Calculate \(\mathrm{E}_{\text {cell }}^{\circ}\) if the equilibrium constant for following reaction is \(1.2 \times 10^6\).
\(
2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{++}+\mathrm{Cu}_{(\mathrm{s})}
\)

\(\begin{aligned} & \mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K} \text { at } 298 \mathrm{~K} \\ & \mathrm{Cu}_{(\mathrm{mq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \\ & \frac{\mathrm{Cu}_{\text {(aq) }}^{+} \longrightarrow \mathrm{Cu}_{\text {(aq) }}^{2+}+\mathrm{e}^{-}}{2 \mathrm{Cu}_{\text {(aq) }}^{+} \longrightarrow \mathrm{Cu}_{(\text {aq) }}^{2+}+\mathrm{Cu}_{(\mathrm{s})}} \\ & \therefore \quad \mathrm{n}=1 \\ & \therefore \quad \mathrm{E}_{\text {eell }}^0=\frac{0.0592}{1} \log _{10}\left(1.2 \times 10^6\right) \\ & =0.0592\left(\log 1.2+\log 10^6\right) \\ & =0.0592(0.079+6) \\ & =0.0592 \times 6.079 \\ & =0.36 \mathrm{~V} \\ & \end{aligned}\)