Question

Balance the redox reaction by inserting the appropriate coefficients.

redox reaction:

H^+ + CrO2^−4 + NO−2⟶Cr3+ + H2O + NO^−3

Answer 1

Cr in CrO4-2 has oxidation state of +6

Cr in Cr+3 has oxidation state of +3

So, Cr in CrO4-2 is reduced to Cr+3

N in NO2- has oxidation state of +3

N in NO3- has oxidation state of +5

So, N in NO2- is oxidised to NO3-

Reduction half cell:

CrO4-2 + 3e- --> Cr+3

Oxidation half cell:

NO2- --> NO3- + 2e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

2 CrO4-2 + 6e- --> 2 Cr+3

Oxidation half cell:

3 NO2- --> 3 NO3- + 6e-

Lets combine both the reactions.

2 CrO4-2 + 3 NO2- --> 2 Cr+3 + 3 NO3-

Balance Oxygen by adding water

2 CrO4-2 + 3 NO2- --> 2 Cr+3 + 3 NO3- + 5 H2O

Balance Hydrogen by adding H+

2 CrO4-2 + 3 NO2- + 10 H+ --> 2 Cr+3 + 3 NO3- + 5 H2O

This is balanced chemical equation in acidic medium

Answer:

10 H^+ + 2 CrO2^−4 + 3 NO−2⟶ 2 Cr3+ + 5 H2O + 3 NO^−3