Question

Use the half-reaction method to balance the following equation in acidic solution. It is not necessary to include any phases of matter for any species.
CN− + MnO4− -------> CNO− + MnO2

Answer 1

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.