In the given reaction, 
           $\begin{array}{c}2{\mathrm{Cu}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)\\ {\mathrm{E}}_{{\mathrm{Cu}}^{+}/\mathrm{Cu}}^{\circ }=0.6\mathrm{V}\text{ and }{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\circ }=0.41\mathrm{V}\end{array}$
Find out the equilibrium constant.

Right hand cell reaction, ${\mathrm{Cu}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Cu}$
Left hand cell reaction, ${\mathrm{Cu}}^{+}⟶{\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}$
Cell reactions, $2{\mathrm{Cu}}^{+}⟶\mathrm{Cu}+{\mathrm{Cu}}^{2+}$
Cell potential ${\mathrm{E}}_{\text{cell }}^{\circ }={\mathrm{E}}_{{\mathrm{Cu}}^{+}/\mathrm{Cu}}^{\circ }-{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{C}}^{\circ }$
            $\begin{array}{l} =0.60-0.41=0.19\mathrm{V}\\ -{\mathrm{nFE}}^{\circ }=-\mathrm{RTln}{\mathrm{K}}_{\mathrm{eq}}\\ \log {\mathrm{K}}_{\mathrm{eq}}=\frac{{\mathrm{nE}}^{\circ }}{(2.303\mathrm{RT}/\mathrm{F})}=\frac{2\times 0.19\mathrm{V}}{0.059\mathrm{V}}=6.44\\ {\mathrm{K}}_{\mathrm{eq}}={10}^{6.44}=2.76\times {10}^6\end{array}$