Given below are the half -cell reactions
`Mn^(2+)+2e^(-) to Mn, E^(@)=-1.18V`
`Mn^(3+)+e^(-) to Mn^(2+), E^(@)=+1.51 V`
The `E^(@)` for `3Mn^(2+)to Mn+2Mn^(3+)` will be _________.

Options

A. `-2.69` V, the reaction will not occur
B. `-2.69` V, the reaction will occur
C. `-0.33` V, the reaction will not occur
D. `0.33` V, the reaction will occur

Answer

A

Explanation

`Mn^(2+) + 2e^(-) to Mn , E^(@) = -1.18 V , " " … (i)`
`Mn^(3+) + e to Mn^(2+) , E^(@) = - 1.51 V , " " … (ii)`
Now multiplying equation (ii) by two and subtracting from equation (i)
`3 Mn^(2+) to Mn^(+) + 2 Mn^(3+)`,
`E^(@) - E_(Ox) + E_("Red")`
`= -1.18 + (-1.51) = -2.69 V`
(-ve value of EMF (i.e. `Delta G = ` +ve ) shows that the reaction is non-spontaneous )