Chemistry - Electrochemistry Question with Solution | TestHub

The reaction $$\begin{array}{l}2{\mathrm{Cu}}^{+}(\mathrm{aq})⟶\mathrm{Cu}(\mathrm{s})+{\mathrm{Cu}}^{2+}(\mathrm{aq})\\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\circ }-\frac{0.0592}{1}\log \frac{\left[{\mathrm{Cu}}^{2+}\right]}{{\left[{\mathrm{Cu}}^{+}\right]}^2}\end{array}$$ At equilibrium ${\mathrm{E}}_{\mathrm{cell}}=0$ $$\therefore {\mathrm{E}}_{\text{cell }}^{\circ }=0.0592\log {\mathrm{K}}_{\mathrm{c}}$$ or, $\log {\mathrm{K}}_{\mathrm{c}}=\frac{0.52-0.16}{0.0592}$ $\therefore {\mathrm{K}}_{\mathrm{c}}=1.2\times 10^6$