For the electrochemical cell, `(M|M^(o+))||(X^(c-)|X)`, `E^(c-)._((M^(o+)|M))=0.44V`, and `E^(c-)._((X|X^(c-))=-0.33V`.
From this data, one can conclude that

Options

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

Explanation

To analyze the electrochemical cell `(M|M^(o+))||(X^(c-)|X)` and draw conclusions based on the given standard reduction potentials, we can follow these steps:

### Step 1: Identify the Given Reduction Potentials
We have the following standard reduction potentials:
- For the half-reaction involving metal M:
\[ E^\circ_{(M^{o+}/M)} = 0.44 \, \text{V} \]
- For the half-reaction involving species X:
\[ E^\circ_{(X/X^{c-})} = -0.33 \, \text{V} \]

### Step 2: Convert the Reduction Potential of M to Oxidation Potential
The oxidation potential is the negative of the reduction potential. Therefore, the oxidation potential for the half-reaction of M is:
\[ E^\circ_{(M/M^{o+})} = -E^\circ_{(M^{o+}/M)} = -0.44 \, \text{V} \]

### Step 3: Calculate the Standard Cell Potential (E°cell)
To find the standard cell potential, we use the formula:
\[ E^\circ_{cell} = E^\circ_{(cathode)} - E^\circ_{(anode)} \]
In this case, X is the cathode (since it has a more positive reduction potential) and M is the anode. Thus:
\[ E^\circ_{cell} = E^\circ_{(X/X^{c-})} - E^\circ_{(M/M^{o+})} \]
Substituting the values:
\[ E^\circ_{cell} = (-0.33 \, \text{V}) - (-0.44 \, \text{V}) = -0.33 + 0.44 = 0.11 \, \text{V} \]

### Step 4: Determine the Spontaneity of the Reaction
A positive E°cell indicates that the reaction is spontaneous. Since we calculated:
\[ E^\circ_{cell} = 0.11 \, \text{V} > 0 \]
This means the reaction is spontaneous under standard conditions.

### Conclusion
From the data provided, we can conclude that the electrochemical reaction in this cell is spontaneous because the standard cell potential (E°cell) is positive.

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