Question 7 of 7: Electrometallurgy (20 marks)

Question 7 — Electrometallurgy (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A galvanic cell couples the Sn²⁺/Sn and Pb²⁺/Pb half-reactions; both standard reduction potentials are supplied at 25 °C, and both are negative — tin and lead are close neighbours in the electrochemical series, unlike the widely separated pairs (e.g. Fe/Cu) usually set for this question.

Find. The standard cell potential, the standard Gibbs free energy, the equilibrium constant, and the cell potential once the ion concentrations depart from standard (1 M) conditions.

Approach. Identify which half-reaction is reduced (cathode) and which is reversed and oxidised (anode) by comparing the two standard potentials, combine them for E°, convert to ΔG° and K via the standard thermodynamic relations, then apply the Nernst equation with the stated non-standard concentrations for part (d).

1. Part (a) — identify electrodes and combine the half-cell potentials. Lead's reduction potential ($-0.125$ V) is higher (less negative) than tin's ($-0.137$ V), so lead is reduced at the cathode and tin is oxidised at the anode (its half-reaction reversed). The cell potential is cathode minus anode, both read as reduction potentials: $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=(-0.125)-(-0.137)$$ $$\boxed{E^\circ_{cell}=0.012\ \text{V}}$$ A small but positive $E^\circ_{cell}$ confirms the reaction as written, $\mathrm{Sn+Pb^{2+}\rightarrow Sn^{2+}+Pb}$, is spontaneous under standard conditions — but only just, because tin and lead sit close together in the electrochemical series, unlike a widely separated pair such as Fe/Cu.
2. Part (b) — standard free energy. Two electrons are transferred per formula unit ($n=2$), so $$\Delta G^\circ=-nFE^\circ_{cell}=-2\times96\,485\times0.012$$ $$\boxed{\Delta G^\circ=-2316\ \text{J/mol}=-2.316\ \text{kJ/mol}}$$ This is roughly 65 times smaller in magnitude than the $-150.5$ kJ/mol found for the Fe/Cu couple at the same $n$ — a direct consequence of $E^\circ_{cell}$ itself being roughly 65 times smaller.
3. Part (c) — equilibrium constant. At equilibrium $\Delta G^\circ=-RT\ln K$, so $$\ln K=\frac{nFE^\circ_{cell}}{RT}=\frac{2\times96\,485\times0.012}{8.314\times298.15}=\frac{2315.6}{2478.8}=0.9342$$ $$K=e^{0.9342}$$ $$\boxed{K\approx2.55}$$ $K$ of order unity is the thermodynamic signature of a weak driving force: the reaction still favours products at equilibrium, but only mildly, in sharp contrast to the $K\approx2.35\times10^{26}$ that a large-$E^\circ_{cell}$ couple like Fe/Cu produces. Practically, this means the Sn/Pb couple cannot be relied on to run to completion the way a strongly favourable couple can.
4. Part (d) — cell potential away from standard conditions. The Nernst equation, $E=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q$, needs the reaction quotient for $\mathrm{Sn+Pb^{2+}\rightarrow Sn^{2+}+Pb}$ (solids excluded): $$Q=\frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}=\frac{1.0}{0.1}=10.0$$ so $$E=0.012-\frac{8.314\times298.15}{2\times96\,485}\ln(10.0)=0.012-0.01285\times2.3026$$ $$\boxed{E\approx-0.0176\ \text{V}}$$ Because $E^\circ_{cell}$ started so small, even this modest departure from standard concentrations (product-favoured $Q=10$) is enough to flip the sign: at these concentrations the reaction as written is no longer spontaneous, and it would in fact run in reverse. This is the genuine teaching point of pairing two electrochemically close metals rather than a widely separated pair — a thermodynamically favourable reaction under standard conditions is not automatically favourable once real, non-standard concentrations are substituted.