The solubiltiy product of silver iodide is `8.3 xx 10^(-17)` and the standard potential (reduction) of `Ag,Ag^(+)` electrode is `+0.800` volts at `25^(@)C`. The standard potential of `Ag,AgI//I^(-)` electrode (reduction) from these data is

Options

A. `-0.30V`
B. `+0.15V`
C. `+0.10V`
D. `-0.15V`

Explanation

To solve the problem, we need to determine the standard potential of the Ag, AgI / I⁻ electrode using the given solubility product (Ksp) of silver iodide (AgI) and the standard reduction potential of the Ag, Ag⁺ electrode.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Solubility product (Ksp) of AgI = \(8.3 \times 10^{-17}\)
- Standard reduction potential of Ag, Ag⁺ = \(E^\circ = +0.800 \, \text{V}\)

2. **Write the Equilibrium Reaction:**
The dissolution of silver iodide can be represented as:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]

3. **Express Ksp in Terms of Concentrations:**
The solubility product expression for AgI is:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]
From the Ksp value, we can say:
\[ K_{sp} = 8.3 \times 10^{-17} \]

4. **Using Nernst Equation:**
The Nernst equation relates the standard potential to the concentrations of the ions:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log K_{sp} \]
Here, \(n = 1\) because one electron is involved in the reduction of Ag⁺ to Ag.

5. **Set Up the Equation:**
At equilibrium, \(E_{cell} = 0\) (since it is a saturated solution):
\[ 0 = E^\circ_{Ag/AgI} - E^\circ_{Ag/Ag^+} - 0.0591 \log K_{sp} \]
Rearranging gives:
\[ E^\circ_{Ag/AgI} = E^\circ_{Ag/Ag^+} + 0.0591 \log K_{sp} \]

6. **Substituting Known Values:**
Substitute \(E^\circ_{Ag/Ag^+} = 0.800 \, \text{V}\) and \(K_{sp} = 8.3 \times 10^{-17}\):
\[ E^\circ_{Ag/AgI} = 0.800 + 0.0591 \log(8.3 \times 10^{-17}) \]

7. **Calculate the Logarithm:**
\[ \log(8.3 \times 10^{-17}) = \log(8.3) + \log(10^{-17}) \approx 0.919 - 17 = -16.081 \]

8. **Substituting Logarithm Value:**
\[ E^\circ_{Ag/AgI} = 0.800 + 0.0591 \times (-16.081) \]
\[ E^\circ_{Ag/AgI} = 0.800 - 0.950 \]
\[ E^\circ_{Ag/AgI} = -0.150 \, \text{V} \]

9. **Final Answer:**
The standard potential of the Ag, AgI / I⁻ electrode is:
\[ E^\circ_{Ag/AgI} = -0.15 \, \text{V} \]