$\mathrm{E}_{0}$ for the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Cu

Question

$\mathrm{E}_{0}$ for the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})ight|\left|\mathrm{Cu}^{2+}(\mathrm{aq})ight| \mathrm{Cu}$ is 1.10 V at $25^{\circ} \mathrm{C}$. The equilibrium constant for the reaction $$ \mathrm{Zn}+\mathrm{Cu}^{2+}(\mathrm{aq}) \leftrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu} $$ is of the order of\

Professor PC Thomas Classes · Answer

$\mathrm{E}^{\mathrm{o}}=\frac{0.059}{\mathrm{n}} \log \mathrm{K}=\frac{0.059}{2} \log \mathrm{~K}$\ $\log \mathrm{K}=\frac{2 	imes \mathrm{E}^{\circ}}{0.059}=\frac{2 	imes 1.1}{0.059}=37.2$\ Therefore, $\mathrm{K}=10^{37 . .2} \approx 10^{37}$.\