If`E^(@)(M^(+)/M)= -0.44V` and `E^(@)(X/X^(-)) = 0.33V`.
From this data one can conclude that:

Options

A. `M+Xrightarrow M^(+)+X^(-)` is the spontaneous reaction
B. `M^(+)+ X^(-) rightarrow M+X` is the spontaneous reaction
C. `E_(cell)^(@) = 0.77V` for the feasible cell reaction
D. `E_(cell)^(@) = -0.77V`

Explanation

To solve the problem, we need to analyze the given standard electrode potentials and draw conclusions about the spontaneity of the reactions involving the species M and X.

### Step-by-Step Solution:

1. **Identify the Given Standard Electrode Potentials**:
- For the half-reaction \( M^{+}/M \): \( E^\circ = -0.44 \, \text{V} \)
- For the half-reaction \( X/X^{-} \): \( E^\circ = 0.33 \, \text{V} \)

2. **Compare the Standard Electrode Potentials**:
- The standard reduction potential for \( X/X^{-} \) (0.33 V) is greater than that for \( M^{+}/M \) (-0.44 V). This indicates that \( X \) has a greater tendency to be reduced than \( M^{+} \).

3. **Determine the Oxidation and Reduction Reactions**:
- Since \( X \) is more likely to be reduced, it will act as the cathode in the electrochemical cell, while \( M \) will be oxidized.
- The oxidation reaction can be written as:
\[ M \rightarrow M^{+} + e^{-} \]
- The reduction reaction can be written as:
\[ X + e^{-} \rightarrow X^{-} \]

4. **Calculate the Standard Cell Potential \( E^\circ_{\text{cell}} \)**:
- The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Here, \( E^\circ_{\text{cathode}} = 0.33 \, \text{V} \) and \( E^\circ_{\text{anode}} = -0.44 \, \text{V} \).
- Substituting the values:
\[ E^\circ_{\text{cell}} = 0.33 \, \text{V} - (-0.44 \, \text{V}) = 0.33 \, \text{V} + 0.44 \, \text{V} = 0.77 \, \text{V} \]

5. **Determine the Spontaneity of the Reaction**:
- A positive \( E^\circ_{\text{cell}} \) indicates that the reaction is spontaneous. Since \( E^\circ_{\text{cell}} = 0.77 \, \text{V} \), the reaction is spontaneous.

6. **Conclusion**:
- The overall reaction is spontaneous, and we can conclude that:
- \( M \) oxidizes to \( M^{+} \) and \( X \) reduces to \( X^{-} \).
- The standard cell potential is \( 0.77 \, \text{V} \).