Which from following expressions is used to find the cell potential of \(\mathrm{Cd}_{\text {(s) }}\left|\mathrm{Cd}_{\text {(aq) }}^{++}\right|\left|\mathrm{Cu}_{\text {(aq) }}^{++}\right| \mathrm{Cu}_{\text {(s) }}\) cell at \(25^{\circ} \mathrm{C}\) ?

Cell reaction:
\(
\begin{aligned}
& \mathrm{Cd}_{(\mathrm{s})} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \quad \text { (oxidation at anode) } \\
& \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \quad \text { (reduction at cathode) } \\
& \hline \mathrm{Cd}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cu}_{(\mathrm{s})} \\
& \text { (overall reaction) }
\end{aligned}
\)
Now,
\(
\begin{aligned}
\mathrm{E}_{\text {cell }} & =\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592 \mathrm{~V}}{\mathrm{n}} \log _{10} \frac{[\text { Product }]}{[\text { Reactant }]} \\
& =0.02-\frac{0.0592 \mathrm{~V}}{2} \log _{10} \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]} \\
& =\mathrm{E}_{\text {cell }}^{\prime}-0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]}
\end{aligned}
\)