Calculate the equilibrium cosntant for the reaction, `Zn+Cd^(2+)hArrZn^(2+)+Cd`,
If `E_(Cd^(2+)//Cd)^(@)=-0.403V and E_(Zn^(2+)//Zn)^(@)=-0.763V`

Answer

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

\[ \text{Zn} + \text{Cd}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cd} \]

we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.

The half-reactions are:
1. For cadmium:
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \quad (E^\circ = -0.403 \, \text{V}) \]
2. For zinc:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad (E^\circ = -0.763 \, \text{V}) \]

### Step 2: Determine the oxidation and reduction processes.

In the reaction:
- Cadmium is reduced (gains electrons).
- Zinc is oxidized (loses electrons).

Thus:
- The cathode (reduction) is cadmium: \( E^\circ_{\text{cathode}} = -0.403 \, \text{V} \)
- The anode (oxidation) is zinc: \( E^\circ_{\text{anode}} = -0.763 \, \text{V} \)

### Step 3: Calculate the standard cell potential (E°cell).

Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.403) - (-0.763) = -0.403 + 0.763 = 0.360 \, \text{V} \]

### Step 4: Relate the cell potential to the Gibbs free energy change (ΔG).

The relationship between the Gibbs free energy change and the cell potential is given by:
\[ \Delta G = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) = number of moles of electrons transferred (2 for this reaction)
- \( F \) = Faraday's constant (approximately \( 96485 \, \text{C/mol} \))

### Step 5: Calculate ΔG.

Substituting the values:
\[ \Delta G = -2 \times 96485 \times 0.360 \]
Calculating this gives:
\[ \Delta G = -69306 \, \text{J/mol} \]

### Step 6: Relate ΔG to the equilibrium constant (Kc).

The relationship between ΔG and Kc is given by:
\[ \Delta G = -RT \ln K_c \]
Rearranging gives:
\[ \ln K_c = -\frac{\Delta G}{RT} \]

### Step 7: Substitute the values for R and T.

Using \( R = 8.314 \, \text{J/(mol K)} \) and assuming \( T = 298 \, \text{K} \):
\[ \ln K_c = -\frac{-69306}{8.314 \times 298} \]
Calculating the denominator:
\[ 8.314 \times 298 \approx 2477.572 \]
Thus:
\[ \ln K_c \approx \frac{69306}{2477.572} \approx 27.95 \]

### Step 8: Calculate Kc.

Exponentiating both sides to solve for Kc:
\[ K_c = e^{27.95} \approx 1.585 \times 10^{12} \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c \approx 1.585 \times 10^{12} \]

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