Calculate `Delta G^(Theta)` and the equilibrium constant for the cell reaction ,
`Cl_2 + 2 I^(-) to 2 Cl^(-) + I_(2)`
Given that `E_((Cl_(2) , Cl^(-)))^(Theta) = 1.36 V , E_((I_(2) , I^(-)))^(Theta) = 0.536V `

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kc) for the cell reaction:

\[ \text{Cl}_2 + 2 \text{I}^- \rightarrow 2 \text{Cl}^- + \text{I}_2 \]

Given:
- \( E^\circ_{\text{(Cl}_2, \text{Cl}^-)} = 1.36 \, \text{V} \)
- \( E^\circ_{\text{(I}_2, \text{I}^-)} = 0.536 \, \text{V} \)

### Step 1: Determine the standard cell potential (E°cell)

1. Identify the half-reactions:
- Reduction: \( \text{Cl}_2 + 2e^- \rightarrow 2 \text{Cl}^- \) (Chlorine is reduced)
- Oxidation: \( 2 \text{I}^- \rightarrow \text{I}_2 + 2e^- \) (Iodide is oxidized)

2. Write the standard cell potential equation:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

3. Substitute the values:
- For reduction of chlorine: \( E^\circ_{\text{reduction}} = 1.36 \, \text{V} \)
- For oxidation of iodide (reverse the sign of reduction potential): \( E^\circ_{\text{oxidation}} = 0.536 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} - 0.536 \, \text{V} = 0.824 \, \text{V} \]

### Step 2: Calculate ΔG°

1. Use the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

2. Determine \( n \) (number of moles of electrons transferred):
- From the balanced equation, \( n = 2 \) (2 electrons are transferred).

3. Use Faraday's constant \( F = 96500 \, \text{C/mol} \).

4. Substitute the values into the equation:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 0.824 \, \text{V} \]
\[ \Delta G^\circ = -159,000 \, \text{J} = -159 \times 10^3 \, \text{J} \]

### Step 3: Calculate the equilibrium constant (Kc)

1. Use the relationship between ΔG° and Kc:
\[ \log K_c = -\frac{\Delta G^\circ}{2.303RT} \]

2. Use \( R = 8.314 \, \text{J/(mol K)} \) and \( T = 298 \, \text{K} \).

3. Substitute the values:
\[ \log K_c = -\frac{-159000 \, \text{J}}{2.303 \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}} \]

4. Calculate:
\[ \log K_c = \frac{159000}{2.303 \times 8.314 \times 298} \]
\[ \log K_c \approx 7.35 \]

5. Convert from logarithmic form to Kc:
\[ K_c = 10^{7.35} \approx 2.24 \times 10^7 \]

### Final Results
- \( \Delta G^\circ = -159 \times 10^3 \, \text{J} \)
- \( K_c \approx 2.24 \times 10^7 \)