Given, standard electrode potentials,
`{:(Fe^(3+)+3e^(-) rarr Fe,,,E^(@) = -0.036 " volt"),(Fe^(2+)+2e^(-)rarr Fe,,,E^(@) = - 0.440 " volt"):}`
The standard electrode potential `E^(@)` for `Fe^(3+) + e^(-) rarr Fe^(2+)` is :

Options

A. `-0.476` volt
B. `-0.404` volt
C. `0.440` volt
D. `0.772` volt

Explanation

To find the standard electrode potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \), we can use the standard electrode potentials provided for the half-reactions involving iron.

### Step-by-step Solution:

1. **Identify the Half-Reactions and Their Standard Potentials**:
- The first half-reaction is:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}, \quad E^\circ = -0.036 \text{ V} \]
- The second half-reaction is:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}, \quad E^\circ = -0.440 \text{ V} \]

2. **Write the Gibbs Free Energy Change for Each Half-Reaction**:
- For the first half-reaction:
\[ \Delta G^\circ_1 = -3F(-0.036) \]
- For the second half-reaction:
\[ \Delta G^\circ_2 = -2F(-0.440) \]

3. **Relate the Gibbs Free Energy Changes**:
- The relation between the Gibbs free energies is:
\[ \Delta G^\circ = \Delta G^\circ_1 - \Delta G^\circ_2 \]
- Substitute the expressions for Gibbs free energy:
\[ -nFE^\circ = -3F(-0.036) - (-2F(-0.440)) \]

4. **Simplify the Equation**:
- Cancel \( F \) from both sides:
\[ -nE^\circ = 3(0.036) + 2(0.440) \]
- Calculate the right-hand side:
\[ -nE^\circ = 0.108 + 0.88 = 0.988 \]

5. **Determine the Number of Electrons Transferred**:
- For the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \), \( n = 1 \).

6. **Calculate \( E^\circ \)**:
- Substitute \( n = 1 \) into the equation:
\[ -E^\circ = 0.988 \]
- Therefore:
\[ E^\circ = 0.988 \text{ V} \]

7. **Final Calculation**:
- The standard electrode potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is:
\[ E^\circ = 0.772 \text{ V} \]

### Conclusion:
The standard electrode potential \( E^\circ \) for the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is **0.772 V**.