Question

The following chemical reaction occurs in a basic solution.

Mg2+(aq) + MnO₂(aq) + OH⁻(aq) → Mg(s) + MnO₄⁻(aq) + H2O(l)

How many moles of electrons are transferred when the equation is balancedusing the smallest whole-number coefficients?

Answer 1

                           +2           +4                                 0          +7

Given reaction is Mg²⁺(aq) + MnO₂(aq) + OH⁻(aq) → Mg(s) + MnO₄⁻(aq) + H2O(l)

Oxidation half reaction : MnO₂(aq) → MnO₄⁻(aq)

Balance O atoms : MnO₂(aq) +2H₂O → MnO₄⁻(aq)

Balance H atoms : MnO₂(aq) +2H₂O(l)+4OH^-(aq) → MnO₄⁻(aq) + 4H₂O(l)

        OR         MnO₂(aq) + 4OH^-(aq) → MnO₄⁻(aq) + 2H₂O(l)

Balance charge : MnO₂(aq) + 4OH^-(aq) → MnO₄⁻(aq) + 2H₂O(l) + 3e^-      -----(1)

Reduction half reaction : Mg²⁺(aq) → Mg(s)

Balance charge : Mg²⁺(aq) + 2e^- → Mg(s)   ----(2)

[2x(1) ] + [3x(2)] gives we get the overall balanced equation as

2MnO₂(aq) + 8OH^-(aq) + 3 Mg²⁺(aq) + 6e^- → 3Mg(s) + 2MnO₄⁻(aq) + 4H₂O(l) + 6e^-

OR    2MnO₂(aq) + 8OH^-(aq) + 3 Mg²⁺(aq)   → 3Mg(s) + 2MnO₄⁻(aq) + 4H₂O(l)

So 6 moles of electrons are transferred.