Zn/Fe 2+ Galvanic Cell — Potential, Free Energy, K and Concentration Dependence: 21-Mat-A1 Thermodynamics, December 2018, Q6

Question 6: Zn/Fe²⁺ Galvanic Cell — Potential, Free Energy, K and Concentration Dependence (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.

Standard reduction potentials at 25°C
Half-reaction	$E^{\circ}$ (V)
Zn²⁺$+2e^-\rightarrow$Zn(s)	−0.76
Fe²⁺$+2e^-\rightarrow$Fe(s)	−0.44

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

Approach. Zn is oxidized (anode) and Fe²⁺ is reduced (cathode); combine $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ with $\Delta G^{\circ}=-nFE^{\circ}$, $\Delta G^{\circ}=-RT\ln K$ and the Nernst equation for part (d).

(a) Standard cell potential. Fe²⁺/Fe is the cathode (more positive/less negative), Zn²⁺/Zn is the anode: $$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=(-0.44)-(-0.76)=\boxed{+0.32\ \text{V}}.$$
(b) Standard free energy. $n=2$ electrons transferred (Zn$\rightarrow$Zn²⁺$+2e^-$; Fe²⁺$+2e^-\rightarrow$Fe): $$\Delta G^{\circ}=-nFE^{\circ}_{cell}=-(2)(96{,}485)(0.32)=\boxed{-61{,}750\ \text{J/mol}\ (-61.75\ \text{kJ/mol})}.$$
(c) Equilibrium constant. $$\ln K=\frac{-\Delta G^{\circ}}{RT}=\frac{61{,}750}{(8.314)(298.15)}=24.91\ \Rightarrow\ K=\boxed{6.59\times10^{10}}.$$
(d) Cell potential at the stated concentrations. For Zn$+$Fe²⁺$\rightarrow$Zn²⁺$+$Fe, $Q=[\text{Zn}^{2+}]/[\text{Fe}^{2+}]$: $$E=E^{\circ}_{cell}-\frac{RT}{nF}\ln Q=0.32-\frac{(8.314)(298.15)}{(2)(96{,}485)}\ln\!\left(\frac{2.0}{0.5}\right)=0.32-(0.01285)(1.386)=\boxed{+0.302\ \text{V}}.$$

Quantity	Value
(a) $E^{\circ}_{cell}$	+0.32 V
(b) $\Delta G^{\circ}$	−61.75 kJ/mol
(c) $K$	6.59×10¹⁰
(d) $E$ ([Fe²⁺]=0.5 M, [Zn²⁺]=2.0 M)	+0.302 V

Concept & Theory

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

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

Why this works. The electrical work available from a cell equals the Gibbs free energy released by its spontaneous redox reaction ($-nFE=\Delta G$), which links the measurable cell voltage directly to the reaction's thermodynamics; the Nernst equation is simply this relation evaluated at non-standard concentrations via $\Delta G=\Delta G^{\circ}+RT\ln Q$.

Common pitfall. Reversing which half-reaction is the cathode — since both potentials are negative, it is tempting to subtract in the wrong order; the cathode is always the LESS negative (more easily reduced) couple, here Fe²⁺/Fe, and getting this backwards flips the sign of $E^{\circ}$, $\Delta G^{\circ}$ and inverts $K$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 14 (electrochemical cells, the Nernst equation).

Question 6 of 7: Zn/Fe 2+ Galvanic Cell — Potential, Free Energy, K and Concentration Dependence

Question 6: Zn/Fe2+ Galvanic Cell — Potential, Free Energy, K and Concentration Dependence (20 marks)

Question 6: Zn/Fe²⁺ Galvanic Cell — Potential, Free Energy, K and Concentration Dependence (20 marks)