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