Chemistry - Electrochemistry Question with Solution | TestHub

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