Balance the redox reactions by ion – electron method :

Step 1

The two half reactions are given below:

Oxidation half reaction: \( I_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) }\)

Reduction half reaction: \( MnO_{ 4 }^{ – } \; \rightarrow \; MnO_{ 2 }\)

Step 2

Balance I in oxidation half reaction:

\( 2 \; I^{ – }_{(aq)} \; \rightarrow \; I_{ 2 \; (s) } \)

Add \( 2 e^{ – }\) to the right hand side of the reaction to balance the charge:

\( 2I^{ – }_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) } \; + \; 2 \; e^{ – } \)

Step 3

The oxidation no. of Mn has decreased from +7 to +4 in the reduction half reaction. Therefore, 3 electrons are added to the left hand side of the reaction.

\( MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \; (aq)} \)

Add \( 4 OH^{ – }\) ions to right hand side of the reaction to balance the charge.

\( MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \;(aq)} \; + \; 4 \; OH^{ – } \)

Step 4

There are 6 oxygen atoms on the right hand side and 4 oxygen atoms on the left hand side. Hence, 2 water molecules are added to the left hand side.

\( MnO^{ – }_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O \; + \; 3\; e^{ – }\; \rightarrow \; MnO_{ 2 \; (aq) } \; + \; 4 OH^{ – } \)

Step 5

Equal the no. of electrons on both the sides by multiplying oxidation half reaction by 3 and reduction half reaction by 2:

\( 6 \; I_{ (aq) }^{ – } \; \rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 6 \; e^{ – } \)

\( 2 \; MnO^{ – }_{ 4 \; (aq) } \; + \; 4 \; H_{ 2 }O \; + \; 6\; e^{ – } \; \)

\( \rightarrow 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }\)

Step 6

After adding both the half reactions, we get the balanced reaction as given below:

\( 6 \; I_{ (aq) }^{ – } \; + \; 2 \; MnO^{ – }_{ 4 \; (aq) } \;+ \; 4 \; H_{ 2 }O_{ (l) } \; \)

\( \rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }\)