Given,`{:(E_(Fe^(3+) //Fe)^o + 3e E^o =- 0.036 V),(E_(Fe^(3+) //Fe)^o =- 0.439 V):}`
The value of standard electrode ptoential for the charge,

Options

A. ` -0.072 V`
B. ` 0.385 V`
C. ` 0.770 V`
D. ` -0. 270 V`

Explanation

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

### Step-by-Step Solution:

1. **Write down the given standard electrode potentials:**
- For the reaction \( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \):
\[ E^\circ = -0.036 \, \text{V} \]
- For the reaction \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \):
\[ E^\circ = -0.439 \, \text{V} \]

2. **Write the half-reactions:**
- Reaction 1 (for \( \text{Fe}^{3+} \)):
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad (1) \]
- Reaction 2 (for \( \text{Fe}^{2+} \)):
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (2) \]

3. **Write the desired reaction:**
- We want to find the standard electrode potential for the reaction:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (3) \]

4. **Subtract Reaction 2 from Reaction 1:**
- By subtracting Reaction 2 from Reaction 1:
\[ \text{Fe}^{3+} + 3e^- - (\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}) = \text{Fe}^{3+} - \text{Fe}^{2+} + e^- \rightarrow 0 \]
- This simplifies to:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (3) \]

5. **Calculate the standard electrode potential for Reaction 3:**
- Using the formula:
\[ E^\circ_{\text{reaction}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}} - E^\circ_{\text{Fe}^{2+}/\text{Fe}} \]
- Substitute the values:
\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = (-0.036 \, \text{V}) - (-0.439 \, \text{V}) \]
- Calculate:
\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = -0.036 + 0.439 = 0.403 \, \text{V} \]

6. **Final Result:**
- The standard electrode potential for the reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is:
\[ E^\circ = 0.403 \, \text{V} \]