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

Answer

B

Explanation

For the given cell `M|M^(+)||X^(-)|X`, the cell reaction is derived as follows:
R.H.S. reduction `X+e^(-)toX^(-)` . . (i)
L.H.S. oxidation `MtoM^(+)+e^(-)` . . . . (ii)
Add (i) and (ii) M+X`toM^(+)+X^(-)`
the cell potential `=-0.11V`
Since, `E_(cell)=-ve`, the cell reaction derived above is not spontaneous. In fact, the reverse reaction will occur spontaneously.