Given the data at $$25^\circ C$$, $$Ag + I^- \longrightarrow AgI + e^-;\ E^\circ = 0.152\,V$$; $$Ag \longrightarrow Ag^+ + e^-;\ E^\circ = -0.800\,V$$. What is the value of $$\log K_{sp}$$ for $$AgI$$? $$\left(2.303\dfrac{RT}{F} = 0.059\,V\right)$$

For the dissolution equilibrium
$$AgI(s) \rightleftharpoons Ag^{+}(aq)+I^{-}(aq)$$
we need its standard Gibbs energy change or, equivalently, the standard potential of a one-electron process that gives the above reaction.

The two given half-reactions (written as oxidations, electrons on the RHS) are

$$Ag + I^- \rightarrow AgI + e^-;\; E^{\circ}_1 = +0.152\,{\rm V}$$
$$Ag \rightarrow Ag^{+} + e^-;\;E^{\circ}_2 = -0.800\,{\rm V}$$

Reverse the first reaction so that $$AgI$$ appears on the left (its potential changes sign):

$$AgI + e^- \rightarrow Ag + I^-;\;E^{\circ}_{1\,\text{rev}} = -0.152\,{\rm V}$$

Add this to the second half-reaction; the electrons and metallic silver cancel:

$$\bigl[AgI + e^- \rightarrow Ag + I^- \bigr]$$
$$\underline{+ \; \bigl[Ag \rightarrow Ag^{+} + e^-\bigr]}$$
$$AgI \rightarrow Ag^{+} + I^-$$

The resulting standard potential is the algebraic sum

$$E^{\circ}_{\text{cell}} = (-0.152) + (-0.800) = -0.952\,{\rm V}$$

For a reaction involving $$n=1$$ electron, the relations between potential, Gibbs energy and equilibrium constant are

$$\Delta G^{\circ} = -n F E^{\circ} \qquad\text{and}\qquad \Delta G^{\circ} = -RT\ln K_{sp}$$

Equating and rearranging,

$$\ln K_{sp} = \frac{n F E^{\circ}_{\text{cell}}}{RT}$$
$$\ln K_{sp} = \frac{1 \times F \times (-0.952)}{RT}$$

Given $$2.303\,\dfrac{RT}{F} = 0.059\,{\rm V}$$, we have $$\dfrac{RT}{F} = \dfrac{0.059}{2.303} = 0.0256\,{\rm V}$$.

Hence

$$\ln K_{sp} = \frac{-0.952}{0.0256} = -37.2$$

Convert to base-10 logarithm:

$$\log K_{sp} = \frac{\ln K_{sp}}{2.303} = \frac{-37.2}{2.303} \approx -16.13$$

Therefore,

Option D which is: $$-16.13$$