Question

Balance the following oxidation-reduction reaction occurring in an acidic solution using the half reaction method:

CN^1-(aq) + MnO₄^-(aq) ---> CNO^1-(aq) + MnO₂(s)

Answer 1

CN^- + MnO₄^- ---> CNO^- +MnO₂

First of all we have to consider two reactions taking place,

CN^- ---> CNO^-

MnO₄^- --> MnO₂

First balance the oxygen atoms,

CN^- + H₂O ---> CNO^- +2H⁺

MnO^-₄ + 4H⁺--> MnO₂ + 2H₂O

For reactions in basic medium,

CN^- + H₂O + 2OH^- --> CNO^- + 2H₂O

MnO₄^- +4H₂O --> MnO₂ +2H₂O + 4OH^-

balancing the charge,

CN^- +H₂O + 2OH^- --> CNO^- + 2H₂O + 2e^-...................................(1)

MnO₄^- + 4H₂O + 3e^- --> MnO₂ + 2H₂O + 4OH^-...........................(2)

Now to make electron gain to electron lost,

Eqⁿ (1) *3 and Eqⁿ (2) * 2 we get,

3CN^- +3H₂O + 6OH^- --> 3CNO^- + 6H₂O + 6e^-

2MnO₄^- + 8H₂O + 6e^- --> 2MnO₂ + 4H₂O + 8 OH^-

overall equation,

3CN^- + 2MnO₄^- + H₂O --> 3CNO^- +2MnO₂ +2OH^-