Calculate the value of equilibrium constant for the reaction :
`2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2)`
Given that `E_(cell)^(@)=0.235" V "`

Explanation

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

\[ 2 \text{Fe}^{3+} + 2 \text{I}^{-} \rightarrow 2 \text{Fe}^{2+} + \text{I}_2 \]

Given that \( E^\circ_{\text{cell}} = 0.235 \, \text{V} \), we can use the Nernst equation to find the equilibrium constant.

### Step-by-Step Solution:

1. **Identify the Reaction and Nernst Equation**:
The Nernst equation relates the standard cell potential to the equilibrium constant. The equation is given by:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K \]
where \( n \) is the number of moles of electrons transferred in the balanced equation.

2. **Determine the Number of Electrons Transferred (n)**:
In the given reaction:
- Each \( \text{Fe}^{3+} \) ion gains 1 electron to become \( \text{Fe}^{2+} \).
- Therefore, for 2 moles of \( \text{Fe}^{3+} \), 2 electrons are gained.
- The \( \text{I}^- \) ions are oxidized to \( \text{I}_2 \), which involves the loss of 2 electrons (1 electron per \( \text{I}^- \) ion, and there are 2 \( \text{I}^- \) ions).
- Thus, the total number of electrons transferred \( n = 2 \).

3. **Substitute Values into the Nernst Equation**:
Now we can substitute the values into the Nernst equation:
\[ 0.235 = \frac{0.0591}{2} \log K \]

4. **Rearranging the Equation**:
To isolate \( \log K \):
\[ \log K = \frac{0.235 \times 2}{0.0591} \]

5. **Calculate \( \log K \)**:
\[ \log K = \frac{0.470}{0.0591} \approx 7.94 \]

6. **Calculate K**:
To find \( K \), we take the antilogarithm:
\[ K = 10^{7.94} \approx 8.76 \times 10^7 \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is approximately \( 8.76 \times 10^7 \).

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