Question

In a particular redox reaction, MnO2 is oxidized to MnO4– and Fe3 is reduced to Fe2 . Complete and balance the equation for this reaction in acidic solution. Phases are optional.

Answer 1

Oxidation number of each element in MnO2 is

Mn=+4

O=-2

Oxidation number of each element in Fe+3 is

Fe=+3

Oxidation number of each element in MnO4-1 is

O=-2

Mn=+7

Oxidation number of each element in Fe+2 is

Fe=+2

Oxidation number of each element in reactant is

Mn=+4

O=-2

Fe=+3

Oxidation number of each element in product is

O=-2

Mn=+7

Fe=+2

Mn in MnO2 has oxidation state of +4

Mn in MnO4- has oxidation state of +7

So, Mn in MnO2 is oxidised to MnO4-

Fe in Fe+3 has oxidation state of +3

Fe in Fe+2 has oxidation state of +2

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

Reduction half cell:

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

Oxidation half cell:

MnO2 --> MnO4- + 3e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

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

Oxidation half cell:

MnO2 --> MnO4- + 3e-

Lets combine both the reactions.

3 Fe+3 + MnO2 --> 3 Fe+2 + MnO4-

Balance Oxygen by adding water

3 Fe+3 + MnO2 + 2 H2O --> 3 Fe+2 + MnO4-

Balance Hydrogen by adding H+

3 Fe+3 + MnO2 + 2 H2O --> 3 Fe+2 + MnO4- + 4 H+

This is balanced chemical equation in acidic medium

Answer:

3 Fe³⁺ + MnO₂ + 2 H₂O --> 3 Fe²⁺ + MnO₄^- + 4 H⁺