Chemistry - Electrochemistry Question with Solution | TestHub

$$\begin{array}{ll}\mathrm{AgI}(\mathrm{s})+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}(\mathrm{s})+{\mathrm{I}}^{-};& {\mathrm{E}}^{\circ }=-0.152\\ \mathrm{Ag}(\mathrm{s})⟶{\mathrm{Ag}}^{+}+{\mathrm{e}}^{-};& {\mathrm{E}}^{\circ }=-0.8\\ \mathrm{AgI}(\mathrm{s})⟶{\mathrm{Ag}}^{+}+{\mathrm{I}}^{-};& {\mathrm{E}}^{\circ }=-0.952\\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }=\frac{0.059}{\mathrm{n}}\mathrm{logK}& \\ -0.952=\frac{0.059}{1}\log {\mathrm{K}}_{\mathrm{sp}}& \\ \log {\mathrm{K}}_{\mathrm{sp}}=-\frac{0.952}{0.059}=-16.135& \end{array}$$