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

Answer

C

Explanation

`{:(Mn^(2+)+2e^(-) rarr Mn,E_(Mn^(2+)//Mn)^(@)=-1.18 V),(2Mn^(2+) rarr 2Mn^(3+)+2e^(-),E_(Mn^(3+)//Mn^(2+))^(@)=+1.51 V),(bar(3Mn^(2+) rarr 2Mn^(3+)+Mn),):}`
`E_("Redox process")^(@)=E_(Mn^(2+)//Mn^(2+))^(@)`
`=-1.81-1.51=-2.69 V`