Consider `Delta G^(@)` for the following cell reaction :
`Zn (s) + Ag_(2)O(s)+H_(2)O(l) hArr Zn^(2+)(aq)+2Ag(s)+2OH^(-)(aq)`
`E_(Ag^(+)//Ag)^(@) = +0.80` and `E_(Zn^(2+)//Zn)^(@) = - 0.76 V`

Options

A. `-305 kJ//mol`
B. `-301 kJ//mol`
C. `+305kJ//mol`
D. `+301kJ//mol`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the cell reaction given, we will follow these steps:

### Step 1: Identify the half-reactions
The overall cell reaction is:
\[ \text{Zn (s)} + \text{Ag}_2\text{O (s)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{Zn}^{2+}(aq) + 2\text{Ag (s)} + 2\text{OH}^-(aq) \]

The half-reactions involved are:
1. Oxidation:
\[ \text{Zn (s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
2. Reduction:
\[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag (s)} \]

### Step 2: Determine the standard reduction potentials
From the problem statement, we have:
- \( E^\circ(\text{Ag}^+/\text{Ag}) = +0.80 \, \text{V} \)
- \( E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \, \text{V} \)

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In this case:
- Cathode (reduction): Ag (s) from Ag+ (aq) with \( E^\circ = +0.80 \, \text{V} \)
- Anode (oxidation): Zn (s) to Zn2+ (aq) with \( E^\circ = -0.76 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 \, \text{V} \]

### Step 4: Calculate ΔG°
Using the relationship between Gibbs free energy and cell potential:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) = number of moles of electrons transferred (2 for Zn)
- \( F \) = Faraday's constant = 96500 C/mol
- \( E^\circ_{\text{cell}} = 1.56 \, \text{V} \)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 1.56 \, \text{V} \]
\[ \Delta G^\circ = -2 \times 96500 \times 1.56 \]
\[ \Delta G^\circ = -301320 \, \text{J} \]
\[ \Delta G^\circ = -301.32 \, \text{kJ} \]

### Final Answer
Thus, the standard Gibbs free energy change for the reaction is approximately:
\[ \Delta G^\circ \approx -301 \, \text{kJ} \]