Half-cell reactions are given below :
`Mn^(2+)+2e^(-) to Mn,E_(0)=-1.18V`
`2(Mn^(3+)+e^(-) to Mn^(2+)),E_(0)=+1.51V` then what is `E_(0)` for `3Mn^(2+) to Mn+2Mn^(3+)` ?

Options

A. `-0.33V`, no reaction
B. `-0.33V`, reaction can be possible
C. `-2.69V`, no reaction
D. `-2.69V`, no reaction

Answer

A, B, C

Explanation

For given reactioin :
`underset(+3)(3Mn^(2+)) to underset(0)(M)n+underset(+3)(2Mn)^(3+)`
So, half oxidation : `2Mn^(2+) to 2Mn^(3+)+2e^(-)("Anode")`
`underline("and half reduction : "Mn^(2+)+2e^(-) to Mn" (Cathode)")`
Total reaction : `3Mn^(2+) to Mn+2Mn^(3+)`
So `E_("reaction")^(@)=E_("reduction")^(@)("Cathode")-E_("red")^(@)("Anode")`
`=E_(Mn^(2+)|Mn)^(@)-E_(Mn^(3+)|Mn^(2+))^(@)`
`=(-1.18-1.51)`
`=-2.69V`
`E_("Reaction")^(@) lt 0` so no reaction is possible and here, `E_(("Reaction"))^(@)=-2.68V=-2.69V`.