Question

Balance the following redox reaction occurring in basic solution. I−(aq)+MnO−4(aq)→I2(aq)+MnO2(s)

Answer 1

I in I- has oxidation state of -1

I in I2 has oxidation state of 0

So, I in I- is oxidised to I2

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:

MnO4- + 3e- --> MnO2

Oxidation half cell:

2 I- --> I2 + 2e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

2 MnO4- + 6e- --> 2 MnO2

Oxidation half cell:

6 I- --> 3 I2 + 6e-

Lets combine both the reactions.

2 MnO4- + 6 I- --> 2 MnO2 + 3 I2

Balance Oxygen by adding water

2 MnO4- + 6 I- --> 2 MnO2 + 3 I2 + 4 H2O

Balance Hydrogen by adding H+

2 MnO4- + 6 I- + 8 H+ --> 2 MnO2 + 3 I2 + 4 H2O

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

2 MnO4- + 6 I- + 8 H+ + 8 OH- --> 2 MnO2 + 3 I2 + 4 H2O + 8 OH-

Combine H+ and OH- to form water

2 MnO4- + 6 I- + 8 H2O --> 2 MnO2 + 3 I2 + 4 H2O + 8 OH-

Remove common H2O from both sides

Balanced Eqn is

2 MnO4- + 6 I- + 4 H2O --> 2 MnO2 + 3 I2 + 8 OH-

This is balanced chemical equation in basic medium

Answer:

2 MnO₄^- + 6 I^- + 4 H₂O --> 2 MnO₂ + 3 I₂ + 8 OH^-