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

Answer

B

Explanation

`M|M^(+)||X^(-)|X`
The two half cell reactions for the cell are
`MrarrM^(+)+e^(-)`
or `X+e^(-)rarrX^(-)`
`M+XrarrM^(+)+X`
`E_("cell")^(@)=E_(X//X^(-))^(@)-E_(M^(+)//M)`
`=0.33V-0.44V=-0.11V`
As `E_("Cell")^(@)` is negative the reaction involved is non spon taneus. the reverse reaction, howev er, will be spontaneous.
`M^(+)+X^(-)rarrM+X`