For the cell reaction:
`2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq)`
`E_(cell)^(ɵ)=0.24V` at `298K`. The rstandard gibbs energy `(triangle,G^(ɵ))` of the cell reaction is
[Given that faraday constnat `F=96400Cmol^(-1)]`

Options

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

Explanation

To calculate the standard Gibbs energy change (ΔG°) for the given cell reaction, we will use the formula:

\[ \Delta G^{\circ} = -nFE^{\circ}_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (96400 C/mol)
- \( E^{\circ}_{cell} \) = standard cell potential (0.24 V)

### Step 1: Determine the number of electrons transferred (n)
In the given reaction:
\[ 2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_{2}(aq) \]
Each \( Fe^{3+} \) ion is reduced to \( Fe^{2+} \), which involves the transfer of 1 electron per iron ion. Since there are 2 moles of \( Fe^{3+} \), the total number of electrons transferred is:
\[ n = 2 \text{ (for 2 moles of } Fe^{3+}\text{)} \]

### Step 2: Substitute the values into the Gibbs energy formula
Now, we can substitute the values into the equation:
\[ \Delta G^{\circ} = -nFE^{\circ}_{cell} \]
\[ \Delta G^{\circ} = -2 \times 96400 \, \text{C/mol} \times 0.24 \, \text{V} \]

### Step 3: Calculate ΔG°
Calculating the above expression:
\[ \Delta G^{\circ} = -2 \times 96400 \times 0.24 \]
\[ \Delta G^{\circ} = -46320 \, \text{J} \]

### Step 4: Convert ΔG° to kJ
To convert from Joules to kilojoules, we divide by 1000:
\[ \Delta G^{\circ} = -46.32 \, \text{kJ/mol} \]

### Final Answer
The standard Gibbs energy change (ΔG°) for the cell reaction is:
\[ \Delta G^{\circ} = -46.32 \, \text{kJ/mol} \]