Given `E_(M^(4+),M^(3+))^(0)=xV, E_(M^(4+),M)^(0)="y V hence "E_(M^(3+),M)^(0)`, is equal to :

Options

A. `(4y-x)/(3)`
B. `(x-3y)/(4)`
C. `(x+4y)/(3)`
D. `(x+2y)/(3)`

Explanation

To solve the problem, we need to find the standard reduction potential \( E_{(M^{3+}, M)}^0 \) given the following information:

1. \( E_{(M^{4+}, M^{3+})}^0 = x \, V \)
2. \( E_{(M^{4+}, M)}^0 = y \, V \)

### Step-by-Step Solution:

1. **Understanding the Given Potentials:**
- The first equation \( E_{(M^{4+}, M^{3+})}^0 = x \) represents the reduction of \( M^{4+} \) to \( M^{3+} \):
\[ M^{4+} + e^- \rightarrow M^{3+} \quad (E^0 = x) \]

- The second equation \( E_{(M^{4+}, M)}^0 = y \) represents the reduction of \( M^{4+} \) to \( M \):
\[ M^{4+} + 4e^- \rightarrow M \quad (E^0 = y) \]

2. **Rearranging the First Reaction:**
- To find the potential for the reaction from \( M^{3+} \) to \( M \), we need to express the first reaction in terms of oxidation:
\[ M^{3+} \rightarrow M^{4+} + e^- \quad (E^0 = -x) \]

3. **Combining the Reactions:**
- Now, we can combine the two reactions:
- The oxidation of \( M^{3+} \) to \( M^{4+} \) (which is \( -x \)) and the reduction of \( M^{4+} \) to \( M \) (which is \( y \)):
\[ M^{3+} \rightarrow M^{4+} + e^- \quad (E^0 = -x) \]
\[ M^{4+} + 4e^- \rightarrow M \quad (E^0 = y) \]

4. **Overall Reaction:**
- The overall reaction combining these two gives:
\[ M^{3+} + 4e^- \rightarrow M \quad (3 \text{ electrons involved}) \]

5. **Finding the Overall Cell Potential:**
- The overall standard cell potential \( E^0 \) can be calculated using the formula:
\[ E^0_{\text{cell}} = E^0_{\text{reduction}} - E^0_{\text{oxidation}} \]
- Substituting the values:
\[ E^0_{\text{cell}} = y - (-x) = y + x \]

6. **Calculating the Final Potential:**
- Since we have 3 electrons involved in the overall reaction, we need to average the potential:
\[ E_{(M^{3+}, M)}^0 = \frac{y - (-x)}{3} = \frac{y + x}{3} \]

### Final Expression:
Thus, the standard reduction potential \( E_{(M^{3+}, M)}^0 \) is:
\[ E_{(M^{3+}, M)}^0 = \frac{4y - x}{3} \]