NCERT Solutions · Class 12 Chemistry Electrochemistry

A metal displaces another from its salt solution only if it is a stronger reducing agent than that metal — and the strength of a metal as a reducing agent is read directly off its standard reduction potential.The relevant half-reaction for any metal \(\displaystyle \mathrm{M} \) is \[\mathrm{M}^{n+}(aq) + n\mathrm{e}^- \rightarrow \mathrm{M}(s), \quad E^\circ \] where \(\displaystyle E^\circ \) is the standard reduction potential — how much the metal "wants" to gain electrons and plate out as solid metal. A more negative \(\displaystyle E^\circ \) means the reverse process, \(\displaystyle \mathrm{M}(s) \rightarrow \mathrm{M}^{n+}(aq) + n\mathrm{e}^- \) (oxidation, losing electrons), is favoured instead — so that metal gives up electrons readily and is the stronger reducing agent.This is the step people get wrong: displacement is governed by which metal is more easily oxidised, so you rank by increasing \(\displaystyle E^\circ_{\text{red}} \) (most negative first), not by the numerical value ignoring sign.The standard reduction potentials (vs. SHE, in volts) for the five metals are:\[\begin{aligned} \mathrm{Mg}^{2+} + 2e^- &\rightarrow \mathrm{Mg}, & E^\circ &= -2.36\ \text{V} \\ \mathrm{Al}^{3+} + 3e^- &\rightarrow \mathrm{Al}, & E^\circ &= -1.66\ \text{V} \\ \mathrm{Zn}^{2+} + 2e^- &\rightarrow \mathrm{Zn}, & E^\circ &= -0.76\ \text{V} \\ \mathrm{Fe}^{2+} + 2e^- &\rightarrow \mathrm{Fe}, & E^\circ &= -0.44\ \text{V} \\ \mathrm{Cu}^{2+} + 2e^- &\rightarrow \mathrm{Cu}, & E^\circ &= +0.34\ \text{V} \end{aligned} \]A metal with a more negative \(\displaystyle E^\circ \) will reduce the ion of a metal with a less negative (or positive) \(\displaystyle E^\circ \), oxidising itself in the process — that is, it displaces that metal from solution. For example, since \(\displaystyle E^\circ_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = -2.36\ \text{V} \) is more negative than \(\displaystyle E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34\ \text{V} \), the spontaneous reaction is \[\mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) \rightarrow \mathrm{Mg}^{2+}(aq) + \mathrm{Cu}(s) \] so magnesium metal displaces copper from copper sulphate solution.Arranging the five values from most negative to least negative: \[-2.36\ (\mathrm{Mg}) \;<\; -1.66\ (\mathrm{Al}) \;<\; -0.76\ (\mathrm{Zn}) \;<\; -0.44\ (\mathrm{Fe}) \;<\; +0.34\ (\mathrm{Cu}) \]This gives the order of decreasing reactivity — decreasing power to displace the others from their salt solutions:\[\mathrm{Mg} > \mathrm{Al} > \mathrm{Zn} > \mathrm{Fe} > \mathrm{Cu} \]So magnesium can displace aluminium, zinc, iron, and copper from their salt solutions; aluminium can displace zinc, iron, and copper (but not magnesium); zinc can displace iron and copper; and iron can displace only copper. Copper, with the least negative (positive) \(\displaystyle E^\circ \), displaces none of the others.Answer: \(\displaystyle \mathrm{Mg} > \mathrm{Al} > \mathrm{Zn} > \mathrm{Fe} > \mathrm{Cu} \) — this is the order of decreasing reactivity, i.e., each metal displaces every metal listed after it from the solution of its salt.

A more negative standard reduction potential means the metal gives up its electrons more easily — it is a stronger reducing agent.The values given are standard reduction potentials \(\displaystyle E^{\circ} \), for the half-reaction written as\[M^{n+} + ne^{-} \rightarrow M \]A large positive \(\displaystyle E^{\circ} \) means that half-reaction happens easily, i.e. the ion \(\displaystyle M^{n+} \) is readily reduced — so the metal \(\displaystyle M \) itself has little tendency to be oxidized (it is a weak reducing agent). A large negative \(\displaystyle E^{\circ} \) means the reverse reaction, \(\displaystyle M \rightarrow M^{n+} + ne^{-} \) (oxidation), is favoured — the metal loses electrons readily, so it is a strong reducing agent.This is the step people get backwards: reducing power runs opposite to the reduction potential, not the same way. The metal with the lowest (most negative) \(\displaystyle E^{\circ} \) has the highest reducing power.List the five values in order from most positive to most negative:\[\begin{aligned} Ag^{+}/Ag &: +0.80\ \text{V} \\ Hg^{2+}/Hg &: +0.79\ \text{V} \\ Cr^{3+}/Cr &: -0.74\ \text{V} \\ Mg^{2+}/Mg &: -2.37\ \text{V} \\ K^{+}/K &: -2.93\ \text{V} \end{aligned} \]Reducing power increases as \(\displaystyle E^{\circ} \) decreases, so reading this same list from most positive (weakest reducing agent) to most negative (strongest reducing agent) already gives the increasing order of reducing power:\[Ag < Hg < Cr < Mg < K \]Ag (\(\displaystyle +0.80\ \text{V}\)) holds on to its electrons most strongly, so it is the weakest reducing agent, while K (\(\displaystyle -2.93\ \text{V}\)) gives up its electron most readily, so it is the strongest reducing agent of the five.Answer: Increasing order of reducing power: \(\displaystyle Ag < Hg < Cr < Mg < K \)

In a galvanic cell, split the overall reaction into its two half-reactions first — the one that loses electrons is oxidation and happens at the anode, the one that gains electrons is reduction and happens at the cathode.The overall cell reaction is \[Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s) \]Splitting this into half-reactions:Zinc goes from the $\displaystyle 0$ oxidation state to \(\displaystyle +2\), so it loses electrons — this is oxidation, and it occurs at the anode: \[Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \quad \text{(anode, oxidation)} \]Silver ion goes from \(\displaystyle +1\) to $\displaystyle 0$, so it gains electrons — this is reduction, and it occurs at the cathode: \[2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s) \quad \text{(cathode, reduction)} \]The two electrons released by one Zn atom exactly supply the two \(\displaystyle \mathrm{Ag^{+}}\) ions that get reduced, so the half-reactions are already balanced against each other.By the IUPAC convention, a galvanic cell is written as anode | anode solution || cathode solution | cathode, so the cell is \[Zn(s)\,|\,Zn^{2+}(aq)\,||\,Ag^{+}(aq)\,|\,Ag(s) \](i) Which electrode is negatively charged?The zinc electrode is the anode. As each Zn atom oxidises, it leaves its two electrons behind on the metal strip before they travel through the external wire. This constant supply of electrons piling up on the zinc makes it the negative electrode. The silver electrode is the cathode, where those electrons are consumed as \(\displaystyle \mathrm{Ag^{+}}\) is reduced, so it is the positive electrode.This is the opposite of an electrolytic cell — there the anode is positive because an external battery is pulling electrons away from it. In a galvanic cell the cell itself is doing the pushing, so the anode is negative.(ii) Carriers of the current in the cellTwo different carriers are at work in two different parts of the circuit: (iii) Individual reaction at each electrodeAt the anode (oxidation): \[Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \]At the cathode (reduction): \[2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s) \]Answer: Cell: \(\displaystyle Zn(s)\,|\,Zn^{2+}(aq)\,||\,Ag^{+}(aq)\,|\,Ag(s)\). (i) The zinc electrode (anode) is negatively charged. (ii) Electrons carry the current in the external circuit; ions carry the current inside the cell and salt bridge. (iii) Anode: \(\displaystyle Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}\); Cathode: \(\displaystyle 2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s)\).

A galvanic cell's \(\displaystyle E^{o}_{cell}\) is always (cathode's reduction potential) − (anode's reduction potential), and the reaction as written tells you which half is which — the species that gets oxidized is the anode.Use the standard reduction potentials from the electrochemical series (all at $\displaystyle 298$ K):\[E^{o}(Cr^{3+}/Cr) = -0.74\ \text{V}, \quad E^{o}(Cd^{2+}/Cd) = -0.40\ \text{V} \] \[E^{o}(Fe^{3+}/Fe^{2+}) = +0.77\ \text{V}, \quad E^{o}(Ag^{+}/Ag) = +0.80\ \text{V} \]Reaction (i): \(\displaystyle 2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)\)Split into half-reactions to see which one is oxidation and which is reduction.Oxidation (anode): \(\displaystyle Cr \rightarrow Cr^{3+} + 3e^{-}\), so \(\displaystyle E^{o}_{anode} = E^{o}(Cr^{3+}/Cr) = -0.74\ \text{V}\)Reduction (cathode): \(\displaystyle Cd^{2+} + 2e^{-} \rightarrow Cd\), so \(\displaystyle E^{o}_{cathode} = E^{o}(Cd^{2+}/Cd) = -0.40\ \text{V}\)The formula for cell potential is \(\displaystyle E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}\) (never anode minus cathode — that sign flip is the most common mistake here):\[E^{o}_{cell} = (-0.40) - (-0.74) = 0.34\ \text{V} \]A positive \(\displaystyle E^{o}_{cell}\) confirms the reaction is spontaneous as written, which is consistent with it being set up as a galvanic cell.To balance electrons, the half-reactions must be multiplied so both sides transfer the same number: \(\displaystyle 2Cr \rightarrow 2Cr^{3+} + 6e^{-}\) and \(\displaystyle 3Cd^{2+} + 6e^{-} \rightarrow 3Cd\), so \(\displaystyle n = 6\) — this is the number of moles of electrons in the overall balanced equation, not in one atom's half-reaction.Free energy change, using \(\displaystyle \Delta_{r}G^{o} = -nFE^{o}_{cell}\), where \(\displaystyle F = 96500\ \text{C mol}^{-1}\) is the Faraday constant:\[\Delta_{r}G^{o} = -(6)(96500\ \text{C mol}^{-1})(0.34\ \text{V}) = -196860\ \text{J mol}^{-1} = -196.86\ \text{kJ mol}^{-1} \]Equilibrium constant, from \(\displaystyle \Delta_{r}G^{o} = -2.303RT\log K\), which combined with the expression above gives\[\log K = \frac{nE^{o}_{cell}}{0.0591\ \text{V}} \](the constant \(\displaystyle 0.0591\ \text{V} = \dfrac{2.303RT}{F}\) at $\displaystyle 298$ K — carry this exact form rather than re-deriving \(\displaystyle RT/F\) each time, since it is easy to drop the $\displaystyle 2.303$).\[\log K = \frac{(6)(0.34)}{0.0591} = \frac{2.04}{0.0591} = 34.52 \]\[K = 10^{34.52} \approx 3.3 \times 10^{34} \]Reaction (ii): \(\displaystyle Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)\)Oxidation (anode): \(\displaystyle Fe^{2+} \rightarrow Fe^{3+} + e^{-}\), so \(\displaystyle E^{o}_{anode} = E^{o}(Fe^{3+}/Fe^{2+}) = 0.77\ \text{V}\)Reduction (cathode): \(\displaystyle Ag^{+} + e^{-} \rightarrow Ag\), so \(\displaystyle E^{o}_{cathode} = E^{o}(Ag^{+}/Ag) = 0.80\ \text{V}\)\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = 0.80 - 0.77 = 0.03\ \text{V} \]Here \(\displaystyle n = 1\), since each half-reaction as written already transfers exactly one electron.Free energy change:\[\Delta_{r}G^{o} = -nFE^{o}_{cell} = -(1)(96500\ \text{C mol}^{-1})(0.03\ \text{V}) = -2895\ \text{J mol}^{-1} = -2.895\ \text{kJ mol}^{-1} \]Equilibrium constant:\[\log K = \frac{nE^{o}_{cell}}{0.0591} = \frac{(1)(0.03)}{0.0591} = 0.508 \]\[K = 10^{0.508} \approx 3.2 \]Notice how small a swing in \(\displaystyle E^{o}_{cell}\) causes: reaction (i)'s cell potential is over ten times larger than reaction (ii)'s, and because \(\displaystyle K\) depends exponentially on \(\displaystyle E^{o}_{cell}\), that turns into the difference between an equilibrium constant of \(\displaystyle 10^{34}\) (reaction essentially goes to completion) and one of only about $\displaystyle 3$ (reaction stops well short of completion, with real amounts of \(\displaystyle \mathrm{Fe^{2+}}\) and \(\displaystyle \mathrm{Ag^{+}}\) left at equilibrium).Answer: (i) \(\displaystyle E^{o}_{cell} = 0.34\ \text{V}\), \(\displaystyle \Delta_{r}G^{o} = -196.86\ \text{kJ mol}^{-1}\), \(\displaystyle K \approx 3.3 \times 10^{34}\); (ii) \(\displaystyle E^{o}_{cell} = 0.03\ \text{V}\), \(\displaystyle \Delta_{r}G^{o} = -2.895\ \text{kJ mol}^{-1}\), \(\displaystyle K \approx 3.2\)

The Nernst equation just corrects the standard potential for the fact that your concentrations aren't the standard $\displaystyle 1$ M / $\displaystyle 1$ bar — get the cell reaction and the electron count \(\displaystyle n\) right first, and the correction term follows on its own.At $\displaystyle 298$ K the Nernst equation is\[E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n}\log Q \]where \(\displaystyle E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\) (cathode is always the right-hand electrode in the cell notation, anode the left), \(\displaystyle n\) is the number of electrons in the balanced overall cell reaction, and \(\displaystyle Q\) is the reaction quotient — products over reactants, each raised to its stoichiometric coefficient, with pure solids and liquids left out (activity $\displaystyle 1$) and gas pressures used in bar.The standard reduction potentials needed here (from the standard electrode potential table) are: \(\displaystyle E^{\circ}(Mg^{2+}/Mg) = -2.36\ V\), \(\displaystyle E^{\circ}(Cu^{2+}/Cu) = +0.34\ V\), \(\displaystyle E^{\circ}(Fe^{2+}/Fe) = -0.44\ V\), \(\displaystyle E^{\circ}(2H^{+}/H_2) = 0.00\ V\), \(\displaystyle E^{\circ}(Sn^{2+}/Sn) = -0.14\ V\), \(\displaystyle E^{\circ}(Br_2/Br^{-}) = +1.09\ V\).(i) Mg(s) | Mg\(\displaystyle ^{2+}\)($\displaystyle 0.001$ M) || Cu\(\displaystyle ^{2+}\)($\displaystyle 0.0001$ M) | Cu(s)Mg is the anode (oxidation), Cu is the cathode (reduction). The overall reaction is\[Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s), \qquad n = 2 \]\[E^{\circ}_{cell} = E^{\circ}(Cu^{2+}/Cu) - E^{\circ}(Mg^{2+}/Mg) = 0.34 - (-2.36) = 2.70\ V \]Solid Mg and solid Cu don't appear in \(\displaystyle Q\):\[E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2}\log\frac{[Mg^{2+}]}{[Cu^{2+}]} = 2.70 - \frac{0.0591}{2}\log\frac{0.001}{0.0001} \]\[= 2.70 - 0.02955\times\log(10) = 2.70 - 0.02955(1) = 2.6705\ V \]Rounding to three significant figures, \(\displaystyle E_{cell} = 2.67\ V\).(ii) Fe(s) | Fe\(\displaystyle ^{2+}\)($\displaystyle 0.001$ M) || H\(\displaystyle ^{+}\)($\displaystyle 1$ M) | H\(\displaystyle _2\)(g)($\displaystyle 1$ bar) | Pt(s)Fe is oxidised at the anode; H\(\displaystyle ^+\) is reduced to H\(\displaystyle _2\) at the Pt cathode.\[Fe(s) + 2H^{+}(aq) \rightarrow Fe^{2+}(aq) + H_2(g), \qquad n = 2 \]\[E^{\circ}_{cell} = E^{\circ}(H^{+}/H_2) - E^{\circ}(Fe^{2+}/Fe) = 0 - (-0.44) = 0.44\ V \]This is the step people get wrong: because $\displaystyle 2$ mol of H\(\displaystyle ^+\) are consumed, \(\displaystyle [H^+]\) enters \(\displaystyle Q\) squared, not to the first power.\[E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2}\log\frac{[Fe^{2+}]\,p_{H_2}}{[H^{+}]^{2}} = 0.44 - \frac{0.0591}{2}\log\frac{(0.001)(1)}{(1)^{2}} \]\[= 0.44 - 0.02955\times\log(10^{-3}) = 0.44 - 0.02955\times(-3) = 0.44 + 0.08865 = 0.5287\ V \]Rounded to three significant figures, \(\displaystyle E_{cell} = 0.529\ V\).(iii) Sn(s) | Sn\(\displaystyle ^{2+}\)($\displaystyle 0.050$ M) || H\(\displaystyle ^{+}\)($\displaystyle 0.020$ M) | H\(\displaystyle _2\)(g)($\displaystyle 1$ bar) | Pt(s)Same skeleton reaction as (ii), with Sn in place of Fe:\[Sn(s) + 2H^{+}(aq) \rightarrow Sn^{2+}(aq) + H_2(g), \qquad n = 2 \]\[E^{\circ}_{cell} = E^{\circ}(H^{+}/H_2) - E^{\circ}(Sn^{2+}/Sn) = 0 - (-0.14) = 0.14\ V \]\[Q = \frac{[Sn^{2+}]\,p_{H_2}}{[H^{+}]^{2}} = \frac{(0.050)(1)}{(0.020)^{2}} = \frac{0.050}{4\times10^{-4}} = 125 \]\[E_{cell} = 0.14 - \frac{0.0591}{2}\log(125) = 0.14 - 0.02955\times2.0969 = 0.14 - 0.06196 = 0.07804\ V \]Rounded to three significant figures, \(\displaystyle E_{cell} = 0.0780\ V\).(iv) Pt(s) | Br\(\displaystyle ^{-}\)($\displaystyle 0.010$ M) | Br\(\displaystyle _2\)(l) || H\(\displaystyle ^{+}\)($\displaystyle 0.030$ M) | H\(\displaystyle _2\)(g)($\displaystyle 1$ bar) | Pt(s)Reading the cell notation left to right, the left half-cell is written as the oxidation: \(\displaystyle 2Br^{-} \rightarrow Br_2 + 2e^{-}\). The right half-cell is the reduction \(\displaystyle 2H^{+} + 2e^{-} \rightarrow H_2\). Adding them:\[2Br^{-}(aq) + 2H^{+}(aq) \rightarrow Br_2(l) + H_2(g), \qquad n = 2 \]The half-cell here is being run as an oxidation, so the anode potential going into \(\displaystyle E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\) is still the reduction potential of \(\displaystyle Br_2/Br^{-}\) — don't flip its sign by hand, the formula already does that.\[E^{\circ}_{cell} = E^{\circ}(H^{+}/H_2) - E^{\circ}(Br_2/Br^{-}) = 0 - 1.09 = -1.09\ V \]\(\displaystyle Br_2\) is a pure liquid and \(\displaystyle H_2\) is at $\displaystyle 1$ bar, so both have activity $\displaystyle 1$ and drop out of \(\displaystyle Q\), leaving only the two ions — each squared, since $\displaystyle 2$ mol of each appear in the balanced equation:\[Q = \frac{[Br_2][H_2]}{[Br^{-}]^{2}[H^{+}]^{2}} = \frac{1}{(0.010)^{2}(0.030)^{2}} = \frac{1}{(1\times10^{-4})(9\times10^{-4})} = \frac{1}{9\times10^{-8}} = 1.111\times10^{7} \]\[E_{cell} = -1.09 - \frac{0.0591}{2}\log(1.111\times10^{7}) = -1.09 - 0.02955\times7.0458 = -1.09 - 0.2082 = -1.2982\ V \]Rounded to three significant figures, \(\displaystyle E_{cell} = -1.30\ V\). The negative sign says this cell reaction is not spontaneous as written at these concentrations — that is a legitimate result of the calculation, not a sign to go back and flip anything.Answer: (i) \(\displaystyle E_{cell} = 2.70 - \dfrac{0.0591}{2}\log\dfrac{[Mg^{2+}]}{[Cu^{2+}]} = 2.67\ V\); (ii) \(\displaystyle E_{cell} = 0.44 - \dfrac{0.0591}{2}\log\dfrac{[Fe^{2+}]}{[H^{+}]^{2}} = 0.529\ V\); (iii) \(\displaystyle E_{cell} = 0.14 - \dfrac{0.0591}{2}\log\dfrac{[Sn^{2+}]}{[H^{+}]^{2}} = 0.0780\ V\); (iv) \(\displaystyle E_{cell} = -1.09 - \dfrac{0.0591}{2}\log\dfrac{1}{[Br^{-}]^{2}[H^{+}]^{2}} = -1.30\ V\).

A cell's standard emf is the cathode's standard potential minus the anode's — and the two must be read off the SAME table, on the same convention.Step $\displaystyle 1$: Split the reaction into its half-reactions.\[Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq) \]Zinc goes from $\displaystyle 0$ to +$\displaystyle 2$, losing two electrons — oxidation, at the anode:\[Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \]Silver goes from +$\displaystyle 1$ (in \(\displaystyle Ag_{2}O\)) to $\displaystyle 0$, gaining those two electrons — reduction, at the cathode. So \(\displaystyle n = 2\) electrons pass per unit of reaction.Step $\displaystyle 2$: Take both standard potentials from the chapter's own table.\[E^{\circ}(Ag^{+}/Ag) = +0.80\ \text{V}, \qquad E^{\circ}(Zn^{2+}/Zn) = -0.76\ \text{V} \]Step $\displaystyle 3$: Combine them.\[E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-0.76) \]\[E^{\circ}_{cell} = 1.56\ \text{V} \]Watch the two minus signs — subtracting a negative anode potential adds to the cell emf. Writing \(\displaystyle 0.80 - 0.76 = 0.04\) V is the usual slip here.Step $\displaystyle 4$: Get the Gibbs energy from the emf.\[\Delta_{r}G^{\circ} = -nFE^{\circ}_{cell} \]where \(\displaystyle n = 2\) is the moles of electrons and \(\displaystyle F = 96500\ \text{C mol}^{-1}\) is the Faraday constant.\[\Delta_{r}G^{\circ} = -(2)(96500)(1.56) = -301080\ \text{J mol}^{-1} \]\[\Delta_{r}G^{\circ} \approx -301\ \text{kJ mol}^{-1} \]The sign is the check worth doing: a positive cell emf must give a negative \(\displaystyle \Delta_{r}G^{\circ}\), because a cell that drives current is a spontaneous reaction.A note on the real battery. In the actual watch cell the electrolyte is alkaline, so the zinc leaves as \(\displaystyle \mathrm{Zn(OH)_{2}}\) rather than as free \(\displaystyle \mathrm{Zn^{2+}}\), and the silver couple is \(\displaystyle Ag_{2}O/Ag\) at \(\displaystyle +0.34\) V against \(\displaystyle Zn(OH)_{2}/Zn\) at \(\displaystyle -1.25\) V. That gives \(\displaystyle 1.59\) V — and a real silver-oxide cell reads about \(\displaystyle 1.55\) V. The two routes land in the same place because the alkali stabilises both metals by nearly the same amount, and the two shifts cancel in the difference.Answer: \(\displaystyle E^{\circ}_{cell} = 1.56\) V and \(\displaystyle \Delta_{r}G^{\circ} = -nFE^{\circ} = -301\) kJ mol\(\displaystyle ^{-1}\)

Conductivity is how well a fixed volume of solution carries current; molar conductivity is how well one mole of the dissolved ions carries current — and these two move in opposite directions as you dilute.Conductivity, \(\displaystyle \kappa \) (kappa)Conductivity is the reciprocal of resistivity: \(\displaystyle \kappa = 1/\rho \). For a solution held between electrodes of area \(\displaystyle A \) and separated by length \(\displaystyle l \), the conductance \(\displaystyle G = 1/R \) is related to \(\displaystyle \kappa \) by\[G = \kappa \frac{A}{l} \quad\Rightarrow\quad \kappa = G \times \frac{l}{A} \]where \(\displaystyle l/A \) is the cell constant \(\displaystyle G^{*} \) of the conductivity cell. So conductivity is the conductance of a column of solution of unit length and unit cross-sectional area — physically, it measures the current-carrying capacity of the ions present per unit volume of solution. Its SI unit is \(\displaystyle \mathrm{S\,m^{-1}} \) (often quoted as \(\displaystyle \mathrm{S\,cm^{-1}} \)), where \(\displaystyle \mathrm{S} \) (siemens) \(\displaystyle = \mathrm{ohm^{-1}} \).A step people skip: conductivity is a property of the solution as a whole (per unit volume), not of the electrolyte itself — that role belongs to molar conductivity.Molar conductivity, \(\displaystyle \Lambda_m \)Molar conductivity is the conducting power of all the ions produced by one mole of electrolyte, spread through however much solution that mole occupies. It is defined as\[\Lambda_m = \frac{\kappa}{c} \]where \(\displaystyle c \) is the molar concentration. If \(\displaystyle \kappa \) is in \(\displaystyle \mathrm{S\,cm^{-1}} \) and \(\displaystyle c \) is in \(\displaystyle \mathrm{mol\,L^{-1}} \), a factor of $\displaystyle 1000$ (to convert litres to cm\(\displaystyle ^3\)) is needed to keep units consistent:\[\Lambda_m = \frac{1000 \,\kappa}{c} \quad \left(\text{units: } \mathrm{S\,cm^{2}\,mol^{-1}}\right) \]In SI units, with \(\displaystyle c \) expressed in \(\displaystyle \mathrm{mol\,m^{-3}} \), \(\displaystyle \Lambda_m = \kappa/c \) directly gives units of \(\displaystyle \mathrm{S\,m^{2}\,mol^{-1}} \).Variation with concentrationConductivity always decreases as the solution is diluted, for both strong and weak electrolytes. This is straightforward: \(\displaystyle \kappa \) counts the number of current-carrying ions per unit volume, and dilution simply reduces the number of ions present in that unit volume, whether or not the electrolyte is fully dissociated.Molar conductivity, in contrast, increases as the solution is diluted. This looks like a contradiction only until you notice what is being divided by what: \(\displaystyle \Lambda_m = \kappa/c \) accounts for the conductance of a fixed amount (one mole) of electrolyte, and as \(\displaystyle c \) falls, the volume containing that one mole grows faster than \(\displaystyle \kappa \) falls, so the quotient rises. Physically, for a strong electrolyte, dissociation is already essentially complete at all concentrations, so the rise in \(\displaystyle \Lambda_m \) on dilution comes from the ions moving apart and interfering less with each other's motion (weaker interionic attraction, so higher ionic mobility). This rise is modest and follows the Debye–Hückel–Onsager relation\[\Lambda_m = \Lambda_m^{\circ} - A\sqrt{c} \]so a plot of \(\displaystyle \Lambda_m \) against \(\displaystyle \sqrt{c} \) is nearly a straight line, and extrapolating it to \(\displaystyle c \to 0 \) gives the limiting molar conductivity \(\displaystyle \Lambda_m^{\circ} \) (also written \(\displaystyle \Lambda_m^{\infty} \)).For a weak electrolyte, the rise in \(\displaystyle \Lambda_m \) on dilution is much sharper, because dilution now also pushes the equilibrium of dissociation further to the right — the degree of dissociation \(\displaystyle \alpha \) itself increases with dilution, so more ions per mole become available to carry current, on top of the mobility effect. As \(\displaystyle c \to 0 \), \(\displaystyle \alpha \to 1 \) and \(\displaystyle \Lambda_m \) rises steeply toward \(\displaystyle \Lambda_m^{\circ} \), but the curve near \(\displaystyle c = 0 \) is too steep to extrapolate reliably from experimental data. \(\displaystyle \Lambda_m^{\circ} \) for a weak electrolyte is therefore obtained not by extrapolation but from Kohlrausch's law of independent migration of ions, using the limiting molar conductivities of a strong acid, a strong base, and the corresponding salt.Answer: Conductivity \(\displaystyle \kappa = 1/\rho = G(l/A) \) (unit \(\displaystyle \mathrm{S\,m^{-1}} \)) falls on dilution because the number of ions per unit volume drops. Molar conductivity \(\displaystyle \Lambda_m = \kappa/c \) (or \(\displaystyle 1000\kappa/c \), unit \(\displaystyle \mathrm{S\,cm^{2}\,mol^{-1}} \)) rises on dilution — modestly for strong electrolytes (via \(\displaystyle \Lambda_m = \Lambda_m^{\circ} - A\sqrt{c} \), from reduced ionic interaction), and sharply for weak electrolytes (from increasing degree of dissociation), both approaching a limiting value \(\displaystyle \Lambda_m^{\circ} \) at infinite dilution.

Molar conductivity tells you how well ONE MOLE of dissolved electrolyte conducts, so a concentration in mol/L must appear in the denominator — it is conductivity per mole, not conductivity per unit volume.The formula connecting the two is\[\Lambda_m = \dfrac{\kappa \times 1000}{M} \]where: This factor of $\displaystyle 1000$ is the step people drop. It comes from writing molar conductivity in its defining form, \(\displaystyle \Lambda_m = \kappa V\), where \(\displaystyle V\) is the volume in \(\displaystyle \text{cm}^3\) that contains $\displaystyle 1$ mole of electrolyte. If the solution has concentration \(\displaystyle M\) mol per litre, then $\displaystyle 1$ mole of solute sits in \(\displaystyle 1/M\) litres, i.e. in \(\displaystyle 1000/M\) cm\(\displaystyle ^3\). Substituting that volume gives the formula above.Now substitute the given values:\[\kappa = 0.0248\ \text{S cm}^{-1}, \qquad M = 0.20\ \text{mol L}^{-1} \]\[\Lambda_m = \frac{0.0248\ \text{S cm}^{-1} \times 1000\ \text{cm}^3\,\text{L}^{-1}}{0.20\ \text{mol L}^{-1}} \]Multiply the numerator first:\[0.0248 \times 1000 = 24.8\ \ \text{S cm}^2\,\text{L}^{-1} \]Then divide by the concentration:\[\Lambda_m = \frac{24.8}{0.20} = 124\ \text{S cm}^2\,\text{mol}^{-1} \]The units work out cleanly: \(\displaystyle \text{S cm}^{-1} \times \text{cm}^3 \div \text{mol} = \text{S cm}^2\,\text{mol}^{-1}\), confirming the answer is a molar conductivity, not a raw conductivity.Answer: \(\displaystyle \Lambda_m = 124\ \text{S cm}^2\,\text{mol}^{-1}\)

The cell constant links a measured resistance to the solution's true conductivity — it does not change with the solution you put in the cell.For any conductivity cell, the conductivity \(\displaystyle \kappa \) (a property of the solution) and the measured resistance \(\displaystyle R \) (a property of the solution and the cell's geometry) are related through the cell constant \(\displaystyle G^{*} \):\[\kappa = G^{*} \times \frac{1}{R} \]Rearranging for the cell constant:\[G^{*} = \kappa \times R \]Here \(\displaystyle \kappa \) is the conductivity of the solution (in \(\displaystyle \text{S cm}^{-1} \)) and \(\displaystyle R \) is the resistance of that same solution measured in the cell (in \(\displaystyle \Omega \)). The cell constant \(\displaystyle G^{*} \) comes out in \(\displaystyle \text{cm}^{-1} \) — it is fixed by the electrode area and separation, so once you know it for a cell, you can find the conductivity of any other solution just by measuring its resistance in that same cell.A step people skip: the cell constant must be worked out using a solution of known conductivity (like the standard KCl solution given here) before the cell can be used to find an unknown conductivity — you cannot invert the formula until \(\displaystyle G^{*} \) is pinned down this way.Substituting the given values, \(\displaystyle \kappa = 0.146 \times 10^{-3}\ \text{S cm}^{-1} \) and \(\displaystyle R = 1500\ \Omega \):\[G^{*} = \left(0.146 \times 10^{-3}\ \text{S cm}^{-1}\right) \times \left(1500\ \Omega\right) \]\[G^{*} = 0.146 \times 1500 \times 10^{-3}\ \text{cm}^{-1} \]\[G^{*} = 219 \times 10^{-3}\ \text{cm}^{-1} = 0.219\ \text{cm}^{-1} \]Answer: The cell constant is \(\displaystyle 0.219\ \text{cm}^{-1} \).

The relation \(\displaystyle Λ_m = κ/c\) only works if \(\displaystyle κ\) and \(\displaystyle c\) are in matching SI units — that's the step people trip on here, because the table gives \(\displaystyle c\) in mol L⁻¹ but \(\displaystyle κ\) in S m⁻¹.Step $\displaystyle 1$ — get the units straight, then compute \(\displaystyle Λ_m\) at each concentrationMolar conductivity is defined as \[Λ_m = \frac{κ}{c} \] where \(\displaystyle κ\) is the conductivity (S m⁻¹) and \(\displaystyle c\) is the molar concentration expressed in mol m⁻³, so that \(\displaystyle Λ_m\) comes out in S m² mol⁻¹.Since \(\displaystyle 1\ \text{L} = 10^{-3}\ \text{m}^3\), a concentration of \(\displaystyle c\) mol L⁻¹ equals \(\displaystyle 1000c\) mol m⁻³. Do the first row explicitly: at \(\displaystyle c = 0.001\ \text{M}\), the table gives \(\displaystyle 10^2κ = 1.237\), so \[κ = 1.237\times10^{-2}\ \text{S m}^{-1}, \qquad c = 0.001\times1000 = 1\ \text{mol m}^{-3} \] \[Λ_m = \frac{1.237\times10^{-2}}{1} = 1.237\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1} \]Repeating this division for every row:

\(\displaystyle c\) / mol L⁻¹	\(\displaystyle κ\) / S m⁻¹	\(\displaystyle c\) / mol m⁻³	\(\displaystyle Λ_m = κ/c\) / \(\displaystyle 10^{-2}\) S m² mol⁻¹	\(\displaystyle \sqrt{c}\) / (mol L⁻¹)^{$\displaystyle 1$/$\displaystyle 2$}
$\displaystyle 0.001$	\(\displaystyle 1.237\times10^{-2}\)	$\displaystyle 1$	$\displaystyle 1.237$	$\displaystyle 0.0316$
$\displaystyle 0.010$	\(\displaystyle 1.185\times10^{-1}\)	$\displaystyle 10$	$\displaystyle 1.185$	$\displaystyle 0.100$
$\displaystyle 0.020$	\(\displaystyle 2.315\times10^{-1}\)	$\displaystyle 20$	$\displaystyle 1.1575$	$\displaystyle 0.1414$
$\displaystyle 0.050$	\(\displaystyle 5.553\times10^{-1}\)	$\displaystyle 50$	$\displaystyle 1.1106$	$\displaystyle 0.2236$
$\displaystyle 0.100$	\(\displaystyle 1.0674\)	$\displaystyle 100$	$\displaystyle 1.0674$	$\displaystyle 0.3162$

Notice \(\displaystyle Λ_m\) falls as concentration rises — more ions crowd each other and mobility drops. This is the opposite of the trend for a weak electrolyte, whose \(\displaystyle Λ_m\) rises sharply on dilution because dissociation increases; NaCl is a strong electrolyte, fully dissociated at every concentration here, so the fall in \(\displaystyle Λ_m\) is only due to ionic interactions, not incomplete dissociation.Step $\displaystyle 2$ — plot \(\displaystyle Λ_m\) against \(\displaystyle \sqrt c\) and read \(\displaystyle Λ^0_m\) as the interceptFor a strong electrolyte, Kohlrausch's law says this plot is (very nearly) a straight line: \[Λ_m = Λ^0_m - A\sqrt{c} \] where \(\displaystyle Λ^0_m\) (the limiting molar conductivity, at infinite dilution) is the y-intercept and \(\displaystyle A\) is the slope. Plotting the five \(\displaystyle (\sqrt c,\,Λ_m)\) pairs from the table above gives points that fall almost on a line running from top-left \(\displaystyle (0.0316,\,1.237)\) down to bottom-right \(\displaystyle (0.3162,\,1.0674)\), curving only slightly at the higher concentrations — exactly what the law predicts, since it is exact only in the dilute limit.To read the intercept precisely (what drawing the best-fit line by eye and extending it to \(\displaystyle \sqrt c = 0\) accomplishes), fit \(\displaystyle Λ_m = Λ^0_m - A\sqrt c\) by least squares. With \(\displaystyle x=\sqrt c\), \(\displaystyle y = Λ_m\) (in units of \(\displaystyle 10^{-2}\ \text{S m}^2\ \text{mol}^{-1}\)), \(\displaystyle n=5\): \[\sum x = 0.8129,\quad \sum y = 5.7575,\quad \sum xy = 0.9072,\quad \sum x^2=\sum c = 0.181 \] (the last equality holds because \(\displaystyle x^2 = c\)). The slope is \[-A=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{5(0.9072)-(0.8129)(5.7575)}{5(0.181)-(0.8129)^2}=\frac{4.536-4.680}{0.905-0.661}=\frac{-0.144}{0.244}=-0.590 \] so \(\displaystyle A \approx 0.590\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1}\)(mol L⁻¹)^{-$\displaystyle 1$/$\displaystyle 2$}. The intercept is \[Λ^0_m=\frac{\sum y-(-A)\sum x}{n}=\frac{5.7575-(-0.590)(0.8129)}{5}=\frac{5.7575+0.4796}{5}=\frac{6.237}{5}=1.247\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1} \]This is the value the line would hit at \(\displaystyle \sqrt c = 0\), i.e. at infinite dilution — where the ions are far enough apart that they no longer hinder each other, which is exactly what \(\displaystyle Λ^0_m\) represents and why it can only be found by extrapolation, never by direct measurement (a truly zero concentration has no ions to carry current).Rounding to three significant figures, matching the precision of the conductivity data: \[Λ^0_m(\text{NaCl}) \approx 1.25\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1} \]Answer: \(\displaystyle Λ_m\) = $\displaystyle 1.237$, $\displaystyle 1.185$, $\displaystyle 1.1575$, $\displaystyle 1.1106$, $\displaystyle 1.0674$ (all \(\displaystyle \times10^{-2}\) S m² mol⁻¹) at c = $\displaystyle 0.001$, $\displaystyle 0.010$, $\displaystyle 0.020$, $\displaystyle 0.050$, $\displaystyle 0.100$ M respectively; extrapolating the \(\displaystyle Λ_m\) vs \(\displaystyle \sqrt c\) plot to \(\displaystyle \sqrt c = 0\) gives \(\displaystyle Λ^0_m \approx 1.25\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1}\) (equivalently $\displaystyle 125$ S cm² mol⁻¹).