For the electrochemical cell, `(M|M^(o+))||(X^(c-)|X),E^(c-)._((M^(o+)|M))=0.44V` and `E^(c-)._((X|X^(c-)))=0.334V`

Options

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

Explanation

To solve the problem, we need to determine the overall cell potential (E_cell) for the given electrochemical cell and identify whether the reaction is spontaneous or non-spontaneous.

### Step-by-Step Solution:

1. **Identify the Half-Cells and Their Potentials:**
- The half-cell reactions are given as:
- \( M^{o+} + e^- \rightarrow M \) with \( E^\circ_{M^{o+}/M} = 0.44 \, V \) (Anode)
- \( X + e^- \rightarrow X^{c-} \) with \( E^\circ_{X/X^{c-}} = 0.334 \, V \) (Cathode)

2. **Determine the Anode and Cathode:**
- In an electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode.
- Since \( M^{o+} \) is being reduced, it will act as the anode (oxidation occurs here).
- \( X^{c-} \) is being reduced, so it will act as the cathode.

3. **Calculate the Cell Potential (E_cell):**
- The formula for the cell potential is:
\[ E_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Substituting the values:
\[ E_{cell} = 0.334 \, V - 0.44 \, V = -0.106 \, V \]

4. **Determine Spontaneity:**
- A positive \( E_{cell} \) indicates a spontaneous reaction, while a negative \( E_{cell} \) indicates a non-spontaneous reaction.
- Since \( E_{cell} = -0.106 \, V \), the reaction is non-spontaneous.

5. **Identify the Overall Reaction:**
- The overall reaction can be written as:
\[ M + X^{c-} \rightarrow M^{o+} + X \]
- Here, \( M \) is oxidized to \( M^{o+} \) and \( X^{c-} \) is reduced to \( X \).

### Final Answer:
The correct option is that the cell reaction is non-spontaneous, as indicated by the negative cell potential.