Step-1: First we write the skeleton ionic equation, equation

\(\overset{+7}MnO_4^-(aq)+2\overset{-1}{I}(aq)\rightarrow \)  \(M\overset{+4}{nO_2}(s) +\overset{o}{I_2}(s)+O_2(s)\)

Step-2: The two half reaction are: Oxidation half reaction

\(\overset{-1}{I}(aq)\) → \(+\overset{o}{I_2}(s)\)

Reduction half reaction,

\(\overset{+7}{MnO_4^-}(aq)\rightarrow Mn\overset{+4}{O_2}(s)\)

Step-3: Balance the iodine atoms in oxidation half reaction

2I (aq) → I₂ (s)

Step-4: To balance O-atoms in oxidation half reaction, we add two water molecules to the right side.

MnO₄^-(aq) → MnO₂(s) + 2H₂O(l)

To balance the H-atoms, we add 4H⁺ ions to the left side.

MnO₄^-(aq) + 4H⁺ → MnO₂(s) + 2H₂O(l)

As the reaction takes place in basic medium, so for four H⁺ ions, we add 4 OH^– ions to both side of the equation.

MnO₄^-(aq) + 4H⁺ + 4OH^-(aq)  → MnO₂(s) 2H₂O(l) + 4OH^-(aq)

Replacing the H⁺ and OH^– ions with H₂O, then equation becomes:

MnO₄^-(aq) + 4H₂O(l)  → MnO₂(s) + 2H₂O(l) + 4OH^-(aq)

or

MnO₄^-(aq) + 2H₂O(l) → MnO₂(s) + 4OH^-(aq)

Step-5: Balance the changes of the two-half reaction

\(\overset{-1}{2I^-}(aq)\) → \(\overset{o}{I_2}(s)\) + 2e^-

\(\overset{+7}{MnO_4}(a) \) + 2H₂O (I) + 3e^- → \(\overset{+4}{MnO_2}(s) \) + 4OH^-(aq)

Now, to equalize the number of electrons, we multiply the oxidation half reaction by 3 and the reduction half reaction by 2, then, we get

6I^-(aq) → 3I₂(s) + 6e^-

2MnO₄^- (aq) + 4H₂O(I) + 6e^- → 132 MnO₂(s) + 8OH^-(aq)

Step-6: Add two half reactions to obtain the net reaction after cancelling the electrons on both sides. The net equation is:

6I^-(aq) + 2MnO₄^- (aq) + 4H₂O(l) → 3I₂ (s) + 2MnO₂(s) + 8OH^-(aq)

A final verification shows that the equation is balanced in respect of the number of atoms and charges in both sides.