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Step-by-Step Solution
Step 1: Identify the Half-Reactions and Their Standard Potentials
We are given the following data at 25°C:
1) $Ag \to Ag^+ + e^-$ with $E^\circ = -0.800\,\text{V}$
2) $Ag + I^- \to AgI + e^-$ with $E^\circ = 0.152\,\text{V}$
Step 2: Determine the Reaction for Dissolution of $AgI$
To find the solubility product $K_{sp}$ of $AgI$, we need the net reaction:
$AgI \to Ag^+ + I^-$
From the two half-reactions given, if we combine them in a way that yields $AgI \to Ag^+ + I^-$, we first reverse the second reaction (to break $AgI$) and then combine appropriately. Effectively, the net standard potential for
$AgI \to Ag^+ + I^-$
is the negative of the sum of the two relevant half-reactions:
$E^\circ(\text{AgI} \to Ag^+ + I^-) = E^\circ(\text{Ag} \to Ag^+) - E^\circ(\text{Ag} + I^- \to AgI)$
Substituting values:
$E^\circ = (-0.800)\,\text{V} - (0.152)\,\text{V} = -0.952\,\text{V}$
Step 3: Apply the Nernst Equation to Relate $K_{sp}$ and $E^\circ$
For a general reaction $\text{Reaction} \leftrightarrow \text{Products}$, the standard potential $E^\circ$ is related to the equilibrium constant $K$ by:
$E^\circ = \frac{0.059}{n} \log K$
Here, $n = 1$ (since only one electron is transferred), $E^\circ = -0.952\,\text{V}$, and $K = K_{sp}$ for our dissolution process. Hence,
$-0.952 = \frac{0.059}{1} \log K_{sp}$
Step 4: Calculate $\log K_{sp}$
Rearrange to solve for $\log K_{sp}$:
$\log K_{sp} = \frac{-0.952}{0.059} = -16.13$
Step 5: Final Answer
Therefore, $\log K_{sp}$ for $AgI$ is $-16.13$.
