Given below are half-cell reaction:
`Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.81 V`
`2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V`
The `E^(@)` for `3Mn^(2+) rarr Mn+2Mn^(3+)` will be:

Options

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

Answer

A

Explanation

Using `DeltaG^(@)=-nFE^(@)`
(1) `Mn^(2+)+2erarrMn,E^(@)=-1.18V,`
`DeltaG_(1)^(@)=-2F(-1.18)=2.36F`
(2) `Mn^(3+)+erarrMn^(2+),E^(@)=+1.51V,`
`DeltaG_(2)^(@)=-F(1.51)=-1.51F`
`(1)-(2)xx(2)`
`3Mn^(2+)rarrMn+2Mn^(3+),`
`DeltaG_(3)^(@)=DeltaG_(1)^(@)-2Delta_(2)^(@)`
`=[2.36-2(-1.51)]F=(2.36+3.02)F=5.8F`
But, `DeltaG_(3)^(@)=-2FE^(@),5.38F=-2FE^(@)`
`E^(@)=-2.69V`
As `E^(@)` value in negative reaction is non-spontaneous.