NCERT Solutions · Class 11 Chemistry Redox Reactions

When the reducing agent is left over after the reaction, it keeps pushing electrons onto the oxidising agent until that oxidising agent has been driven all the way down to its most reduced (lowest oxidation number) product. When the oxidising agent is left over instead, there is only a fixed, limited supply of electrons coming from the small amount of reducing agent present, so the oxidising agent is reduced only part of the way, and it settles into a product with a higher oxidation number.Every redox reaction runs on one piece of bookkeeping: electrons lost by the reducing agent = electrons gained by the oxidising agent. That equality always holds — what changes when you change the ratio of the two reactants is how far each individual atom's oxidation number moves before the electron supply runs out. A common mistake here is to think "more oxidising agent must mean more oxidation happening everywhere" — but it is the relative amount, reducing agent to oxidising agent, that decides how deep the electron transfer goes, not the absolute quantity of either one.Illustration (i): zinc and nitric acidTake the pair zinc (reducing agent, \(\displaystyle \text{Zn} \to \text{Zn}^{2+} + 2e^- \)) and nitric acid (oxidising agent, nitrogen starts at \(\displaystyle +5 \) in \(\displaystyle \text{HNO}_3 \)).With dilute nitric acid and zinc present in excess (the reducing agent has more electrons to give than the small amount of \(\displaystyle \text{HNO}_3\) needs for a shallow reduction), nitrogen is driven all the way down to \(\displaystyle -3 \), forming ammonium nitrate, \(\displaystyle \text{NH}_4\text{NO}_3 \): \[4\,\text{Zn} + 10\,\text{HNO}_3 \,(\text{dilute}) \longrightarrow 4\,\text{Zn(NO}_3)_2 + \text{NH}_4\text{NO}_3 + 3\,\text{H}_2\text{O} \] To balance this: one nitrogen atom is reduced, \(\displaystyle \overset{+5}{\text{N}} \to \overset{-3}{\text{N}} \), an $\displaystyle 8$-electron gain. Each zinc atom supplies $\displaystyle 2$ electrons on oxidation, \(\displaystyle \overset{0}{\text{Zn}} \to \overset{+2}{\text{Zn}} \), so $\displaystyle 4$ zinc atoms are needed to supply the $\displaystyle 8$ electrons — that fixes the "$\displaystyle 4$" in front of \(\displaystyle \text{Zn}\). The other $\displaystyle 8$ nitrogen atoms just carry \(\displaystyle -1\) charge as spectator nitrate ions in \(\displaystyle \text{Zn(NO}_3)_2\) and in \(\displaystyle \text{NH}_4\text{NO}_3\); counting total N ($\displaystyle 10$), H ($\displaystyle 10$), and O ($\displaystyle 30$) on each side confirms the rest of the coefficients.With concentrated nitric acid in large excess (now it is the oxidising agent that is oversupplied relative to zinc, so the small amount of zinc's electrons get shared thinly among many nitrate ions), nitrogen is reduced only one step, to \(\displaystyle +4 \), forming nitrogen dioxide, \(\displaystyle \text{NO}_2 \): \[\text{Zn} + 4\,\text{HNO}_3\,(\text{concentrated}) \longrightarrow \text{Zn(NO}_3)_2 + 2\,\text{NO}_2 + 2\,\text{H}_2\text{O} \] Here one zinc atom gives up $\displaystyle 2$ electrons; each nitrogen reduced to \(\displaystyle \text{NO}_2\) only takes \(\displaystyle \overset{+5}{\text{N}} \to \overset{+4}{\text{N}} \), a $\displaystyle 1$-electron gain, so $\displaystyle 2$ such nitrogens absorb the $\displaystyle 2$ electrons zinc released — that fixes the "$\displaystyle 2$" in front of \(\displaystyle \text{NO}_2\). Nitrogen ($\displaystyle 4$), hydrogen ($\displaystyle 4$) and oxygen ($\displaystyle 12$) balance on both sides.The same metal, the same acid — but the product's oxidation state moves from \(\displaystyle -3 \) (reducing agent zinc in excess) to \(\displaystyle +4 \) (oxidising agent \(\displaystyle \text{HNO}_3\) in excess), exactly as the statement claims.Illustration (ii): sodium thiosulphate with a mild versus a strong, excess oxidising agentIn sodium thiosulphate, \(\displaystyle \text{Na}_2\text{S}_2\text{O}_3 \), sulphur sits at an average oxidation number of \(\displaystyle +2 \) (each oxygen is \(\displaystyle -2 \), so \(\displaystyle 2S - 6 = -2 \Rightarrow S = +2 \)).Iodine is a mild oxidising agent, so even without any deliberate "excess" of thiosulphate, iodine simply cannot pull sulphur very far — the reducing agent's capacity outlasts iodine's oxidising power, and the reaction stops at sodium tetrathionate, \(\displaystyle \text{Na}_2\text{S}_4\text{O}_6 \), where sulphur is only at \(\displaystyle +2.5 \) on average: \[2\,\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \longrightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\,\text{NaI} \] Two thiosulphate ions contain $\displaystyle 4$ sulphur atoms, total oxidation number \(\displaystyle +8 \); in tetrathionate those same $\displaystyle 4$ sulphur atoms total \(\displaystyle +10 \) — a loss of only $\displaystyle 2$ electrons overall, exactly matched by \(\displaystyle \text{I}_2 + 2e^- \to 2\,\text{I}^- \). Sodium ($\displaystyle 4$), sulphur ($\displaystyle 4$), oxygen ($\displaystyle 6$) and iodine ($\displaystyle 2$) all check out on both sides.Chlorine is a much stronger oxidising agent and is used here in clear excess ($\displaystyle 4$ moles of it per mole of thiosulphate); with that much oxidising capacity available, sulphur is driven all the way up to \(\displaystyle +6 \), forming sulfuric acid, \(\displaystyle \text{H}_2\text{SO}_4 \): \[\text{Na}_2\text{S}_2\text{O}_3 + 4\,\text{Cl}_2 + 5\,\text{H}_2\text{O} \longrightarrow 2\,\text{H}_2\text{SO}_4 + 2\,\text{NaCl} + 6\,\text{HCl} \] Here the $\displaystyle 2$ sulphur atoms go from a total of \(\displaystyle +4 \) to a total of \(\displaystyle +12 \), an $\displaystyle 8$-electron loss; four \(\displaystyle \text{Cl}_2\) molecules each gaining $\displaystyle 2$ electrons (\(\displaystyle \text{Cl}_2 + 2e^- \to 2\,\text{Cl}^- \)) accounts for exactly $\displaystyle 8$ electrons gained. Sodium ($\displaystyle 2$), sulphur ($\displaystyle 2$), oxygen (\(\displaystyle 3+5=8\) on the left, \(\displaystyle 8\) on the right), hydrogen ($\displaystyle 10$) and chlorine ($\displaystyle 8$) all balance.The mild, comparatively "reducing-agent-favoured" reaction with iodine stops at the lower oxidation state \(\displaystyle (+2.5) \); the reaction run with oxidising agent in real excess reaches the higher oxidation state \(\displaystyle (+6) \) — this is exactly why iodometric titrations are done with iodine and never with chlorine or bromine: a stronger oxidant, especially in excess, would over-oxidise the thiosulphate past the clean, reproducible tetrathionate endpoint.Illustration (iii): hydrogen sulphide with the same pair of halogensHydrogen sulphide, \(\displaystyle \text{H}_2\text{S} \), has sulphur at \(\displaystyle -2 \). With a limited amount of iodine, sulphur is only oxidised to elemental sulphur, \(\displaystyle \text{S} \) (oxidation number \(\displaystyle 0 \)): \[\text{H}_2\text{S} + \text{I}_2 \longrightarrow \text{S} + 2\,\text{HI} \] Sulphur loses $\displaystyle 2$ electrons (\(\displaystyle -2 \to 0 \)); iodine gains exactly $\displaystyle 2$ electrons (\(\displaystyle \text{I}_2 + 2e^- \to 2\,\text{I}^- \)). Hydrogen ($\displaystyle 2$) and iodine ($\displaystyle 2$) balance directly.With chlorine present in excess, sulphur is oxidised all the way to \(\displaystyle +6 \), again as sulfuric acid: \[\text{H}_2\text{S} + 4\,\text{Cl}_2 + 4\,\text{H}_2\text{O} \longrightarrow \text{H}_2\text{SO}_4 + 8\,\text{HCl} \] Sulphur loses $\displaystyle 8$ electrons (\(\displaystyle -2 \to +6 \)); four \(\displaystyle \text{Cl}_2\) molecules gaining $\displaystyle 2$ electrons each supplies the $\displaystyle 8$ electrons needed. Hydrogen (\(\displaystyle 2+8=10\) left, \(\displaystyle 2+8=10\) right), sulphur ($\displaystyle 1$), oxygen ($\displaystyle 4$) and chlorine ($\displaystyle 8$) all balance.Once more: limited oxidising agent (iodine) → lower oxidation state product (\(\displaystyle S\), \(\displaystyle 0\)); oxidising agent in excess (chlorine) → higher oxidation state product (\(\displaystyle \text{H}_2\text{SO}_4\), \(\displaystyle +6\)).A short aside worth keeping in mind while balancing any of these: only the atom that is actually changing oxidation state (nitrogen in illustration (i), sulphur in (ii) and (iii)) needs its electron change tracked — oxygen stays at \(\displaystyle -2 \) and hydrogen at \(\displaystyle +1 \) throughout, and mixing those into the electron count is the most common place this kind of balancing goes wrong.All three pairs of reactants tell the same story: whichever reagent is left in surplus after the reaction controls how deep the electron transfer went, because the electron budget is fixed by whichever reagent runs out first. Reducing agent in surplus → the oxidising agent (or the element being oxidised, depending on which side you're tracking) is pushed to its lowest oxidation state; oxidising agent in surplus → the reaction stops at a higher, only partially changed, oxidation state.Answer: The statement is justified by (i) Zn + dilute HNO₃ (Zn in excess) giving NH₄NO₃ with N at −$\displaystyle 3$, versus Zn + concentrated HNO₃ (HNO₃ in excess) giving NO₂ with N at +$\displaystyle 4$; (ii) Na₂S₂O₃ + I₂ (mild, limited oxidant) giving Na₂S₄O₆ with S at +$\displaystyle 2.5$, versus Na₂S₂O₃ + excess Cl₂ giving H₂SO₄ with S at +$\displaystyle 6$; and (iii) H₂S + I₂ (limited) giving elemental S at $\displaystyle 0$, versus H₂S + excess Cl₂ giving H₂SO₄ with S at +$\displaystyle 6$ — in every case, reducing agent in excess drives the product to the lower oxidation state, and oxidising agent in excess drives it to the higher one.

Alkaline potassium permanganate is a weaker, gentler oxidant than acidic potassium permanganate — that mildness is exactly why it can turn the side-chain of toluene into a carboxylate without also tearing apart the benzene ring, while the chloride/bromide difference in part (b) comes down to which halide ion gives up an electron more easily.(a) Toluene, \(\displaystyle C_6H_5CH_3\), needs its \(\displaystyle -CH_3\) group oxidised all the way to a carboxyl group to give benzoic acid, \(\displaystyle C_6H_5COOH\). Two forms of \(\displaystyle KMnO_4\) could, in principle, do this: acidic \(\displaystyle KMnO_4\) (\(\displaystyle MnO_4^- \rightarrow Mn^{2+}\), a five-electron change, \(\displaystyle E^\circ = +1.51\ \text{V}\)) and alkaline \(\displaystyle KMnO_4\) (\(\displaystyle MnO_4^- \rightarrow MnO_2\), a three-electron change, \(\displaystyle E^\circ = +0.59\ \text{V}\)). The lower \(\displaystyle E^\circ\) tells you alkaline permanganate is the weaker oxidant of the two — and a weaker, more controllable oxidant is what this job needs. Acidic \(\displaystyle KMnO_4\) is vigorous enough to keep oxidising past the carboxyl stage and attack the aromatic ring itself, breaking it into a mixture of smaller degradation products; alkaline \(\displaystyle KMnO_4\), used hot, is just strong enough to lift the methyl group up to a carboxylate and then stop, leaving the ring untouched.Aside — "alkaline" sounds gentle, and it is, but not because alkaline conditions protect the ring directly. It is gentle because the alkaline half-reaction has the smaller reduction potential; that lower oxidising power is precisely why it is the safer, more selective choice here.Assign oxidation numbers to see the electron count. In the \(\displaystyle -CH_3\) carbon of toluene, that carbon carries three C–H bonds (\(\displaystyle -1\) each, since carbon is more electronegative than hydrogen) and one C–C bond to the ring (\(\displaystyle 0\)), so its oxidation number is \(\displaystyle -3\). In the \(\displaystyle -COOK\) carbon of the product, that same carbon carries one C–C bond to the ring (\(\displaystyle 0\)) and, in effect, two bonds to oxygen — one double, one single, worth \(\displaystyle +2\) and \(\displaystyle +1\) — giving \(\displaystyle +3\). So the oxidation number of that one carbon rises by \(\displaystyle 6\), from \(\displaystyle -3\) to \(\displaystyle +3\): it loses \(\displaystyle 6\) electrons. Manganese falls from \(\displaystyle +7\) in \(\displaystyle KMnO_4\) to \(\displaystyle +4\) in \(\displaystyle MnO_2\), a drop of \(\displaystyle 3\), i.e. a gain of \(\displaystyle 3\) electrons per manganese. To make electrons lost equal electrons gained, one toluene (losing \(\displaystyle 6\)) needs two permanganate ions (gaining \(\displaystyle 3\) each, \(\displaystyle 2\times3=6\)):\[C_6H_5CH_3 + 2KMnO_4 \xrightarrow{\ \Delta\ } C_6H_5COOK + 2MnO_2 + KOH + H_2O \]Checking atoms confirms the coefficients: carbon \(\displaystyle 7=7\); hydrogen \(\displaystyle 8 = 5+1+2\); oxygen \(\displaystyle 8 = 2+4+1+1\); potassium \(\displaystyle 2 = 1+1\); manganese \(\displaystyle 2=2\). The organic product here is potassium benzoate, \(\displaystyle C_6H_5COOK\) — a salt, not the acid itself. A mineral acid finishes the job by protonating the carboxylate:\[C_6H_5COOK + HCl \rightarrow C_6H_5COOH + KCl \]giving benzoic acid, the actual manufacturing target, alongside potassium chloride.(b) Concentrated \(\displaystyle H_2SO_4\) is a strong, non-volatile acid, so it can always push a more volatile acid out of its salt by simple double displacement — no electron transfer required. With a chloride, that is all that happens:\[NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\uparrow \]\(\displaystyle HCl\) escapes as the colourless, sharp-smelling gas described in the question, and neither chlorine nor sulphur changes oxidation state — chlorine stays \(\displaystyle -1\) throughout, sulphur stays \(\displaystyle +6\) throughout. This step is acid–base chemistry, not redox.Bromide goes one step further, because \(\displaystyle Br^-\) is a noticeably better reducing agent than \(\displaystyle Cl^-\): being the larger ion, with its outermost electrons held less tightly, it gives one up more readily (reducing power of the halide ions runs \(\displaystyle F^- \ll Cl^- < Br^- < I^-\)). Concentrated \(\displaystyle H_2SO_4\) is not a strong enough oxidant to pull an electron off \(\displaystyle Cl^-\) — chlorine resists oxidation to \(\displaystyle Cl_2\) here — but it is strong enough to pull one off \(\displaystyle Br^-\). So the \(\displaystyle HBr\) formed by the same displacement step does not survive: it is oxidised on the spot, and the acid itself acts as the oxidising agent:\[2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O \]Here bromine rises from \(\displaystyle -1\) (in \(\displaystyle HBr\)) to \(\displaystyle 0\) (in \(\displaystyle Br_2\)): two bromine atoms together lose \(\displaystyle 2\) electrons. Sulphur falls from \(\displaystyle +6\) (in \(\displaystyle H_2SO_4\)) to \(\displaystyle +4\) (in \(\displaystyle SO_2\)), gaining those same \(\displaystyle 2\) electrons, so the electron count matches and the equation balances in atoms too (\(\displaystyle H:4=4\), \(\displaystyle Br:2=2\), \(\displaystyle S:1=1\), \(\displaystyle O:4=4\)). Folding the two steps into one, for a bromide salt reacting with excess concentrated acid:\[2NaBr + 3H_2SO_4 \rightarrow 2NaHSO_4 + SO_2 + Br_2 + 2H_2O \]The red vapour seen is bromine, \(\displaystyle Br_2\), escaping together with the colourless, choking gas sulphur dioxide, \(\displaystyle SO_2\).Aside — the chloride and bromide reactions look like the "same" step written with a different halogen, but they are not the same kind of chemistry: the chloride case never changes anyone's oxidation number, while the bromide case is a full oxidation–reduction, with sulphur playing the oxidising agent that accepts the electrons bromide gives up.Answer: (a) Alkaline \(\displaystyle KMnO_4\) (\(\displaystyle MnO_4^-\to MnO_2\), \(\displaystyle E^\circ=+0.59\,V\), a $\displaystyle 3$-electron change) is the weaker, more selective oxidant and stops cleanly at the carboxylate stage without attacking the ring, whereas acidic \(\displaystyle KMnO_4\) (\(\displaystyle MnO_4^-\to Mn^{2+}\), \(\displaystyle E^\circ=+1.51\,V\), a $\displaystyle 5$-electron change) is vigorous enough to over-oxidise and degrade the aromatic ring: \(\displaystyle C_6H_5CH_3 + 2KMnO_4 \xrightarrow{\Delta} C_6H_5COOK + 2MnO_2 + KOH + H_2O\), and acidifying the potassium benzoate gives benzoic acid. (b) \(\displaystyle NaCl+H_2SO_4\to NaHSO_4+HCl\) is a non-redox displacement because chloride resists oxidation, while \(\displaystyle Br^-\) is a stronger reducing agent than \(\displaystyle Cl^-\) and is oxidised by concentrated \(\displaystyle H_2SO_4\): \(\displaystyle 2NaBr+3H_2SO_4\to 2NaHSO_4+SO_2+Br_2+2H_2O\), which is why a bromide gives red bromine vapour (with \(\displaystyle SO_2\)) instead of just \(\displaystyle HBr\) gas.

The whole question is "who lost electrons and who gained them" — the species whose oxidation number goes up is oxidised (and is called the reducing agent), and the one whose oxidation number goes down is reduced (and is called the oxidising agent). The one detail students slip on: the oxidising/reducing agent is named as the whole reactant (\(\displaystyle \text{AgBr}\), \(\displaystyle [\text{Ag(NH}_3)_2]^+\), \(\displaystyle \text{PbO}_2\) …), while "oxidised/reduced" describes the atom inside it whose number actually changed.(a) \(\displaystyle 2\text{AgBr(s)} + \text{C}_6\text{H}_6\text{O}_2\text{(aq)} \rightarrow 2\text{Ag(s)} + 2\text{HBr(aq)} + \text{C}_6\text{H}_4\text{O}_2\text{(aq)}\)In \(\displaystyle \text{AgBr}\), bromine is fixed at \(\displaystyle -1\) (it stays \(\displaystyle -1\) in \(\displaystyle \text{HBr}\) too, so bromine is a spectator here — not every atom in a redox equation has to change). Since the compound is neutral, \[\text{Ag} + (-1) = 0 \implies \text{Ag} = +1 \text{ in AgBr.} \] In \(\displaystyle \text{Ag(s)}\) the oxidation number is \(\displaystyle 0\) (element in its free state). So \[\overset{+1}{\text{Ag}}\text{Br} \longrightarrow \overset{0}{\text{Ag}}, \] a fall from \(\displaystyle +1\) to \(\displaystyle 0\): each \(\displaystyle \text{Ag}^+\) gains one electron, and there are two of them, so \(\displaystyle 2\) electrons are gained overall — silver is reduced.\(\displaystyle \text{C}_6\text{H}_6\text{O}_2\) is hydroquinone (quinol, $\displaystyle 1,4$-dihydroxybenzene); \(\displaystyle \text{C}_6\text{H}_4\text{O}_2\) is the compound it turns into, benzoquinone. Going from hydroquinone to quinone, two \(\displaystyle \text{C–OH}\) groups become two \(\displaystyle \text{C=O}\) groups, and the molecule throws off exactly the two hydrogen atoms that show up on the product side as \(\displaystyle 2\text{HBr}\). By the classical definition — oxidation is loss of hydrogen — hydroquinone is oxidised, and that loss of \(\displaystyle 2\) H (as \(\displaystyle 2\text{H}^+ + 2e^-\)) supplies precisely the \(\displaystyle 2\) electrons that the \(\displaystyle 2\text{Ag}^+\) needed. The electron count matches, so the equation is balanced as a redox process.Substance oxidised: \(\displaystyle \text{C}_6\text{H}_6\text{O}_2\) (hydroquinone). Substance reduced: the \(\displaystyle \text{Ag}^+\) in \(\displaystyle \text{AgBr}\). Oxidising agent: \(\displaystyle \text{AgBr}\). Reducing agent: \(\displaystyle \text{C}_6\text{H}_6\text{O}_2\) (hydroquinone).(b) \(\displaystyle \text{HCHO(l)} + 2[\text{Ag(NH}_3)_2]^+\text{(aq)} + 3\text{OH}^-\text{(aq)} \rightarrow 2\text{Ag(s)} + \text{HCOO}^-\text{(aq)} + 4\text{NH}_3\text{(aq)} + 2\text{H}_2\text{O(l)}\)} — this is Tollens' silver-mirror test.\(\displaystyle \text{NH}_3\) is a neutral ligand, so in \(\displaystyle [\text{Ag(NH}_3)_2]^+\) all of the complex's \(\displaystyle +1\) charge sits on silver: \(\displaystyle \text{Ag} = +1\). In \(\displaystyle \text{Ag(s)}\) it is \(\displaystyle 0\). So \[\overset{+1}{\text{Ag}}(\text{NH}_3)_2^+ \longrightarrow \overset{0}{\text{Ag}} \] is a gain of \(\displaystyle 1\) electron per silver, \(\displaystyle 2\) electrons for the two silvers in the equation — silver is reduced.Now the carbon in methanal, \(\displaystyle \text{HCHO}\) (\(\displaystyle \text{H–CHO}\)): it carries two bonds to hydrogen (each hydrogen is less electronegative than carbon, so each contributes \(\displaystyle -1\) to carbon) and a double bond to oxygen (oxygen is more electronegative, so it contributes \(\displaystyle +2\)): \[\text{C in HCHO} = (-1) + (-1) + (+2) = 0. \] In the formate ion, \(\displaystyle \text{HCOO}^-\), carbon has one bond to H (\(\displaystyle -1\)), one double bond to O (\(\displaystyle +2\)), and one single bond to a second, negatively-charged O (single bond to the more electronegative atom, \(\displaystyle +1\)): \[\text{C in HCOO}^- = (-1) + (+2) + (+1) = +2. \] Carbon rises from \(\displaystyle 0\) to \(\displaystyle +2\): a loss of \(\displaystyle 2\) electrons, so methanal is oxidised to formate. That matches the \(\displaystyle 2\) electrons the two silver ions gained — balanced.Substance oxidised: \(\displaystyle \text{HCHO}\) (methanal/formaldehyde). Substance reduced: \(\displaystyle [\text{Ag(NH}_3)_2]^+\) (the silver in Tollens' reagent). Oxidising agent: \(\displaystyle [\text{Ag(NH}_3)_2]^+\). Reducing agent: \(\displaystyle \text{HCHO}\).(c) \(\displaystyle \text{HCHO(l)} + 2\text{Cu}^{2+}\text{(aq)} + 5\text{OH}^-\text{(aq)} \rightarrow \text{Cu}_2\text{O(s)} + \text{HCOO}^-\text{(aq)} + 3\text{H}_2\text{O(l)}\) — Fehling's test.\(\displaystyle \text{Cu}_2\text{O}\) is a neutral compound with two oxygens at \(\displaystyle -2\) each: \[2(\text{Cu}) + 2(-2) = 0 \implies \text{Cu} = +1 \text{ in } \text{Cu}_2\text{O.} \] So copper goes from \(\displaystyle +2\) (as \(\displaystyle \text{Cu}^{2+}\)) to \(\displaystyle +1\): each copper gains \(\displaystyle 1\) electron, and with \(\displaystyle 2\) coppers that is \(\displaystyle 2\) electrons gained — copper is reduced (this is the same numbered result as blue \(\displaystyle \text{Cu}^{2+}\) forming the brick-red \(\displaystyle \text{Cu}_2\text{O}\) precipitate you may already know from the test).The carbon math is identical to part (b): \(\displaystyle \text{HCHO}\) (carbon \(\displaystyle 0\)) becomes \(\displaystyle \text{HCOO}^-\) (carbon \(\displaystyle +2\)), losing \(\displaystyle 2\) electrons — oxidised. The \(\displaystyle 2\) electrons lost by carbon equal the \(\displaystyle 2\) electrons gained by the two \(\displaystyle \text{Cu}^{2+}\) ions, so the equation balances as a redox change even though it looks unfamiliar with \(\displaystyle \text{OH}^-\) floating around (those hydroxide ions are just neutralising the \(\displaystyle \text{H}^+\) that would otherwise appear, they don't change oxidation state — the trap here is assuming every species in a long equation must be part of the electron transfer).Substance oxidised: \(\displaystyle \text{HCHO}\). Substance reduced: \(\displaystyle \text{Cu}^{2+}\). Oxidising agent: \(\displaystyle \text{Cu}^{2+}\) (copper(II) ion). Reducing agent: \(\displaystyle \text{HCHO}\).(d) \(\displaystyle \text{N}_2\text{H}_4\text{(l)} + 2\text{H}_2\text{O}_2\text{(l)} \rightarrow \text{N}_2\text{(g)} + 4\text{H}_2\text{O(l)}\)Hydrazine, \(\displaystyle \text{N}_2\text{H}_4\), is neutral with \(\displaystyle 4\) hydrogens at \(\displaystyle +1\): \[2(\text{N}) + 4(+1) = 0 \implies \text{N} = -2 \text{ in } \text{N}_2\text{H}_4. \] In elemental \(\displaystyle \text{N}_2(g)\), nitrogen is \(\displaystyle 0\). So nitrogen rises from \(\displaystyle -2\) to \(\displaystyle 0\): each nitrogen loses \(\displaystyle 2\) electrons, and there are \(\displaystyle 2\) nitrogens, so \(\displaystyle 4\) electrons are lost overall — hydrazine is oxidised.In hydrogen peroxide, \(\displaystyle \text{H}_2\text{O}_2\), the O–O peroxide linkage puts oxygen at \(\displaystyle -1\) (not the usual \(\displaystyle -2\)) — this is exactly the step people get wrong, forgetting peroxides break the "oxygen is \(\displaystyle -2\)" rule: \[2(+1) + 2(\text{O}) = 0 \implies \text{O} = -1 \text{ in } \text{H}_2\text{O}_2. \] In water, \(\displaystyle \text{H}_2\text{O}\), oxygen is back to \(\displaystyle -2\). So oxygen falls from \(\displaystyle -1\) to \(\displaystyle -2\), gaining \(\displaystyle 1\) electron per oxygen atom; with \(\displaystyle 2\) molecules of \(\displaystyle \text{H}_2\text{O}_2\) (\(\displaystyle 4\) oxygen atoms total), that is \(\displaystyle 4\) electrons gained — hydrogen peroxide is reduced. The \(\displaystyle 4\) electrons lost by nitrogen match the \(\displaystyle 4\) gained by oxygen, so the equation balances.Substance oxidised: \(\displaystyle \text{N}_2\text{H}_4\) (hydrazine — nitrogen is oxidised). Substance reduced: \(\displaystyle \text{H}_2\text{O}_2\) (hydrogen peroxide — oxygen is reduced). Oxidising agent: \(\displaystyle \text{H}_2\text{O}_2\). Reducing agent: \(\displaystyle \text{N}_2\text{H}_4\).(e) \(\displaystyle \text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \rightarrow 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)}\) — this is the lead-acid storage battery discharging.Metallic \(\displaystyle \text{Pb(s)}\) is \(\displaystyle 0\) (element in its free state). In \(\displaystyle \text{PbO}_2\), with two oxygens at \(\displaystyle -2\): \[\text{Pb} + 2(-2) = 0 \implies \text{Pb} = +4 \text{ in } \text{PbO}_2. \] In \(\displaystyle \text{PbSO}_4\), the sulfate ion \(\displaystyle \text{SO}_4^{2-}\) carries \(\displaystyle -2\), so \[\text{Pb} + (-2) = 0 \implies \text{Pb} = +2 \text{ in } \text{PbSO}_4. \] Now watch both lead atoms land on the same product: the metallic lead rises from \(\displaystyle 0\) to \(\displaystyle +2\) (loses \(\displaystyle 2\) electrons — oxidised), while the lead in \(\displaystyle \text{PbO}_2\) falls from \(\displaystyle +4\) to \(\displaystyle +2\) (gains \(\displaystyle 2\) electrons — reduced). One reactant supplies the electrons the other consumes, \(\displaystyle 2\) for \(\displaystyle 2\), so the equation balances — this is the case where the same element sits on both the oxidised and reduced side of one reaction, which is exactly why a lead-acid cell can run on two lead electrodes.Substance oxidised: \(\displaystyle \text{Pb}\) (metallic lead, the anode). Substance reduced: \(\displaystyle \text{PbO}_2\) (lead dioxide, the cathode). Oxidising agent: \(\displaystyle \text{PbO}_2\). Reducing agent: \(\displaystyle \text{Pb}\).Answer: (a) oxidised: \(\displaystyle \text{C}_6\text{H}_6\text{O}_2\) (hydroquinone); reduced: \(\displaystyle \text{AgBr}\); oxidising agent: \(\displaystyle \text{AgBr}\); reducing agent: \(\displaystyle \text{C}_6\text{H}_6\text{O}_2\). (b) oxidised: \(\displaystyle \text{HCHO}\); reduced: \(\displaystyle [\text{Ag(NH}_3)_2]^+\); oxidising agent: \(\displaystyle [\text{Ag(NH}_3)_2]^+\); reducing agent: \(\displaystyle \text{HCHO}\). (c) oxidised: \(\displaystyle \text{HCHO}\); reduced: \(\displaystyle \text{Cu}^{2+}\); oxidising agent: \(\displaystyle \text{Cu}^{2+}\); reducing agent: \(\displaystyle \text{HCHO}\). (d) oxidised: \(\displaystyle \text{N}_2\text{H}_4\); reduced: \(\displaystyle \text{H}_2\text{O}_2\); oxidising agent: \(\displaystyle \text{H}_2\text{O}_2\); reducing agent: \(\displaystyle \text{N}_2\text{H}_4\). (e) oxidised: \(\displaystyle \text{Pb}\); reduced: \(\displaystyle \text{PbO}_2\); oxidising agent: \(\displaystyle \text{PbO}_2\); reducing agent: \(\displaystyle \text{Pb}\).

The same reductant can end up more or less oxidized depending on how strong an oxidant it meets — a weak oxidant only pulls electrons partway, a strong one pulls them all the way.Thiosulphate, \(\displaystyle \text{S}_2\text{O}_3^{2-} \), is the reductant in both reactions. To see "differently" precisely, track the oxidation number of sulphur, using the rule that the oxidation numbers of all atoms in an ion must add up to the ion's charge.Step $\displaystyle 1$ — oxidation number of S in each speciesIn \(\displaystyle \text{S}_2\text{O}_3^{2-} \), let \(\displaystyle x\) be the (average) oxidation number of S. Oxygen is \(\displaystyle -2\): \[2x + 3(-2) = -2 \;\;\Rightarrow\;\; 2x = 4 \;\;\Rightarrow\;\; x = +2 \]In tetrathionate, \(\displaystyle \text{S}_4\text{O}_6^{2-} \): \[4x + 6(-2) = -2 \;\;\Rightarrow\;\; 4x = 10 \;\;\Rightarrow\;\; x = +2.5 \]In sulphate, \(\displaystyle \text{SO}_4^{2-} \): \[x + 4(-2) = -2 \;\;\Rightarrow\;\; x = +6 \]So with iodine, sulphur only creeps up from \(\displaystyle +2\) to \(\displaystyle +2.5\); with bromine, it is driven all the way from \(\displaystyle +2\) to \(\displaystyle +6\). That gap is the whole answer — now check it against balanced electron counts.Step $\displaystyle 2$ — the reaction with iodine (mild oxidation)Total S-oxidation-number on each side of \(\displaystyle 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} \): left = \(\displaystyle 2\times(+2)=4\) (that's for one ion; two ions give \(\displaystyle 2\times4=8\) counting both S atoms — equivalently, $\displaystyle 4$ S atoms at \(\displaystyle +2\) each), right = $\displaystyle 4$ S atoms at \(\displaystyle +2.5\) each = \(\displaystyle 10\). The rise of \(\displaystyle 2\) units means $\displaystyle 2$ electrons are lost: \[2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2e^- \] Iodine, the oxidant, is reduced and takes exactly those $\displaystyle 2$ electrons: \[\text{I}_2 + 2e^- \rightarrow 2\text{I}^- \] Adding the two half-reactions (electrons cancel) reproduces the reaction as given: \[2\text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(s) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2\text{I}^-(aq) \] Charge check: left \(\displaystyle 2(-2)+0=-4\); right \(\displaystyle -2+2(-1)=-4\). Balanced. Only a small, $\displaystyle 1$-electron-per-thiosulphate change — the two sulphurs stay linked (as they are in tetrathionate), they're just joined into a bigger ion.Step $\displaystyle 3$ — the reaction with bromine (full oxidation) — and a balance check that mattersHere sulphur is oxidized all the way to sulphate. The oxidation half-reaction, built atom by atom (balance S, then O with \(\displaystyle \text{H}_2\text{O} \), then H with \(\displaystyle \text{H}^+ \), then charge with electrons), is: \[\text{S}_2\text{O}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 2\text{SO}_4^{2-} + 10\text{H}^+ + 8e^- \] Check it: O, \(\displaystyle 3+5=8\) both sides; H, \(\displaystyle 10\) both sides; charge, left \(\displaystyle -2\), right \(\displaystyle 2(-2)+10(1)-8(-1)\text{-worth}= -4+10-8=-2\) — balanced, and it needs $\displaystyle 8$ electrons, matching sulphur's jump from \(\displaystyle +2\) to \(\displaystyle +6\) on two S atoms (\(\displaystyle 2\times4=8\)).The reduction half-reaction is \[\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \] which supplies only $\displaystyle 2$ electrons per \(\displaystyle \text{Br}_2 \). To soak up $\displaystyle 8$ electrons you need four \(\displaystyle \text{Br}_2 \), giving eight \(\displaystyle \text{Br}^- \): \[4\text{Br}_2 + 8e^- \rightarrow 8\text{Br}^- \] Adding the two half-reactions (the \(\displaystyle 8e^-\) cancel): \[\text{S}_2\text{O}_3^{2-}(aq) + 4\text{Br}_2(l) + 5\text{H}_2\text{O}(l) \rightarrow 2\text{SO}_4^{2-}(aq) + 8\text{Br}^-(aq) + 10\text{H}^+(aq) \] Charge check: left \(\displaystyle -2\); right \(\displaystyle 2(-2)+8(-1)+10(+1) = -4-8+10=-2\). Balanced.This is the step people skip: matching atom counts is not the same as balancing an equation. If you only count Br atoms ($\displaystyle 2$ mol \(\displaystyle \text{Br}_2\) does give $\displaystyle 4$ mol Br, matching "$\displaystyle 4$ Br\(\displaystyle ^-\)"), the equation looks fine — but its charge comes out to \(\displaystyle -2\) on the left against \(\displaystyle +2\) on the right, off by 4. Electrons, not just atoms, have to balance, and the fix is the same factor of two throughout: $\displaystyle 4$ \(\displaystyle \text{Br}_2 \) and $\displaystyle 8$ \(\displaystyle \text{Br}^- \), not $\displaystyle 2$ and 4.Step $\displaystyle 4$ — why the depth of oxidation differsThe reductant is identical in both cases; what differs is how strongly the oxidant pulls electrons off it, measured by standard reduction potential: \[\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq), \quad E^{\circ} = +0.54\ \text{V} \] \[\text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq), \quad E^{\circ} = +1.09\ \text{V} \] The higher \(\displaystyle E^{\circ}\) for bromine means \(\displaystyle \text{Br}_2 \) is the stronger oxidizing agent. Iodine is too weak an oxidant to break the S–S bond inside thiosulphate; it can only strip one electron pair off the pair of ions, coupling two thiosulphate units into tetrathionate (S rises just from \(\displaystyle +2\) to \(\displaystyle +2.5\), S–S bond kept). Bromine is a strong enough oxidant to break that S–S bond completely and push both sulphur atoms up to their maximum common oxidation state, \(\displaystyle +6\), giving two separate sulphate ions.Answer: Thiosulphate is the reductant in both cases, but the extent to which it is oxidized is set by the oxidizing strength of the other reactant. \(\displaystyle \text{I}_2 \) (\(\displaystyle E^{\circ}=+0.54\ \text{V}\)) is a weak oxidant and can raise sulphur's oxidation number only from \(\displaystyle +2\) (in \(\displaystyle \text{S}_2\text{O}_3^{2-} \)) to \(\displaystyle +2.5\) (in \(\displaystyle \text{S}_4\text{O}_6^{2-} \)), so \(\displaystyle 2\text{S}_2\text{O}_3^{2-}(aq)+\text{I}_2(s)\rightarrow \text{S}_4\text{O}_6^{2-}(aq)+2\text{I}^-(aq)\), a $\displaystyle 2$-electron change with the S–S linkage retained. \(\displaystyle \text{Br}_2 \) (\(\displaystyle E^{\circ}=+1.09\ \text{V}\)) is a much stronger oxidant and drives sulphur all the way to \(\displaystyle +6\) (sulphate), breaking the S–S bond entirely: \(\displaystyle \text{S}_2\text{O}_3^{2-}(aq)+4\text{Br}_2(l)+5\text{H}_2\text{O}(l)\rightarrow 2\text{SO}_4^{2-}(aq)+8\text{Br}^-(aq)+10\text{H}^+(aq) \), an $\displaystyle 8$-electron change (the coefficients must be $\displaystyle 4$ mol \(\displaystyle \text{Br}_2 \) and $\displaystyle 8$ mol \(\displaystyle \text{Br}^- \) for the equation to balance in charge, not $\displaystyle 2$ and $\displaystyle 4$).

Every one of these reactions comes down to who wants an electron more (the oxidant) and who lets go of one most easily (the reductant).(a) Fluorine is the best oxidant among the halogensAn oxidizing agent is the species that gets reduced — it pulls electrons away from something else. The strength of an oxidant is measured by its standard reduction potential \(\displaystyle E^\circ\), for the half‑reaction \(\displaystyle X_2 + 2e^- \rightarrow 2X^-\). For the halogens,\[E^\circ(F_2/F^-) = +2.87\ \text{V}, \quad E^\circ(Cl_2/Cl^-) = +1.36\ \text{V}, \quad E^\circ(Br_2/Br^-) = +1.09\ \text{V}, \quad E^\circ(I_2/I^-) = +0.54\ \text{V} \]\(\displaystyle E^\circ\) is the standard reduction potential in volts — the more positive it is, the more strongly that species pulls in electrons under standard conditions. Fluorine's value is the highest of any common element, so \(\displaystyle F_2\) is reduced more readily than any other halogen. That is exactly what "best oxidant" means, and two reactions put it into practice.Reaction $\displaystyle 1$ — fluorine oxidizes water itself, something no other halogen does at room temperature:\[2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g) \]giving hydrogen fluoride (\(\displaystyle HF\)) and oxygen gas (\(\displaystyle O_2\)). Balance it by tracking oxidation numbers: fluorine goes \(\displaystyle \overset{0}{F_2} \rightarrow \overset{-1}{F^-} \), gaining \(\displaystyle 1\,e^-\) per atom — $\displaystyle 4$ F atoms in \(\displaystyle 2F_2\), so $\displaystyle 4$ electrons gained. Oxygen in water goes \(\displaystyle \overset{-2}{O} \rightarrow \overset{0}{O_2} \), losing \(\displaystyle 2\,e^-\) per atom — $\displaystyle 2$ O atoms in \(\displaystyle 2H_2O\), so $\displaystyle 4$ electrons lost. Electrons gained equal electrons lost (\(\displaystyle 4=4\)), and the atom counts already match on both sides ($\displaystyle 4$ H, $\displaystyle 2$ O), so the equation is balanced as written. This is the step people forget: it's why fluorine is never made by chemically oxidizing \(\displaystyle F^-\) in water — the water gets oxidized first.Reaction $\displaystyle 2$ — fluorine displaces every other halogen from its salt, and nothing displaces fluorine back:\[F_2(g) + 2X^-(aq) \rightarrow 2F^-(aq) + X_2 \qquad (X = Cl,\ Br,\ I) \]for example \(\displaystyle F_2(g) + 2Cl^-(aq) \rightarrow 2F^-(aq) + Cl_2(g)\), giving chlorine gas; with \(\displaystyle Br^-\) it gives liquid bromine, \(\displaystyle Br_2(l)\); with \(\displaystyle I^-\) it gives solid iodine, \(\displaystyle I_2(s)\). Oxidation-number check: \(\displaystyle F_2\) goes \(\displaystyle 0\to -1\) ($\displaystyle 2$ F atoms gain \(\displaystyle 1\,e^-\) each = $\displaystyle 2$ gained), \(\displaystyle X^-\) goes \(\displaystyle -1\to 0\) ($\displaystyle 2$ ions lose \(\displaystyle 1\,e^-\) each = $\displaystyle 2$ lost); \(\displaystyle 2=2\), and the atom counts ($\displaystyle 2$ F, $\displaystyle 2$ X on each side) need no further adjustment. No halogen below fluorine can run this in reverse on \(\displaystyle F^-\) — fluoride is the smallest, most tightly-held halide ion in existence, and nothing here can strip an electron back off it. A displacement that only ever runs one way is the signature of the strongest oxidant in the family.(b) Hydroiodic acid is the best reductant among the hydrohalic acidsA reducing agent is the species that gets oxidized — it gives an electron away. For an acid \(\displaystyle HX\), the electron on offer sits on the halide ion \(\displaystyle X^-\); the \(\displaystyle H^+\) is just a spectator. So the real question is: which of \(\displaystyle F^-, Cl^-, Br^-, I^-\) lets go of an electron most easily?This is where the two trends of the chapter get mixed up. Going down the group, ionic radius increases (\(\displaystyle F^- < Cl^- < Br^- < I^-\)), so the outermost electrons of \(\displaystyle I^-\) sit farthest from the nucleus and are held the most loosely — making \(\displaystyle I^-\) the easiest of the four to oxidize. That gives the reducing-power order \(\displaystyle HI > HBr > HCl > HF\), which runs in the opposite direction to the halogens' own oxidizing power (\(\displaystyle F_2>Cl_2>Br_2>I_2\)) — one trend is about how tightly a neutral atom grabs a spare electron, the other about how loosely an ion that already has one holds it, and size increasing down the group drives both, just in opposite senses.The proof is each halide's reaction with concentrated sulphuric acid, an oxidant only strong enough to be reduced when the halide pushes hard enough.Chloride — no redox at all:\[NaCl(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HCl(g) \]giving sodium hydrogen sulphate and hydrogen chloride gas. Checking oxidation numbers: Na stays \(\displaystyle +1\), Cl stays \(\displaystyle -1\), S stays \(\displaystyle +6\), O stays \(\displaystyle -2\), H stays \(\displaystyle +1\) — nothing changes state, so despite the arrow this is only an acid–base (proton-transfer) step: \(\displaystyle H_2SO_4\) is the stronger, less volatile acid, so it simply displaces \(\displaystyle HCl\) gas out of the solid chloride. \(\displaystyle HCl\) forms, but chloride is too poor a reductant to react with \(\displaystyle H_2SO_4\) any further.Bromide — reduces sulphur partway, to sulphur dioxide:\[NaBr(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HBr(g) \] \[2HBr(g) + H_2SO_4(l) \rightarrow Br_2(l) + SO_2(g) + 2H_2O(l) \]giving reddish-brown bromine vapour, choking sulphur dioxide gas, and water. In the second step, sulphur drops from \(\displaystyle \overset{+6}{S} \) in \(\displaystyle H_2SO_4\) to \(\displaystyle \overset{+4}{S} \) in \(\displaystyle SO_2\) — a gain of \(\displaystyle 2\,e^-\). Bromine goes \(\displaystyle \overset{-1}{Br} \rightarrow \overset{0}{Br_2}\): each of the $\displaystyle 2$ Br atoms loses \(\displaystyle 1\,e^-\), so \(\displaystyle 2\,e^-\) lost in total. \(\displaystyle 2\) gained \(\displaystyle =2\) lost fixes those coefficients; the rest follows by inspection — H is \(\displaystyle 2+2=4\) on the left and \(\displaystyle 2\times2=4\) on the right ($\displaystyle 2$ \(\displaystyle H_2O\)), O is \(\displaystyle 4\) on the left and \(\displaystyle 2+2=4\) on the right (\(\displaystyle SO_2\) plus $\displaystyle 2$ \(\displaystyle H_2O\)). Fully balanced.Iodide — reduces sulphur all the way to hydrogen sulfide:\[NaI(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HI(g) \] \[8HI(g) + H_2SO_4(l) \rightarrow 4I_2(s) + H_2S(g) + 4H_2O(l) \]giving violet solid iodine, rotten-egg-smelling hydrogen sulfide gas, and water. Sulphur is pulled all the way from \(\displaystyle \overset{+6}{S} \) to \(\displaystyle \overset{-2}{S} \) in \(\displaystyle H_2S\) — a gain of \(\displaystyle 8\,e^-\) on that one atom. Iodine goes \(\displaystyle \overset{-1}{I}\rightarrow \overset{0}{I_2}\), each of the $\displaystyle 8$ I atoms losing \(\displaystyle 1\,e^-\), for \(\displaystyle 8\,e^-\) lost — matching the $\displaystyle 8$ gained, which is why \(\displaystyle HI\) needs the coefficient $\displaystyle 8$ and not 2. Atom check: H is \(\displaystyle 8+2=10\) on the left and \(\displaystyle 2+4\times2=10\) on the right (\(\displaystyle H_2S\) plus $\displaystyle 4$ \(\displaystyle H_2O\)); O is \(\displaystyle 4\) on the left and \(\displaystyle 4\) on the right ($\displaystyle 4$ \(\displaystyle H_2O\)); balanced.That sulphur can be dragged all the way down to \(\displaystyle -2\) — something neither chloride nor bromide manages — is the direct experimental proof that \(\displaystyle HI\) is the strongest reductant of the four: chloride does no redox chemistry with \(\displaystyle H_2SO_4\) at all, bromide reduces it only as far as \(\displaystyle +4\), and only iodide has electrons to spare all the way down to \(\displaystyle -2\).Answer: Fluorine is the best oxidant among the halogens because \(\displaystyle E^\circ(F_2/F^-)=+2.87\ \text{V}\) is the highest in the group; it oxidizes water itself, \(\displaystyle 2F_2(g)+2H_2O(l)\rightarrow 4HF(aq)+O_2(g)\), and displaces every other halide from solution, \(\displaystyle F_2(g)+2X^-(aq)\rightarrow 2F^-(aq)+X_2\) (\(\displaystyle X=Cl,Br,I\)), a displacement no other halogen can ever run in reverse on \(\displaystyle F^-\). Hydroiodic acid is the best reductant among the hydrohalic acids because \(\displaystyle I^-\), the largest and most loosely-held halide ion, gives up its electron most easily: with concentrated \(\displaystyle H_2SO_4\), \(\displaystyle Cl^-\) shows no redox at all (\(\displaystyle NaCl+H_2SO_4\rightarrow NaHSO_4+HCl\)), \(\displaystyle Br^-\) reduces sulphur only to \(\displaystyle SO_2\) (\(\displaystyle 2HBr+H_2SO_4\rightarrow Br_2+SO_2+2H_2O\)), while \(\displaystyle I^-\) alone reduces it all the way to \(\displaystyle H_2S\) (\(\displaystyle 8HI+H_2SO_4\rightarrow 4I_2+H_2S+4H_2O\)) — giving the reducing-power order \(\displaystyle HI>HBr>HCl>HF\).

Xenon is reduced and fluoride is oxidised here — that electron transfer is the whole reaction, and it tells you exactly how powerful Na₄XeO₆ is.Work out the oxidation number of xenon on each side. Oxidation number is the charge an atom would carry if every bond to it were treated as fully ionic; for an ion, the oxidation numbers of all atoms must add up to the ion's charge, and for a neutral molecule they must add up to zero.Start with \(\displaystyle \text{XeO}_6^{4-} \) (the perxenate ion — this is the anion that Na₄XeO₆ releases in water). Oxygen is taken as \(\displaystyle -2\) here — that's the normal rule, and it's safe to use because there are no O–O bonds in this ion (the exception, peroxide oxygen at \(\displaystyle -1\), doesn't apply). Let \(\displaystyle x\) be the oxidation number of Xe: \[x + 6(-2) = -4 \] \[x = -4 + 12 = +8 \] So xenon sits at \(\displaystyle +8\) in \(\displaystyle \text{XeO}_6^{4-} \). This is worth pausing on: most Xe–O compounds you meet (XeO₃, XeOF₄, XeO₂F₂) cap out at \(\displaystyle +6\); \(\displaystyle +8\) is xenon's highest known oxidation state, and \(\displaystyle \text{XeO}_6^{4-} \) is one of only two places it turns up (the other is \(\displaystyle \text{XeO}_4 \)).Now do the same for \(\displaystyle \text{XeO}_3 \) (xenon trioxide, the neutral product gas). Let \(\displaystyle y\) be Xe's oxidation number: \[y + 3(-2) = 0 \] \[y = +6 \]So xenon goes from \(\displaystyle +8\) to \(\displaystyle +6\): each Xe atom gains $\displaystyle 2$ electrons. Gaining electrons is reduction, so \(\displaystyle \text{XeO}_6^{4-} \) is the species reduced — it is the oxidising agent. As a half-reaction (balancing O with \(\displaystyle H_2O\) and charge with \(\displaystyle H^+\), the standard way to balance a half-reaction in acid): \[\text{XeO}_6^{4-} + 6H^+ + 2e^- \rightarrow \text{XeO}_3 + 3H_2O \]Next, fluorine. In \(\displaystyle F^-\) (fluoride ion) the oxidation number is \(\displaystyle -1\) — a monatomic ion's oxidation number is just its charge. In \(\displaystyle F_2\) (fluorine gas, the free element) the oxidation number is \(\displaystyle 0\) — any element in its own elemental form is defined as oxidation state zero, by convention, regardless of how reactive it is. So fluorine goes from \(\displaystyle -1\) to \(\displaystyle 0\): each F atom loses $\displaystyle 1$ electron, and two F atoms lose $\displaystyle 2$ electrons between them. Losing electrons is oxidation, so \(\displaystyle F^-\) is oxidised — it is the reducing agent: \[2F^- \rightarrow F_2 + 2e^- \]To combine two half-reactions you normally have to scale each one so the electron counts match before adding — that's the step people skip and then wonder why their final equation won't balance. Here it's already a match: $\displaystyle 2$ electrons gained equals $\displaystyle 2$ electrons lost, so the half-reactions add directly with nothing left over: \[\text{XeO}_6^{4-} + 2F^- + 6H^+ \rightarrow \text{XeO}_3 + F_2 + 3H_2O \]Check it the way you'd check any balanced equation — atoms of each element, then charge, on both sides: Xe: \(\displaystyle 1 = 1\). O: \(\displaystyle 6\) on the left, and \(\displaystyle 3\;(\text{from XeO}_3) + 3\;(\text{from } 3H_2O) = 6\) on the right. F: \(\displaystyle 2 = 2\). H: \(\displaystyle 6 = 6\;(\text{from } 3H_2O)\). Charge on the left: \(\displaystyle (-4) + 2(-1) + 6(+1) = -4-2+6 = 0\); charge on the right: all three products are neutral, so \(\displaystyle 0\). Mass balances and charge balances, so this is exactly the equation given, and it is a genuine redox reaction, not just a rearrangement of formulas.Now the "why" the question is actually asking for. Fluoride ion is notoriously hard to oxidise — fluorine is the most electronegative element there is, so \(\displaystyle F^-\) clings to its extra electron more tightly than almost any other ion, which is precisely why \(\displaystyle F_2\)/\(\displaystyle F^-\) is normally used as one of the strongest oxidisers available (turning \(\displaystyle F^-\) back into \(\displaystyle F_2\) usually takes something extreme, like electrolysis). Here the roles are reversed: \(\displaystyle \text{XeO}_6^{4-} \) is the one doing the oxidising, pulling electrons off \(\displaystyle F^-\) hard enough to release \(\displaystyle F_2\) gas. For that to happen, the \(\displaystyle \text{XeO}_6^{4-}/\text{XeO}_3 \) couple has to be an even stronger oxidiser than the \(\displaystyle F_2/F^-\) couple itself.That is the conclusion the reaction hands you about Na₄XeO₆: it isn't just "an oxidising agent" in some mild sense — the fact that it can strip electrons from an ion as reluctant as \(\displaystyle F^-\) marks it as one of the most powerful oxidising agents known, stronger even than elemental fluorine under these conditions.**Answer: In this reaction, Xe in \(\displaystyle \text{XeO}_6^{4-} \) is reduced from \(\displaystyle +8\) to \(\displaystyle +6\) (in \(\displaystyle \text{XeO}_3 \)), gaining $\displaystyle 2$ electrons, while $\displaystyle 2$ \(\displaystyle F^-\) ions are oxidised to \(\displaystyle F_2\), losing $\displaystyle 2$ electrons — the electron counts match, which is why the equation balances as written. Because \(\displaystyle \text{XeO}_6^{4-} \) is able to oxidise \(\displaystyle F^-\) (normally very hard to oxidise) all the way to \(\displaystyle F_2\), the reaction shows that Na₄XeO₆ (sodium perxenate, containing Xe in its rare \(\displaystyle +8\) oxidation state) is an exceptionally powerful oxidising agent — one of the strongest chemical oxidants known.

Both \(\displaystyle \mathrm{AgNO_3}\) and \(\displaystyle \mathrm{CuSO_4}\) oxidise \(\displaystyle \mathrm{H_3PO_2}\) equally well — the difference between silver and copper only shows up when you test them on an aromatic aldehyde, and that is what tells you which one is the stronger oxidising agent.Step $\displaystyle 1$ — find out who is oxidised, using oxidation numbers.For a neutral molecule, the oxidation numbers of all the atoms, each multiplied by how many times it occurs, must add to zero: \[\sum(\text{no. of atoms})\times(\text{oxidation number}) = 0 \] Take \(\displaystyle \mathrm{H}=+1\) and \(\displaystyle \mathrm{O}=-2\) by the usual convention.In \(\displaystyle \mathrm{H_3PO_2}\), with P's oxidation number as \(\displaystyle x\): \[3(+1)+x+2(-2)=0 \ \Rightarrow\ x=+1 \] In \(\displaystyle \mathrm{H_3PO_4}\): \[3(+1)+x+4(-2)=0 \ \Rightarrow\ x=+5 \] Aside — the step people get wrong here: \(\displaystyle \mathrm{H_3PO_2}\) is actually a monobasic acid — only one of its three hydrogens (the P–O–H proton) is acidic; the other two sit directly on phosphorus. That doesn't matter for this bookkeeping method: you still put in all three H's as \(\displaystyle +1\) when solving for phosphorus's number. The "+$\displaystyle 1$" you get for P is a formal average, not a claim about which bonds are which.So phosphorus goes from \(\displaystyle +1\) to \(\displaystyle +5\) — a loss of $\displaystyle 4$ electrons per \(\displaystyle \mathrm{H_3PO_2}\) molecule, i.e. \(\displaystyle \mathrm{H_3PO_2}\) is the species being oxidised in both (a) and (b), and its oxidation half-reaction is \[\mathrm{H_3PO_2 + 2H_2O \rightarrow H_3PO_4 + 4H^+ + 4e^-} \] (check: $\displaystyle 7$ H, $\displaystyle 1$ P, $\displaystyle 4$ O on each side; charge $\displaystyle 0$ on the left, \(\displaystyle 4(+1)+4(-1)=0\) on the right — balanced.)Step $\displaystyle 2$ — reaction (a): silver as the oxidant.In \(\displaystyle \mathrm{AgNO_3}\), the nitrate ion \(\displaystyle \mathrm{NO_3^-}\) carries charge \(\displaystyle -1\), so silver must be \(\displaystyle \mathrm{Ag^+}\), oxidation number \(\displaystyle +1\). Reduced to the metal, \(\displaystyle \mathrm{Ag(s)}\), its oxidation number is $\displaystyle 0$ — a gain of $\displaystyle 1$ electron per Ag atom: \[\mathrm{Ag^+ + e^- \rightarrow Ag} \] Since the oxidation step releases $\displaystyle 4$ electrons, you need $\displaystyle 4$ of these reduction steps to soak them up: \[\mathrm{4Ag^+ + 4e^- \rightarrow 4Ag} \] Adding this to the oxidation half-reaction (the \(\displaystyle 4e^-\) cancel) and dressing the ions back up with their spectator counter-ions (\(\displaystyle \mathrm{NO_3^-}\) pairing with the \(\displaystyle 4\mathrm{H^+}\) to give \(\displaystyle 4\mathrm{HNO_3}\)) reproduces exactly the equation you were given: \[\mathrm{H_3PO_2(aq) + 4AgNO_3(aq) + 2H_2O(l) \rightarrow H_3PO_4(aq) + 4Ag(s) + 4HNO_3(aq)} \] Atom count both sides: H = $\displaystyle 7$, P = $\displaystyle 1$, O = $\displaystyle 16$, Ag = $\displaystyle 4$, N = $\displaystyle 4$ — balanced.Step $\displaystyle 3$ — reaction (b): copper as the oxidant.In \(\displaystyle \mathrm{CuSO_4}\), sulfate \(\displaystyle \mathrm{SO_4^{2-}}\) carries \(\displaystyle -2\), so copper is \(\displaystyle \mathrm{Cu^{2+}}\). Reduced to \(\displaystyle \mathrm{Cu(s)}\), each copper atom gains $\displaystyle 2$ electrons: \[\mathrm{Cu^{2+}+2e^- \rightarrow Cu} \] To absorb the same $\displaystyle 4$ electrons you now need only $\displaystyle 2$ copper ions: \[\mathrm{2Cu^{2+}+4e^-\rightarrow 2Cu} \] Add this to the oxidation half-reaction (the \(\displaystyle 4e^-\) cancel again): \[\mathrm{H_3PO_2+2H_2O+2Cu^{2+}\rightarrow H_3PO_4+4H^++2Cu} \] Now dress the ions: the $\displaystyle 2$ \(\displaystyle \mathrm{CuSO_4}\) bring in two sulfate ions, and each \(\displaystyle \mathrm{H_2SO_4}\) needs one sulfate plus two \(\displaystyle \mathrm{H^+}\) — so those $\displaystyle 2$ sulfates and $\displaystyle 4$ \(\displaystyle \mathrm{H^+}\) pair up into two molecules of sulfuric acid, not one: \[\mathrm{H_3PO_2(aq) + 2CuSO_4(aq) + 2H_2O(l) \rightarrow H_3PO_4(aq) + 2Cu(s) + 2H_2SO_4(aq)} \] Aside — the step people get wrong: matching only the electron count ($\displaystyle 4$ lost = $\displaystyle 4$ gained) is not enough to balance a full molecular equation; you still have to carry every spectator atom through. Here the sulfur count forces the coefficient $\displaystyle 2$ in front of \(\displaystyle \mathrm{H_2SO_4}\) — check: H = $\displaystyle 3$ + $\displaystyle 4$ = $\displaystyle 7$ on the left, $\displaystyle 3$ + $\displaystyle 4$ = $\displaystyle 7$ on the right; O = $\displaystyle 2$ + $\displaystyle 8$ + $\displaystyle 2$ = $\displaystyle 12$ on the left, $\displaystyle 4$ + $\displaystyle 8$ = $\displaystyle 12$ on the right; S = $\displaystyle 2$ on both sides.So far \(\displaystyle \mathrm{Ag^+}\) and \(\displaystyle \mathrm{Cu^{2+}}\) look interchangeable: both fully oxidise \(\displaystyle \mathrm{H_3PO_2}\) to \(\displaystyle \mathrm{H_3PO_4}\), each being reduced all the way to the free metal.Step $\displaystyle 4$ — reaction (c): the test that tells them apart.Assign the oxidation number of the carbonyl carbon by counting its bonds: a bond to a more electronegative atom (O) contributes \(\displaystyle +1\) per bond (a double bond counts twice), a bond to a less electronegative atom (H) contributes \(\displaystyle -1\), and a bond to another carbon contributes 0.In the aldehyde group \(\displaystyle -\mathrm{CHO}\) of benzaldehyde, \(\displaystyle \mathrm{C_6H_5CHO}\): one C=O double bond (\(\displaystyle +2\)), one C–H bond (\(\displaystyle -1\)), one C–C bond to the ring ($\displaystyle 0$). Total \(\displaystyle =+1\).In the carboxylate group \(\displaystyle -\mathrm{COO^-}\) of the benzoate ion, \(\displaystyle \mathrm{C_6H_5COO^-}\): one C=O double bond (\(\displaystyle +2\)), one C–O(single) bond (\(\displaystyle +1\)), one C–C bond to the ring ($\displaystyle 0$). Total \(\displaystyle =+3\).So the carbonyl carbon goes from \(\displaystyle +1\) to \(\displaystyle +3\) — a loss of $\displaystyle 2$ electrons: benzaldehyde is oxidised to the benzoate ion (the carboxylate salt of benzoic acid). In basic solution: \[\mathrm{C_6H_5CHO + 3OH^- \rightarrow C_6H_5COO^- + 2H_2O + 2e^-} \] (check: C = $\displaystyle 7$, H = $\displaystyle 9$, O = $\displaystyle 4$ on both sides; charge \(\displaystyle -3\) on the left, \(\displaystyle (-1)+0+(-2)=-3\) on the right.)In \(\displaystyle \mathrm{[Ag(NH_3)_2]^+}\), the two ammonia ligands are neutral (an aside worth remembering: a neutral ligand contributes nothing to the metal's oxidation number, so the whole \(\displaystyle +1\) charge of the complex sits on silver — complexing Ag⁺ with ammonia changes how reactive it is, not what oxidation state it's in). Reduced to \(\displaystyle \mathrm{Ag(s)}\), each silver gains $\displaystyle 1$ electron, releasing its ammonias: \[\mathrm{[Ag(NH_3)_2]^+ + e^- \rightarrow Ag + 2NH_3} \] Two of these absorb the $\displaystyle 2$ electrons released by the aldehyde: \[\mathrm{2[Ag(NH_3)_2]^+ + 2e^- \rightarrow 2Ag + 4NH_3} \] Adding the two half-reactions (the \(\displaystyle 2e^-\) cancel) reproduces exactly: \[\mathrm{C_6H_5CHO(l) + 2[Ag(NH_3)_2]^+(aq) + 3OH^-(aq) \rightarrow C_6H_5COO^-(aq) + 2Ag(s) + 4NH_3(aq) + 2H_2O(l)} \] Atom and charge count both sides: C = $\displaystyle 7$, H = $\displaystyle 21$, O = $\displaystyle 4$, N = $\displaystyle 4$, Ag = $\displaystyle 2$; charge \(\displaystyle -1\) on both sides — balanced. This is the classic Tollens' test (silver-mirror test): the metallic silver deposited is literally the \(\displaystyle 2\mathrm{Ag(s)}\) on the right.Now try the same aldehyde with \(\displaystyle \mathrm{Cu^{2+}}\) in base: \[\mathrm{C_6H_5CHO(l) + 2Cu^{2+}(aq) + 5OH^-(aq) \rightarrow \text{no observable reaction}} \] No electrons move, so there is nothing to balance — the carbonyl carbon stays at \(\displaystyle +1\), copper stays at \(\displaystyle +2\). Copper simply cannot pull electrons off this particular aldehyde.Step $\displaystyle 5$ — the inference.\(\displaystyle \mathrm{Ag^+}\) and \(\displaystyle \mathrm{Cu^{2+}}\) are both perfectly capable oxidising agents against \(\displaystyle \mathrm{H_3PO_2}\) — they oxidise it to \(\displaystyle \mathrm{H_3PO_4}\) with the identical $\displaystyle 4$-electron transfer, being reduced to the free metal in each case. But when the substrate is changed to benzaldehyde, an aromatic aldehyde, only \(\displaystyle \mathrm{Ag^+}\) (as the diammine complex) can still strip $\displaystyle 2$ electrons off the carbonyl carbon to give the benzoate ion and a silver mirror; \(\displaystyle \mathrm{Cu^{2+}}\) cannot touch it. The benzene ring donates electron density into the carbonyl by resonance, making an aromatic aldehyde a weaker reducing agent than an aliphatic one — so it takes a stronger oxidant to pull electrons out of it. Since \(\displaystyle \mathrm{Ag^+}\) succeeds where \(\displaystyle \mathrm{Cu^{2+}}\) fails, \(\displaystyle \mathrm{Ag^+}\) must be the stronger of the two oxidising agents. (This is exactly why Tollens' reagent, not a Cu²⁺-based test, is used to identify aromatic aldehydes.)Answer: Ag⁺ is the stronger oxidising agent of the two. Both Ag⁺ (in \(\displaystyle \mathrm{AgNO_3}\)) and Cu²⁺ (in \(\displaystyle \mathrm{CuSO_4}\)) oxidise \(\displaystyle \mathrm{H_3PO_2}\) (P: \(\displaystyle +1\to+5\), $\displaystyle 4$ electrons) to \(\displaystyle \mathrm{H_3PO_4}\) equally well, being reduced to the free metal — $\displaystyle 4$ Ag⁺ each gaining $\displaystyle 1$ electron, or $\displaystyle 2$ Cu²⁺ each gaining $\displaystyle 2$ electrons. But with the aromatic aldehyde benzaldehyde, only Ag⁺ (as \(\displaystyle \mathrm{[Ag(NH_3)_2]^+}\)) can oxidise the carbonyl carbon (\(\displaystyle +1\to+3\)) to the benzoate ion, depositing metallic silver (the Tollens'/silver-mirror test); Cu²⁺ produces no reaction at all with benzaldehyde. So Ag⁺ can oxidise both aliphatic (\(\displaystyle \mathrm{H_3PO_2}\)) and aromatic (\(\displaystyle \mathrm{C_6H_5CHO}\)) reducing agents, while Cu²⁺ manages only the aliphatic one — Ag⁺ is therefore the stronger, more broadly capable oxidising agent.

The ion–electron (half-reaction) method never asks you to guess coefficients — it splits the reaction into a loss-of-electrons half and a gain-of-electrons half, balances each one completely on its own, and only then adds them together.For every part below the recipe is the same five moves on each half-reaction: 1. Balance every atom except O and H. 2. Balance O by adding \(\displaystyle \text{H}_2\text{O}\) to the side short of oxygen. 3. Balance H by adding \(\displaystyle \text{H}^+\) to the side short of hydrogen (this is done first as if the medium were acidic). 4. Balance charge by adding electrons \(\displaystyle e^-\) to the more positive side. 5. If the medium is basic, add enough \(\displaystyle \text{OH}^-\) to both sides to cancel every \(\displaystyle \text{H}^+\) you introduced (since \(\displaystyle \text{H}^+ + \text{OH}^- \to \text{H}_2\text{O}\)), then merge any \(\displaystyle \text{H}_2\text{O}\) that appears on both sides.A step people get wrong: the number of electrons you add in step $\displaystyle 4$ must make the total charge equal on both sides — check it by adding up every ion's charge, not just "does it look balanced."---(a) \(\displaystyle \text{MnO}_4^-(aq) + \text{I}^-(aq) \rightarrow \text{MnO}_2(s) + \text{I}_2(s)\) — basic mediumReduction half (Mn goes from \(\displaystyle +7\) to \(\displaystyle +4\), so it gains \(\displaystyle 3e^-\)): \[\text{MnO}_4^- \rightarrow \text{MnO}_2 \] Balance O with water ($\displaystyle 4$ O on the left, $\displaystyle 2$ on the right, so add $\displaystyle 2$ \(\displaystyle \text{H}_2\text{O}\) to the right): \[\text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \] Balance H with \(\displaystyle \text{H}^+\) (right side has $\displaystyle 4$ H, so add $\displaystyle 4$ \(\displaystyle \text{H}^+\) to the left): \[\text{MnO}_4^- + 4\text{H}^+ \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \] Balance charge (left is \(\displaystyle -1+4=+3\), right is \(\displaystyle 0\), so add \(\displaystyle 3e^-\) to the left): \[\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \] Now convert to basic medium — add $\displaystyle 4$ \(\displaystyle \text{OH}^-\) to both sides to soak up the $\displaystyle 4$ \(\displaystyle \text{H}^+\): \[\text{MnO}_4^- + 4\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} + 4\text{OH}^- \] Cancel the $\displaystyle 2$ \(\displaystyle \text{H}_2\text{O}\) common to both sides: \[\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \quad \text{...(i)} \]Oxidation half (I goes from \(\displaystyle -1\) to \(\displaystyle 0\), losing \(\displaystyle 1e^-\) per atom, and iodine leaves as the diatomic molecule \(\displaystyle \text{I}_2\)): \[2\text{I}^- \rightarrow \text{I}_2 + 2e^- \quad \text{...(ii)} \]Equation (i) carries $\displaystyle 3$ electrons, (ii) carries 2. The lowest common multiple is $\displaystyle 6$, so multiply (i) by $\displaystyle 2$ and (ii) by $\displaystyle 3$: \[2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- \] \[6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \] Add the two (the \(\displaystyle 6e^-\) cancel): \[2\text{MnO}_4^-(aq) + 4\text{H}_2\text{O}(l) + 6\text{I}^-(aq) \rightarrow 2\text{MnO}_2(s) + 8\text{OH}^-(aq) + 3\text{I}_2(s) \] Check: Mn \(\displaystyle 2=2\); O \(\displaystyle 8+4=12\) both sides; H \(\displaystyle 8=8\); I \(\displaystyle 6=6\); charge left \(\displaystyle 2(-1)+6(-1)=-8\), right \(\displaystyle 8(-1)=-8\). The product manganese(IV) oxide and molecular iodine are formed.---(b) \(\displaystyle \text{MnO}_4^-(aq) + \text{SO}_2(g) \rightarrow \text{Mn}^{2+}(aq) + \text{HSO}_4^-(aq)\) — acidic solutionReduction half (Mn: \(\displaystyle +7 \to +2\), gain \(\displaystyle 5e^-\)): \[\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad (\text{4 O on the left needs 4 } \text{H}_2\text{O}) \] Balance H with $\displaystyle 8$ \(\displaystyle \text{H}^+\) on the left: \[\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \] Charge left \(\displaystyle =-1+8=+7\), right \(\displaystyle =+2\); add \(\displaystyle 5e^-\) to the left: \[\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \text{...(iii)} \]Oxidation half (S: \(\displaystyle +4 \to +6\) inside \(\displaystyle \text{HSO}_4^-\), lose \(\displaystyle 2e^-\)): \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- \] ($\displaystyle 2$ O in \(\displaystyle \text{SO}_2\) plus $\displaystyle 2$ from the water gives the $\displaystyle 4$ O needed in \(\displaystyle \text{HSO}_4^-\)). Balance H — left has $\displaystyle 4$ H, right has $\displaystyle 1$, so put $\displaystyle 3$ \(\displaystyle \text{H}^+\) on the right: \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ \] Charge left \(\displaystyle =0\), right \(\displaystyle =-1+3=+2\); add \(\displaystyle 2e^-\) to the right: \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^- \quad \text{...(iv)} \]LCM of $\displaystyle 5$ and $\displaystyle 2$ is $\displaystyle 10$, so multiply (iii) by $\displaystyle 2$ and (iv) by $\displaystyle 5$: \[2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \] \[5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 15\text{H}^+ + 10e^- \] Adding and cancelling the electrons, and then cancelling the \(\displaystyle \text{H}^+\) and \(\displaystyle \text{H}_2\text{O}\) that appear on both sides ($\displaystyle 16$ \(\displaystyle \text{H}^+\) left minus $\displaystyle 15$ \(\displaystyle \text{H}^+\) right leaves $\displaystyle 1$ \(\displaystyle \text{H}^+\) on the left; $\displaystyle 10$ \(\displaystyle \text{H}_2\text{O}\) left minus $\displaystyle 8$ \(\displaystyle \text{H}_2\text{O}\) right leaves $\displaystyle 2$ \(\displaystyle \text{H}_2\text{O}\) on the left): \[2\text{MnO}_4^-(aq) + 5\text{SO}_2(g) + 2\text{H}_2\text{O}(l) + \text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 5\text{HSO}_4^-(aq) \] Check: Mn \(\displaystyle 2=2\); S \(\displaystyle 5=5\); O: left \(\displaystyle 8+10+2=20\), right \(\displaystyle 5\times4=20\); H: left \(\displaystyle 1+4=5\), right \(\displaystyle 5\times1=5\); charge: left \(\displaystyle -2+1=-1\), right \(\displaystyle 4-5=-1\). Manganese(II) ion and the hydrogensulfate ion are the products — do not simplify \(\displaystyle \text{HSO}_4^-\) to \(\displaystyle \text{SO}_4^{2-}\), the question names \(\displaystyle \text{HSO}_4^-\) explicitly because the medium is acidic and one H stays attached.---(c) \(\displaystyle \text{H}_2\text{O}_2(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{H}_2\text{O}(l)\) — acidic solutionReduction half (each O in \(\displaystyle \text{H}_2\text{O}_2\) is at \(\displaystyle -1\); in \(\displaystyle \text{H}_2\text{O}\) it is at \(\displaystyle -2\), so the $\displaystyle 2$ oxygens together gain \(\displaystyle 2e^-\)): \[\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} \] O is already balanced ($\displaystyle 2$ = $\displaystyle 2$). Balance H — right has $\displaystyle 4$ H, left has $\displaystyle 2$, so add $\displaystyle 2$ \(\displaystyle \text{H}^+\) to the left: \[\text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} \] Charge left \(\displaystyle =0+2=+2\), right \(\displaystyle =0\); add \(\displaystyle 2e^-\) to the left: \[\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O} \quad \text{...(v)} \]Oxidation half (Fe: \(\displaystyle +2 \to +3\), loses \(\displaystyle 1e^-\); no O or H involved): \[\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \] Multiply by $\displaystyle 2$ so its electrons match (v)'s $\displaystyle 2$ electrons: \[2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^- \quad \text{...(vi)} \]Add (v) and (vi): \[\text{H}_2\text{O}_2(aq) + 2\text{H}^+(aq) + 2\text{Fe}^{2+}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2\text{H}_2\text{O}(l) \] Check: H: left \(\displaystyle 2+2=4\), right \(\displaystyle 2\times2=4\); O: left \(\displaystyle 2\), right \(\displaystyle 2\); Fe \(\displaystyle 2=2\); charge: left \(\displaystyle 2+4=6\), right \(\displaystyle 6\). Iron(II) is oxidised to iron(III) while hydrogen peroxide is reduced to water — the H\(\displaystyle _2\)O\(\displaystyle _2\) here is the oxidising agent, the opposite of its usual textbook role as a reducing agent, which is exactly why this equation is worth checking by oxidation number, not by memory.---(d) \(\displaystyle \text{Cr}_2\text{O}_7^{2-}(aq) + \text{SO}_2(g) \rightarrow \text{Cr}^{3+}(aq) + \text{SO}_4^{2-}(aq)\) — acidic solutionReduction half (Cr: \(\displaystyle +6 \to +3\) for each of the $\displaystyle 2$ Cr atoms, so \(\displaystyle 6e^-\) total): \[\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \] (Cr already balanced, $\displaystyle 2$ = 2.) Balance O with $\displaystyle 7$ \(\displaystyle \text{H}_2\text{O}\) on the right: \[\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Balance H with $\displaystyle 14$ \(\displaystyle \text{H}^+\) on the left: \[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Charge left \(\displaystyle =-2+14=+12\), right \(\displaystyle =2(+3)=+6\); add \(\displaystyle 6e^-\) to the left: \[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \quad \text{...(vii)} \]Oxidation half (S: \(\displaystyle +4 \to +6\), loses \(\displaystyle 2e^-\)): \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} \] ($\displaystyle 2$ O from \(\displaystyle \text{SO}_2\) plus $\displaystyle 2$ from water give the $\displaystyle 4$ O in \(\displaystyle \text{SO}_4^{2-}\)). Balance H with $\displaystyle 4$ \(\displaystyle \text{H}^+\) on the right: \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ \] Charge left \(\displaystyle =0\), right \(\displaystyle =-2+4=+2\); add \(\displaystyle 2e^-\) to the right: \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \quad \text{...(viii)} \]LCM of $\displaystyle 6$ and $\displaystyle 2$ is $\displaystyle 6$, so keep (vii) as is and multiply (viii) by $\displaystyle 3$: \[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] \[3\text{SO}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 6e^- \] Add and cancel electrons; then cancel \(\displaystyle \text{H}^+\) ($\displaystyle 14$ left minus $\displaystyle 12$ right leaves $\displaystyle 2$ \(\displaystyle \text{H}^+\) on the left) and \(\displaystyle \text{H}_2\text{O}\) ($\displaystyle 7$ right minus $\displaystyle 6$ left leaves $\displaystyle 1$ \(\displaystyle \text{H}_2\text{O}\) on the right): \[\text{Cr}_2\text{O}_7^{2-}(aq) + 3\text{SO}_2(g) + 2\text{H}^+(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{SO}_4^{2-}(aq) + \text{H}_2\text{O}(l) \] Check: Cr \(\displaystyle 2=2\); S \(\displaystyle 3=3\); O: left \(\displaystyle 7+6=13\), right \(\displaystyle 12+1=13\); H: left \(\displaystyle 2\), right \(\displaystyle 2\); charge: left \(\displaystyle -2+2=0\), right \(\displaystyle 6-6=0\). The dichromate ion oxidises sulfur dioxide to the sulfate ion while itself being reduced to chromium(III) ion.**Answer: (a) \(\displaystyle 2\text{MnO}_4^-+4\text{H}_2\text{O}+6\text{I}^- \rightarrow 2\text{MnO}_2+8\text{OH}^-+3\text{I}_2\); (b) \(\displaystyle 2\text{MnO}_4^-+5\text{SO}_2+2\text{H}_2\text{O}+\text{H}^+ \rightarrow 2\text{Mn}^{2+}+5\text{HSO}_4^-\); (c) \(\displaystyle \text{H}_2\text{O}_2+2\text{H}^++2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+}+2\text{H}_2\text{O}\); (d) \(\displaystyle \text{Cr}_2\text{O}_7^{2-}+3\text{SO}_2+2\text{H}^+ \rightarrow 2\text{Cr}^{3+}+3\text{SO}_4^{2-}+\text{H}_2\text{O}\).

A disproportionation and two more redox changes — balance each by tracking electrons twice, once per half-reaction (ion-electron method) and once per atom's oxidation number, and both routes must land on the same equation.Part (a): \(\displaystyle \mathrm{P_4(s) + OH^-(aq) \rightarrow PH_3(g) + H_2PO_2^-(aq)} \)(The hypophosphite ion needs two hydrogens to balance its charge, \(\displaystyle \mathrm{H_2PO_2^-} \), not \(\displaystyle \mathrm{HPO_2^-} \) — the oxidation-number check below confirms this.)First find every oxidation number, so you know which phosphorus atoms are oxidised and which are reduced. So phosphorus in \(\displaystyle \mathrm{P_4}\) splits two ways: some atoms go \(\displaystyle 0 \rightarrow -3\) (gain \(\displaystyle 3e^-\), reduced) and some go \(\displaystyle 0 \rightarrow +1\) (lose \(\displaystyle 1e^-\), oxidised). This is a disproportionation — the same element, starting in the same molecule, is both oxidised and reduced — so \(\displaystyle \mathrm{P_4} \) is its own oxidising and its own reducing agent.Ion-electron method. Balance each half-reaction on its own first, using \(\displaystyle \mathrm{H_2O}\) to supply oxygen, \(\displaystyle \mathrm{H^+}\) to supply hydrogen, and electrons to balance charge (the acidic-medium routine), then add \(\displaystyle \mathrm{OH^-}\) to both sides to turn every \(\displaystyle \mathrm{H^+}\) into \(\displaystyle \mathrm{H_2O}\) — that conversion is what "basic medium" means for this method.Oxidation half (take a full \(\displaystyle \mathrm{P_4}\) as the reference; the true split gets fixed when the halves are combined below): \[\mathrm{P_4 + 8H_2O \rightarrow 4H_2PO_2^- + 8H^+ + 4e^-} \] Adding \(\displaystyle 8\mathrm{OH^-}\) to both sides turns the \(\displaystyle 8\mathrm{H^+}\) into \(\displaystyle 8\mathrm{H_2O}\), which exactly cancels the \(\displaystyle 8\mathrm{H_2O}\) already on the left: \[\mathrm{P_4 + 8OH^- \rightarrow 4H_2PO_2^- + 4e^-} \]Reduction half: \[\mathrm{P_4 + 12H^+ + 12e^- \rightarrow 4PH_3} \] Adding \(\displaystyle 12\mathrm{OH^-}\) to both sides converts the \(\displaystyle 12\mathrm{H^+}\) into \(\displaystyle 12\mathrm{H_2O}\): \[\mathrm{P_4 + 12H_2O + 12e^- \rightarrow 4PH_3 + 12OH^-} \]The oxidation step releases \(\displaystyle 4e^-\); the reduction step needs \(\displaystyle 12e^-\). Multiply the oxidation half by $\displaystyle 3$ so both sides move the same $\displaystyle 12$ electrons — this is the step people skip: every half-reaction must be scaled so the electron counts match exactly before adding them, or the final equation is left holding loose electrons. \[\mathrm{3P_4 + 24OH^- \rightarrow 12H_2PO_2^- + 12e^-} \] Adding this to the reduction half, the \(\displaystyle 12e^-\) cancel, \(\displaystyle 12\mathrm{OH^-}\) common to both sides cancel, and every coefficient divides by $\displaystyle 4$ (the \(\displaystyle 4P_4\) on the left, from \(\displaystyle 3+1\), becomes \(\displaystyle 1P_4\)): \[\mathrm{P_4(s) + 3OH^-(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3H_2PO_2^-(aq)} \]Check it: phosphorus \(\displaystyle 4=1+3\); oxygen \(\displaystyle 3+3=6\) on the left, \(\displaystyle 3\times2=6\) on the right; hydrogen \(\displaystyle 3+6=9\) on the left, \(\displaystyle 3+3\times2=9\) on the right; charge \(\displaystyle 3\times(-1)=-3\) on the left, \(\displaystyle 3\times(-1)=-3\) on the right.Oxidation number method. Phosphorus reduced: \(\displaystyle 0\rightarrow-3\), a gain of \(\displaystyle 3e^-\) per atom. Phosphorus oxidised: \(\displaystyle 0\rightarrow+1\), a loss of \(\displaystyle 1e^-\) per atom. For electrons lost to equal electrons gained you need $\displaystyle 3$ oxidised atoms for every $\displaystyle 1$ reduced atom, and \(\displaystyle 3+1=4\) is exactly the atom count of one \(\displaystyle \mathrm{P_4}\) molecule — so one whole \(\displaystyle \mathrm{P_4}\) supplies both products: $\displaystyle 1$ atom becomes \(\displaystyle \mathrm{PH_3}\), $\displaystyle 3$ atoms become \(\displaystyle \mathrm{H_2PO_2^-}\). Filling in \(\displaystyle \mathrm{OH^-}\) and \(\displaystyle \mathrm{H_2O}\) to balance the remaining oxygen, hydrogen and charge — exactly as above — gives the same equation: \[\mathrm{P_4(s) + 3OH^-(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3H_2PO_2^-(aq)} \]Part (b): \(\displaystyle \mathrm{N_2H_4(l) + ClO_3^-(aq) \rightarrow NO(g) + Cl^-(aq)} \)(The chloride product is dissolved in solution, \(\displaystyle \mathrm{Cl^-(aq)}\), not a gas.)Oxidation numbers: in \(\displaystyle \mathrm{N_2H_4}\) (hydrazine), \(\displaystyle 2x + 4(+1) = 0\) gives nitrogen \(\displaystyle =-2\). In \(\displaystyle \mathrm{NO}\) (nitric oxide), \(\displaystyle x+(-2) = 0\) gives nitrogen \(\displaystyle =+2\). Nitrogen goes \(\displaystyle -2\rightarrow+2\), losing \(\displaystyle 4e^-\) per atom — hydrazine is oxidised, so it is the reducing agent. In \(\displaystyle \mathrm{ClO_3^-}\) (chlorate ion), \(\displaystyle x+3(-2)=-1\) gives chlorine \(\displaystyle =+5\); in \(\displaystyle \mathrm{Cl^-}\) (chloride ion), chlorine \(\displaystyle =-1\). Chlorine goes \(\displaystyle +5\rightarrow-1\), gaining \(\displaystyle 6e^-\) per atom — chlorate is reduced, so it is the oxidising agent.Ion-electron method. Balance each half in acid form first, then add \(\displaystyle \mathrm{OH^-}\) to both sides to remove every \(\displaystyle \mathrm{H^+}\).Oxidation half: \[\mathrm{N_2H_4 + 2H_2O \rightarrow 2NO + 8H^+ + 8e^-} \] Adding \(\displaystyle 8\mathrm{OH^-}\) to both sides turns the \(\displaystyle 8\mathrm{H^+}\) into \(\displaystyle 8\mathrm{H_2O}\), leaving $\displaystyle 6$ water molecules net on the product side: \[\mathrm{N_2H_4 + 8OH^- \rightarrow 2NO + 6H_2O + 8e^-} \]Reduction half: \[\mathrm{ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O} \] Adding \(\displaystyle 6\mathrm{OH^-}\) to both sides turns the \(\displaystyle 6\mathrm{H^+}\) into \(\displaystyle 6\mathrm{H_2O}\), leaving $\displaystyle 3$ water molecules net on the reactant side: \[\mathrm{ClO_3^- + 3H_2O + 6e^- \rightarrow Cl^- + 6OH^-} \]One \(\displaystyle \mathrm{N_2H_4}\) releases \(\displaystyle 8e^-\) ($\displaystyle 2$ nitrogens \(\displaystyle \times4\)); one \(\displaystyle \mathrm{ClO_3^-}\) absorbs \(\displaystyle 6e^-\). The lowest common multiple of $\displaystyle 8$ and $\displaystyle 6$ is $\displaystyle 24$, so multiply the oxidation half by $\displaystyle 3$ and the reduction half by $\displaystyle 4$: \[\mathrm{3N_2H_4 + 24OH^- \rightarrow 6NO + 18H_2O + 24e^-} \] \[\mathrm{4ClO_3^- + 12H_2O + 24e^- \rightarrow 4Cl^- + 24OH^-} \] Adding these, the \(\displaystyle 24e^-\) cancel, the \(\displaystyle 24\mathrm{OH^-}\) cancel, and $\displaystyle 12$ of the $\displaystyle 18$ water molecules on the right cancel against the $\displaystyle 12$ on the left, leaving $\displaystyle 6$ net: \[\mathrm{3N_2H_4(l) + 4ClO_3^-(aq) \rightarrow 6NO(g) + 4Cl^-(aq) + 6H_2O(l)} \]Check it: nitrogen \(\displaystyle 6=6\); chlorine \(\displaystyle 4=4\); oxygen \(\displaystyle 4\times3=12\) on the left, \(\displaystyle 6+6=12\) on the right; hydrogen \(\displaystyle 3\times4=12\) on the left, \(\displaystyle 6\times2=12\) on the right; charge \(\displaystyle 4\times(-1)=-4\) on both sides.Oxidation number method. Nitrogen loses \(\displaystyle 4e^-\) per atom, and \(\displaystyle \mathrm{N_2H_4}\) carries $\displaystyle 2$ nitrogens, so one hydrazine molecule releases \(\displaystyle 8e^-\). Chlorine gains \(\displaystyle 6e^-\) per atom. Equalising $\displaystyle 8$ and $\displaystyle 6$ needs multiplier $\displaystyle 3$ on hydrazine and $\displaystyle 4$ on chlorate (\(\displaystyle 3\times8=4\times6=24\)) — the same \(\displaystyle 3:4\) ratio found above. Balancing the leftover oxygen and hydrogen with water (no \(\displaystyle \mathrm{OH^-}\) or \(\displaystyle \mathrm{H^+}\) survives in the finished equation here) reproduces: \[\mathrm{3N_2H_4(l) + 4ClO_3^-(aq) \rightarrow 6NO(g) + 4Cl^-(aq) + 6H_2O(l)} \]Part (c): \(\displaystyle \mathrm{Cl_2O_7(g) + H_2O_2(aq) \rightarrow ClO_2^-(aq) + O_2(g)} \)(The question's skeleton shows a stray \(\displaystyle \mathrm{H^+}\); a true basic-medium balance carries no free \(\displaystyle \mathrm{H^+}\) in the finished equation — every \(\displaystyle \mathrm{H^+}\) is converted to \(\displaystyle \mathrm{OH^-}\)/\(\displaystyle \mathrm{H_2O}\), by the same step used in (a) and (b), shown below. This is the step to watch for: don't leave the printed \(\displaystyle \mathrm{H^+}\) sitting in a "basic medium" answer.)Oxidation numbers: in \(\displaystyle \mathrm{Cl_2O_7}\) (dichlorine heptoxide), \(\displaystyle 2x+7(-2)=0\) gives chlorine \(\displaystyle =+7\). In \(\displaystyle \mathrm{ClO_2^-}\) (chlorite ion), \(\displaystyle x+2(-2)=-1\) gives chlorine \(\displaystyle =+3\). Chlorine goes \(\displaystyle +7\rightarrow+3\), gaining \(\displaystyle 4e^-\) per atom — \(\displaystyle \mathrm{Cl_2O_7}\) is reduced, so it is the oxidising agent. In \(\displaystyle \mathrm{H_2O_2}\) (hydrogen peroxide) the oxygen is peroxide oxygen: \(\displaystyle 2(+1)+2x=0\) gives \(\displaystyle x=-1\); in \(\displaystyle \mathrm{O_2}\) oxygen is \(\displaystyle 0\). Oxygen goes \(\displaystyle -1\rightarrow0\), losing \(\displaystyle 1e^-\) per atom — \(\displaystyle \mathrm{H_2O_2}\) is oxidised, so it is the reducing agent (hydrogen peroxide handing electrons to a stronger oxidiser, the reverse of its more familiar role).Ion-electron method.Reduction half: \[\mathrm{Cl_2O_7 + 6H^+ + 8e^- \rightarrow 2ClO_2^- + 3H_2O} \] Adding \(\displaystyle 6\mathrm{OH^-}\) to both sides turns the \(\displaystyle 6\mathrm{H^+}\) into \(\displaystyle 6\mathrm{H_2O}\), leaving $\displaystyle 3$ water molecules net on the reactant side: \[\mathrm{Cl_2O_7 + 3H_2O + 8e^- \rightarrow 2ClO_2^- + 6OH^-} \]Oxidation half: \[\mathrm{H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-} \] Adding \(\displaystyle 2\mathrm{OH^-}\) to both sides turns the \(\displaystyle 2\mathrm{H^+}\) into \(\displaystyle 2\mathrm{H_2O}\): \[\mathrm{H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-} \]Reduction needs \(\displaystyle 8e^-\); oxidation supplies only \(\displaystyle 2e^-\), so multiply the oxidation half by $\displaystyle 4$: \[\mathrm{4H_2O_2 + 8OH^- \rightarrow 4O_2 + 8H_2O + 8e^-} \] Adding this to the reduction half and cancelling the \(\displaystyle 8e^-\): \(\displaystyle \mathrm{OH^-}\) stands at $\displaystyle 8$ on one side against $\displaystyle 6$ on the other, leaving $\displaystyle 2$ net; \(\displaystyle \mathrm{H_2O}\) stands at $\displaystyle 8$ on one side against $\displaystyle 3$ on the other, leaving $\displaystyle 5$ net on the product side: \[\mathrm{Cl_2O_7(g) + 4H_2O_2(aq) + 2OH^-(aq) \rightarrow 2ClO_2^-(aq) + 4O_2(g) + 5H_2O(l)} \]Check it: chlorine \(\displaystyle 2=2\); oxygen, left \(\displaystyle 7+8+2=17\), right \(\displaystyle 4+8+5=17\); hydrogen, left \(\displaystyle 8+2=10\), right \(\displaystyle 10\); charge, left \(\displaystyle -2\), right \(\displaystyle -2\).Oxidation number method. Chlorine gains \(\displaystyle 4e^-\) per atom, times $\displaystyle 2$ atoms per \(\displaystyle \mathrm{Cl_2O_7}\), so \(\displaystyle 8e^-\) gained per molecule. Oxygen loses \(\displaystyle 1e^-\) per atom, times $\displaystyle 2$ atoms per \(\displaystyle \mathrm{H_2O_2}\), so \(\displaystyle 2e^-\) lost per molecule. Matching $\displaystyle 8$ and $\displaystyle 2$ needs $\displaystyle 4$ molecules of \(\displaystyle \mathrm{H_2O_2}\) for every $\displaystyle 1$ molecule of \(\displaystyle \mathrm{Cl_2O_7}\) — the same \(\displaystyle 1:4\) ratio used above — and balancing the remaining oxygen and hydrogen with \(\displaystyle \mathrm{OH^-}\)/\(\displaystyle \mathrm{H_2O}\) (never \(\displaystyle \mathrm{H^+}\), since the medium is basic) gives the identical equation: \[\mathrm{Cl_2O_7(g) + 4H_2O_2(aq) + 2OH^-(aq) \rightarrow 2ClO_2^-(aq) + 4O_2(g) + 5H_2O(l)} \]Answer: (a) \(\displaystyle \mathrm{P_4(s) + 3OH^-(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3H_2PO_2^-(aq)} \), with \(\displaystyle \mathrm{P_4}\) as both oxidising and reducing agent (disproportionation); (b) \(\displaystyle \mathrm{3N_2H_4(l) + 4ClO_3^-(aq) \rightarrow 6NO(g) + 4Cl^-(aq) + 6H_2O(l)} \), oxidising agent \(\displaystyle \mathrm{ClO_3^-}\), reducing agent \(\displaystyle \mathrm{N_2H_4}\); (c) \(\displaystyle \mathrm{Cl_2O_7(g) + 4H_2O_2(aq) + 2OH^-(aq) \rightarrow 2ClO_2^-(aq) + 4O_2(g) + 5H_2O(l)} \), oxidising agent \(\displaystyle \mathrm{Cl_2O_7}\), reducing agent \(\displaystyle \mathrm{H_2O_2}\).

Assign oxidation numbers atom-by-atom before deciding what kind of reaction this is — a formula alone hides which atom is actually being oxidised or reduced.Step $\displaystyle 1$: Fix nitrogen and oxygen at their usual values. In every nitrogen-containing species here, nitrogen keeps its familiar value of \(\displaystyle -3 \) (as in \(\displaystyle NH_3\)), and oxygen keeps its familiar value of \(\displaystyle -2 \) (as in \(\displaystyle H_2O\)). Fixing these two lets the rule "the oxidation numbers in a species add up to that species' overall charge" solve for carbon in each species.Step $\displaystyle 2$: Find carbon's oxidation number in each species.For cyanogen, \(\displaystyle (CN)_2 \), a neutral molecule containing two carbon and two nitrogen atoms: \[2(C) + 2(-3) = 0 \implies C = +3 \]For the cyanide ion, \(\displaystyle CN^{-} \), overall charge \(\displaystyle -1\): \[C + (-3) = -1 \implies C = +2 \]For the cyanate ion, \(\displaystyle CNO^{-} \), overall charge \(\displaystyle -1\): \[C + (-3) + (-2) = -1 \implies C = +4 \]Aside: it is tempting to treat "CN" as one unbroken unit and assume nothing changes across the arrow. Oxidation numbers are worked out atom by atom, and here it is carbon alone that moves — nitrogen sits fixed at \(\displaystyle -3\) on both sides of the equation.Step $\displaystyle 3$: Read off what happened to carbon. One carbon starts at \(\displaystyle +3\) in \(\displaystyle (CN)_2\) and ends at \(\displaystyle +2\) in \(\displaystyle \mathrm{CN^{-}}\) — a gain of one electron, a reduction. The other carbon also starts at \(\displaystyle +3\) in \(\displaystyle (CN)_2\) and ends at \(\displaystyle +4\) in \(\displaystyle \mathrm{CNO^{-}}\) — a loss of one electron, an oxidation.Because the same starting species, \(\displaystyle (CN)_2\), supplies both the atom that is reduced and the atom that is oxidised, this is a disproportionation reaction (also called an auto-oxidation-reduction): \(\displaystyle (CN)_2\) acts as its own oxidising agent and its own reducing agent.Step $\displaystyle 4$: Confirm the equation is balanced, using electron gain = electron loss. Electrons gained (reduction, one carbon \(\displaystyle +3 \to +2\)) \(\displaystyle = 1\). Electrons lost (oxidation, one carbon \(\displaystyle +3 \to +4\)) \(\displaystyle = 1\). Since these are equal, one formula unit each of \(\displaystyle \mathrm{CN^{-}}\) and \(\displaystyle \mathrm{CNO^{-}}\) already balances the electron transfer, so the equation as printed, \[(CN)_2(g) + 2OH^{-}(aq) \rightarrow CN^{-}(aq) + CNO^{-}(aq) + H_2O(l) \] needs no further scaling. Checking every atom and the charge confirms it: Carbon: \(\displaystyle 2\) on the left \(\displaystyle = 1+1 = 2\) on the right. Nitrogen: \(\displaystyle 2\) on the left \(\displaystyle = 1+1 = 2\) on the right. Oxygen: \(\displaystyle 2\) on the left (from \(\displaystyle 2OH^{-}\)) \(\displaystyle = 1\) (in \(\displaystyle CNO^{-}\)) \(\displaystyle + 1\) (in \(\displaystyle H_2O\)) \(\displaystyle = 2\) on the right. Hydrogen: \(\displaystyle 2\) on the left (from \(\displaystyle 2OH^{-}\)) \(\displaystyle = 2\) (in \(\displaystyle H_2O\)) on the right. Charge: left \(\displaystyle = 0 + 2(-1) = -2\); right \(\displaystyle = (-1)+(-1)+0 = -2\). All four atom counts and the charge match on both sides, so the equation is balanced.Step $\displaystyle 5$: What this reaction tells us. Answer: The reaction is a disproportionation (auto-oxidation-reduction) of cyanogen, \(\displaystyle (CN)_2\). Carbon's oxidation number, \(\displaystyle +3\) in \(\displaystyle (CN)_2\), falls to \(\displaystyle +2\) in cyanide ion, \(\displaystyle \mathrm{CN^{-}}\) (reduction), and rises to \(\displaystyle +4\) in cyanate ion, \(\displaystyle \mathrm{CNO^{-}}\) (oxidation), while nitrogen (\(\displaystyle -3\)) and oxygen (\(\displaystyle -2\)) stay unchanged. This also identifies \(\displaystyle (CN)_2\) as a pseudohalogen, since it disproportionates in hydroxide exactly as a halogen such as \(\displaystyle Cl_2\) does.

In a disproportionation reaction the same element is simultaneously reduced and oxidised — one \(\displaystyle \text{Mn}^{3+} \) ion picks up the electron that another \(\displaystyle \text{Mn}^{3+} \) ion gives away, so \(\displaystyle \text{Mn}^{3+} \) never appears alone on the product side.Step $\displaystyle 1$ — find what happens to the oxidation number of Mn in each species.In \(\displaystyle \text{Mn}^{3+} \), manganese is \(\displaystyle +3 \) (the ion's own charge).In \(\displaystyle \text{Mn}^{2+} \), manganese is \(\displaystyle +2 \).In \(\displaystyle \text{MnO}_2 \), oxygen is \(\displaystyle -2 \) (its usual value), so if \(\displaystyle x \) is Mn's oxidation number, \[x + 2(-2) = 0 \quad\Rightarrow\quad x = +4 . \]So one \(\displaystyle \text{Mn}^{3+} \) is reduced ( \(\displaystyle +3 \to +2\), gains $\displaystyle 1$ electron) and a second \(\displaystyle \text{Mn}^{3+} \) is oxidised ( \(\displaystyle +3 \to +4\), loses $\displaystyle 1$ electron). Both changes are exactly one electron, which is the reason this equation balances so neatly — that equality of electron count is what people usually forget to check before adding half-reactions.Step $\displaystyle 2$ — write the two half-reactions.Reduction half: \[\text{Mn}^{3+} + e^- \longrightarrow \text{Mn}^{2+} \] Charge check: left \(\displaystyle +3-1=+2 \); right \(\displaystyle +2 \). Balanced.Oxidation half: going from \(\displaystyle \text{Mn}^{3+} \) to \(\displaystyle \text{MnO}_2 \) adds two oxygen atoms that must come from water (this is why \(\displaystyle \text{H}^+ \) shows up as a product — it's the medium being acidic, exactly as the question states). Two water molecules supply the two oxygens and release four \(\displaystyle \text{H}^+ \): \[\text{Mn}^{3+} + 2\text{H}_2\text{O} \longrightarrow \text{MnO}_2 + 4\text{H}^+ + e^- \] Check atoms: Mn — $\displaystyle 1$ = $\displaystyle 1$; O — $\displaystyle 2$ = $\displaystyle 2$; H — $\displaystyle 4$ = 4. ✓ Check charge: left \(\displaystyle +3 \); right \(\displaystyle 0 + 4(+1) + (-1) = +3 \). ✓ (This is the step people rush past — balance mass first, oxygen with \(\displaystyle \text{H}_2\text{O} \) and hydrogen with \(\displaystyle \text{H}^+ \), and only then confirm charge; if you try to force charge balance before the atoms are right, the coefficients come out wrong.)Step $\displaystyle 3$ — add the half-reactions.Since each half-reaction already involves exactly one electron, they can be added directly without multiplying either one: \[\text{Mn}^{3+} + e^- \longrightarrow \text{Mn}^{2+} \] \[\text{Mn}^{3+} + 2\text{H}_2\text{O} \longrightarrow \text{MnO}_2 + 4\text{H}^+ + e^- \]Adding (the \(\displaystyle e^- \) on each side cancels): \[2\text{Mn}^{3+} + 2\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+ \]Step $\displaystyle 4$ — verify.Mn atoms: left \(\displaystyle 2 \); right \(\displaystyle 1+1=2 \). ✓ O atoms: left \(\displaystyle 2 \); right \(\displaystyle 2 \) (in \(\displaystyle \text{MnO}_2 \)). ✓ H atoms: left \(\displaystyle 4 \); right \(\displaystyle 4 \). ✓ Charge: left \(\displaystyle 2(+3) = +6 \); right \(\displaystyle +2 + 0 + 4(+1) = +6 \). ✓Both mass and charge balance, so the equation is complete as written — no extra electrons or ions are needed.Answer: \(\displaystyle 2\text{Mn}^{3+} + 2\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+ \) (one \(\displaystyle \text{Mn}^{3+} \) is reduced to \(\displaystyle \text{Mn}^{2+} \), a second \(\displaystyle \text{Mn}^{3+} \) is oxidised to manganese dioxide, \(\displaystyle \text{MnO}_2 \))

Whether an element shows a positive or a negative oxidation state comes down to electronegativity: the more electronegative atom in a bond is always assigned the negative number, the less electronegative one the positive number — and an atom with a full octet already (a noble gas) usually forms no bonds at all, so it gets no oxidation number to speak of.Look at each of the four elements in terms of its electron configuration and its electronegativity relative to the other atoms it bonds with. → Fluorine shows only the negative oxidation state, \(\displaystyle -1\). → Cesium shows only the positive oxidation state, \(\displaystyle +1\). (Careful here: "loses an electron" always means the oxidation number goes up, i.e., positive — that direction is easy to get backwards.) → Iodine shows both signs of oxidation state, ranging from \(\displaystyle -1\) up to \(\displaystyle +7\). → Neon shows neither a positive nor a negative oxidation state — its oxidation number stays \(\displaystyle 0\), the same as it is in the free, uncombined atom.The one pattern tying all four together: an atom's oxidation-state range is decided by where it sits relative to a full octet (how many electrons it is "short" or "over") and by how electronegative it is compared with whatever it bonds to — not by any single fixed property of the atom alone.**Answer: (a) Fluorine (F) — only the negative state, \(\displaystyle -1\). (b) Cesium (Cs) — only the positive state, \(\displaystyle +1\). (c) Iodine (I) — both, ranging from \(\displaystyle -1\) to \(\displaystyle +7\) (e.g. \(\displaystyle -1\) in HI, \(\displaystyle +7\) in \(\displaystyle \text{HIO}_4\)). (d) Neon (Ne) — neither; it stays at oxidation number \(\displaystyle 0\) because it forms no compounds.

Excess chlorine is destroyed by making it react with sulphur dioxide dissolved in the water — chlorine gets reduced, sulphur dioxide gets oxidised, and the two electrons each side handles must match.The two gases dissolved in water react to give two acids: hydrochloric acid, \(\displaystyle \text{HCl}\), and sulphuric acid, \(\displaystyle \text{H}_2\text{SO}_4\). To balance this correctly you track the oxidation number of chlorine and of sulphur across the reaction, because that is what tells you how many electrons move — the step people skip and then get the wrong ratio of the two reactants.Step $\displaystyle 1$ — assign oxidation numbers. In \(\displaystyle \text{Cl}_2\) (an element in its free form) chlorine is at oxidation number \(\displaystyle 0\). In \(\displaystyle \text{HCl}\), hydrogen is \(\displaystyle +1\), and since the molecule is neutral, chlorine must be \(\displaystyle -1\). So chlorine goes from \(\displaystyle 0\) to \(\displaystyle -1\): each Cl atom gains one electron.In \(\displaystyle \text{SO}_2\), oxygen is \(\displaystyle -2\) each, so for the neutral molecule, \[S + 2(-2) = 0 \implies S = +4. \] In \(\displaystyle \text{H}_2\text{SO}_4\), hydrogen is \(\displaystyle +1\) each and oxygen is \(\displaystyle -2\) each, so \[2(+1) + S + 4(-2) = 0 \implies S = +6. \] So sulphur goes from \(\displaystyle +4\) to \(\displaystyle +6\): each S atom loses two electrons.Step $\displaystyle 2$ — write the two half-reactions and balance electrons.Reduction (chlorine gains electrons): \[\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \] Two electrons are needed because \(\displaystyle \text{Cl}_2\) supplies two Cl atoms, each gaining one electron. Charge check: left side \(\displaystyle 0 + (-2) = -2\); right side \(\displaystyle 2(-1) = -2\). Balanced.Oxidation (sulphur loses electrons, and in water \(\displaystyle \text{SO}_2\) is oxidised all the way to sulphate): \[\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \] Here the two water molecules supply the extra oxygens needed to turn \(\displaystyle \text{SO}_2\) ($\displaystyle 2$ oxygens) into \(\displaystyle \text{SO}_4^{2-}\) ($\displaystyle 4$ oxygens), and the $\displaystyle 4$ hydrogens released end up as \(\displaystyle 4\text{H}^+\). Charge check: left side is neutral (\(\displaystyle 0\)); right side is \(\displaystyle (-2) + 4(+1) - 2 = 0\). Balanced, and the $\displaystyle 2$ electrons on the product side match the $\displaystyle 2$-electron loss found in Step 1.Step $\displaystyle 3$ — add the half-reactions. Since both involve exactly $\displaystyle 2$ electrons, they combine with no further scaling: \[\text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} + 2e^- \rightarrow 2\text{Cl}^- + \text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \] The \(\displaystyle 2e^-\) cancels from both sides, leaving the overall ionic equation \[\text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Cl}^- + \text{SO}_4^{2-} + 4\text{H}^+. \]Step $\displaystyle 4$ — write it in molecular form. Group the \(\displaystyle 4\text{H}^+\) with the anions to form the neutral acids: \(\displaystyle 2\text{H}^+\) with \(\displaystyle \text{SO}_4^{2-}\) gives one \(\displaystyle \text{H}_2\text{SO}_4\), and the remaining \(\displaystyle 2\text{H}^+\) with \(\displaystyle 2\text{Cl}^-\) gives \(\displaystyle 2\text{HCl}\): \[\text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{HCl} + \text{H}_2\text{SO}_4 \]Step $\displaystyle 5$ — verify the atom count, which is the final check that catches most balancing slips: Every element balances, and the charge is already zero throughout, so this is the fully balanced molecular equation.Chlorine (oxidation number \(\displaystyle 0 \to -1\)) is the oxidising agent, getting reduced; sulphur dioxide (oxidation number \(\displaystyle +4 \to +6\)) is the reducing agent, getting oxidised. This is exactly why treating excess chlorine with sulphur dioxide works: the chlorine is consumed as it is reduced to harmless chloride ion (as \(\displaystyle \text{HCl}\)), instead of remaining as toxic \(\displaystyle \text{Cl}_2\) gas in the water.Answer: \(\displaystyle \text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{HCl} + \text{H}_2\text{SO}_4\), in which chlorine (oxidation number \(\displaystyle 0 \to -1\)) is reduced and sulphur (oxidation number \(\displaystyle +4 \to +6\)) is oxidised, each side involving a transfer of $\displaystyle 2$ electrons.

Disproportionation needs an element sitting in the MIDDLE of its own oxidation-state range — one part of it goes up, another part goes down, in the same reaction.Formally: a species undergoes disproportionation when the same element, starting in one oxidation state, ends up in a higher oxidation state in one product (that part is oxidised) and a lower oxidation state in another product (that part is reduced). For this to be chemically possible, the element must have at least three oxidation states available to it — the "middle" one it starts from, plus a higher one and a lower one it can reach. Scanning the periodic table for elements with a wide spread of oxidation states, and elements with a very narrow one, is exactly how you answer both parts of this question.(a) Non-metals that can disproportionateGo group by group among the non-metals.Group $\displaystyle 17$ (halogens). Chlorine, bromine and iodine each show oxidation states from \(\displaystyle -1\) all the way up to \(\displaystyle +7\) (think \(\displaystyle Cl^-, Cl_2, ClO^-, ClO_2^-, ClO_3^-, ClO_4^-\)). Sitting at \(\displaystyle 0\) in the free element, they have both a lower state (\(\displaystyle -1\)) and higher states (\(\displaystyle +1, +3, +5,\dots\)) to fall into, so \(\displaystyle Cl_2\), \(\displaystyle Br_2\) and \(\displaystyle I_2\) all disproportionate readily in alkali. For chlorine: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O \] Balance check: $\displaystyle 2$ Cl atoms on each side, $\displaystyle 2$ Na, $\displaystyle 2$ O (one in \(\displaystyle NaOCl\), one in \(\displaystyle H_2O\)), $\displaystyle 2$ H — every atom count matches, so the equation is balanced as written, with no extra coefficients needed. Oxidation numbers: chlorine starts at \(\displaystyle 0\) in \(\displaystyle Cl_2\), and ends at \(\displaystyle -1\) in sodium chloride (reduction) and \(\displaystyle +1\) in sodium hypochlorite (oxidation) — the same element split two ways. Bromine and iodine behave the same way with \(\displaystyle NaOBr\)/\(\displaystyle NaOI\) forming alongside \(\displaystyle NaBr\)/\(\displaystyle NaI\).Fluorine is the one halogen that is excluded. Fluorine is the most electronegative element there is, so no other atom can ever pull electron density away from it — it can never be pushed to a positive oxidation state. \(\displaystyle F_2\) only ever has the states \(\displaystyle 0\) (element) and \(\displaystyle -1\) (fluoride); with no positive state reachable, there is nowhere "up" for part of it to go, so \(\displaystyle F_2\) cannot disproportionate.Group 15. Phosphorus spans \(\displaystyle -3\) to \(\displaystyle +5\). White phosphorus (\(\displaystyle P_4\), oxidation state \(\displaystyle 0\)) disproportionates in hot concentrated alkali: \[P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2 \] Balance check: P — $\displaystyle 4$ on the left, \(\displaystyle 1+3=4\) on the right; Na — $\displaystyle 3$ and $\displaystyle 3$; H — \(\displaystyle (3+6)=9\) on the left from the \(\displaystyle NaOH\) and the \(\displaystyle H_2O\), and \(\displaystyle (3+6)=9\) on the right from \(\displaystyle PH_3\) and the three \(\displaystyle NaH_2PO_2\); O — $\displaystyle 3$ (from \(\displaystyle NaOH\)) \(\displaystyle +3\) (from \(\displaystyle H_2O\)) \(\displaystyle =6\) on the left, and \(\displaystyle 3\times2=6\) on the right from the hypophosphite. All four elements balance. Phosphorus goes from \(\displaystyle 0\) in \(\displaystyle P_4\) to \(\displaystyle -3\) in phosphine gas, \(\displaystyle PH_3\) (reduction), and to \(\displaystyle +1\) in sodium hypophosphite, \(\displaystyle NaH_2PO_2\) (oxidation).Group 16. Sulfur spans \(\displaystyle -2\) to \(\displaystyle +6\), and elemental sulfur (\(\displaystyle S_8\), oxidation state \(\displaystyle 0\)) disproportionates in hot concentrated alkali too: \[S_8 + 12NaOH \rightarrow 4Na_2S + 2Na_2S_2O_3 + 6H_2O \] Balance check: S — $\displaystyle 8$ on the left, \(\displaystyle 4+4=8\) on the right; Na — $\displaystyle 12$ and \(\displaystyle 8+4=12\); O — $\displaystyle 12$ (from the \(\displaystyle NaOH\)) on the left, \(\displaystyle 6+6=12\) on the right; H — $\displaystyle 12$ and 12. Sulfur goes from \(\displaystyle 0\) to \(\displaystyle -2\) in sodium sulfide (reduction) and to an average of \(\displaystyle +2\) in sodium thiosulfate (oxidation). Oxygen itself is the exception here: it only ever reaches \(\displaystyle -2\), \(\displaystyle -1\) (peroxide) or \(\displaystyle 0\), with a positive state possible only when forced by bonding to fluorine specifically — elemental \(\displaystyle O_2\) has no accessible higher state on its own, so it does not disproportionate.Group $\displaystyle 15$'s other member, nitrogen, does have a huge range of oxidation states in its compounds (\(\displaystyle NO_2\), for instance, does disproportionate: \(\displaystyle 3NO_2 + H_2O \rightarrow 2HNO_3 + NO\)), but elemental \(\displaystyle N_2\) itself does not — the \(\displaystyle N\equiv N\) triple bond is so strong that dinitrogen gas is essentially chemically inert under ordinary conditions and never gets pulled into a redox change at all, let alone a self-redox one.(b) Three metals that can disproportionateThe clearest metal examples are the ones whose lower oxidation state is thermodynamically unstable in aqueous solution relative to the metal and the higher state together.Copper: copper(I) is unstable in water and converts to copper(II) and copper metal, \[2Cu^{+} \rightarrow Cu^{2+} + Cu \] Balance check: $\displaystyle 2$ copper atoms on each side, and charge balances too — \(\displaystyle 2(+1) = +2\) on the left equals \(\displaystyle +2 + 0\) on the right. \(\displaystyle Cu^+\) is oxidised to \(\displaystyle \mathrm{Cu^{2+}}\) and simultaneously reduced to metallic copper.Gold: gold(I) behaves the same way, converting to gold(III) and gold metal, \[3Au^{+} \rightarrow Au^{3+} + 2Au \] Balance check: $\displaystyle 3$ gold atoms each side; charge \(\displaystyle 3(+1)=+3\) on the left equals \(\displaystyle +3+0\) on the right.Mercury: the mercurous ion is the dimer \(\displaystyle Hg_2^{2+}\), with each mercury formally at \(\displaystyle +1\); when something stabilises \(\displaystyle \mathrm{Hg^{2+}}\) strongly (a good complexing or precipitating ion), it disproportionates to mercury(II) and liquid mercury metal, \[Hg_2^{2+} \rightarrow Hg^{2+} + Hg \] Balance check: $\displaystyle 2$ mercury atoms each side; charge \(\displaystyle +2\) on the left equals \(\displaystyle +2+0\) on the right.In each case the metal's \(\displaystyle +1\) ion is the "middle" oxidation state, splitting into a \(\displaystyle +2\)/\(\displaystyle +3\) product (oxidation) and the neutral metal (reduction) — exactly the disproportionation pattern from part (a), just applied to metals instead of non-metals.The step people get wrong here is checking only the identity of the atoms and skipping the oxidation-number bookkeeping — a reaction can look like a single element forming two different products and still not be disproportionation unless one product genuinely sits at a higher oxidation number and the other at a lower one than the reactant.Answer: (a) The halogens chlorine, bromine and iodine (\(\displaystyle Cl_2\), \(\displaystyle Br_2\), \(\displaystyle I_2\)), together with phosphorus (\(\displaystyle P_4\)) and sulfur (\(\displaystyle S_8\)), are non-metals that disproportionate — fluorine, elemental oxygen and elemental nitrogen cannot, because each is stuck with too narrow a range of oxidation states. (b) Copper, gold and mercury are three metals that disproportionate: \(\displaystyle 2Cu^{+} \rightarrow Cu^{2+} + Cu\), \(\displaystyle 3Au^{+} \rightarrow Au^{3+} + 2Au\), and \(\displaystyle Hg_2^{2+} \rightarrow Hg^{2+} + Hg\).

When two reactants are both given amounts, the smaller mass is not automatically the limiting reagent — you convert both to moles and compare each one against its own coefficient in the balanced equation.Step $\displaystyle 1$ — write and balance the equation.The skeleton reaction (ammonia oxidized by oxygen to nitric oxide and water vapour) is \[\text{NH}_3(g) + \text{O}_2(g) \rightarrow \text{NO}(g) + \text{H}_2\text{O}(g) \]Nitrogen is already $\displaystyle 1$:$\displaystyle 1$ between \(\displaystyle \text{NH}_3\) and \(\displaystyle \text{NO}\), so give both the same coefficient, \(\displaystyle a\). Let \(\displaystyle b\) be the coefficient of \(\displaystyle \text{O}_2\) and \(\displaystyle c\) that of \(\displaystyle \text{H}_2\text{O}\). Taking \(\displaystyle a = 4\) clears both fractions: \(\displaystyle b = 5\), \(\displaystyle c = 6\). The balanced equation is \[4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g) \](Checking: N, $\displaystyle 4$ = $\displaystyle 4$; H, \(\displaystyle 4\times3=12\) on the left, \(\displaystyle 6\times2=12\) on the right; O, \(\displaystyle 5\times2=10\) on the left, \(\displaystyle 4\times1+6\times1=10\) on the right — balanced. This is also a redox step: nitrogen goes from the \(\displaystyle -3\) oxidation state in \(\displaystyle \text{NH}_3\) to \(\displaystyle +2\) in NO, so ammonia is oxidized and \(\displaystyle \text{O}_2\) is the oxidizing agent.)Step $\displaystyle 2$ — convert both given masses to moles.Use \(\displaystyle n = \dfrac{m}{M}\), where \(\displaystyle n\) is moles, \(\displaystyle m\) is the given mass, and \(\displaystyle M\) is the molar mass of that substance.Molar mass of \(\displaystyle \text{NH}_3\): \(\displaystyle M = 14 + 3(1) = 17\ \text{g mol}^{-1}\) (N = $\displaystyle 14$, H = $\displaystyle 1$ each, three H atoms). \[n(\text{NH}_3) = \frac{10.00\ \text{g}}{17\ \text{g mol}^{-1}} = 0.588\ \text{mol} \]Molar mass of \(\displaystyle \text{O}_2\): \(\displaystyle M = 2(16) = 32\ \text{g mol}^{-1}\). \[n(\text{O}_2) = \frac{20.00\ \text{g}}{32\ \text{g mol}^{-1}} = 0.625\ \text{mol} \]Step $\displaystyle 3$ — find the limiting reagent.The trap here is comparing the two masses directly ($\displaystyle 20.00$ g of \(\displaystyle \text{O}_2\) looks like "plenty" next to $\displaystyle 10.00$ g of \(\displaystyle \text{NH}_3\)) — but the reaction does not consume the two reactants in a $\displaystyle 1$:$\displaystyle 1$ mass ratio, or even a $\displaystyle 1$:$\displaystyle 1$ mole ratio. It consumes them in the $\displaystyle 4$:$\displaystyle 5$ mole ratio fixed by the balanced equation. The correct test is to divide each reactant's moles by its own coefficient and see which gives the smaller number — that reactant runs out first.\[\frac{n(\text{NH}_3)}{4} = \frac{0.588}{4} = 0.147, \qquad \frac{n(\text{O}_2)}{5} = \frac{0.625}{5} = 0.125 \]Since \(\displaystyle 0.125 < 0.147\), \(\displaystyle \text{O}_2\) is used up first: \(\displaystyle \text{O}_2\) is the limiting reagent, and \(\displaystyle \text{NH}_3\) is in excess (some ammonia will be left over unreacted).Step $\displaystyle 4$ — get moles of NO from the limiting reagent.Every $\displaystyle 5$ mol of \(\displaystyle \text{O}_2\) that reacts produces $\displaystyle 4$ mol of NO, so \[n(\text{NO}) = \frac{4}{5} \times n(\text{O}_2) = \frac{4}{5} \times 0.625\ \text{mol} = 0.500\ \text{mol} \]Step $\displaystyle 5$ — convert moles of NO to mass.Molar mass of NO: \(\displaystyle M = 14 + 16 = 30\ \text{g mol}^{-1}\) (N = $\displaystyle 14$, O = $\displaystyle 16$). Using \(\displaystyle m = n \times M\): \[m(\text{NO}) = 0.500\ \text{mol} \times 30\ \text{g mol}^{-1} = 15.0\ \text{g} \]So starting from $\displaystyle 10.00$ g of ammonia and $\displaystyle 20.00$ g of oxygen, oxygen runs out first, and the reaction can produce at most $\displaystyle 15.0$ g of nitric oxide gas (\(\displaystyle \text{NO}\)), with some ammonia left unreacted.Answer: The maximum mass of nitric oxide obtainable is $\displaystyle 15.0$ g (oxygen is the limiting reagent, giving $\displaystyle 0.500$ mol of NO from the reaction \(\displaystyle 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}\)).

A redox reaction runs on its own only when the cell potential you get from pairing the two half-reactions comes out positive. The rule for building that pairing:\[E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]Here \(\displaystyle E^\circ_{cathode}\) and \(\displaystyle E^\circ_{anode}\) are both taken straight from the table of standard reduction potentials — never flip a sign by hand. The species with the higher (more positive) reduction potential is the one that actually gets reduced (it sits at the cathode); the other species is forced to run backwards, giving up electrons, at the anode. This is the step people trip on: if you reverse the anode half-reaction on paper and flip the sign of its E° and then add, you've double-counted the sign flip — subtracting the two tabulated reduction potentials already does that job for you.From Table $\displaystyle 8.1$, the standard reduction potentials needed here are:\[\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s), \quad E^\circ = +0.80\ \text{V} \] \[\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq), \quad E^\circ = +0.77\ \text{V} \] \[\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s), \quad E^\circ = +0.34\ \text{V} \] \[\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq), \quad E^\circ = +0.54\ \text{V} \] \[\text{Br}_2(aq) + 2e^- \rightarrow 2\text{Br}^-(aq), \quad E^\circ = +1.09\ \text{V} \](a) Fe³⁺(aq) and I⁻(aq). Compare \(\displaystyle E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\ \text{V}\) with \(\displaystyle E^\circ(\text{I}_2/\text{I}^-) = 0.54\ \text{V}\). The iron couple is higher, so Fe³⁺ is reduced (cathode) and I⁻ is oxidised (anode). Iodide loses one electron per ion but the iron half-reaction only accepts one electron per ion, so double the iron equation to match electrons before adding:\[2\text{Fe}^{3+}(aq) + 2e^- \rightarrow 2\text{Fe}^{2+}(aq) \] \[2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^- \] \[2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(s) \]Check: $\displaystyle 2$ Fe and $\displaystyle 2$ I on each side; charge is \(\displaystyle 2(+3) + 2(-1) = +4\) on the left and \(\displaystyle 2(+2) + 0 = +4\) on the right — balanced.\[E^\circ_{cell} = 0.77 - 0.54 = +0.23\ \text{V} \;(>0) \Rightarrow \textbf{feasible} \](b) Ag⁺(aq) and Cu(s). \(\displaystyle E^\circ(\text{Ag}^+/\text{Ag}) = 0.80\ \text{V}\) beats \(\displaystyle E^\circ(\text{Cu}^{2+}/\text{Cu}) = 0.34\ \text{V}\), so Ag⁺ is reduced and copper metal is oxidised. Copper releases $\displaystyle 2$ electrons per atom, so the silver half-reaction is doubled to match:\[2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s) \] \[\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^- \] \[2\text{Ag}^+(aq) + \text{Cu}(s) \rightarrow 2\text{Ag}(s) + \text{Cu}^{2+}(aq) \]Check: $\displaystyle 2$ Ag and $\displaystyle 1$ Cu on each side; charge \(\displaystyle 2(+1) + 0 = +2\) on the left, \(\displaystyle 0 + (+2) = +2\) on the right — balanced.\[E^\circ_{cell} = 0.80 - 0.34 = +0.46\ \text{V} \;(>0) \Rightarrow \textbf{feasible} \](c) Fe³⁺(aq) and Cu(s). \(\displaystyle E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\ \text{V}\) is higher than \(\displaystyle E^\circ(\text{Cu}^{2+}/\text{Cu}) = 0.34\ \text{V}\), so Fe³⁺ is reduced and copper metal is oxidised. Copper gives up $\displaystyle 2$ electrons per atom, so the iron half-reaction is doubled:\[2\text{Fe}^{3+}(aq) + 2e^- \rightarrow 2\text{Fe}^{2+}(aq) \] \[\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^- \] \[2\text{Fe}^{3+}(aq) + \text{Cu}(s) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq) \]Check: $\displaystyle 2$ Fe and $\displaystyle 1$ Cu on each side; charge \(\displaystyle 2(+3) + 0 = +6\) on the left, \(\displaystyle 2(+2) + (+2) = +6\) on the right — balanced.\[E^\circ_{cell} = 0.77 - 0.34 = +0.43\ \text{V} \;(>0) \Rightarrow \textbf{feasible} \](d) Ag(s) and Fe³⁺(aq). For silver metal to be oxidised here, Fe³⁺ has to be the one reduced, which needs \(\displaystyle E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+})\) to exceed \(\displaystyle E^\circ(\text{Ag}^+/\text{Ag})\). But \(\displaystyle 0.77\ \text{V} < 0.80\ \text{V}\) — silver's own reduction potential is (barely) the higher one, meaning Ag⁺ is a slightly stronger electron-grabber than Fe³⁺. Running the subtraction the same way as before, with Fe³⁺/Fe²⁺ forced into the cathode role it doesn't win:\[E^\circ_{cell} = E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\circ(\text{Ag}^+/\text{Ag}) = 0.77 - 0.80 = -0.03\ \text{V} \;(<0) \Rightarrow \textbf{not feasible} \]A negative \(\displaystyle E^\circ_{cell}\) means \(\displaystyle \Delta G^\circ = -nFE^\circ_{cell}\) (\(\displaystyle n\) = moles of electrons transferred, \(\displaystyle F\) = Faraday constant) comes out positive, so the reaction as written does not proceed. It's a close call — only $\displaystyle 0.03$ V — which is exactly why solid silver does not dissolve in a solution of iron(III) salts.(e) Br₂(aq) and Fe²⁺(aq). \(\displaystyle E^\circ(\text{Br}_2/\text{Br}^-) = 1.09\ \text{V}\) is well above \(\displaystyle E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\ \text{V}\), so bromine is reduced and Fe²⁺ is oxidised to Fe³⁺. Iron loses $\displaystyle 1$ electron per ion while bromine's half-reaction needs $\displaystyle 2$, so the iron half-reaction is doubled:\[\text{Br}_2(aq) + 2e^- \rightarrow 2\text{Br}^-(aq) \] \[2\text{Fe}^{2+}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2e^- \] \[\text{Br}_2(aq) + 2\text{Fe}^{2+}(aq) \rightarrow 2\text{Br}^-(aq) + 2\text{Fe}^{3+}(aq) \]Check: $\displaystyle 2$ Br and $\displaystyle 2$ Fe on each side; charge \(\displaystyle 0 + 2(+2) = +4\) on the left, \(\displaystyle 2(-1) + 2(+3) = +4\) on the right — balanced.\[E^\circ_{cell} = 1.09 - 0.77 = +0.32\ \text{V} \;(>0) \Rightarrow \textbf{feasible} \]Answer: (a) feasible, \(\displaystyle E^\circ_{cell}=+0.23\) V, giving Fe²⁺(aq) and I₂(s); (b) feasible, \(\displaystyle E^\circ_{cell}=+0.46\) V, giving Ag(s) and Cu²⁺(aq); (c) feasible, \(\displaystyle E^\circ_{cell}=+0.43\) V, giving Fe²⁺(aq) and Cu²⁺(aq); (d) not feasible, \(\displaystyle E^\circ_{cell}=-0.03\) V; (e) feasible, \(\displaystyle E^\circ_{cell}=+0.32\) V, giving Br⁻(aq) and Fe³⁺(aq).

Electrolysis sends the current through the solution to force a redox reaction that would not happen on its own — cations go to the cathode and get reduced, anions (or water) go to the anode and get oxidized. Which species actually reacts is decided by ease of reduction/oxidation, not by which ion "belongs" to the salt.Two ideas run through all four parts:1. At the cathode, whichever species is easiest to reduce (the more positive standard reduction potential, \(\displaystyle E^\circ\)) is discharged first. Relevant values: \[\text{Ag}^+ + e^- \rightarrow \text{Ag}(s), \quad E^\circ = +0.80\ \text{V} \] \[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s), \quad E^\circ = +0.34\ \text{V} \] \[2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g), \quad E^\circ = 0.00\ \text{V} \] Since \(\displaystyle \text{Ag}^+\) and \(\displaystyle \text{Cu}^{2+}\) are both easier to reduce than \(\displaystyle \text{H}^+\), they are deposited as metal before any \(\displaystyle \text{H}_2\) forms — a metal below hydrogen in the electrochemical series always wins at the cathode.2. At the anode, whichever species is easiest to oxidize is discharged first — normally the one with the lowest \(\displaystyle E^\circ\) — but there is a catch: gas evolution (especially \(\displaystyle \text{O}_2\)) needs extra voltage in practice ("overvoltage") beyond what the tables predict, so a poorly-oxidizable ion like \(\displaystyle \text{Cl}^-\) can still beat water to the anode. Also, if the electrode metal itself is easily oxidized (a reactive, "active" electrode such as Ag or Cu, as opposed to an inert one like Pt), the electrode dissolves instead of the solution reacting.A step people skip: \(\displaystyle \text{NO}_3^-\) and \(\displaystyle \text{SO}_4^{2-}\) are never discharged at an inert anode in dilute aqueous solution — nitrogen and sulfur are already in their highest common oxidation states (+$\displaystyle 5$ and +$\displaystyle 6$), so these ions have nothing further to give up. Water is oxidized in their place.(i) Aqueous AgNO₃ with silver electrodesHere the electrodes are active — made of the same metal as the cation in solution. \(\displaystyle \text{Ag}(s)\) is oxidized far more easily (\(\displaystyle E^\circ = +0.80\ \text{V}\)) than water (\(\displaystyle E^\circ = +1.23\ \text{V}\) for \(\displaystyle \text{O}_2/\text{H}_2\text{O}\)), so the silver anode itself dissolves instead of water being oxidized:Anode (oxidation): \(\displaystyle \text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-\)Cathode (reduction): \(\displaystyle \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)Each equation already balances one silver atom and one electron on each side. Silver metal dissolves off the anode as \(\displaystyle \text{Ag}^+\) ions and an equal amount of silver metal plates back out at the cathode — the \(\displaystyle \text{AgNO}_3\) concentration in solution stays essentially unchanged. This is exactly the principle used to electrorefine silver.(ii) Aqueous AgNO₃ with platinum electrodesPlatinum is inert, so it cannot dissolve. At the cathode, \(\displaystyle \text{Ag}^+\) is still easier to reduce than \(\displaystyle \text{H}^+\), so silver still plates out:Cathode: \(\displaystyle \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)At the anode, \(\displaystyle \text{NO}_3^-\) cannot be oxidized further (N is already +$\displaystyle 5$), so water is oxidized instead, releasing oxygen gas:Anode: \(\displaystyle 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\)To combine them, the cathode step must supply the same $\displaystyle 4$ electrons the anode step releases, so multiply it by $\displaystyle 4$: \[4\text{Ag}^+(aq) + 4e^- \rightarrow 4\text{Ag}(s) \] \[2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \] Adding and cancelling the \(\displaystyle 4e^-\): \[4\text{Ag}^+(aq) + 2\text{H}_2\text{O}(l) \rightarrow 4\text{Ag}(s) + \text{O}_2(g) + 4\text{H}^+(aq) \] Check the balance: $\displaystyle 4$ Ag, $\displaystyle 4$ H, $\displaystyle 2$ O on each side, and charge \(\displaystyle +4\) on both sides. Silver metal deposits at the cathode, oxygen gas is evolved at the anode, and the solution is left with the \(\displaystyle \text{H}^+\) produced (it slowly turns into dilute \(\displaystyle \text{HNO}_3\) as the \(\displaystyle \text{Ag}^+\) is used up).(iii) Dilute H₂SO₄ with platinum electrodes\(\displaystyle \text{SO}_4^{2-}\) cannot be oxidized (sulfur is already +$\displaystyle 6$), so, just as with \(\displaystyle \text{NO}_3^-\) above, water is oxidized at the anode. At the cathode, the only cation present besides water is \(\displaystyle \text{H}^+\), so hydrogen gas is evolved:Cathode: \(\displaystyle 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\)Anode: \(\displaystyle 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\)Multiply the cathode step by $\displaystyle 2$ so both steps carry $\displaystyle 4$ electrons, then add: \[4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2(g) \] \[2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \] The \(\displaystyle 4\text{H}^+\) and \(\displaystyle 4e^-\) cancel exactly, leaving \[2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g) \] This is simply the electrolysis of water — the \(\displaystyle \text{H}_2\text{SO}_4\) is regenerated, not consumed, and only serves to carry the current. Hydrogen gas collects at the cathode and oxygen gas at the anode, in the $\displaystyle 2$ : $\displaystyle 1$ mole ratio that matches water's own composition.(iv) Aqueous CuCl₂ with platinum electrodesAt the cathode, \(\displaystyle \text{Cu}^{2+}\) (\(\displaystyle E^\circ = +0.34\ \text{V}\)) is easier to reduce than \(\displaystyle \text{H}^+\) (\(\displaystyle E^\circ = 0.00\ \text{V}\)), so copper metal deposits:Cathode: \(\displaystyle \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\)At the anode there is a genuine competition between \(\displaystyle \text{Cl}^-\) and water. By the tabulated potentials, water should be easier to oxidize (\(\displaystyle E^\circ = +1.23\ \text{V}\) for \(\displaystyle \text{O}_2/\text{H}_2\text{O}\), against \(\displaystyle +1.36\ \text{V}\) for \(\displaystyle \text{Cl}_2/\text{Cl}^-\)) — this is the step where it's easy to guess "oxygen" and be wrong. In practice, oxygen gas evolution on an electrode surface carries a large overvoltage (an extra energy cost beyond the thermodynamic value), so with an appreciable concentration of \(\displaystyle \text{Cl}^-\) present, chlorine gas is discharged instead:Anode: \(\displaystyle 2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\)The cathode and anode steps both involve exactly $\displaystyle 2$ electrons, so they add directly: \[\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \] \[2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^- \] \[\text{Cu}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{Cu}(s) + \text{Cl}_2(g) \] i.e. \(\displaystyle \text{CuCl}_2(aq) \rightarrow \text{Cu}(s) + \text{Cl}_2(g)\) — one copper and two chlorines on each side, charge zero on each side. Copper metal deposits at the cathode and chlorine gas is liberated at the anode.Answer: (i) With Ag electrodes, the silver anode dissolves (\(\displaystyle \text{Ag} \rightarrow \text{Ag}^+ + e^-\)) and an equal amount of silver metal deposits at the cathode — the electrode metal itself reacts, and the solution composition is unchanged. (ii) With Pt electrodes, silver metal deposits at the cathode while oxygen gas is evolved at the anode (\(\displaystyle 4\text{Ag}^+ + 2\text{H}_2\text{O} \rightarrow 4\text{Ag} + \text{O}_2 + 4\text{H}^+\)). (iii) Dilute \(\displaystyle \text{H}_2\text{SO}_4\) with Pt electrodes simply electrolyses water: hydrogen gas at the cathode and oxygen gas at the anode, in a $\displaystyle 2$ : $\displaystyle 1$ mole ratio (\(\displaystyle 2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2\)). (iv) With CuCl₂ and Pt electrodes, copper metal deposits at the cathode and chlorine gas is liberated at the anode (\(\displaystyle \text{CuCl}_2 \rightarrow \text{Cu} + \text{Cl}_2\)), chlorine winning out over oxygen because of oxygen's overvoltage.

A metal displaces another metal from its salt solution only if it is oxidised more easily than the metal already in solution — that is, only if it has a more negative standard reduction potential.Every displacement reaction here has the same shape: a solid metal \(\displaystyle A\) goes into solution as \(\displaystyle A^{n+}\) (it is oxidised, loses electrons) while the metal ion \(\displaystyle \mathrm{B^{n+}}\) already dissolved comes out as solid \(\displaystyle B\) (it is reduced, gains electrons): \[A(s) + B^{n+}(aq) \rightarrow A^{n+}(aq) + B(s) \]Whether this runs forward or not is decided by the standard reduction potential, \(\displaystyle E^\circ\) — the voltage, measured against the standard hydrogen electrode, for the half-reaction \(\displaystyle M^{n+} + ne^- \rightarrow M\). A more negative \(\displaystyle E^\circ\) means that metal resists being reduced, i.e. it would much rather sit as the ion and hand its electrons to something else — so it is the stronger reducing agent and the more easily oxidised metal.The cell potential for the overall reaction is \[E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] where \(\displaystyle E^\circ_{\text{cathode}}\) is the reduction potential of the ion being reduced (\(\displaystyle B^{n+}\), the one that gets displaced) and \(\displaystyle E^\circ_{\text{anode}}\) is the reduction potential of the metal being oxidised (\(\displaystyle A\), the one added). A displacement happens on its own only when \(\displaystyle E^\circ_{cell}\) is positive — which means \(\displaystyle A\) must have the smaller (more negative) \(\displaystyle E^\circ\) of the two.This is exactly the step people skip: it is tempting to rank metals by "how shiny" or "how commonly used" they are, but the only thing that decides who displaces whom is the sign and size of \(\displaystyle E^\circ\), not intuition about reactivity.The standard reduction potentials (\(\displaystyle M^{n+}/M\), $\displaystyle 298$ K) are: \[E^\circ(\mathrm{Mg^{2+}/Mg}) = -2.36\ \mathrm{V},\quad E^\circ(\mathrm{Al^{3+}/Al}) = -1.66\ \mathrm{V},\quad E^\circ(\mathrm{Zn^{2+}/Zn}) = -0.76\ \mathrm{V}, \] \[E^\circ(\mathrm{Fe^{2+}/Fe}) = -0.44\ \mathrm{V},\quad E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\ \mathrm{V} \]Reading these from most negative to most positive gives the order in which each metal can push the next one out of solution: \[\mathrm{Mg} \;(-2.36) < \mathrm{Al}\;(-1.66) < \mathrm{Zn}\;(-0.76) < \mathrm{Fe}\;(-0.44) < \mathrm{Cu}\;(+0.34) \]So Mg is the strongest reducing agent and Cu the weakest: every metal on the left of this list displaces every metal to its right from a solution of that metal's salt, but never the reverse. Concretely:Magnesium displaces aluminium from aluminium sulfate solution (balanced by putting $\displaystyle 3$ in front of Mg and MgSO\(\displaystyle _4\) so that $\displaystyle 3$ Mg atoms supply the $\displaystyle 6$ electrons needed to reduce $\displaystyle 2$ Al\(\displaystyle ^{3+}\)): \[3\,\mathrm{Mg}(s) + \mathrm{Al_2(SO_4)_3}(aq) \rightarrow 3\,\mathrm{MgSO_4}(aq) + 2\,\mathrm{Al}(s) \] Here Mg goes from oxidation number \(\displaystyle 0\) to \(\displaystyle +2\) (oxidised), and Al goes from \(\displaystyle +3\) to \(\displaystyle 0\) (reduced) — solid magnesium (a grey metal) dissolves while solid aluminium metal is deposited.Aluminium in turn displaces zinc from zinc sulfate solution ($\displaystyle 2$ Al atoms give up $\displaystyle 6$ electrons, enough to reduce $\displaystyle 3$ Zn\(\displaystyle ^{2+}\)): \[2\,\mathrm{Al}(s) + 3\,\mathrm{ZnSO_4}(aq) \rightarrow \mathrm{Al_2(SO_4)_3}(aq) + 3\,\mathrm{Zn}(s) \] Al: \(\displaystyle 0 \rightarrow +3\); Zn: \(\displaystyle +2 \rightarrow 0\).Zinc displaces iron from iron(II) sulfate solution (already balanced $\displaystyle 1$:$\displaystyle 1$:$\displaystyle 1$:$\displaystyle 1$, since both metals form \(\displaystyle 2+\) ions here): \[\mathrm{Zn}(s) + \mathrm{FeSO_4}(aq) \rightarrow \mathrm{ZnSO_4}(aq) + \mathrm{Fe}(s) \] Zn: \(\displaystyle 0 \rightarrow +2\); Fe: \(\displaystyle +2 \rightarrow 0\).And iron displaces copper from copper(II) sulfate solution — the classic "iron nail in blue vitriol solution" reaction, where the blue colour fades as reddish-brown copper metal is deposited: \[\mathrm{Fe}(s) + \mathrm{CuSO_4}(aq) \rightarrow \mathrm{FeSO_4}(aq) + \mathrm{Cu}(s) \] Fe: \(\displaystyle 0 \rightarrow +2\) (oxidised); Cu: \(\displaystyle +2 \rightarrow 0\) (reduced).Copper, sitting at the bottom with the only positive \(\displaystyle E^\circ\), cannot push any of the other four metals out of their salt solutions — copper metal placed in magnesium, aluminium, zinc, or iron sulfate solution does nothing, because copper is a weaker reducing agent than all four.Answer: Mg > Al > Zn > Fe > Cu — each metal in this order displaces every metal to its right from a solution of that metal's salt (e.g. Mg displaces Al, Zn, Fe, Cu; Fe displaces only Cu; Cu displaces none of them), because standard reduction potential increases in exactly this order (\(\displaystyle -2.36\) V to \(\displaystyle +0.34\) V), and the metal with the more negative \(\displaystyle E^\circ\) is always the one oxidised.

The reducing power of a metal is its tendency to lose electrons (get oxidised), and that is the exact opposite of what a standard reduction potential measures — so the most negative reduction potential belongs to the strongest reducing agent.Each value you were given is a standard reduction potential, \(\displaystyle E^{\circ} \), for the half-reaction\[M^{n+}(aq) + n e^{-} \rightarrow M(s) \]Here \(\displaystyle M^{n+}\) is the metal ion, \(\displaystyle M(s)\) is the solid metal, \(\displaystyle n\) is the number of electrons transferred, and \(\displaystyle e^{-}\) is the electron. Each equation above is already balanced — the charge on the left, \(\displaystyle n+\) from the ion plus \(\displaystyle n\) negative charges from the \(\displaystyle n\) electrons, adds to zero, matching the neutral metal atom on the right. Written out for the five metals here:\[\mathrm{Ag^{+}(aq) + e^{-} \rightarrow Ag(s)}, \quad E^{\circ} = 0.80\ \text{V} \] \[\mathrm{Hg^{2+}(aq) + 2e^{-} \rightarrow Hg(l)}, \quad E^{\circ} = 0.79\ \text{V} \] \[\mathrm{Cr^{3+}(aq) + 3e^{-} \rightarrow Cr(s)}, \quad E^{\circ} = -0.74\ \text{V} \] \[\mathrm{Mg^{2+}(aq) + 2e^{-} \rightarrow Mg(s)}, \quad E^{\circ} = -2.37\ \text{V} \] \[\mathrm{K^{+}(aq) + e^{-} \rightarrow K(s)}, \quad E^{\circ} = -2.93\ \text{V} \]A large positive \(\displaystyle E^{\circ}\) means the forward reaction (gaining electrons, i.e. reduction) is favourable — the ion "wants" to be reduced, so the metal has little tendency to give up electrons again. A large negative \(\displaystyle E^{\circn}\) means the reverse reaction is favourable instead: the metal atom \(\displaystyle M(s)\) readily loses its electrons and turns into \(\displaystyle M^{n+}\). Losing electrons is oxidation, and a substance that is readily oxidised is, by definition, a good reducing agent. This is the step people mix up: they rank the metals by the size of \(\displaystyle E^{\circ}\) as printed, forgetting that reducing power runs in the opposite direction to reduction potential, not the same direction.So arranging the five reduction-potential values from most positive to most negative:\[0.80\ (\mathrm{Ag}) \;>\; 0.79\ (\mathrm{Hg}) \;>\; -0.74\ (\mathrm{Cr}) \;>\; -2.37\ (\mathrm{Mg}) \;>\; -2.93\ (\mathrm{K}) \]gives the metals in exactly increasing order of reducing power, because the metal with the lowest (most negative) reduction potential — potassium — is the one most eager to lose its electron, making it the strongest reducing agent, while silver, with the highest (most positive) value, is the least eager to lose an electron and so is the weakest reducing agent of the five.Answer: Increasing order of reducing power: \(\displaystyle \mathrm{Ag} < \mathrm{Hg} < \mathrm{Cr} < \mathrm{Mg} < \mathrm{K} \) (silver is the weakest reducing agent, potassium is the strongest).

A galvanic cell is built by splitting one redox reaction into its two half‑reactions — oxidation at the anode, reduction at the cathode — and the anode is always the negative electrode, because it is the one pumping electrons out into the wire.Step $\displaystyle 1$ — split the reaction using oxidation numbersThe given reaction is \[\text{Zn(s)} + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag(s)} \]Track the oxidation number of each element on both sides: Oxidation happens at the electrode called the anode; reduction happens at the cathode. So the zinc electrode is the anode and the silver electrode is the cathode.Step $\displaystyle 2$ — write and balance each half-reaction separatelyOxidation (anode), zinc metal losing $\displaystyle 2$ electrons to become the +$\displaystyle 2$ ion: \[\text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \] This is already balanced: $\displaystyle 1$ zinc atom on each side, and the charge on the right \(\displaystyle (+2) + (-2\times1) = 0\) matches the neutral charge on the left.Reduction (cathode), silver ion gaining $\displaystyle 1$ electron to become silver metal: \[\text{Ag}^+(aq) + e^- \rightarrow \text{Ag(s)} \] This too is balanced atom-for-atom and charge-for-charge \(\displaystyle (+1 - 1 = 0)\) on both sides.The step people skip: matching the electron count before adding the halves back together. The zinc half-reaction releases $\displaystyle 2$ electrons, but the silver half-reaction as written only absorbs 1. Electrons cannot be left over or borrowed — the number given up at the anode must exactly equal the number taken up at the cathode. So the silver half-reaction is multiplied through by $\displaystyle 2$: \[2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag(s)} \] Adding this to the zinc half-reaction, the \(\displaystyle 2e^-\) cancel from both sides and the two halves reconstruct exactly the original overall equation — confirming the split is correct.Step $\displaystyle 3$ — depict the cellThe standard shorthand for a galvanic cell lists, from left to right: anode metal | anode solution || cathode solution | cathode metal, with a single vertical line for a phase boundary (metal–solution) and a double vertical line for the salt bridge joining the two half-cells: \[\text{Zn(s)} \mid \text{Zn}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag(s)} \] Physically this is a zinc rod dipped in a solution of a zinc salt (e.g. \(\displaystyle \text{Zn(NO}_3)_2\)) in one beaker, a silver rod dipped in a solution of a silver salt (e.g. \(\displaystyle \text{AgNO}_3\)) in a second beaker, the two rods joined by a wire (with a voltmeter/external load), and the two solutions connected by a salt bridge (a tube of an inert electrolyte gel, such as \(\displaystyle \text{KNO}_3\), that completes the internal circuit without mixing the two solutions).Step $\displaystyle 4$ — answer each part(i) Which electrode is negatively charged? The zinc electrode is losing electrons into the external wire as it dissolves \(\displaystyle (\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-)\), so it is building up a surplus of electrons relative to the circuit — it is the anode, and it is the negative electrode. The silver electrode is consuming electrons and is the positive electrode (cathode).(ii) Carriers of the current in the cell? Two different carriers operate in two different parts of the circuit, and conflating them is the usual mistake. In the external wire (through the voltmeter/load) the current is carried by electrons, flowing from the zinc anode to the silver cathode. Inside the cell — through the two solutions and the salt bridge — there are no free electrons; the current is carried by the ions: \(\displaystyle \text{Zn}^{2+}\) and \(\displaystyle \text{NO}_3^-\) (or whatever anion is present) move away from the anode compartment, and \(\displaystyle \text{Ag}^+\) is consumed at the cathode while anions drift in to replace the charge, with the salt bridge's own ions maintaining electrical neutrality on both sides.(iii) Individual reaction at each electrode? At the anode (zinc electrode, oxidation): \[\text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \] At the cathode (silver electrode, reduction): \[2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag(s)} \]Answer: Cell: \(\displaystyle \text{Zn(s)} \mid \text{Zn}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag(s)}\). (i) The zinc electrode (anode) is negatively charged. (ii) Electrons carry the current through the external wire; ions (e.g. \(\displaystyle \text{Zn}^{2+}\), \(\displaystyle \text{Ag}^+\), \(\displaystyle \text{NO}_3^-\), and the salt-bridge ions) carry the current through the solutions and salt bridge. (iii) Anode: \(\displaystyle \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^-\); Cathode: \(\displaystyle 2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag(s)}\).