For the cell reaction
`2Fe^(3+)(aq) + 2l^(-1) (aq) to 2Fe^(2+) (aq) + I_2(aq)`
`E_("cell")^(Theta) = 0.24 V` at 298 K. The standard Gibbs energy `(Delta_r G^Theta)` of the cell reaction is
[Given that Faraday constnat `F = 96500 C mol^(-1)`]

Options

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

Explanation

To calculate the standard Gibbs energy change (Δ_rG^Θ) for the given cell reaction, we can use the following formula:

\[ \Delta_r G^Θ = -nFE_{cell}^{Θ} \]

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

### Step 1: Determine the number of electrons transferred (n)

For the given reaction:
\[ 2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq) \]

We can break this down into half-reactions:

1. Reduction half-reaction:
\[ Fe^{3+} + e^{-} \rightarrow Fe^{2+} \]
(Each \( Fe^{3+} \) gains 1 electron, and since there are 2 \( Fe^{3+} \), the total is 2 electrons.)

2. Oxidation half-reaction:
\[ 2I^{-} \rightarrow I_2 + 2e^{-} \]
(2 \( I^{-} \) lose 2 electrons.)

Thus, the total number of electrons transferred in the balanced reaction is:
\[ n = 2 \]

### Step 2: Substitute values into the Gibbs energy formula

Now that we have \( n \), \( F \), and \( E_{cell}^{Θ} \), we can substitute these values into the formula:

\[ \Delta_r G^Θ = -nFE_{cell}^{Θ} \]
\[ \Delta_r G^Θ = - (2)(96500 \, \text{C/mol})(0.24 \, \text{V}) \]

### Step 3: Calculate Δ_rG^Θ

Calculating the above expression:

\[ \Delta_r G^Θ = - (2)(96500)(0.24) \]
\[ \Delta_r G^Θ = - (2)(96500)(0.24) = - 46320 \, \text{J/mol} \]

### Step 4: Convert Joules to Kilojoules

To convert Joules to Kilojoules, we divide by 1000:

\[ \Delta_r G^Θ = - 46.32 \, \text{kJ/mol} \]

### Final Answer

The standard Gibbs energy change (Δ_rG^Θ) of the cell reaction is:

\[ \Delta_r G^Θ = -46.32 \, \text{kJ/mol} \]

---