Given the data at 25⁰C  
$\begin{array}{r}\mathrm{Ag}+{\mathrm{I}}^{-}⟶\mathrm{AgI}+{\mathrm{e}}^{-};{\mathrm{E}}^0=0.152\mathrm{V}\\ \mathrm{Ag}⟶{\mathrm{Ag}}^{+}+{\mathrm{e}}^{-};{\mathrm{E}}^0=-0.800\mathrm{V}\end{array}$
What is the value of log Ksp for AgI? (2.303 RT /F = 0.059 V) 

$$\begin{gathered} \mathrm{Ag} +\hspace{0.17em}{\mathrm{I}}^{-}\to \mathrm{AgI}\hspace{0.17em}+\hspace{0.17em}{\mathrm{e}}^{-} {\mathrm{E}}^{\mathrm{o}}= 0.152\mathrm{V}\hspace{0.17em} \\ \mathrm{Ag} \to {\mathrm{Ag}}^{+}+\hspace{0.17em}{\mathrm{e}}^{-} {\mathrm{E}}^{\mathrm{o}}= -0.8 \mathrm{V}\hspace{0.17em} \\ {\mathrm{Ag}}^{+} +\hspace{0.17em}{\mathrm{e}}^{-}\to \mathrm{Ag} {\mathrm{E}}^{\mathrm{o}} = 0.8\mathrm{V}\hspace{0.17em} \\ {\mathrm{Ag}}^{+} +\hspace{0.17em}{\mathrm{I}}^{-} \to \mathrm{AgI}\hspace{0.17em}+\hspace{0.17em}{\mathrm{e}}^{-} {\mathrm{E}}_{ }^{\mathrm{o}}= 0.152\mathrm{V}\hspace{0.17em} \\ {\mathrm{Ag}}^{+}+\hspace{0.17em}{\mathrm{I}}^{-}\to \mathrm{AgI} {\mathrm{E}}^{\mathrm{o}}= 0.952 \mathrm{V} \end{gathered}$$

$$\begin{gathered} ∆{\mathrm{G}}^{\mathrm{o}}=-\mathrm{nFE} \\ △{\mathrm{G}}^{\mathrm{o}}=+\mathrm{RT} \ln {\mathrm{K}}_{\mathrm{sp}} \\ -1 \times 96500\times 0.952=8.314 \times 298\times 2.303\times \log {\mathrm{K}}_{\mathrm{sp}} \\ \log {\mathrm{K}}_{\mathrm{sp}}= -16.1 \end{gathered}$$