Galvanic Cell — Sn/Pb Concentration and Potential: 21-Mat-A1 Thermodynamics, December 2017, Q6

Question 6: Galvanic Cell — Sn/Pb Concentration and Potential (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. $E^\circ(Sn^{2+}/Sn)=-0.137$ V; $E^\circ(Pb^{2+}/Pb)=-0.125$ V; $n=2$ electrons transferred; $T=298.15$ K; $F=96{,}485$ C mol⁻¹; $R=8.314$ J K⁻¹mol⁻¹. Non-standard part (d): $[Pb^{2+}]=0.1$ M, $[Sn^{2+}]=1.0$ M.

Find. (a) $E^\circ_{cell}$; (b) $\Delta G^\circ$; (c) $K$; (d) $E_{cell}$ at the stated concentrations.

Approach. Identify Sn as the anode (oxidized, per the written cell reaction) and Pb as the cathode (reduced); $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ using the two given reduction potentials as-is, then chain into $\Delta G^\circ=-nFE^\circ$, $K=\exp(-\Delta G^\circ/RT)$, and the Nernst equation.

(a) Standard cell potential. The written reaction reduces $Pb^{2+}$ (cathode) and oxidizes Sn (anode), so $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=E^\circ(Pb^{2+}/Pb)-E^\circ(Sn^{2+}/Sn)=-0.125-(-0.137)=\boxed{+0.012\ \text{V}}.$$
(b) Standard free energy. With $n=2$: $$\Delta G^\circ=-nFE^\circ_{cell}=-(2)(96{,}485)(0.012)=\boxed{-2316\ \text{J}\,\text{mol}^{-1}}\ (-2.32\ \text{kJ}\,\text{mol}^{-1}).$$
(c) Equilibrium constant. From $\Delta G^\circ=-RT\ln K$: $$K=\exp\!\left(\dfrac{-\Delta G^\circ}{RT}\right)=\exp\!\left(\dfrac{2316}{(8.314)(298.15)}\right)=\exp(0.9342)=\boxed{2.55}.$$ $K$ close to 1 reflects how small $E^\circ_{cell}$ is — Sn and Pb are nearly equally (un)reactive, consistent with their reduction potentials differing by only 12 mV.
(d) Nernst equation. $Q=\dfrac{[Sn^{2+}]}{[Pb^{2+}]}=\dfrac{1.0}{0.1}=10$, and $$E_{cell}=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q=0.012-\dfrac{(8.314)(298.15)}{(2)(96{,}485)}\ln(10)=0.012-(0.01285)(2.3026)=\boxed{-0.0176\ \text{V}}.$$ Lowering $[Pb^{2+}]$ to 0.1 M pulls the cell below its (already small) standard potential and reverses its sign — under these concentrations the spontaneous direction is actually the reverse of reaction (1).

Quantity	Value
(a) $E^\circ_{cell}$	+0.012 V
(b) $\Delta G^\circ$	−2316 J mol⁻¹ (−2.32 kJ mol⁻¹)
(c) $K$	2.55
(d) $E_{cell}$ ($[Pb^{2+}]=0.1$ M, $[Sn^{2+}]=1.0$ M)	−0.0176 V

Concept & Theory

Topic. Galvanic cell thermodynamics — linking $E^\circ_{cell}$, $\Delta G^\circ$, $K$ and the Nernst equation.

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

Why this works. A cell's standard potential is simply the reduction potential of the reaction actually written as reduction (cathode) minus that of the reaction written as reduction but occurring as oxidation (anode); the resulting $E^\circ_{cell}$ is a direct electrical measure of $\Delta G^\circ$ per electron transferred, and the Nernst equation extends this to any non-standard concentration via the reaction quotient $Q$.

Common pitfall. Flipping the sign of $E^\circ(Sn^{2+}/Sn)$ when identifying the anode — both half-reactions are given as REDUCTION potentials, so the cell potential is a straight subtraction, never an addition of a "reversed" sign.

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

Question 6 of 7: Galvanic Cell — Sn/Pb Concentration and Potential

Question 6: Galvanic Cell — Sn/Pb Concentration and Potential (20 marks)