Question

Write a balanced equation for the oxidation of methanol with KMnO₄ in acidic solution; show how you balanced the reaction using half-reactions.

Answer 1

When methanol(CH3OH) reacts with acidic solution of KMnO4, redox reaction takes place in which methanol is oxidised to HCOOH and MnO₄^- (aq) is reduced to Mn²⁺(aq).

To balanced this redx reaction we need to write oxidation and reduction half reaction separately and balance them. After balancing them we need to add both reaction to get the final balanced reaction.

Unbalanced reduction - half reaction:  MnO₄^-(aq) ----- > Mn²⁺(aq)

To balance this reaction in acidic medium, add 1 H2O molecule in opposite side for each excess oxygen atom and 2H⁺ on the same side. Finally the number of electrons need to be balanced on both sides. Hence

Balanced reduction - half reaction:

MnO₄^-(aq) + 8 H⁺(aq) + 5 e^- ----- > Mn²⁺(aq) + 4 H2O --------- (1)

Unbalanced oxidation - half reaction: CH3OH(l) ----- > HCOOH(l)

To balance this reaction in acidic medium, add 1 H2O molecule in opposite side for each excess oxygen atom and 2H⁺ on the same side. Finally the number of electrons need to be balanced on both sides. Hence

Balanced oxidation - half reaction:

CH3OH(l) + H2O ----- > HCOOH(l) + 4 H⁺(aq) + 4 e^- -------(2)

Multiplying reaction (1) by 4 and reactin (2) by 5 and then adding, we will get final balanced reaction.

4MnO₄^-(aq) + 32 H⁺(aq) + 20 e^- ----- > 4 Mn²⁺(aq) + 16 H2O --------- (3)

5 CH3OH(l) + 5 H2O ----- > 5 HCOOH(l) + 20 H⁺(aq) + 20 e^- -------(2)

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4MnO₄^-(aq) + 5 CH3OH(l) + 12 H⁺(aq) ------ >  4 Mn²⁺(aq) +5 HCOOH(l) + 11 H2O (answer)