Calculate the equilibrium constant for the following reaction at 298K.
`Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)`
`R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)`

Explanation

To calculate the equilibrium constant (Kc) for the reaction:

\[ \text{Cu(s)} + \text{Cl}_2(g) \rightarrow \text{CuCl}_2(aq) \]

at 298 K, we will follow these steps:

### Step 1: Identify the standard reduction potentials
We have the following standard reduction potentials:
- \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \)
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \)

### Step 2: Determine the half-reactions
The half-reactions for the given reaction are:
1. Reduction at the cathode:
\[ \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^- \]
2. Oxidation at the anode:
\[ \text{Cu(s)} \rightarrow \text{Cu}^{2+}(aq) + 2e^- \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cl}_2/\text{Cl}^-} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} \]
\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} - 0.34 \, \text{V} = 1.02 \, \text{V} \]

### Step 4: Calculate the Gibbs free energy change (ΔG°)
The relationship between the Gibbs free energy change and the cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, \text{C/mol} \)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 1.02 \, \text{V} \]
\[ \Delta G^\circ = -196,020 \, \text{J/mol} \]

### Step 5: Relate ΔG° to the equilibrium constant (Kc)
Using the relation:
\[ \Delta G^\circ = -RT \ln K_c \]
We can rearrange it to find Kc:
\[ \ln K_c = -\frac{\Delta G^\circ}{RT} \]
Substituting the values:
- \( R = 8.314 \, \text{J/(K·mol)} \)
- \( T = 298 \, \text{K} \)

Calculating:
\[ \ln K_c = -\frac{-196020 \, \text{J/mol}}{8.314 \, \text{J/(K·mol)} \times 298 \, \text{K}} \]
\[ \ln K_c = \frac{196020}{2477.572} \approx 79.06 \]

### Step 6: Calculate Kc
To find Kc, we exponentiate the result:
\[ K_c = e^{79.06} \approx 3.715 \times 10^{34} \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at 298 K is approximately:
\[ K_c \approx 3.715 \times 10^{34} \]

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