Calculate the `DeltaG^(@)` of the following reaction :-
`Fe^(+2)(aq)+Ag^(+)(aq)toFe^(+2)(aq)+Ag(s)`
`E_(Ag^(+)//Ag)=0.8V" "E_(Fe^(+3)//Fe^(+2))^(0)=0.77V`

Options

A. `-2895 J mol^(-1)`
B. `-3845 J mol^(-1)`
C. `-1874 J mol^(-1)`
D. `-375 J mol^(-1)`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the given reaction:

**Reaction:**
\[ \text{Fe}^{2+}(aq) + \text{Ag}^{+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag}(s) \]

**Given:**
- Standard reduction potential for \( \text{Ag}^{+}/\text{Ag} = 0.80 \, \text{V} \)
- Standard reduction potential for \( \text{Fe}^{3+}/\text{Fe}^{2+} = 0.77 \, \text{V} \)

### Step 1: Identify the half-reactions and their potentials

1. **Reduction half-reaction for Silver:**
\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag}(s) \quad E^{\circ} = 0.80 \, \text{V} \]

2. **Reduction half-reaction for Iron:**
\[ \text{Fe}^{3+} + e^{-} \rightarrow \text{Fe}^{2+} \quad E^{\circ} = 0.77 \, \text{V} \]

### Step 2: Determine the anode and cathode

- The half-reaction with the higher reduction potential will occur at the cathode. Here, silver has a higher reduction potential (0.80 V) than iron (0.77 V).
- Therefore, the cathode is where silver is reduced, and the anode is where iron is oxidized.

### Step 3: Calculate the standard cell potential (E°cell)

Using the formula:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]

Substituting the values:
\[ E^{\circ}_{cell} = 0.80 \, \text{V} - 0.77 \, \text{V} = 0.03 \, \text{V} \]

### Step 4: Calculate ΔG°

The relationship between Gibbs free energy change and cell potential is given by:
\[ \Delta G^{\circ} = -nFE^{\circ}_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred (1 for this reaction)
- \( F \) = Faraday's constant \( (96500 \, \text{C/mol}) \)
- \( E^{\circ}_{cell} = 0.03 \, \text{V} \)

Substituting the values:
\[ \Delta G^{\circ} = -1 \times 96500 \, \text{C/mol} \times 0.03 \, \text{V} \]
\[ \Delta G^{\circ} = -2895 \, \text{J/mol} \]

### Final Answer
\[ \Delta G^{\circ} = -2895 \, \text{J/mol} \]