Electrometallurgy (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2016, 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. A galvanic (Daniell-type) cell couples the Fe²⁺/Fe and Cu²⁺/Cu half-reactions; iron is oxidised and copper is deposited, and both standard reduction potentials are supplied at 25 °C.

Given data
Quantity	Symbol	Value
Cathode half-reaction (reduction)	Cu²⁺ + 2e⁻ → Cu	E° = +0.34 V
Anode half-reaction (as reduction)	Fe²⁺ + 2e⁻ → Fe	E° = −0.44 V
Electrons transferred	n	2
Temperature	T	298.15 K (25 °C)
Faraday constant	F	96 485 C/mol
Gas constant	R	8.314 J/(mol·K)
Part (d) concentrations	[Cu²⁺], [Fe²⁺]	0.5 M, 1.5 M

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

Figure 7.1 — The Fe/Cu galvanic cell. Iron, the less noble metal, is oxidised at the anode and supplies electrons through the external circuit to reduce Cu²⁺ at the cathode; the concentrations shown are those used in part (d).

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).

Part (a) — identify electrodes and combine the half-cell potentials. Copper's reduction potential (+0.34 V) is higher (more positive) than iron's (−0.44 V), so copper is reduced at the cathode and iron is oxidised at the anode (its half-reaction is reversed). The cell potential is cathode minus anode, both read as reduction potentials: $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-0.44)$$ $$\boxed{E^\circ_{cell}=0.78\ \text{V}}$$ A positive $E^\circ_{cell}$ confirms the reaction as written, $\mathrm{Fe+Cu^{2+}\rightarrow Fe^{2+}+Cu}$, is spontaneous under standard conditions — consistent with iron being the more reactive (less noble) metal.
Part (b) — standard free energy. Two electrons are transferred per formula unit ($n=2$), so $$\Delta G^\circ=-nFE^\circ_{cell}=-2\times96\,485\times0.78$$ $$\boxed{\Delta G^\circ=-150\,517\ \text{J/mol}=-150.5\ \text{kJ/mol}}$$ The negative sign is the thermodynamic statement of the same spontaneity already read off the positive $E^\circ_{cell}$ in part (a) — the two are not independent checks so much as the same fact expressed twice.
Part (c) — equilibrium constant. At equilibrium $\Delta G^\circ=-RT\ln K$, so $$\ln K=\frac{nFE^\circ_{cell}}{RT}=\frac{2\times96\,485\times0.78}{8.314\times298.15}=\frac{150\,517}{2478.8}=60.72$$ $$K=e^{60.72}\approx2.35\times10^{26}$$ $$\boxed{K\approx2.35\times10^{26}}$$ Equivalently, using $\log_{10}K=nE^\circ_{cell}/0.0592=2\times0.78/0.0592=26.37$ at 25 °C gives the same result. A $K$ this large means the reaction runs essentially to completion — consistent with iron being routinely used as a cheap sacrificial reductant to cement copper out of dilute leach solutions in industrial practice.
Part (d) — cell potential away from standard conditions. The Nernst equation adjusts $E^\circ_{cell}$ for the actual ion concentrations, $E=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q$, with the reaction quotient for $\mathrm{Fe+Cu^{2+}\rightarrow Fe^{2+}+Cu}$ (solids excluded) written as $$Q=\frac{[\text{Fe}^{2+}]}{[\text{Cu}^{2+}]}=\frac{1.5}{0.5}=3.0$$ so $$E=0.78-\frac{8.314\times298.15}{2\times96\,485}\ln(3.0)=0.78-0.01285\times1.0986$$ $$\boxed{E=0.7659\ \text{V}\approx0.766\ \text{V}}$$ Raising the product-side concentration (Fe²⁺) relative to the reactant-side concentration (Cu²⁺) pushes $Q$ above 1 and, by Le Châtelier's principle applied through the Nernst equation, pulls $E$ slightly below $E^\circ_{cell}$ — exactly the small downward shift computed here.

Final results — Question 7
Part	Quantity	Result
(a)	Standard cell potential, E°	0.78 V
(b)	Standard free energy, ΔG°	−150.5 kJ/mol
(c)	Equilibrium constant, K	2.35 × 10²⁶
(d)	Cell potential at [Cu²⁺]=0.5 M, [Fe²⁺]=1.5 M	0.766 V

Topic. Electrochemistry of a galvanic cell: cell potential, Gibbs free energy, equilibrium constant and the Nernst equation.

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

Why this works. All four parts are one piece of thermodynamics viewed through different lenses. The cell potential is simply the difference in electrochemical driving force between the two half-reactions; multiplying that potential by the charge moved ($nF$) converts it to the same free-energy currency used everywhere else in thermodynamics, and $\Delta G^\circ=-RT\ln K$ is the standard bridge from free energy to an equilibrium constant — the same relation used for any reaction, electrochemical or not. The Nernst equation is nothing more than the general free-energy-of-reaction relation $\Delta G=\Delta G^\circ+RT\ln Q$ divided through by $-nF$ to convert it back into a potential; it is the tool that lets a cell's potential be predicted at whatever concentrations actually apply, not just at the standard 1 M reference state the tabulated E° values assume.

Common pitfall. The most common slip is subtracting the potentials the wrong way round (anode minus cathode), which silently flips the sign of $E^\circ_{cell}$ and, downstream, the sign of $\Delta G^\circ$ and the direction implied for K. The second is misidentifying which species belongs in the numerator of Q in the Nernst equation for part (d) — Q follows the reaction as written (products over reactants, solids omitted), not an arbitrary ratio of "the two given concentrations."

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 13 (electrochemistry); Rosenqvist, Principles of Extractive Metallurgy, Ch. 16 (electrometallurgy and cementation).

Question 7 of 7: Electrometallurgy (20 marks)

Question 7 — Electrometallurgy (20 marks)