Given standard `E^(c-):`
`Fe^(3+)+3e^(-)rarrFe," "E^(c-)=-0.036`
`Fe^(2+)+2e^(-)rarr Fe," "E^(c-)=-0.440V`
The `E^(c-)` of `Fe^(3+)+e^(-) rarr Fe^(2+) ` is

Options

A. `-0.476V`
B. `-0.404V`
C. `0.404V`
D. `0.772V`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Potentials**:
- For the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \), the standard reduction potential \( E^{\circ}_1 = -0.036 \, \text{V} \).
- For the reaction \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \), the standard reduction potential \( E^{\circ}_2 = -0.440 \, \text{V} \).

2. **Write the Target Reaction**:
- We want to find \( E^{\circ} \) for the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \).

3. **Use Gibbs Free Energy Relation**:
- The Gibbs free energy change \( \Delta G \) is related to the cell potential by the equation:
\[ \Delta G = -nFE \]
- Here, \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E \) is the cell potential.

4. **Express \( \Delta G \) for Each Reaction**:
- For the first reaction:
\[ \Delta G_1 = -3F(-0.036) = 0.108F \]
- For the second reaction:
\[ \Delta G_2 = -2F(-0.440) = 0.880F \]

5. **Calculate \( \Delta G \) for the Target Reaction**:
- The Gibbs free energy change for the target reaction can be expressed as:
\[ \Delta G_3 = \Delta G_1 - \Delta G_2 \]
- Substituting the values:
\[ \Delta G_3 = 0.108F - 0.880F = -0.772F \]

6. **Relate \( \Delta G_3 \) to the Standard Potential \( E^{\circ}_3 \)**:
- Using the relation \( \Delta G_3 = -F E^{\circ}_3 \):
\[ -0.772F = -F E^{\circ}_3 \]
- Canceling \( F \) from both sides gives:
\[ E^{\circ}_3 = 0.772 \, \text{V} \]

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

### Final Answer:
\[ E^{\circ}(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) = 0.772 \, \text{V} \]