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

Options

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

Explanation

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 Standard Reduction Potentials:**
- For the half-reaction \( M^{+} + e^{-} \rightarrow M \), the standard reduction potential \( E^{\circ} (M^{+}/M) = 0.44 \, V \).
- For the half-reaction \( X + e^{-} \rightarrow X^{-} \), the standard reduction potential \( E^{\circ} (X/X^{-}) = 0.33 \, V \).

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