Answer:
2H⁺ + 2MnO4⁻ + 3CN⁻ → 2MnO₂ + H₂O + 3CNO⁻
Explanation:
CN⁻ + MnO4⁻ → CNO⁻ + MnO₂
In right side permangante, with Mn element has +7 as oxidation number.
In the MnO₂, Mn acts with +4.
This is the half reaction of reduction, where the Mn has gained 3 electrons.
in right side, cianide with C element has +2 as oxidation number. In the anion cianate, C acts with +4.
The oxidation number has increased in this half reaction. It's oxidation where the C, has lost 2 elecontrons
( 4H⁺ + MnO4⁻ + 3e⁻ → MnO₂ + 2H₂O ) .2
(H₂O + CN⁻ → CNO⁻ + 2e⁻ + 2H⁺) .3
I have to add water, to ballance the amount of oxygens and protons to ballance H, in the opposite side
To ballance the half reactions, I have to multiply x2 (reduction) and x3 (oxidation) so I can cancel, the electrons.
2H⁺ + 2MnO4⁻ + 3CN⁻ → 2MnO₂ + H₂O + 3CNO⁻
The electrons are now cancelled, and I can also modify water. In reactant side I have 3H₂O and in product side, I have 4H₂O so, H₂O in right side is gone so I finally obtained 1 H₂O on left. I had 8H⁺ on right, and 6H⁺ on left, so finally I obtained 2H⁺ on right, and the protons of product side are gone.