For the cell reaction:
`2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq)`
`E_(cell)^(ɵ)=0.24V` at `298K`. The standard gibbs energy `(triangle,G^(ɵ))` of the cell reaction is
[Given that faraday constnat `F=96500Cmol^(-1)]`

Options

A. `23.16kJmol^(-1)`
B. `-46.32kJmol^(-1)`
C. `-23.16kJmol^(-1)`
D. `46.32kJmol^(-1)`

Explanation

To find the standard Gibbs free energy change (ΔG°) for the given cell reaction, we can use the relationship between Gibbs free energy, the number of electrons transferred, Faraday's constant, and the standard cell potential. The equation we will use is:

\[ \Delta G^{\circ} = -n \cdot F \cdot E^{\circ}_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (96500 C/mol)
- \( E^{\circ}_{cell} \) = standard cell potential (0.24 V)

### Step-by-Step Solution:

1. **Identify the number of electrons transferred (n)**:
- The cell reaction is:
\[ 2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_{2}(aq) \]
- The half-reactions are:
- Reduction: \( 2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+} \)
- Oxidation: \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)
- From the half-reactions, we see that 2 electrons are transferred in total.
- Therefore, \( n = 2 \).

2. **Substitute the values into the Gibbs free energy equation**:
- Now we can substitute \( n \), \( F \), and \( E^{\circ}_{cell} \) into the equation:
\[ \Delta G^{\circ} = -n \cdot F \cdot E^{\circ}_{cell} \]
\[ \Delta G^{\circ} = -2 \cdot 96500 \, \text{C/mol} \cdot 0.24 \, \text{V} \]

3. **Calculate the value**:
\[ \Delta G^{\circ} = -2 \cdot 96500 \cdot 0.24 \]
\[ \Delta G^{\circ} = -2 \cdot 23160 = -46320 \, \text{J/mol} \]

4. **Convert Joules to Kilojoules**:
- Since we need the answer in kilojoules per mole:
\[ \Delta G^{\circ} = -46320 \, \text{J/mol} \div 1000 = -46.32 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G^{\circ} = -46.32 \, \text{kJ/mol} \]