Given below are the half-cell reactions Mn2++2e−→Mn,             Eo=−1.18 V2(Mn3++e−→Mn2+),      Eo=+1.51 VThe Eo for 3 Mn2+→Mn+2Mn3+ will be

Mn2++2e−→Mn,              Eo=−1.18V2(Mn3++e−→Mn2+),       Eo=+1.51VSubtracting Eq. (ii) from Eq. (i), we get 3Mn2+→Mn+2Mn3+Eo=−1.18−(+1.51)=−2.69 VSince, the value of Eo is -ve, therefore the reaction is non-spontaneous. Alternate method Mn2++2e−→Mn,Eo=−1.18V          ...(i)ΔGo=−nFEo [here, n = number of e− inoved in the reaction]ΔG1o=−2F(−1.18)=2.36 F2Mn3++2e−→2Mn2+,Eo=+1.51 V       ...(ii)ΔG2o=−2F[−1.51]=−3.02FSubtracting Eq. (ii) from Eq. (i), we get 3Mn2+→Mn+2Mn3+,[n=2]ΔG3o=ΔG1o−ΔG2o=5.38F⇒−2FEo=5.38F⇒Eo=−2.69 V