Electrometallurgy (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2019, Q7

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. The overall reaction $\mathrm{Zn(s)+Cd^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cd(s)}$ transfers $n=2$ electrons; zinc is oxidised (anode) and cadmium is reduced (cathode), each with the standard reduction potential listed.

Given data
Quantity	Symbol	Value
Reduction potential, Cd²⁺/Cd (cathode)	E°_Cd	−0.40 V
Reduction potential, Zn²⁺/Zn (anode)	E°_Zn	−0.76 V
Electrons transferred	n	2
Temperature	T	298.15 K (25 °C)
Faraday constant	F	96 485 C/mol
Part (d) concentrations	[Cd²⁺], [Zn²⁺]	0.5 M, 1.5 M

Find. The standard cell potential, standard free energy, equilibrium constant, and the actual cell potential under the stated non-standard concentrations.

Figure 7.1 — The Zn–Cd galvanic cell. Zinc is the more negative (more easily oxidised) electrode and is the anode; cadmium, less negative, is the cathode where reduction occurs. Electrons flow anode → external wire → cathode; conventional current and ion flow through the salt bridge complete the circuit.

Approach. Identify the cathode as whichever half-reaction has the higher (less negative) reduction potential, subtract to get E°, then apply $\Delta G^\circ=-nFE^\circ$, $\Delta G^\circ=-RT\ln K$, and the Nernst equation in turn.

Part (a) — identify electrodes and compute E°. Comparing the two reduction potentials, Cd²⁺/Cd (−0.40 V) is less negative than Zn²⁺/Zn (−0.76 V), so cadmium is reduced (cathode) and zinc is oxidised (anode), consistent with the reaction as written. The standard cell potential is cathode minus anode: $$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=(-0.40)-(-0.76)$$ $$\boxed{E^\circ_{\text{cell}}=0.36\ \text{V}}$$ Positive, confirming the reaction as written is spontaneous under standard conditions.
Part (b) — standard free energy. With $n=2$ and $F=96\,485$ C/mol, $$\Delta G^\circ=-nFE^\circ_{\text{cell}}=-2\times96\,485\times0.36=-69\,469\ \text{J/mol}$$ $$\boxed{\Delta G^\circ=-69.47\ \text{kJ/mol}}$$ The negative sign confirms spontaneity, consistent with the positive E° in part (a) — the two must always agree in sign.
Part (c) — equilibrium constant. At equilibrium $\Delta G^\circ=-RT\ln K$, so combining with part (b), $$\ln K=\frac{nFE^\circ_{\text{cell}}}{RT}=\frac{2\times96\,485\times0.36}{8.314\times298.15}=28.03$$ $$\boxed{K=e^{28.03}\approx1.48\times10^{12}}$$ A very large $K$ is expected from a cell potential this far above zero — the reaction goes essentially to completion, which is exactly what "spontaneous with $\Delta G^\circ\ll0$" means at equilibrium.
Part (d) — cell potential under the stated concentrations. The reaction quotient for $\mathrm{Zn(s)+Cd^{2+}\rightarrow Zn^{2+}+Cd(s)}$ (pure solids omitted) is $$Q=\frac{[Zn^{2+}]}{[Cd^{2+}]}=\frac{1.5}{0.5}=3.0$$ The Nernst equation at 298.15 K gives $$E=E^\circ_{\text{cell}}-\frac{RT}{nF}\ln Q=0.36-\frac{8.314\times298.15}{2\times96\,485}\ln(3.0)$$ $$E=0.36-0.012\,85\times1.0986=0.36-0.0141$$ $$\boxed{E=0.346\ \text{V}}$$ Raising the product-ion concentration (Zn²⁺) relative to the reactant ion (Cd²⁺) pushes $Q$ above 1 and pulls the cell potential down slightly from its standard value, exactly as Le Châtelier's principle predicts for a reaction quotient moved toward products.

Final results — Question 7
Part	Quantity	Result
(a)	Standard cell potential E°	0.36 V
(b)	Standard free energy ΔG°	−69.47 kJ/mol
(c)	Equilibrium constant K	≈ 1.48 × 10¹²
(d)	Cell potential at [Cd²⁺]=0.5 M, [Zn²⁺]=1.5 M	0.346 V

Topic. Galvanic-cell thermodynamics: standard potential, free energy, equilibrium constant and the Nernst equation.

Key relations. $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$; $\Delta G^\circ=-nFE^\circ$; $\Delta G^\circ=-RT\ln K$; Nernst equation $E=E^\circ-\frac{RT}{nF}\ln Q$.

Why this works. All four quantities are restatements of the same underlying free-energy change viewed through different lenses — a cell potential, a molar energy, an equilibrium position, and a concentration-corrected potential — so any one of them, once known, fixes the other three through the bridging relations above. The sign convention (cathode minus anode, using the tabulated REDUCTION potentials directly, without flipping the anode's sign by hand) is what most often goes wrong; the anode's sign is already correctly handled by the subtraction itself.

Common pitfall. A frequent error is reversing which half-reaction is the cathode by picking the more negative potential, which silently flips the sign of every downstream answer. On part (d), the reciprocal error — writing $Q=[Cd^{2+}]/[Zn^{2+}]$ instead of products over reactants — changes the sign of the correction term and pushes E the wrong way.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 13 (electrochemistry); Rosenqvist, Principles of Extractive Metallurgy, Ch. 4 (electrometallurgy).

Question 7 of 7: Electrometallurgy (20 marks)

Question 7 — Electrometallurgy (20 marks)