Let's use these steps to find the standard cell potential for an electrochemical cell with the following cell reaction.
Write the Half redox reactions as follows:
$$
\begin{aligned}
\mathrm{Cu} \rightarrow \mathrm{Cu}^{+}+\mathrm{e}^{-} \quad ; & E^{0}=-0.521 \mathrm{~V} \\
\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \quad ; \quad E^{0}=0.7996 \mathrm{~V} \\
\hline \mathrm{Cu}+\mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{+}+\mathrm{Ag} \quad ; \quad \begin{aligned}
E_{\mathrm{cen}}^{0} &=0.7996+(-0.521 \mathrm{~V}) \\
&=0.28 \mathrm{~V}
\end{aligned}
\end{aligned}
$$
Hence the \(E_{\text {cel }}^{\circ}\) of the reaction is \(0.28 \mathrm{~V}\).
