At `25^(@)C, {:(Ag+I^(-)rarr,AgI+e,,E^(@) =0.152V),(Agrarr,Ag.^(+)+e,,E^(@) =- 0.80V):}`
The `log K_(sp)` of `AgI` is: `((2.303RT)/(F)=0.059)`

Options

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

Explanation

To find the log Ksp of AgI, we will follow these steps:

### Step 1: Write the solubility equilibrium reaction for AgI
The solubility equilibrium reaction for silver iodide (AgI) can be expressed as:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]

### Step 2: Identify the half-reactions and their standard electrode potentials
From the question, we have the following half-reactions and their standard electrode potentials (E°):
1. \( \text{Ag} + \text{e}^- \rightleftharpoons \text{Ag}^+ \) with \( E° = -0.80 \, \text{V} \) (Equation 1)
2. \( \text{Ag}^+ + \text{I}^- \rightleftharpoons \text{AgI} + \text{e}^- \) with \( E° = 0.152 \, \text{V} \) (Equation 2)

### Step 3: Combine the half-reactions to find the overall cell reaction
To find the overall cell reaction, we can subtract Equation 1 from Equation 2:
- Reverse Equation 1:
\[ \text{Ag}^+ + \text{e}^- \rightleftharpoons \text{Ag} \quad (E° = +0.80 \, \text{V}) \]
- Now add this to Equation 2:
\[ \text{Ag}^+ + \text{I}^- \rightleftharpoons \text{AgI} + \text{e}^- \quad (E° = 0.152 \, \text{V}) \]

### Step 4: Calculate the standard cell potential (E°cell)
The standard cell potential (E°cell) is given by:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]
Substituting the values:
\[ E°_{\text{cell}} = 0.152 \, \text{V} - (-0.80 \, \text{V}) = 0.152 + 0.80 = 0.952 \, \text{V} \]

### Step 5: Use the Nernst equation to find log Ksp
The Nernst equation at equilibrium (where Ecell = 0) is:
\[ 0 = E°_{\text{cell}} - \frac{0.0591}{n} \log Ksp \]
Where \( n = 1 \) (one electron is involved in the reaction). Rearranging gives:
\[ \log Ksp = \frac{E°_{\text{cell}} \cdot n}{0.0591} \]
Substituting the values:
\[ \log Ksp = \frac{0.952 \, \text{V}}{0.0591} \]

### Step 6: Calculate log Ksp
Calculating the above expression:
\[ \log Ksp = \frac{0.952}{0.0591} \approx 16.1 \]
Since we are looking for the negative value:
\[ \log Ksp \approx -16.13 \]

### Final Answer
Thus, the log Ksp of AgI is:
\[ \log Ksp = -16.13 \]