Question

Balance redox reaction in basic solution: Fe(OH)2(s) + MnO4-(aq) -> MnO2 (s) + Fe(OH)3

Answer 1

Fe in Fe(OH)2 has oxidation state of +2

Fe in Fe(OH)3 has oxidation state of +3

So, Fe in Fe(OH)2 is oxidised to Fe(OH)3

Mn in MnO4- has oxidation state of +7

Mn in MnO2 has oxidation state of +4

So, Mn in MnO4- is reduced to MnO2

Reduction half cell:

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

Oxidation half cell:

1 Fe(OH)2 --> 1 Fe(OH)3 + 1e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

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

Oxidation half cell:

3 Fe(OH)2 --> 3 Fe(OH)3 + 3e-

Lets combine both the reactions.

1 MnO4- + 3 Fe(OH)2 --> 1 MnO2 + 3 Fe(OH)3

Balance Oxygen by adding water

1 MnO4- + 3 Fe(OH)2 + 1 H2O --> 1 MnO2 + 3 Fe(OH)3

Balance Hydrogen by adding H+

1 MnO4- + 3 Fe(OH)2 + 1 H2O + 1 H+ --> 1 MnO2 + 3 Fe(OH)3

Add equal number of OH- on both sides as the number of H+

1 MnO4- + 3 Fe(OH)2 + 1 H2O + 1 H+ + 1 OH- --> 1 MnO2 + 3 Fe(OH)3 + 1 OH-

Combine H+ and OH- to form water

1 MnO4- + 3 Fe(OH)2 + 1 H2O + 1 H2O --> 1 MnO2 + 3 Fe(OH)3 + 1 OH-

Remove common H2O from both sides

Balanced Eqn is

1 MnO4- + 3 Fe(OH)2 + 2 H2O --> 1 MnO2 + 3 Fe(OH)3 + 1 OH-

This is balanced chemical equation in basic medium

MnO4- + 3 Fe(OH)2 + 2 H2O --> MnO2 + 3 Fe(OH)3 + OH-