Question

Balance the following reaction equations in ACIDIC condition.

a,MnO4- (aq) + Fe2+ (aq) → Mn2+ (aq)+ Fe3+ (aq)

b,MnO4- (aq) + C2O4-2 (aq) → Mn2+ (aq)+ CO2 (aq)

Answer 1

a)

Mn in MnO4- has oxidation state of +7

Mn in Mn+2 has oxidation state of +2

So, Mn in MnO4- is reduced to Mn+2

Fe in Fe+2 has oxidation state of +2

Fe in Fe+3 has oxidation state of +3

So, Fe in Fe+2 is oxidised to Fe+3

Reduction half cell:

MnO4- + 5e- --> Mn+2

Oxidation half cell:

Fe+2 --> Fe+3 + 1e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

MnO4- + 5e- --> Mn+2

Oxidation half cell:

5 Fe+2 --> 5 Fe+3 + 5e-

Lets combine both the reactions.

MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3

Balance Oxygen by adding water

MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3 + 4 H2O

Balance Hydrogen by adding H+

MnO4- + 5 Fe+2 + 8 H+ --> Mn+2 + 5 Fe+3 + 4 H2O

This is balanced chemical equation in acidic medium

MnO₄^- + 5 Fe²⁺ + 8 H⁺ --> Mn²⁺ + 5 Fe³⁺ + 4 H₂O

b)

Mn in MnO4- has oxidation state of +7

Mn in Mn+2 has oxidation state of +2

So, Mn in MnO4- is reduced to Mn+2

C in C2O4-2 has oxidation state of +3

C in CO2 has oxidation state of +4

So, C in C2O4-2 is oxidised to CO2

Reduction half cell:

MnO4- + 5e- --> Mn+2

Oxidation half cell:

C2O4-2 --> 2 CO2 + 2e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

2 MnO4- + 10e- --> 2 Mn+2

Oxidation half cell:

5 C2O4-2 --> 10 CO2 + 10e-

Lets combine both the reactions.

2 MnO4- + 5 C2O4-2 --> 2 Mn+2 + 10 CO2

Balance Oxygen by adding water

2 MnO4- + 5 C2O4-2 --> 2 Mn+2 + 10 CO2 + 8 H2O

Balance Hydrogen by adding H+

2 MnO4- + 5 C2O4-2 + 16 H+ --> 2 Mn+2 + 10 CO2 + 8 H2O

This is balanced chemical equation in acidic medium

2 MnO₄^- + 5 C₂O₄^2- + 16 H⁺ --> 2 Mn²⁺ + 10 CO₂ + 8 H₂O