Calculate the equilibrium constant for the reaction
Fe+ `CuSO_4`→ `FeSO_4`+ Cu at `25^(@)C`
Given `E_(Fe//Fe^(2+))^0=0.44V, E^0(Cu//Cu^(2+))=-0.337V`

Answer

To calculate the equilibrium constant for the reaction:

\[ \text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu} \]

at \( 25^\circ C \), we will follow these steps:

### Step 1: Identify the half-reactions
The first step is to identify the oxidation and reduction half-reactions.

- **Oxidation half-reaction**:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]

- **Reduction half-reaction**:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 2: Determine the standard electrode potentials
We are given the standard electrode potentials:
- \( E^\circ(\text{Fe}^{2+}/\text{Fe}) = 0.44 \, \text{V} \)
- \( E^\circ(\text{Cu}^{2+}/\text{Cu}) = -0.337 \, \text{V} \)

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

Here, the cathode is where reduction occurs (Cu) and the anode is where oxidation occurs (Fe).

Thus,
\[ E^\circ_{\text{cell}} = E^\circ(\text{Cu}^{2+}/\text{Cu}) - E^\circ(\text{Fe}^{2+}/\text{Fe}) \]
\[ E^\circ_{\text{cell}} = (-0.337 \, \text{V}) - (0.44 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.777 \, \text{V} \]

### Step 4: Calculate \( \Delta G^\circ \)
Using the relationship between Gibbs free energy and cell potential:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 in this case)
- \( F \) = Faraday's constant \( (96500 \, \text{C/mol}) \)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times (-0.777 \, \text{V}) \]
\[ \Delta G^\circ = 2 \times 96500 \times 0.777 \]
\[ \Delta G^\circ = 150,000 \, \text{J/mol} \]

### Step 5: Relate \( \Delta G^\circ \) to the equilibrium constant \( K \)
Using the equation:
\[ \Delta G^\circ = -RT \ln K \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \) (since \( 25^\circ C = 298 \, \text{K} \))

Rearranging gives:
\[ \ln K = -\frac{\Delta G^\circ}{RT} \]

Substituting the values:
\[ \ln K = -\frac{150000}{8.314 \times 298} \]
\[ \ln K = -60.24 \]

### Step 6: Calculate \( K \)
To find \( K \):
\[ K = e^{-60.24} \]
Calculating this gives:
\[ K \approx 0.000327 \]

### Final Answer
The equilibrium constant \( K \) for the reaction at \( 25^\circ C \) is approximately:
\[ K \approx 0.000327 \]

---