Given the data at `25^(@)C`
`Ag+I^(-)rarrAgl+e^(-)" "E^(@)=0.152V`
`Ag rarrAg^(+)+e^(-)E^(@)=-0.800V`
What is the value of `log K_(sp)` for AgI?

Options

A. `-37.83`
B. `-16.13`
C. `-8.12`
D. `+8.612`

Explanation

To find the value of \( \log K_{sp} \) for AgI, we will follow these steps:

### Step 1: Write down the given half-reactions and their standard electrode potentials.
We have the following reactions:
1. \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) with \( E^\circ = -0.800 \, \text{V} \)
2. \( \text{Ag} + \text{I}^- \rightarrow \text{AgI} + e^- \) with \( E^\circ = 0.152 \, \text{V} \)

### Step 2: Write the reaction for the dissolution of AgI.
The dissolution of AgI can be represented as:
\[ \text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^- \]

### Step 3: Reverse the second reaction to match the dissolution reaction.
To find the \( K_{sp} \) for AgI, we need to reverse the second reaction:
\[ \text{AgI} + e^- \rightarrow \text{Ag} + \text{I}^- \]
The standard potential for the reversed reaction will be the negative of the given potential:
\[ E^\circ = -0.152 \, \text{V} \]

### Step 4: Combine the half-reactions.
Now, we need to combine the two half-reactions:
1. \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) (from the first reaction)
2. \( \text{AgI} + e^- \rightarrow \text{Ag} + \text{I}^- \) (reversed second reaction)

When we add these two reactions, the electrons will cancel out:
\[ \text{AgI} \rightarrow \text{Ag}^+ + \text{I}^- \]

### Step 5: Calculate the overall standard potential.
The overall standard potential \( E^\circ_{cell} \) can be calculated by adding the potentials of the two half-reactions:
\[ E^\circ_{cell} = E^\circ_{Ag^+/Ag} + E^\circ_{AgI/Ag + I^-} \]
\[ E^\circ_{cell} = (-0.800 \, \text{V}) + (0.152 \, \text{V}) \]
\[ E^\circ_{cell} = -0.800 + 0.152 = -0.648 \, \text{V} \]

### Step 6: Use the Nernst equation to find \( K_{sp} \).
The Nernst equation relates the standard potential to the equilibrium constant:
\[ E^\circ_{cell} = \frac{0.0591}{n} \log K_{sp} \]
Here, \( n = 1 \) (the number of electrons transferred).

Substituting the values:
\[ -0.648 = \frac{0.0591}{1} \log K_{sp} \]

### Step 7: Solve for \( \log K_{sp} \).
Rearranging gives:
\[ \log K_{sp} = \frac{-0.648}{0.0591} \]
Calculating this gives:
\[ \log K_{sp} \approx -10.96 \]

### Step 8: Final answer.
Thus, the value of \( \log K_{sp} \) for AgI is approximately:
\[ \log K_{sp} \approx -10.96 \]

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