Calculate the equilibrium constant for the reaction : `Fe^(2+)+Ce^(4+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(Fe^(3+)//Fe^(2+))^(@)=0.68V`

Explanation

To calculate the equilibrium constant for the reaction \( \text{Fe}^{2+} + \text{Ce}^{4+} \rightleftharpoons \text{Fe}^{3+} + \text{Ce}^{3+} \), we can follow these steps:

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

The half-reactions involved are:
1. Reduction of cerium:
\[ \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+} \quad E^\circ = 1.44 \, \text{V} \]
2. Oxidation of iron:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = 0.68 \, \text{V} \]

### Step 2: Determine the cell potential \( E_{\text{cell}} \).

The cell potential can be calculated using the formula:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Here, the cathode is the reduction half-reaction (cerium) and the anode is the oxidation half-reaction (iron):
\[ E_{\text{cell}} = E_{\text{Ce}^{4+}/\text{Ce}^{3+}} - E_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 1.44 \, \text{V} - 0.68 \, \text{V} = 0.76 \, \text{V} \]

### Step 3: Relate cell potential to the equilibrium constant.

At equilibrium, the relationship between the standard cell potential \( E^\circ \) and the equilibrium constant \( K \) is given by the Nernst equation:
\[ E^\circ = \frac{0.059}{n} \log K \]
Where \( n \) is the number of moles of electrons transferred in the balanced equation. In this case, \( n = 1 \) since one electron is transferred in both half-reactions.

### Step 4: Substitute the values into the equation.

Substituting \( E^\circ = 0.76 \, \text{V} \) and \( n = 1 \):
\[ 0.76 = \frac{0.059}{1} \log K \]

### Step 5: Solve for \( \log K \).

Rearranging the equation gives:
\[ \log K = \frac{0.76}{0.059} \]
Calculating this gives:
\[ \log K \approx 12.88 \]

### Step 6: Calculate \( K \).

To find \( K \), we take the antilogarithm:
\[ K = 10^{12.88} \approx 7.6 \times 10^{12} \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is approximately \( 7.6 \times 10^{12} \).

---