The cell, $\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(1\mathrm{M}\right)\parallel {\mathrm{Cu}}^{2+}\left(1\mathrm{M}\right)\right|\mathrm{Cu}\left(E_{\mathrm{cell}}^{\circ }=1.10\mathrm{V}\right)$ was allowed to be completely discharged at 298 K. The relative concentration of ${\mathrm{Zn}}^{2+}$ to ${\mathrm{Cu}}^{2+}$, i.e. $\left(\left[{\mathrm{Zn}}^{2+}\right]/\left[{\mathrm{Cu}}^{2+}\right]\right)$ is

When the cell is completely discharged, the cell reaction is reached its equilibrium state. For the given cell $\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(1\mathrm{M}\right)\parallel {\mathrm{Cu}}^{2+}\left(1\mathrm{M}\right)\right|\mathrm{Cu}$ the cell reaction is $\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)$ The equilibrium constant of the reaction is $K_{\mathrm{eq}}=\frac{\left[{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)\right]}{\left[{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\right]}$

The equilibrium constant is related to ${\mathrm{E}}_{\mathrm{cell}}^{°}$ by the relation

 $\mathrm{\Delta }{G}^{\circ }=-nF{E}_{\mathrm{cell}}^{\circ }=-RT\ln K_{\mathrm{eq}}^{\circ }$ or $\log K_{\mathrm{eq}}^{\circ }=\frac{n{E}_{\mathrm{cell}}}{(2.303RT/F)}$

At 298 K, 2.303 RT/F = 0.059 V. Hence $\log K_{\mathrm{eq}}^{\circ }=\frac{\left(2\right)\left(1.10\mathrm{V}\right)}{\left(0.059\mathrm{V}\right)}=37.3$ or $K_{\mathrm{eq}}^{\circ }={10}^{37.3}$