Question

A cell in which the following reaction occurs:
\[ \text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)} \] has standard electrode potential \(E^\circ_{\text{cell}} = 1.1~\text{V}\) at \(298~\text{K}\). Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Hint:

Use \(\Delta G^\circ = -n F E^\circ\) first to find the energy change, then connect to the equilibrium constant via \(\Delta G^\circ = -RT\ln K\). For quick order-of-magnitude checks, convert to base-10 using \(\log_{10}K = \dfrac{n F E^\circ}{2.303\,RT}\).

Answer 1

For a galvanic cell at standard state, \(\Delta G^\circ = -n F E^\circ_{\text{cell}}\) and \(\Delta G^\circ = -RT \ln K\). Hence, \(K = \exp\!\left(\dfrac{n F E^\circ_{\text{cell}}}{RT}\right)\).
Given: \(E^\circ_{\text{cell}} = 1.1~\text{V},~T = 298~\text{K},~F = 96485~\text{C mol}^{-1},~R = 8.314~\text{J mol}^{-1}\text{K}^{-1}\).
Overall reaction involves transfer of \(n = 2\) electrons.

Step 1: Calculate standard Gibbs energy change.
\[ \Delta G^\circ = -n F E^\circ_{\text{cell}} = -(2)(96485)(1.1)~\text{J mol}^{-1} = -2.12267\times 10^{5}~\text{J mol}^{-1} = \boxed{-2.12\times 10^{5}~\text{J mol}^{-1} \; (= -2.12\times 10^{2}~\text{kJ mol}^{-1})} \]

Step 2: Calculate equilibrium constant.
\[ K = \exp\!\left(\frac{n F E^\circ_{\text{cell}}}{RT}\right) = \exp\!\left(\frac{(2)(96485)(1.1)}{(8.314)(298)}\right) = \exp(85.65) \] \[ $\Rightarrow$ \ \log_{10}K = \frac{85.65}{2.3026} = 37.20 \ $\Rightarrow$ \ \boxed{K \approx 1.6 \times 10^{37}} \] Answer: \(\boxed{\Delta G^\circ \approx -2.12 \times 10^{5}~\text{J mol}^{-1} = -212~\text{kJ mol}^{-1}, K \approx 1.6\times 10^{37}}\).