The `pK_(sp)` of `AgI` is `16.07` if the `E^(@)` value for `Ag^(+)//Ag` is `0.7991V`, find the `E^(@)` for the half cell reaction `AgI(s) +e^(-) rarr Ag+I^(-)`

Explanation

To find the standard electrode potential \( E^\circ \) for the half-cell reaction \( \text{AgI(s)} + e^- \rightarrow \text{Ag}^+ + \text{I}^- \), we can use the relationship between the solubility product constant \( K_{sp} \) and the standard electrode potentials. Here are the steps to solve the problem:

### Step 1: Understand the relationship
The standard electrode potential for the half-cell reaction can be calculated using the equation:
\[ E^\circ_{\text{AgI/Ag}^+} = E^\circ_{\text{Ag}^+/Ag} + \frac{0.0591}{n} \log K_{sp} \]
where:
- \( E^\circ_{\text{Ag}^+/Ag} \) is the standard reduction potential for the silver half-cell.
- \( K_{sp} \) is the solubility product of silver iodide (AgI).
- \( n \) is the number of electrons transferred in the half-reaction.

### Step 2: Convert \( pK_{sp} \) to \( K_{sp} \)
Given that \( pK_{sp} = 16.07 \), we can find \( K_{sp} \) using the formula:
\[ K_{sp} = 10^{-pK_{sp}} = 10^{-16.07} \]

### Step 3: Substitute the values
We know:
- \( E^\circ_{\text{Ag}^+/Ag} = 0.7991 \, V \)
- \( n = 1 \) (since one electron is involved in the reaction)

Now we can substitute these values into the equation:
\[ E^\circ_{\text{AgI/Ag}^+} = 0.7991 + \frac{0.0591}{1} \log(10^{-16.07}) \]

### Step 4: Calculate \( \log(10^{-16.07}) \)
Using logarithmic properties:
\[ \log(10^{-16.07}) = -16.07 \]

### Step 5: Substitute and solve
Now substituting back into the equation:
\[ E^\circ_{\text{AgI/Ag}^+} = 0.7991 + 0.0591 \times (-16.07) \]
Calculating the second term:
\[ 0.0591 \times (-16.07) = -0.948 \]
Now substituting this value:
\[ E^\circ_{\text{AgI/Ag}^+} = 0.7991 - 0.948 \]
\[ E^\circ_{\text{AgI/Ag}^+} = -0.1489 \, V \]

### Step 6: Round the answer
Rounding to three significant figures, we get:
\[ E^\circ_{\text{AgI/Ag}^+} \approx -0.149 \, V \]

### Final Answer
The standard electrode potential for the half-cell reaction \( \text{AgI(s)} + e^- \rightarrow \text{Ag}^+ + \text{I}^- \) is approximately:
\[ E^\circ \approx -0.149 \, V \]