Galvanic Cell Thermodynamics — Zn/Cu²⁺ and the Nernst Equation: 21-Mat-A1 Thermodynamics, December 2015, Q6

Question 6: Galvanic Cell Thermodynamics — Zn/Cu²⁺ and the Nernst Equation (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.

Half-reaction	$E^\circ$ (V)
Cu²⁺ + 2e⁻ → Cu(s) (cathode, reduction)	+0.34
Zn²⁺ + 2e⁻ → Zn(s) (anode, reverse of oxidation)	−0.76

Find. (a) $E^\circ_{cell}$. (b) $\Delta G^\circ$. (c) $K$. (d) $E_{cell}$ at $[\text{Cu}^{2+}]=0.5$ M, $[\text{Zn}^{2+}]=2.0$ M.

Approach. Zn is oxidized (anode) and Cu²⁺ is reduced (cathode), so $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$; then $\Delta G^\circ=-nFE^\circ$, $\Delta G^\circ=-RT\ln K$, and finally the Nernst equation for the non-standard concentrations.

(a) Standard cell potential. $n=2$ electrons transferred. $$E^\circ_{cell}=E^\circ_{Cu^{2+}/Cu}-E^\circ_{Zn^{2+}/Zn}=0.34-(-0.76)=\boxed{1.10\ \text{V}}.$$
(b) Standard free energy. $$\Delta G^\circ=-nFE^\circ=-(2)(96{,}485)(1.10)=\boxed{-212{,}267\ \text{J/mol}}=-212.3\ \text{kJ/mol}.$$ Strongly negative, confirming the reaction is spontaneous as written under standard conditions.
(c) Equilibrium constant. From $\Delta G^\circ=-RT\ln K$: $$\ln K=\frac{-\Delta G^\circ}{RT}=\frac{212{,}267}{(8.314)(298.15)}=85.63\ \Longrightarrow\ K=\boxed{1.5\times10^{37}}.$$ Such a huge $K$ reflects the large $E^\circ_{cell}$ — the reaction runs essentially to completion.
(d) Cell potential at the stated concentrations. Reaction quotient $Q=[\text{Zn}^{2+}]/[\text{Cu}^{2+}]=2.0/0.5=4.0$ (solids omitted). Nernst equation: $$E=E^\circ-\frac{RT}{nF}\ln Q=1.10-\frac{(8.314)(298.15)}{(2)(96{,}485)}\ln(4.0)=1.10-(0.01285)(1.386)=\boxed{1.082\ \text{V}}.$$ Moving both concentrations away from 1 M in the direction that favours the forward reaction ($\text{Cu}^{2+}$ depleted, $\text{Zn}^{2+}$ built up) drops $Q$ below what standard conditions would give at equilibrium, but here $Q>1$ still pulls $E$ slightly below $E^\circ$.

Part	Result
(a) $E^\circ_{cell}$	1.10 V
(b) $\Delta G^\circ$	−212.3 kJ/mol
(c) $K$	1.5 × 10³⁷
(d) $E_{cell}$ (Nernst)	1.082 V

Concept & Theory

Topic. Electrochemical cell thermodynamics: standard cell potential, its link to $\Delta G^\circ$ and $K$, and the Nernst equation for non-standard conditions.

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

Why this works. The maximum non-expansion (electrical) work a cell can deliver equals $-\Delta G$, and for a cell that work is $nFE$ — equating the two links electrochemistry directly to chemical thermodynamics; the Nernst equation is just this same $\Delta G=\Delta G^\circ+RT\ln Q$ relation divided through by $-nF$.

Common pitfall. Reversing which half-reaction is the cathode vs. anode (the more positive $E^\circ$ is always the reduction/cathode in a spontaneous cell), and forgetting that solids (Zn(s), Cu(s)) do not appear in $Q$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 13 (electrochemistry, the Nernst equation).

Question 6 of 7: Galvanic Cell Thermodynamics — Zn/Cu²⁺ and the Nernst Equation

Question 6: Galvanic Cell Thermodynamics — Zn/Cu²⁺ and the Nernst Equation (20 marks)