For the electrochemical cell, `M|M^(+) ||X^(-)|X`, `E_(M^(+)//M)^(@) = 0.44 V` and `E_(X//X^(-))^(@) = 0.33 V`. From this data we 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") = 0.77 V`
D. `E_("cell") = - 0.77 V`

Answer

To solve the problem, we need to analyze the given electrochemical cell and the standard reduction potentials provided.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard reduction potentials**:
- The half-reaction for M to M⁺ is:
\[ M + e^- \leftrightarrow M^+ \quad (E^\circ = +0.44 \, V) \]
- The half-reaction for X⁻ to X is:
\[ X^- + e^- \leftrightarrow X \quad (E^\circ = +0.33 \, V) \]

2. **Determine the oxidation and reduction processes**:
- In the electrochemical cell notation \( M|M^+ || X^-|X \):
- M is oxidized to M⁺ (loses electrons).
- X⁻ is reduced to X (gains electrons).

3. **Calculate the standard cell potential (E°cell)**:
- The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \]
- Here, M is oxidized, so we take the negative of its reduction potential:
\[ E^\circ_{cell} = E^\circ_{X/X^-} - E^\circ_{M^+/M} \]
\[ E^\circ_{cell} = 0.33 \, V - 0.44 \, V \]
\[ E^\circ_{cell} = -0.11 \, V \]

4. **Analyze the spontaneity of the reaction**:
- A positive cell potential indicates a spontaneous reaction. Since \( E^\circ_{cell} = -0.11 \, V \), the reaction is non-spontaneous.
- According to the Gibbs free energy equation:
\[ \Delta G = -nFE^\circ_{cell} \]
- Since \( E^\circ_{cell} < 0 \), it implies \( \Delta G > 0 \), indicating that the reaction is not spontaneous.

5. **Conclusion**:
- From the calculations, we can conclude that the reaction where M is oxidized and X⁻ is reduced is not spontaneous. Therefore, the correct deduction from the given data is that the reaction is non-spontaneous.