(a) Mn2++2e−→Mn;E∘=−1.18V;...(i) (b) Mn3++e−→Mn2+;E0=−1.51V;... (ii) Now multiplying equation (ii) by two and subtracting from equation (i) 3Mn2+→Mn++2Mn3+ E0=EOx.+ERed =−1.18+(−1.51)=−2.69V (-ve value of EMF (i.e. ∆G=+ve ) shows that the reaction is non-spontaneous)