NCERT Solutions · Class 12 Chemistry Electrochemistry

Molar conductivity scales conductivity up from "per cm³" to "per mole of solute," and that scaling factor is $\displaystyle 1000$/c.Conductivity \(\displaystyle \kappa\) tells you how well a full cubic centimetre of solution conducts. Molar conductivity \(\displaystyle \Lambda_m\) asks a different question: how well does all the acetic acid dissolved in one litre conduct? Since \(\displaystyle \kappa\) is measured per cm\(\displaystyle ^3\) and molarity \(\displaystyle c\) is measured per litre ($\displaystyle 1000$ cm\(\displaystyle ^3\)), you convert with\[\Lambda_m = \frac{\kappa \times 1000 \ \text{cm}^3\text{L}^{-1}}{c} \]where \(\displaystyle \kappa\) is the conductivity in S cm\(\displaystyle ^{-1}\) and \(\displaystyle c\) is the molarity in mol L\(\displaystyle ^{-1}\).Substituting \(\displaystyle \kappa = 7.896\times10^{-5}\) S cm\(\displaystyle ^{-1}\) and \(\displaystyle c = 0.00241\) mol L\(\displaystyle ^{-1}\):\[\Lambda_m = \frac{7.896\times10^{-5}\ \text{S cm}^{-1}\times 1000\ \text{cm}^3\text{L}^{-1}}{0.00241\ \text{mol L}^{-1}} = \frac{7.896\times10^{-2}}{2.41\times10^{-3}}\ \text{S cm}^2\text{mol}^{-1} \]\[\Lambda_m = 32.76 \ \text{S cm}^2\text{mol}^{-1} \]This measured \(\displaystyle \Lambda_m\) is well below \(\displaystyle \Lambda_m^{\circ}\) because acetic acid is a weak electrolyte — most of it sits undissociated in solution, and only the dissociated fraction carries current. For a weak acid, the ratio of the actual (measured) molar conductivity to the molar conductivity at infinite dilution (where every molecule is dissociated) IS the degree of dissociation:\[\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} \]With \(\displaystyle \Lambda_m^{\circ} = 390.5\) S cm\(\displaystyle ^2\)mol\(\displaystyle ^{-1}\):\[\alpha = \frac{32.76}{390.5} = 0.08389 \]So only about $\displaystyle 8.4$% of the acetic acid molecules have dissociated at this concentration.Now apply Ostwald's dilution law to get the dissociation constant — don't just report \(\displaystyle \alpha\), the question asks for \(\displaystyle K_a\). For the equilibrium \(\displaystyle \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+\), starting from concentration \(\displaystyle c\) with degree of dissociation \(\displaystyle \alpha\), the equilibrium concentrations are \(\displaystyle c(1-\alpha)\) for the undissociated acid and \(\displaystyle c\alpha\) for each ion, giving\[K_a = \frac{c\alpha^2}{1-\alpha} \]Substituting \(\displaystyle c = 0.00241\) mol L\(\displaystyle ^{-1}\) and \(\displaystyle \alpha = 0.08389\):\[\alpha^2 = (0.08389)^2 = 7.038\times10^{-3} \]\[c\alpha^2 = 0.00241 \times 7.038\times10^{-3} = 1.696\times10^{-5} \]\[1-\alpha = 1-0.08389 = 0.9161 \]\[K_a = \frac{1.696\times10^{-5}}{0.9161} = 1.851\times10^{-5}\ \text{mol L}^{-1} \]Carrying the numbers through without rounding until this last step (rounding \(\displaystyle \alpha\) too early is where this problem usually goes wrong) gives a dissociation constant of about \(\displaystyle 1.85\times10^{-5}\) mol L\(\displaystyle ^{-1}\), consistent with acetic acid being a weak acid.Answer: \(\displaystyle \Lambda_m = 32.76\ \text{S cm}^2\text{mol}^{-1}\), degree of dissociation \(\displaystyle \alpha = 0.0839\), and \(\displaystyle K_a = 1.85\times10^{-5}\ \text{mol L}^{-1}\).

The charge needed is the number of electrons in the half-reaction, multiplied by the Faraday constant.For any reduction \(\displaystyle M^{n+} + ne^- \rightarrow M \), $\displaystyle 1$ mole of the metal ion needs \(\displaystyle n\) moles of electrons. Since $\displaystyle 1$ mole of electrons carries a charge of $\displaystyle 1$ Faraday,\[Q = n \times F \]where \(\displaystyle Q\) is the charge in coulombs, \(\displaystyle n\) is the number of moles of electrons transferred per mole of product, and \(\displaystyle F = 96500\ \text{C mol}^{-1}\) is the charge on one mole of electrons.The step people skip is writing out the balanced half-reaction first — the "$\displaystyle 1$ mol" given in the question is $\displaystyle 1$ mol of the ion, not $\displaystyle 1$ mol of electrons, and \(\displaystyle n\) has to be read off the balanced equation, not guessed from the ionic charge alone (it usually matches, but for \(\displaystyle MnO_4^-\) you need the full half-reaction to see it).(i) \(\displaystyle \mathrm{Al^{3+}}\) to \(\displaystyle Al\)Half-reaction: \[Al^{3+} + 3e^- \rightarrow Al \]Each mole of \(\displaystyle \mathrm{Al^{3+}}\) needs $\displaystyle 3$ mol of electrons, so \(\displaystyle n = 3\).\[Q = n \times F = 3 \times 96500\ \text{C mol}^{-1} = 289500\ \text{C} \](ii) \(\displaystyle \mathrm{Cu^{2+}}\) to \(\displaystyle Cu\)Half-reaction: \[Cu^{2+} + 2e^- \rightarrow Cu \]Each mole of \(\displaystyle \mathrm{Cu^{2+}}\) needs $\displaystyle 2$ mol of electrons, so \(\displaystyle n = 2\).\[Q = n \times F = 2 \times 96500\ \text{C mol}^{-1} = 193000\ \text{C} \](iii) \(\displaystyle MnO_4^-\) to \(\displaystyle \mathrm{Mn^{2+}}\)Here manganese goes from oxidation state \(\displaystyle +7\) (in \(\displaystyle MnO_4^-\)) to \(\displaystyle +2\) (in \(\displaystyle Mn^{2+}\)), a drop of $\displaystyle 5$ units, so the half-reaction needs $\displaystyle 5$ electrons. Balancing it fully in acid:\[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]Each mole of \(\displaystyle MnO_4^-\) needs $\displaystyle 5$ mol of electrons, so \(\displaystyle n = 5\).\[Q = n \times F = 5 \times 96500\ \text{C mol}^{-1} = 482500\ \text{C} \]Answer: (i) $\displaystyle 289500$ C for $\displaystyle 1$ mol \(\displaystyle Al^{3+} \to Al\); (ii) $\displaystyle 193000$ C for $\displaystyle 1$ mol \(\displaystyle Cu^{2+} \to Cu\); (iii) $\displaystyle 482500$ C for $\displaystyle 1$ mol \(\displaystyle MnO_4^- \to Mn^{2+}\).

Faraday's law: the charge needed is (moles of metal) × (electrons transferred per ion), in units of Faraday ($\displaystyle 1$ F = $\displaystyle 1$ mole of electrons).Write the reduction half-reaction first — it tells you how many electrons (how many Faradays) deposit one mole of metal.(i) Calcium from molten \(\displaystyle CaCl_2\)The cation is \(\displaystyle \mathrm{Ca^{2+}}\), so the half-reaction is\[Ca^{2+} + 2e^- \rightarrow Ca \]One mole of \(\displaystyle Ca\) needs $\displaystyle 2$ mol of electrons, i.e. $\displaystyle 2$ F.Moles of Ca in $\displaystyle 20.0$ g (molar mass of Ca = $\displaystyle 40.0$ g mol⁻¹):\[n(Ca) = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0\ \text{g}}{40.0\ \text{g mol}^{-1}} = 0.500\ \text{mol} \]Electricity required:\[\text{Faraday required} = n(Ca) \times 2 = 0.500\ \text{mol} \times 2\ \text{F mol}^{-1} = 1.00\ \text{F} \](ii) Aluminium from molten \(\displaystyle Al_2O_3\)The cation here is \(\displaystyle \mathrm{Al^{3+}}\) (aluminium is in the +$\displaystyle 3$ state in \(\displaystyle Al_2O_3\)), so\[Al^{3+} + 3e^- \rightarrow Al \]This is the step people slip on: it is the charge on the individual metal ion \(\displaystyle \mathrm{(Al^{3+})}\) that fixes the electron count, not the "$\displaystyle 2$" or "$\displaystyle 3$" subscript sitting in the oxide formula \(\displaystyle Al_2O_3\). One mole of \(\displaystyle Al\) needs $\displaystyle 3$ mol of electrons, i.e. $\displaystyle 3$ F.Moles of Al in $\displaystyle 40.0$ g (molar mass of Al = $\displaystyle 27.0$ g mol⁻¹):\[n(Al) = \frac{40.0\ \text{g}}{27.0\ \text{g mol}^{-1}} = 1.4815\ \text{mol} \]Electricity required:\[\text{Faraday required} = n(Al) \times 3 = 1.4815\ \text{mol} \times 3\ \text{F mol}^{-1} = 4.444\ \text{F} \]Rounding to three significant figures (matching the three-sig-fig data given):Answer: (i) $\displaystyle 1.00$ F of electricity is needed to produce $\displaystyle 20.0$ g of Ca; (ii) $\displaystyle 4.44$ F of electricity is needed to produce $\displaystyle 40.0$ g of Al.

The charge needed is Faraday's constant times the number of MOLES OF ELECTRONS the balanced half-reaction actually transfers — and that number is not always "$\displaystyle 1$ per formula unit." Work out the electron count from oxidation-state change (or from the half-reaction), then apply\[Q = nF \]where \(\displaystyle n\) is the moles of electrons transferred and \(\displaystyle F\) is the Faraday constant, \(\displaystyle F = 96500\ \text{C mol}^{-1}\) (the charge on one mole of electrons).(i) Oxidation of \(\displaystyle H_2O\) to \(\displaystyle O_2\)Write the balanced oxidation half-reaction:\[2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-} \]This says $\displaystyle 2$ mol of \(\displaystyle H_2O\) release $\displaystyle 4$ mol of electrons on oxidation. So per $\displaystyle 1$ mol of \(\displaystyle H_2O\), the electrons released are\[n = \frac{4\ \text{mol } e^-}{2\ \text{mol } H_2O} \times 1\ \text{mol } H_2O = 2\ \text{mol } e^- \]This is the step people rush past: it is tempting to read "4e⁻" off the equation and use \(\displaystyle n=4\), but that $\displaystyle 4$ belongs to $\displaystyle 2$ mol of water, not $\displaystyle 1$ mol — halve it.Substituting into \(\displaystyle Q = nF\):\[Q = 2\ \text{mol} \times 96500\ \text{C mol}^{-1} = 193000\ \text{C} = 1.93 \times 10^{5}\ \text{C} \](ii) Oxidation of \(\displaystyle FeO\) to \(\displaystyle Fe_2O_3\)Here the electron count comes from the change in oxidation number of iron, not from balancing oxygen. In \(\displaystyle FeO\), iron is \(\displaystyle \mathrm{Fe^{2+}}\); in \(\displaystyle Fe_2O_3\), iron is \(\displaystyle Fe^{3+}\). Oxidation is the loss of electrons:\[Fe^{2+} \rightarrow Fe^{3+} + e^{-} \]Each iron atom loses exactly $\displaystyle 1$ electron. Since $\displaystyle 1$ mol of \(\displaystyle FeO\) contains $\displaystyle 1$ mol of \(\displaystyle Fe\) atoms,\[n = 1\ \text{mol } e^- \]Substituting into \(\displaystyle Q = nF\):\[Q = 1\ \text{mol} \times 96500\ \text{C mol}^{-1} = 96500\ \text{C} = 9.65 \times 10^{4}\ \text{C} \]Answer: (i) \(\displaystyle 1.93 \times 10^{5}\ \text{C}\) ($\displaystyle 193000$ C); (ii) \(\displaystyle 9.65 \times 10^{4}\ \text{C}\) ($\displaystyle 96500$ C)

The mass deposited at an electrode is fixed by the total charge passed and the number of electrons each ion needs to be discharged — Faraday's laws, not the current alone.The cathode half-reaction for nickel is\[Ni^{2+} + 2e^- \rightarrow Ni \]so depositing $\displaystyle 1$ mole of Ni requires $\displaystyle 2$ moles of electrons, i.e. \(\displaystyle 2F\) of charge (\(\displaystyle F\) = Faraday constant = $\displaystyle 96500$ C/mol e\(\displaystyle ^-\)).Step $\displaystyle 1$: Find the total charge passed.\[Q = I \times t \]where \(\displaystyle I\) is the current in amperes and \(\displaystyle t\) is the time in seconds (not minutes — this is the step people slip on).\[t = 20\ \text{min} = 20 \times 60\ \text{s} = 1200\ \text{s} \]\[Q = 5\ \text{A} \times 1200\ \text{s} = 6000\ \text{C} \]Step $\displaystyle 2$: Convert charge to moles of electrons.\[n(e^-) = \frac{Q}{F} = \frac{6000\ \text{C}}{96500\ \text{C mol}^{-1}} = 0.06218\ \text{mol}\ e^- \]Step $\displaystyle 3$: Convert moles of electrons to moles of Ni.Since each \(\displaystyle \mathrm{Ni^{2+}}\) ion needs $\displaystyle 2$ electrons,\[n(Ni) = \frac{n(e^-)}{2} = \frac{0.06218}{2} = 0.03109\ \text{mol} \]Step $\displaystyle 4$: Convert moles of Ni to mass.Using the molar mass of nickel, \(\displaystyle M(Ni) = 58.7\ \text{g mol}^{-1}\):\[m(Ni) = n(Ni) \times M(Ni) = 0.03109\ \text{mol} \times 58.7\ \text{g mol}^{-1} = 1.825\ \text{g} \]The nitrate ion and the fact that the solution is aqueous do not enter the calculation at all — only the charge passed and the number of electrons the cathode reaction consumes per ion decide how much metal comes out.Answer: $\displaystyle 1.825$ g of Ni is deposited at the cathode.

In cells wired in series, the same current flows through all three — which means the same charge (and so the same number of equivalents) is deposited in each cell, never the same mass. Ag, Cu, and Zn have different molar masses and pick up different numbers of electrons per ion, so their masses come out different even though the charge passing through them is identical.Step $\displaystyle 1$: Find the charge that passed, from cell B.Cell B contains \(\displaystyle AgNO_3\); silver is deposited by\[Ag^{+} + e^{-} \rightarrow Ag \]so \(\displaystyle n = 1\) electron per Ag atom deposited. Using Faraday's first law of electrolysis,\[m = \frac{M}{nF}\,Q \]where \(\displaystyle m\) = mass deposited (g), \(\displaystyle M\) = molar mass (g/mol), \(\displaystyle n\) = electrons transferred per ion, \(\displaystyle F = 96500\ \text{C/mol}\) is Faraday's constant, and \(\displaystyle Q\) = charge passed (C). Solving for \(\displaystyle Q\), with \(\displaystyle M(Ag) = 108\ \text{g/mol}\) and \(\displaystyle m = 1.45\ \text{g}\):\[Q = \frac{m\,nF}{M} = \frac{1.45\ \text{g} \times 1 \times 96500\ \text{C/mol}}{108\ \text{g/mol}} = 1295.6\ \text{C} \]Step $\displaystyle 2$: Convert charge to time using the cell's current.Charge and current are related by \(\displaystyle Q = I t\), so\[t = \frac{Q}{I} = \frac{1295.6\ \text{C}}{1.5\ \text{A}} = 863.7\ \text{s} \]Converting to minutes: \(\displaystyle 863.7\ \text{s} \div 60 = 14.40\ \text{min}\), i.e. about $\displaystyle 14$ min $\displaystyle 24$ s.This is the step people skip: since the cells are in series, the same \(\displaystyle t\) and the same \(\displaystyle Q\) apply to cells A and C too — you never need to solve them separately.Step $\displaystyle 3$: Use the shared charge to get moles of electrons.\[\text{moles of } e^{-} = \frac{Q}{F} = \frac{1295.6\ \text{C}}{96500\ \text{C/mol}} = 0.013426\ \text{mol } e^{-} \]This is the one quantity that is identical in all three cells — not the mass, the moles of electrons transferred.Step $\displaystyle 4$: Find the mass of copper (cell C).Copper is deposited by \(\displaystyle Cu^{2+} + 2e^{-} \rightarrow Cu\), so \(\displaystyle n = 2\). Each mole of Cu needs $\displaystyle 2$ moles of electrons:\[\text{moles of Cu} = \frac{\text{moles of } e^{-}}{n} = \frac{0.013426}{2} = 0.0067130\ \text{mol} \]With \(\displaystyle M(Cu) = 63.5\ \text{g/mol}\):\[m(Cu) = 0.0067130\ \text{mol} \times 63.5\ \text{g/mol} = 0.4263\ \text{g} \]Step $\displaystyle 5$: Find the mass of zinc (cell A).Zinc is deposited by \(\displaystyle Zn^{2+} + 2e^{-} \rightarrow Zn\), also \(\displaystyle n = 2\):\[\text{moles of Zn} = \frac{0.013426}{2} = 0.0067130\ \text{mol} \]With \(\displaystyle M(Zn) = 65.4\ \text{g/mol}\):\[m(Zn) = 0.0067130\ \text{mol} \times 65.4\ \text{g/mol} = 0.4390\ \text{g} \]Notice copper and zinc deposit almost the same number of moles (both take $\displaystyle 2$ electrons per ion) — the mass difference between them comes only from their different molar masses, not from unequal charge.Rounding each result to three significant figures, matching the precision of the given mass ($\displaystyle 1.45$ g) and current ($\displaystyle 1.5$ A):Answer: the current flowed for \(\displaystyle t \approx 863.7\ \text{s} \approx 14.4\ \text{min}\) (about $\displaystyle 14$ min $\displaystyle 24$ s); the mass of copper deposited was \(\displaystyle \approx 0.426\ \text{g}\), and the mass of zinc deposited was \(\displaystyle \approx 0.439\ \text{g}\).

A reaction between a metal ion/species and another species is feasible only if the species being reduced has the higher (more positive) standard reduction potential — the one with the higher \(\displaystyle E^\circ\) gets reduced, the one with the lower \(\displaystyle E^\circ\) gets oxidised, and the resulting \(\displaystyle E^\circ_{cell}\) must come out positive.The rule to apply each time: for the pair given, write both half-reactions as reductions, look up their standard reduction potentials \(\displaystyle E^\circ\), and compute \[E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] where the cathode is the half-reaction with the higher \(\displaystyle E^\circ\) (it actually gets reduced) and the anode is the one with the lower \(\displaystyle E^\circ\) (it gets oxidised, running in reverse of the tabulated reduction). If \(\displaystyle E^\circ_{cell} > 0\), the reaction is feasible; if \(\displaystyle E^\circ_{cell} < 0\), it is not.The standard reduction potentials needed, from Table $\displaystyle 3.1$, are: \[Fe^{3+} + e^- \rightarrow Fe^{2+}, \quad E^\circ = 0.77\ \text{V} \] \[I_2 + 2e^- \rightarrow 2I^-, \quad E^\circ = 0.54\ \text{V} \] \[Ag^+ + e^- \rightarrow Ag, \quad E^\circ = 0.80\ \text{V} \] \[Cu^{2+} + 2e^- \rightarrow Cu, \quad E^\circ = 0.34\ \text{V} \] \[Br_2 + 2e^- \rightarrow 2Br^-, \quad E^\circ = 1.09\ \text{V} \]A step people get wrong here: the number of electrons in a half-reaction ($\displaystyle 1$ for \(\displaystyle Ag^+/Ag\), $\displaystyle 2$ for \(\displaystyle Cu^{2+}/Cu\)) never enters the \(\displaystyle E^\circ_{cell}\) subtraction — \(\displaystyle E^\circ\) is an intensive quantity, so you never multiply it up to "balance" electrons before subtracting.(i) \(\displaystyle \mathrm{Fe^{3+}(aq)}\) and \(\displaystyle I^-\)(aq)The possible reaction is \(\displaystyle 2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\).Comparing the two \(\displaystyle E^\circ\) values, \(\displaystyle Fe^{3+}/Fe^{2+}\) ($\displaystyle 0.77$ V) is higher than \(\displaystyle I_2/I^-\) ($\displaystyle 0.54$ V), so \(\displaystyle \mathrm{Fe^{3+}}\) is reduced (cathode) and \(\displaystyle I^-\) is oxidised (anode): \[E^\circ_{cell} = 0.77\ \text{V} - 0.54\ \text{V} = +0.23\ \text{V} \] Positive, so the reaction is feasible.(ii) \(\displaystyle Ag^+\)(aq) and Cu(s)The possible reaction is \(\displaystyle 2Ag^+ + Cu \rightarrow 2Ag + Cu^{2+}\).\(\displaystyle Ag^+/Ag\) ($\displaystyle 0.80$ V) is higher than \(\displaystyle Cu^{2+}/Cu\) ($\displaystyle 0.34$ V), so \(\displaystyle Ag^+\) is reduced (cathode) and Cu is oxidised (anode): \[E^\circ_{cell} = 0.80\ \text{V} - 0.34\ \text{V} = +0.46\ \text{V} \] Positive, so the reaction is feasible.(iii) \(\displaystyle \mathrm{Fe^{3+}(aq)}\) and \(\displaystyle Br^-\)(aq)The possible reaction is \(\displaystyle 2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2\).Here \(\displaystyle Br_2/Br^-\) ($\displaystyle 1.09$ V) is higher than \(\displaystyle Fe^{3+}/Fe^{2+}\) ($\displaystyle 0.77$ V) — so if anything, \(\displaystyle Br_2\) would be reduced, not \(\displaystyle Fe^{3+}\). Taking \(\displaystyle Fe^{3+}/Fe^{2+}\) as the (attempted) cathode and \(\displaystyle Br_2/Br^-\) as the (attempted) anode: \[E^\circ_{cell} = 0.77\ \text{V} - 1.09\ \text{V} = -0.32\ \text{V} \] Negative, so the reaction is not feasible.(iv) Ag(s) and \(\displaystyle \mathrm{Fe^{3+}(aq)}\)The possible reaction is \(\displaystyle Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}\).\(\displaystyle Ag^+/Ag\) ($\displaystyle 0.80$ V) is higher than \(\displaystyle Fe^{3+}/Fe^{2+}\) ($\displaystyle 0.77$ V), which means Ag would need to be reduced and \(\displaystyle \mathrm{Fe^{3+}}\) oxidised for a spontaneous cell — but the proposed reaction asks for the opposite (Ag oxidised, \(\displaystyle \mathrm{Fe^{3+}}\) reduced). Taking \(\displaystyle Fe^{3+}/Fe^{2+}\) as the (attempted) cathode: \[E^\circ_{cell} = 0.77\ \text{V} - 0.80\ \text{V} = -0.03\ \text{V} \] Negative, so the reaction is not feasible. This margin is the one people wave away as "close enough" — but even a small negative \(\displaystyle E^\circ_{cell}\) still means the reaction does not proceed as written.(v) \(\displaystyle Br_2\)(aq) and \(\displaystyle \mathrm{Fe^{2+}(aq)}\)The possible reaction is \(\displaystyle Br_2 + 2Fe^{2+} \rightarrow 2Br^- + 2Fe^{3+}\).\(\displaystyle Br_2/Br^-\) ($\displaystyle 1.09$ V) is higher than \(\displaystyle Fe^{3+}/Fe^{2+}\) ($\displaystyle 0.77$ V), so \(\displaystyle Br_2\) is reduced (cathode) and \(\displaystyle \mathrm{Fe^{2+}}\) is oxidised to \(\displaystyle \mathrm{Fe^{3+}}\) (anode): \[E^\circ_{cell} = 1.09\ \text{V} - 0.77\ \text{V} = +0.32\ \text{V} \] Positive, so the reaction is feasible.**Answer: Feasible — (i) \(\displaystyle Fe^{3+} + I^- \) (\(\displaystyle E^\circ_{cell} = +0.23\) V), (ii) \(\displaystyle Ag^+ + Cu\) (\(\displaystyle E^\circ_{cell} = +0.46\) V), (v) \(\displaystyle Br_2 + Fe^{2+}\) (\(\displaystyle E^\circ_{cell} = +0.32\) V). Not feasible — (iii) \(\displaystyle Fe^{3+} + Br^-\) (\(\displaystyle E^\circ_{cell} = -0.32\) V), (iv) \(\displaystyle Ag + Fe^{3+}\) (\(\displaystyle E^\circ_{cell} = -0.03\) V).

At the cathode, the species with the highest (most positive) standard reduction potential is reduced first; at an inert anode, the species with the lowest reduction potential is oxidised first — except that an active metal electrode oxidises itself before anything in the solution, and overvoltage can override the plain \(\displaystyle E^\circ\) ranking for gases evolving on platinum.The relevant standard reduction potentials (vs SHE) are:\[\mathrm{Ag^+ + e^- \rightarrow Ag} \qquad E^\circ = +0.80\ \text{V} \] \[\mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \qquad E^\circ = +0.34\ \text{V} \] \[\mathrm{2H^+ + 2e^- \rightarrow H_2} \qquad E^\circ = 0.00\ \text{V} \] \[\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O} \qquad E^\circ = +1.23\ \text{V} \] \[\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-} \qquad E^\circ = +1.36\ \text{V} \]A species is reduced more easily the more positive its \(\displaystyle E^\circ\) is; it is oxidised more easily the less positive (more negative) its \(\displaystyle E^\circ\) is, because oxidation is just the reverse of that same half-reaction.(i) Aqueous \(\displaystyle AgNO_3\), silver electrodesAt the cathode the only reducible cation present in any quantity is \(\displaystyle Ag^+\), and its \(\displaystyle E^\circ = +0.80\) V sits far above \(\displaystyle H^+/H_2\) at \(\displaystyle 0.00\) V, so silver plates out: \[\mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)} \] The anode here is silver metal itself, not an inert electrode — this is the step people get wrong: an active electrode dissolves in preference to oxidising water or \(\displaystyle NO_3^-\), so the electrode material is almost always what actually gets oxidised: \[\mathrm{Ag(s) \rightarrow Ag^+(aq) + e^-} \] Net result: silver simply moves from the anode to the cathode and the solution's \(\displaystyle Ag^+\) concentration stays essentially constant — this is exactly how silver is electrorefined.(ii) Aqueous \(\displaystyle AgNO_3\), platinum electrodesThe cathode reaction is unchanged: \(\displaystyle Ag^+\) is still the easiest species present to reduce, so \(\displaystyle Ag(s)\) deposits there.The anode is now inert platinum and cannot dissolve, so the choice is between oxidising \(\displaystyle H_2O\) or \(\displaystyle NO_3^-\). Nitrogen in \(\displaystyle NO_3^-\) is already at its highest oxidation state, \(\displaystyle +5\), so it cannot be oxidised any further; water is oxidised instead: \[\mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-} \] Oxygen gas is evolved at the anode.(iii) Dilute \(\displaystyle H_2SO_4\), platinum electrodesAt the cathode, \(\displaystyle H^+\) is reduced rather than \(\displaystyle SO_4^{2-}\), because sulphate cannot accept electrons under these conditions: \[\mathrm{2H^+(aq) + 2e^- \rightarrow H_2(g)} \] At the anode, sulphur in \(\displaystyle SO_4^{2-}\) is already at its highest oxidation state, \(\displaystyle +6\) — the same situation as \(\displaystyle NO_3^-\) above — so water is oxidised instead: \[\mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-} \] So dilute \(\displaystyle H_2SO_4\) between inert electrodes simply electrolyses water: \(\displaystyle H_2\) collects at the cathode and \(\displaystyle O_2\) at the anode, in the usual $\displaystyle 2$:$\displaystyle 1$ volume ratio.(iv) Aqueous \(\displaystyle CuCl_2\), platinum electrodesAt the cathode, \(\displaystyle \mathrm{Cu^{2+}}\) (\(\displaystyle E^\circ = +0.34\) V) is reduced in preference to \(\displaystyle H^+/H_2O\) (\(\displaystyle E^\circ = 0.00\) V): \[\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)} \] Copper metal deposits on the cathode.At the anode there are two real competitors: \(\displaystyle Cl^-\) (\(\displaystyle E^\circ = +1.36\) V for \(\displaystyle Cl_2/Cl^-\)) and \(\displaystyle H_2O\) (\(\displaystyle E^\circ = +1.23\) V for \(\displaystyle O_2/H_2O\)). Reading the plain table, water has the lower \(\displaystyle E^\circ\) and should be oxidised first — but this is exactly where the plain ranking fails: oxygen evolution on a platinum surface carries a large overvoltage, needing noticeably more than the thermodynamic \(\displaystyle 1.23\) V to actually proceed, while chlorine evolution has very little overvoltage. Once that overvoltage is accounted for, chlorine is oxidised first in practice: \[\mathrm{2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-} \] Chlorine gas is evolved at the anode, and the solution's blue-green colour fades as \(\displaystyle \mathrm{Cu^{2+}}\) is consumed.Answer: (i) Ag deposits at the cathode while the Ag anode dissolves (silver just moves electrode to electrode). (ii) Ag deposits at the cathode; \(\displaystyle O_2\) gas evolves at the anode. (iii) \(\displaystyle H_2\) gas evolves at the cathode; \(\displaystyle O_2\) gas evolves at the anode — net electrolysis of water. (iv) Cu deposits at the cathode; \(\displaystyle Cl_2\) gas evolves at the anode (because of the high overvoltage of \(\displaystyle O_2\) evolution on platinum).