Calculate at `25^(@)` C , the equilibrium constant for the reaction :
`2 Fe^(3+) + Sn^(2+) hArr 2 Fe^(2+) + Sn^(4+)`
Given that `E_((Fe^(3+)|Fe^(2+)))^(Theta) = 0.771 V , E_((Sn^(4+) |Sn^(2+)))^(Theta) = 0.150 V`

Explanation

To calculate the equilibrium constant for the reaction

\[ 2 \text{Fe}^{3+} + \text{Sn}^{2+} \rightleftharpoons 2 \text{Fe}^{2+} + \text{Sn}^{4+} \]

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

### Step 1: Identify the half-reactions and their standard reduction potentials
The given standard reduction potentials are:
- For the reduction of iron:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = 0.771 \, \text{V} \]
- For the reduction of tin:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad E^\circ = 0.150 \, \text{V} \]

### Step 2: Write the oxidation half-reaction for tin
Since tin is being oxidized in the reaction, we need to reverse the reduction potential:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \quad E^\circ = -0.150 \, \text{V} \]

### Step 3: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
The overall cell potential is calculated by adding the reduction potential of iron and the oxidation potential of tin:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction (Fe)}} + E^\circ_{\text{oxidation (Sn)}} \]
\[ E^\circ_{\text{cell}} = 0.771 \, \text{V} + (-0.150 \, \text{V}) = 0.621 \, \text{V} \]

### Step 4: Use the Nernst equation to relate \( E^\circ_{\text{cell}} \) to the equilibrium constant \( K \)
At equilibrium, the cell potential \( E_{\text{cell}} \) is 0. The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K \]
Setting \( E_{\text{cell}} = 0 \):
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K \]
Rearranging gives:
\[ \log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591} \]

### Step 5: Determine the number of electrons transferred \( n \)
In the balanced reaction, 2 moles of electrons are transferred (1 electron for each Fe and 2 electrons for Sn). Thus, \( n = 2 \).

### Step 6: Substitute values into the equation
Substituting \( E^\circ_{\text{cell}} = 0.621 \, \text{V} \) and \( n = 2 \):
\[ \log K = \frac{2 \cdot 0.621}{0.0591} \]
Calculating this gives:
\[ \log K = \frac{1.242}{0.0591} \approx 21.03 \]

### Step 7: Calculate \( K \)
To find \( K \), we take the antilogarithm:
\[ K \approx 10^{21.03} \approx 1.07 \times 10^{21} \]

### Final Answer
The equilibrium constant \( K \) for the reaction at \( 25^\circ C \) is approximately:
\[ K \approx 1.07 \times 10^{21} \]

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