`Delta_rG^@` for the cell with the cell reaction : `Zn_((s))+ Ag_2O_((s)) + H_2O_((l)) to Zn_((aq))^(2+) + 2Ag_((s)) + 2OH_((aq))^(-)`
`[E_(Ag_2O // Ag)^(@)=0.344 V, E_(Zn^(2+)//Zn)^@=-0.76 V]`

Options

A. `2.13xx10^5 "J mol"^(-1)`
B. `-2.13xx10^5 "J mol"^(-1)`
C. `1.06xx10^5 "J mol"^(-1)`
D. `-1.06xx10^5 "J mol"^(-1)`

Explanation

To find the standard Gibbs free energy change (Δ_rG^@) for the given cell reaction, we can follow these steps:

### Step 1: Identify the Cell Reaction
The cell reaction provided is:
\[ \text{Zn}_{(s)} + \text{Ag}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2\text{Ag}_{(s)} + 2\text{OH}^-_{(aq)} \]

### Step 2: Identify the Standard Reduction Potentials
We are given the standard reduction potentials:
- \( E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} = 0.344 \, \text{V} \)
- \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \)

### Step 3: Determine the Cathode and Anode
In the cell reaction:
- The reduction occurs at the cathode. Here, \( \text{Ag}_2\text{O} \) is reduced to \( \text{Ag} \).
- The oxidation occurs at the anode. Here, \( \text{Zn} \) is oxidized to \( \text{Zn}^{2+} \).

### Step 4: Calculate the Standard Cell Potential (E°_cell)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \]
\[ E^\circ_{\text{cell}} = 0.344 \, \text{V} - (-0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.344 \, \text{V} + 0.76 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 1.104 \, \text{V} \]

### Step 5: Calculate Δ_rG^@
Using the formula:
\[ \Delta_rG^@ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) = number of moles of electrons transferred (from the balanced reaction, \( n = 2 \))
- \( F \) = Faraday's constant \( \approx 96500 \, \text{C/mol} \)

Substituting the values:
\[ \Delta_rG^@ = -2 \times 96500 \, \text{C/mol} \times 1.104 \, \text{V} \]
\[ \Delta_rG^@ = -2 \times 96500 \times 1.104 \]
\[ \Delta_rG^@ = -213131.2 \, \text{J/mol} \]

### Step 6: Convert to Standard Form
Expressing the answer in scientific notation:
\[ \Delta_rG^@ \approx -2.1313 \times 10^5 \, \text{J/mol} \]

### Final Answer
The value of \( \Delta_rG^@ \) for the cell reaction is:
\[ \Delta_rG^@ \approx -2.1313 \times 10^5 \, \text{J/mol} \]