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 reduction potentials and determine the spontaneity of the reactions involving the species M and X.

### Step-by-Step Solution:

1. **Identify the Given Potentials:**
- The standard reduction potential for the half-reaction \( M^+ + e^- \rightarrow M \) is given as:
\[ E^\circ(M^+/M) = -0.44 \, \text{V} \]
- The standard reduction potential for the half-reaction \( X + e^- \rightarrow X^- \) is given as:
\[ E^\circ(X/X^-) = 0.33 \, \text{V} \]

2. **Determine the Oxidation and Reduction Reactions:**
- Since the reduction potential for \( M^+/M \) is negative, it indicates that \( M^+ \) is a weaker oxidizing agent compared to \( X \). Therefore, \( M \) will undergo oxidation:
\[ M \rightarrow M^+ + e^- \]
- The reduction will occur with \( X \):
\[ X + e^- \rightarrow X^- \]

3. **Calculate the Standard Cell Potential (\( E^\circ_{cell} \)):**
- The standard cell potential is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Here, \( E^\circ_{cathode} \) (for reduction) is \( E^\circ(X/X^-) = 0.33 \, \text{V} \) and \( E^\circ_{anode} \) (for oxidation) is \( E^\circ(M^+/M) = -0.44 \, \text{V} \).
- Therefore:
\[ E^\circ_{cell} = 0.33 - (-0.44) = 0.33 + 0.44 = 0.77 \, \text{V} \]

4. **Determine Spontaneity of the Reaction:**
- A positive \( E^\circ_{cell} \) indicates that the reaction is spontaneous. Since \( E^\circ_{cell} = 0.77 \, \text{V} > 0 \), the reaction is spontaneous.

5. **Conclusion:**
- From the analysis, we conclude that the reaction involving \( M \) being oxidized and \( X \) being reduced is spontaneous, and the calculated standard cell potential is \( 0.77 \, \text{V} \).

### Summary of Options:
- Based on the calculations, we can conclude:
- Option A: Correct (spontaneous reaction)
- Option C: Correct (cell potential is positive)
- Option B: Incorrect (reverse reaction is non-spontaneous)
- Option D: Incorrect (not relevant based on our calculations)