Question

\(E^\circ = 1.10\ \text{V}\) for the cell: \(\mathrm{Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s)}\). At \(298\ \text{K}\), the equilibrium constant is \(y \times 10^{37}\). Find \(y\) (to two decimals). (Given: \(F = 96485\ \text{C mol}^{-1}\), \(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\)).

Hint:

Use \(\ln K=\dfrac{nFE^\circ}{RT}\) at \(298\ \text{K}\). For quick estimates, \(\log_{10}K=\dfrac{nE^\circ}{0.05916}\).

Answer 1

Step 1: Relate \(E^\circ\) and \(K\). \(\Delta G^\circ=-nFE^\circ=-RT\ln K ⇒ \ln K=\dfrac{nFE^\circ}{RT}\). For this cell, \(n=2\).
Step 2: Substitute numbers. \[ \ln K=\frac{(2)(96485)(1.10)}{(8.314)(298)}=\frac{212{,}267}{2477.57}\approx 85.675. \] \[ ⇒\ K=e^{85.675}=10^{\,85.675/\ln 10}\approx 10^{37.208}=1.6157\times 10^{37}. \] Step 3: Express as \(y\times 10^{37}\). \(K\approx 1.6157\times 10^{37}⇒ y\approx 1.62\) (to two decimals).