Consider the following half cell reactions
`Mn^(2+) +2e^(-) rarr Mn E = -1.18 v `
`Mn^(2+) rarr Mn^(3+) + e E^(-) = -1.51 V `
The `E^(-)` for the reaction `3 Mn^(2+) rarr Mn+2Mn^(3+)` and the possibility of the forward reaction are respectively .

Options

A. 2.69V and spontaneous
B. `-2.69` and non spontaneous
C. 0.33V and Spontaneous
D. 4.18V and non spontaneous

Answer

B

Explanation

`Mn^(2+) +2e^(-) rarr Mn (E_("red")^(0)) = -1.18 V `
`2[Mn^(2+) rarr Mn^(3+) +e^(-) ] (E_(ox)^(@)) = -1.51 V `
`3 Mn^(2+) rarr Mn^(3+) + 2Mn^(3+) +2Mn^(3+) E_("cell")^(@) `=?
`E_("cell")^(@) = (E_(ox)^(@)) +(E_("red")^(@))`
`= -1.51 -1.18 ` and non spontaneous
`= -2.69 V `
Since `E^(@) ` is = - ve `DeltaG` is +ve and the given forward cell reaction is non - spontaneous.