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 determine the standard electrode potential \( E^{0}_{(M^{3+}, M)} \) using the given values \( E^{0}_{(M^{4+}, M^{3+})} = x \) volts and \( E^{0}_{(M^{4+}, M)} = y \) volts.

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- The half-reaction for the conversion of \( M^{4+} \) to \( M^{3+} \) is:
\[ M^{4+} + e^- \rightarrow M^{3+} \quad (1) \]
This corresponds to the standard potential \( E^{0}_{(M^{4+}, M^{3+})} = x \).

- The half-reaction for the conversion of \( M^{4+} \) to \( M \) is:
\[ M^{4+} + 4e^- \rightarrow M \quad (2) \]
This corresponds to the standard potential \( E^{0}_{(M^{4+}, M)} = y \).

- The half-reaction for the conversion of \( M^{3+} \) to \( M \) is:
\[ M^{3+} + 3e^- \rightarrow M \quad (3) \]
We need to find \( E^{0}_{(M^{3+}, M)} \).

2. **Using the Nernst Equation**:
- The relationship between the potentials can be derived from the overall reaction:
\[ M^{3+} + 3e^- \rightarrow M \quad (3) \]
can be expressed in terms of reactions (1) and (2).

3. **Setting Up the Equation**:
- We can express the potential for reaction (3) as:
\[ E^{0}_{(M^{3+}, M)} = E^{0}_{(M^{4+}, M)} - E^{0}_{(M^{4+}, M^{3+})} \]
- Substituting the known values:
\[ E^{0}_{(M^{3+}, M)} = y - x \]

4. **Final Expression**:
- To find \( E^{0}_{(M^{3+}, M)} \) in terms of \( x \) and \( y \), we can rearrange the equation:
\[ E^{0}_{(M^{3+}, M)} = \frac{4y + x}{3} \]

### Conclusion:
Thus, the standard electrode potential \( E^{0}_{(M^{3+}, M)} \) is given by:
\[ E^{0}_{(M^{3+}, M)} = \frac{4y + x}{3} \]