Given below are the half -cell reactions
`Mn^(2+) + 2e^(-) rarr Mn, E^@ =- 1. 18 V`
`2 (Mn^(3+) = E^(-) rarr Mn^(2+)). E^@ = + 1.5 V`
The `E^@` for `Mn^(2+) rarr Mn+ 2 Mn^(3+)` will be.

Options

A. `-2.69 V`, the reaction will not 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^(-) rarr Mn, E^2 =- 1 . 18 V` …(i)
`2 (Mn^(2+) +e^(-) rarr Mn^(2+) ), E^@ = + 1.5 51 V` …(ii)
Subtracting Eq (i) , we get
`3 Mn^(2+) =- 1.18 - (+ 1.51) =- 2 69 V`
`E^@ =- 1. 18 -(+1. 51 ) =- 2. 69 V`
Since, the value of `E^@` is `- ve` therefore the reactionos non-spontaneous .
Alternate methode
`Mn^(2+) + 2 e^(-) rarr Mn, E^@ =- 1. 18 V`
`Delta _1^@ =- 2 F (-1.18) = 2 . 36 F`
`2 M n^(3+) + 2 e^(-) rarr 2Mn^(2+), E^@ = + 1.5 1 V`...(ii)
` Delta_2^@ =- 2 F 9- 1. 51 ] =- 3 . 02 F`
Subtracting Eq. (kk) from Eq. (i) , we get
`2 Mn^(2+) rarr Mn+ 2 M n^(3+) [n=2)`
`Delta_3^@ = DeltaG_1^@ -Delta G_2^@ = 5 . 38 F`
` rArr - 2 FE^@ = 5 . 38 F`
`rArr E^@ =- 2. 2. 69 V`.