The standard half-cell reduction potential for `Ag+|Ag` is 0.7991 V at `25^(@)C`. Given the experimental value `K_(sp) =1.56xx10^(-10)` for AgCl, calculate the standard half-cell reduction potential for the AglAgCI electrode.

Options

A. `0.2192 V `
B. ` - 0.2192V `
C. ` -1.2192 ` V
D. `1.2192 V `

Explanation

To calculate the standard half-cell reduction potential for the Ag|AgCl electrode, we will follow these steps:

### Step 1: Identify the given data
- Standard half-cell reduction potential for Ag⁺|Ag: \( E^\circ = 0.7991 \, \text{V} \)
- Solubility product constant for AgCl: \( K_{sp} = 1.56 \times 10^{-10} \)

### Step 2: Write the relevant half-reactions
1. For the reduction of silver ions:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (1) \]
2. For the dissociation of AgCl:
\[ \text{AgCl} \rightarrow \text{Ag}^+ + \text{Cl}^- \quad (2) \]

### Step 3: Calculate the Gibbs free energy change (\( \Delta G^\circ \)) for the reduction of Ag⁺
Using the formula:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( n = 1 \) (number of electrons)
- \( F = 96.48 \, \text{kJ/mol/V} \)
- \( E^\circ = 0.7991 \, \text{V} \)

Calculating \( \Delta G^\circ \):
\[ \Delta G^\circ = -1 \times 96.48 \times 0.7991 = -77.097 \, \text{kJ/mol} \]

### Step 4: Calculate the Gibbs free energy change (\( \Delta G^\circ \)) for the dissociation of AgCl
Using the formula:
\[ \Delta G^\circ = -RT \ln K_{sp} \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( K_{sp} = 1.56 \times 10^{-10} \)

Calculating \( \Delta G^\circ \):
\[ \Delta G^\circ = -0.008314 \times 298 \times \ln(1.56 \times 10^{-10}) \]
Calculating \( \ln(1.56 \times 10^{-10}) \):
\[ \ln(1.56 \times 10^{-10}) \approx -18.4 \]
Thus,
\[ \Delta G^\circ \approx -0.008314 \times 298 \times (-18.4) \approx 55.95 \, \text{kJ/mol} \]

### Step 5: Calculate the overall \( \Delta G^\circ \) for the combined reaction
Adding the two Gibbs free energies:
\[ \Delta G^\circ_{\text{total}} = -77.097 + 55.95 = -21.147 \, \text{kJ/mol} \]

### Step 6: Calculate the standard half-cell reduction potential for Ag|AgCl
Using the formula:
\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]
Substituting in the values:
\[ E^\circ = -\frac{-21.147 \times 10^3}{96.48} \approx 0.2192 \, \text{V} \]

### Final Answer:
The standard half-cell reduction potential for the Ag|AgCl electrode is approximately \( 0.2192 \, \text{V} \).

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