For the electrochemicl cell, `M|M^(+)||X^(-)|X E_((M^(+)//M))^(@) = 0.44 V` and `E_((X//X^(-)))^(@) = 0.33 V`
From this data one can deduce that :

Options

A. `M +X rarr M^(+) +X^(-)` is the spontaneous reaction
B. `M^(+) +X^(-) rarr M +X` is the spontaneous reaction
C. `E_(cell) = 0.77V`
D. `E_(cell) = - 0.77V`

Explanation

To solve the question, we need to analyze the given electrochemical cell data and determine the spontaneity of the reactions based on the standard electrode potentials.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Standard reduction potential for \( M^{+}/M \) is \( E^{\circ} = 0.44 \, V \)
- Standard reduction potential for \( X/X^{-} \) is \( E^{\circ} = 0.33 \, V \)

2. **Determine the Reactions:**
- For \( M^{+} + e^{-} \rightarrow M \) (Reduction at cathode)
- For \( X + e^{-} \rightarrow X^{-} \) (Reduction at cathode)

3. **Identify Anode and Cathode:**
- In the first reaction, \( M \) is reduced, so it acts as the cathode.
- In the second reaction, \( X \) is reduced to \( X^{-} \), so it acts as the anode.

4. **Calculate the Cell Potential (E°cell):**
- The cell potential can be calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
- Here, we consider the overall reaction \( M + X^{-} \rightarrow M^{+} + X \):
- \( E^{\circ}_{\text{cathode}} = E^{\circ}_{M^{+}/M} = 0.44 \, V \)
- \( E^{\circ}_{\text{anode}} = E^{\circ}_{X/X^{-}} = 0.33 \, V \)
- Therefore,
\[ E^{\circ}_{\text{cell}} = 0.44 \, V - 0.33 \, V = 0.11 \, V \]

5. **Determine Spontaneity:**
- A positive \( E^{\circ}_{\text{cell}} \) indicates that the reaction is spontaneous.
- Since \( E^{\circ}_{\text{cell}} = 0.11 \, V \) is positive, the reaction \( M + X^{-} \rightarrow M^{+} + X \) is spontaneous.

6. **Evaluate the Options:**
- **Option A:** \( M + X \rightarrow M^{+} + X^{-} \) - Non-spontaneous (as calculated, \( E^{\circ}_{\text{cell}} < 0 \))
- **Option B:** \( M^{+} + X^{-} \rightarrow M + X \) - Spontaneous (as calculated, \( E^{\circ}_{\text{cell}} > 0 \))
- **Option C:** \( E^{\circ}_{\text{cell}} = 0.77 \, V \) - Incorrect
- **Option D:** \( E^{\circ}_{\text{cell}} = -0.77 \, V \) - Incorrect

### Conclusion:
The correct answer is **Option B**: \( M^{+} + X^{-} \rightarrow M + X \) is a spontaneous reaction.