Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \operatorname{AgI}(s)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\text {sp }}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

The given reaction can be broken down into the following reduction and oxidation half-reactions: Reduction: $$ \operatorname{AgI}(s)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)+\mathrm{I}^{-}(a q) $$ Oxidation: $$ \operatorname{Ag}(s) \longrightarrow \operatorname{Ag}^{+}(a q)+\mathrm{e}^{-} $$ Right now we mostly care about the reduction half-reaction.

The reaction quotient (Q) can be written for the reduction half-reaction as: $$ Q = [\mathrm{I}^{-}] $$

The solubility product for AgI is given by: $$ K_{sp} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] $$ Since Ag⁺ concentration is equal to I⁻ concentration due to the 1:1 stoichiometry of the dissociation, we can write: $$ K_{sp} = [\mathrm{I}^{-}]^2 = (\sqrt{K_{sp}})^2 $$ Thus, the equilibrium constant (K) for the reduction half-reaction is: $$ K = \frac{Q}{[\mathrm{Ag}^{+}]} = \frac{[\mathrm{I}^{-}]}{[\mathrm{Ag}^{+}]} = \sqrt{K_{sp}} $$ (Keep in mind that the actual value of Ksp for AgI is needed for actual calculations)

The Nernst equation is: $$ \mathscr{E}^{\circ} = \frac{RT}{nF} \ln{K} $$ Where, - R represents the ideal gas constant: \( R = 8.314 \frac{J}{mol \cdot K} \) - T represents the temperature in Kelvin - n represents the number of electrons transferred in the half-reaction (in this case, n=1) - F represents Faraday's constant: \( F = 96,485 \frac{C}{mol} \) - K represents the equilibrium constant for the half-reaction

Substitute the values in the Nernst equation: $$ \mathscr{E}_{Red}^{\circ} = \frac{8.314 \cdot 298}{1 \cdot 96,485} \ln{\sqrt{K_{sp}}} $$ Let's denote standard reduction potential for Ag⁺ as 𝔼°(Ag⁺) = 0.7996V (as given in standard reduction potential tables) Now, we have to consider both the reduction and the oxidation half-reactions. Since the standard reduction potential for the oxidation half-reaction is the opposite of the reduction potential for Ag⁺, $$ \mathscr{E}_{Ox}^{\circ} = -\mathscr{E}^{\circ}\text{(Ag⁺)} = -0.7996V $$ Finally, the standard cell potential for the given redox reaction is the difference between the standard reduction and oxidation potentials: $$ \mathscr{E}_{cell}^{\circ} = \mathscr{E}_{Red}^{\circ} - \mathscr{E}_{Ox}^{\circ} = \frac{8.314 \cdot 298}{1 \cdot 96,485} \ln{\sqrt{K_{sp}}} + 0.7996V $$ By plugging in the Ksp value of AgI, we can calculate the numerical value of \(\mathscr{E}^{\circ}\).