Question

Balance the reactions in acid using the half-reactions method. (could you also explain this process in words please. Thank you)

a. MnO2 + BiO3– → MnO4– + Bi3+

b. CH3CH2OH + Cr2O72– → CH3COOH + Cr3+

c. NO3– + I2 → IO3– + NO2–

Answer 1

a)

Mn in MnO2 has oxidation state of +4

Mn in MnO4- has oxidation state of +7

So, Mn in MnO2 is oxidised to MnO4-

Bi in BiO3 has oxidation state of +6

Bi in Bi+3 has oxidation state of +3

So, Bi in BiO3 is reduced to Bi+3

Reduction half cell:

1 BiO3 + 3e- --> 1 Bi+3

Oxidation half cell:

1 MnO2 --> 1 MnO4- + 3e-

Number of electrons is same in both half reactions.

So, balancing of electrons is not required. It's already balanced

Lets combine both the reactions.

1 BiO3 + 1 MnO2 --> 1 Bi+3 + 1 MnO4-

Balance Oxygen by adding water

1 BiO3 + 1 MnO2 --> 1 Bi+3 + 1 MnO4- + 1 H2O

Balance Hydrogen by adding H+

1 BiO3 + 1 MnO2 + 2 H+ --> 1 Bi+3 + 1 MnO4- + 1 H2O

This is balanced chemical equation in acidic medium

BiO₃ + MnO₂ + 2 H⁺ --> Bi⁺³ + MnO₄^- + H₂O

I am allowed to answer only 1 question at a time