`E_(M^(3+)//(M))^(@) = -0.036V` , `E_(M^(2+)//M)^(@)= -0.439V`. The value of standard electrode potential for the change, `M^(3+)(aq) + e^(-)rightarrow M^(2+) (aq)` will be :

Options

A. `(-0.072V)`
B. `(0.385V)`
C. `(0.770V)`
D. `0.270V`

Explanation

To find the standard electrode potential for the reaction \( M^{3+} + e^- \rightarrow M^{2+} \), we can use the given standard electrode potentials for the half-reactions involving \( M^{3+} \) and \( M^{2+} \).

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( E^\circ_{M^{3+}/M} = -0.036 \, \text{V} \)
- \( E^\circ_{M^{2+}/M} = -0.439 \, \text{V} \)

2. **Write the Reactions**:
- The first reaction for \( M^{3+} \) to \( M \):
\[ M^{3+} + 3e^- \rightarrow M \quad (E^\circ = -0.036 \, \text{V}) \]
- The second reaction for \( M^{2+} \) to \( M \):
\[ M^{2+} + 2e^- \rightarrow M \quad (E^\circ = -0.439 \, \text{V}) \]

3. **Determine the Reaction of Interest**:
- We want to find the standard electrode potential for the reaction:
\[ M^{3+} + e^- \rightarrow M^{2+} \]

4. **Calculate Gibbs Free Energy Changes**:
- For the first reaction:
\[ \Delta G_1 = -nF E^\circ_{M^{3+}/M} = -3F(-0.036) = 0.108F \]
- For the second reaction:
\[ \Delta G_2 = -nF E^\circ_{M^{2+}/M} = -2F(-0.439) = 0.878F \]

5. **Set Up the Equation for the Desired Reaction**:
- The Gibbs free energy change for the desired reaction can be expressed as:
\[ \Delta G_3 = \Delta G_1 - \Delta G_2 \]
- Therefore:
\[ \Delta G_3 = 0.108F - 0.878F = -0.770F \]

6. **Relate Gibbs Free Energy to Electrode Potential**:
- We know that:
\[ \Delta G_3 = -F E^\circ \]
- Thus:
\[ -F E^\circ = -0.770F \]
- Dividing both sides by \( -F \):
\[ E^\circ = 0.770 \, \text{V} \]

7. **Final Answer**:
- The standard electrode potential for the reaction \( M^{3+} + e^- \rightarrow M^{2+} \) is:
\[ \boxed{0.770 \, \text{V}} \]