The \(\mathrm{E}_{\text {cell }}^{\circ}\) of \(\mathrm{Cu}_{(\mathrm{s})}\left|\mathrm{Cu}_{(\mathrm{IM})}^{++}\right|\left|\mathrm{Ag}_{(\mathrm{IM})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\) is 0.647 volt. Calculate the \(\mathrm{E}_{\mathrm{Ag}}^{\circ}\) if \(\mathrm{E}_{\mathrm{Cu}_{\mathrm{u}}^{\circ}}\) is 0.153 V .

The given cell is composed of copper electrode as the anode and silver electrode as the cathode. The standard cell potential is given by
\(\begin{aligned}
& \mathrm{E}_{\text {cell }}^{\mathrm{o}}=\mathrm{E}_{\text {cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^0 \\
& \therefore \quad \mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}^0}^0-\mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^{\mathrm{o}} \\
& \therefore \quad \mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^{\mathrm{o}}=\mathrm{E}_{\mathrm{cell}}^0+\mathrm{E}_{\mathrm{Ou}^2+\mathrm{Clu}^{\prime}}^{\mathrm{o}} \\
& =(0.647 \mathrm{~V})+(0.153 \mathrm{~V})=0.8 \mathrm{~V}
\end{aligned}\)