Calculate the \(\mathrm{E}_{\text {cell }}\) for \(\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(0.1 \mathrm{M})}^{++}\right|\left|\mathrm{Cr}_{(0.1 \mathrm{M})}^{+++}\right| \mathrm{Cr}_{(\mathrm{s})}\) at \(25^{\circ} \mathrm{C}\) if \(\mathrm{E}_{\text {cell }}^{\circ}\) is \(0.02 \mathrm{~V}\)

The reaction is
\(
\begin{aligned}
& 3 \mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{Cr}_{(0.1 \mathrm{M})}^{3+} \longrightarrow 3 \mathrm{Zn}_{(0.1)}^{2+}+2 \mathrm{Cr}_{(\mathrm{s})} \\
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0592 \mathrm{~V}}{\mathrm{n}} \log _{10} \frac{\left[\mathrm{Zn}^{2+}\right]^3}{\left[\mathrm{Cr}^{3+}\right]^2}
\end{aligned}
\)
\(\begin{aligned} \therefore \mathrm{E}_{\text {cell }} & =0.02 \mathrm{~V}-\frac{0.0592 \mathrm{~V}}{6} \log _{10} \frac{(0.1)^3}{(0.1)^2} \\ & =0.02 \mathrm{~V}-\frac{0.0592 \mathrm{~V}}{6} \log _{10} 0.1 \\ & =0.02 \mathrm{~V}+\frac{0.0592 \mathrm{~V}}{6} \times 1 \\ & =0.02 \mathrm{~V}+0.0099 \mathrm{~V} \\ & =0.03 \mathrm{~V}\end{aligned}\)