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

Explanation

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

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

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

### Step 1: Identify the Anode and Cathode
In the reaction, chlorine (Cl2) is being reduced (gaining electrons) and iodine (I2) is being oxidized (losing electrons). Therefore:
- **Cathode:** Cl2 (reduction)
- **Anode:** I2 (oxidation)

### Step 2: Calculate the Standard Cell Potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ(\text{Cl}_2, \text{Cl}^-) - E^\circ(\text{I}_2, \text{I}^-) \]
\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} - 0.536 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.824 \, \text{V} \]

### Step 3: Calculate ΔG°
The Gibbs free energy change can be calculated using the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (2 for this reaction)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 0.824 \, \text{V} \]
\[ \Delta G^\circ = -159032 \, \text{J} \]
(Converting to kJ, we get \( \Delta G^\circ = -159.032 \, \text{kJ} \))

### Step 4: Calculate the Equilibrium Constant (K)
We can also relate ΔG° to the equilibrium constant (K) using the formula:
\[ \Delta G^\circ = -2.303RT \log K \]
Where:
- \( R \) = gas constant \( (8.314 \, \text{J/(mol K)}) \)
- \( T \) = temperature in Kelvin (298 K)

Rearranging the formula to find K:
\[ \log K = -\frac{\Delta G^\circ}{2.303RT} \]

Substituting the values:
\[ \log K = -\frac{-159032 \, \text{J}}{2.303 \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}} \]
Calculating the denominator:
\[ \log K = \frac{159032}{5730.78} \]
\[ \log K \approx 27.871 \]

Now, taking the antilog to find K:
\[ K = 10^{27.871} \approx 7.43 \times 10^{27} \]

### Final Results
- **ΔG° = -159032 J (or -159.032 kJ)**
- **Equilibrium Constant, K ≈ 7.43 × 10²⁷**