$\begin{array}{l}\mathrm{Ag}+{\mathrm{I}}^{-}⟶\mathrm{AgI}+{\mathrm{e}}^{-};{\mathrm{E}}^0=0.152\mathrm{V}\\ \mathrm{Ag}⟶{\mathrm{Ag}}^{+}{\mathrm{e}}^{-};{\mathrm{E}}^0=-0.800\mathrm{V}\end{array}$
What is the value of log ${\mathrm{K}}_{\mathrm{sp}}$ for AgI ? (2.303 RT/F = 0.059 V)

Given that 

$$\begin{gathered} \mathrm{Ag}+\hspace{0.17em}{\mathrm{I}}^{-}\to \mathrm{AgI}\hspace{0.17em}+\hspace{0.17em}{\mathrm{e}}^{- } ; {\mathrm{E}}^{\mathrm{o}}= +\hspace{0.17em}0.152\mathrm{V} \left(\mathrm{cathode}\right)\hspace{0.17em} \\ \mathrm{at} \mathrm{cathode} \mathrm{gaon} \mathrm{of} \mathrm{electrons} \mathrm{takes} \mathrm{place} \mathrm{so} \mathrm{the} \mathrm{equation} \mathrm{i}s \mathrm{written} \mathrm{as} \\ \mathrm{AgI}\hspace{0.17em}+\hspace{0.17em}{\mathrm{e}}^{-}\to \mathrm{Ag}+\hspace{0.17em}{\mathrm{I}}^{-}; {\mathrm{E}}^{\mathrm{o}}= -0.152\mathrm{V} \\ \mathrm{Ag}\to {\mathrm{Ag}}^{+}+\hspace{0.17em}{\mathrm{e}}^{-} ; {\mathrm{E}}^{\mathrm{o}}=+0.800\mathrm{V} \left(\mathrm{anode}\right) \\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}= {\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}}- {\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}}= (-0.152)-(0.800)=-0.952\mathrm{V} \end{gathered}$$

The Nernst equation is 

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \mathrm{Qc}$

Solubility product is considered for a saturated solution at equilibrium. 

At equilibrium , Qc= Kc = Ksp  then E_cell = 0

$0= {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{1}\mathrm{Ksp}$

$$\begin{gathered} 0=-0.952-\frac{0.0591}{1}\log \mathrm{Ksp} \\ \log \mathrm{Ksp} =\frac{-0.952}{0.0591}=-16.135 \end{gathered}$$