The equilibrium constant of the following redox rection at 298 K is `1 xx 10^(8)`
`2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s)`
If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of `Fe^(3+)//Fe^(2+)` ?

Options

A. `+ 1.006 V`
B. `-1.006 V`
C. `+ 0.77 V`
D. `-0.77 V`

Explanation

To find the standard reduction potential of the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \), we can use the relationship between the equilibrium constant \( K \) and the standard cell potential \( E^\circ \).

### Step-by-Step Solution:

1. **Identify the given data:**
- Equilibrium constant \( K = 1 \times 10^8 \)
- Standard reduction potential of iodine \( \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \) is \( +0.54 \, \text{V} \).

2. **Use the Nernst equation to relate \( E^\circ \) and \( K \):**
\[ E^\circ = \frac{0.059}{n} \log K \]
Here, \( n \) is the number of moles of electrons transferred in the balanced redox reaction. For the given reaction, \( n = 2 \).

3. **Substitute the values into the equation:**
\[ E^\circ = \frac{0.059}{2} \log(1 \times 10^8) \]

4. **Calculate \( \log(1 \times 10^8) \):**
\[ \log(1 \times 10^8) = 8 \]

5. **Now calculate \( E^\circ \):**
\[ E^\circ = \frac{0.059}{2} \times 8 = 0.236 \, \text{V} \]

6. **Set up the equation for the overall reaction:**
The overall redox reaction can be expressed as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]
Here, the reduction potential for iodine is given as \( +0.54 \, \text{V} \).

7. **Substituting the known values:**
\[ 0.236 = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - 0.54 \]

8. **Solve for \( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} \):**
\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.236 + 0.54 = 0.776 \, \text{V} \]

9. **Round to two decimal places:**
\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} \approx 0.77 \, \text{V} \]

### Final Answer:
The standard reduction potential of \( \text{Fe}^{3+}/\text{Fe}^{2+} \) is \( +0.77 \, \text{V} \).

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