Galvanic Cell Thermodynamics for Ba/Cu 2+: 21-Mat-A1 Thermodynamics, May 2017, Q6

Question 6: Galvanic Cell Thermodynamics for Ba/Cu²⁺ (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. Cell reaction $Ba(s)+Cu^{2+}(aq)\to Ba^{2+}(aq)+Cu(s)$; $E^\circ(Cu^{2+}/Cu)=+0.34$ V, $E^\circ(Ba^{2+}/Ba)=-2.92$ V; $T=298$ K; part (d) concentrations $[Cu^{2+}]=0.5$ M, $[Ba^{2+}]=2.0$ M.

Find. $E^\circ_{cell}$, $\Delta G^\circ$, $K$, and the concentration-adjusted cell potential $E$.

Approach. Identify Cu²⁺/Cu as the reduction (cathode) and Ba/Ba²⁺ as the oxidation (anode), so $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$; convert to $\Delta G^\circ=-nFE^\circ$ and $K$ via $\Delta G^\circ=-RT\ln K$; then apply the Nernst equation for part (d).

(a) Standard cell potential. Cu²⁺ is reduced at the cathode, Ba is oxidised at the anode: $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-2.92)=\boxed{+3.26\ \text{V}}.$$
(b) Standard free energy. $n=2$ electrons transfer per formula unit ($F=96{,}485$ C/mol): $$\Delta G^\circ=-nFE^\circ_{cell}=-(2)(96{,}485)(3.26)=\boxed{-629.1\ \text{kJ}}.$$
(c) Equilibrium constant. $\Delta G^\circ=-RT\ln K$ at $T=298.15$ K: $$\ln K=\dfrac{-\Delta G^\circ}{RT}=\dfrac{629{,}100}{(8.314)(298.15)}=253.8\ \Rightarrow\ \log_{10}K=\boxed{110.2}.$$ $K$ is astronomically large, as expected from a strongly positive $E^\circ_{cell}$ — the reaction runs essentially to completion.
(d) Nernst equation at the stated concentrations. $Ba(s)$ and $Cu(s)$ are pure solids (activity 1), so $Q=[Ba^{2+}]/[Cu^{2+}]=2.0/0.5=4.0$: $$E=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q=3.26-\dfrac{(8.314)(298.15)}{(2)(96{,}485)}\ln(4.0)=3.26-0.0128(1.386)=\boxed{3.242\ \text{V}}.$$ Away from standard concentrations the cell potential barely moves (−18 mV) because the $\ln Q$ term is scaled by the small factor $RT/nF\approx13$ mV, while $E^\circ_{cell}$ itself is over 3 V.

Quantity	Value
$E^\circ_{cell}$	+3.26 V
$\Delta G^\circ$	−629.1 kJ
$\log_{10}K$	110.2
$E$ ($[Cu^{2+}]=0.5$ M, $[Ba^{2+}]=2.0$ M)	+3.242 V

Concept & Theory

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

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

Why this works. The electrical work a cell can deliver at reversible (zero-current) conditions equals $-\Delta G$, which is why $E^\circ$, $\Delta G^\circ$ and $K$ are three faces of the same quantity; the Nernst equation is simply this relation applied to an arbitrary (non-standard) reaction quotient $Q$ instead of the standard-state $Q=1$.

Common pitfall. Writing $E^\circ_{cell}=E^\circ_{anode}-E^\circ_{cathode}$ (reversed) or using the wrong sign/species in $Q$ — only the aqueous ions appear in $Q$, since $Ba(s)$ and $Cu(s)$ have unit activity.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 15 (electrochemistry, Nernst equation, relation between $E^\circ$, $\Delta G^\circ$ and $K$).

Question 6 of 7: Galvanic Cell Thermodynamics for Ba/Cu 2+

Question 6: Galvanic Cell Thermodynamics for Ba/Cu2+ (20 marks)

Question 6: Galvanic Cell Thermodynamics for Ba/Cu²⁺ (20 marks)