`Delta G^@` of the cell reaction
`AgCl (s) = 1/2 H_2 (g) =Ag(s) H^+ + Cl^(-)` is `-21 . 52 kJ`.
` Delta G^2` of `2 AgCl (s) + H_2 (g) = 2 Ag(s) + 2 H^(+) 2 Cl^(-)` is .

Options

A. ` -21. 52 k J`
B. ` - 10 . 76 k J`
C. ` - 43. 04 k J`
D. ` 43 . 04 k J`

Explanation

To find the value of \( \Delta G^{\circ} \) for the reaction \( 2 \text{AgCl (s)} + \text{H}_2 (g) \rightarrow 2 \text{Ag (s)} + 2 \text{H}^+ + 2 \text{Cl}^- \), we can use the relationship between the Gibbs free energy change of a reaction and the stoichiometry of the reaction.

### Step-by-Step Solution:

1. **Identify the given data**:
We know that for the reaction:
\[ \text{AgCl (s)} \rightarrow \frac{1}{2} \text{H}_2 (g) + \text{Ag (s)} + \text{H}^+ + \text{Cl}^- \]
the standard Gibbs free energy change \( \Delta G^{\circ} = -21.52 \, \text{kJ} \).

2. **Understand the stoichiometry**:
The second reaction is simply double the first reaction:
\[ 2 \text{AgCl (s)} + \text{H}_2 (g) \rightarrow 2 \text{Ag (s)} + 2 \text{H}^+ + 2 \text{Cl}^- \]
This means that we are doubling the number of moles of all reactants and products.

3. **Apply the relationship for Gibbs free energy**:
The relationship between the Gibbs free energy change of a reaction and its stoichiometry is:
\[ \Delta G^{\circ}_{\text{new}} = n \cdot \Delta G^{\circ}_{\text{original}} \]
where \( n \) is the factor by which the reaction is multiplied. Here, \( n = 2 \).

4. **Calculate \( \Delta G^{\circ} \) for the new reaction**:
\[ \Delta G^{\circ}_{\text{new}} = 2 \cdot (-21.52 \, \text{kJ}) = -43.04 \, \text{kJ} \]

5. **Final answer**:
Thus, the value of \( \Delta G^{\circ} \) for the reaction \( 2 \text{AgCl (s)} + \text{H}_2 (g) \rightarrow 2 \text{Ag (s)} + 2 \text{H}^+ + 2 \text{Cl}^- \) is:
\[ \Delta G^{\circ} = -43.04 \, \text{kJ} \]