The cell $\mathrm{Zn} / \mathrm{Zn}^{2}(1 \mathrm{M}) / / \mathrm{Cu}^{2+}(1 \mathrm{M}) / \mathrm{C

Question

The cell $\mathrm{Zn} / \mathrm{Zn}^{2}(1 \mathrm{M}) / / \mathrm{Cu}^{2+}(1 \mathrm{M}) / \mathrm{Cu}\left(\mathrm{E}^{0}=1.1 \mathrm{~V}\right)$ was allowed to discharge completely at 298 K . The ratio concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}$ is _____?

Professor PC Thomas Classes · Answer

The cell reaction is $\mathrm{Zn}+\mathrm{Cu}^{2+} ightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$\ when the cell is completely discharged $\mathrm{E}=0$\ $	herefore \mathrm{E}=\mathrm{E}^{0}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}$.\ $0=1.1-\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}$\ $\log \frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=\frac{1.1 	imes 2}{0.0591}=37.3$\ $	herefore \frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=10^{37.3}$