Step 1: The two half reactions involved in the given reaction are:

The oxidation half reaction is as below :

I^-_(aq)→ I_2(s)

The reduction half reaction is as below :

MnO^-_4(aq) → MnO_2(aq)

Step 2: Balancing I in the oxidation half reaction, we have:

2I^-_(aq)→ I_2(aq)

Now, to balance the charge, we add 2 e^- to the RHS of the reaction.

2I^-_(aq)→ I_2(aq) + 2 e^-

Step 3: In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.

MnO^-_4(aq) + 3e^-→ MnO_2(aq)

Step 4:To balance reduction half reaction first balance oxidation number by writing required number of electrons to LHS and then balance charge by adding OH^- ions because reaction occurs in basic medium.

MnO^-_4(aq) + 3e^-→ MnO_2(aq) + 4OH^-

Step 5: Balance oxygen atoms by adding H₂O to get :-

MnO^-_4(aq) + 3e^- + 2H₂O → MnO_2(aq) + 4OH^-

Step 6: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have the following equations:-

6I^-_(aq)→ 3I_2(aq) + 6 e^-

2MnO^-_4(aq) + 6e^- + 4H₂O → 2MnO_2(aq) + 8OH^-

Step 7: Adding the two half reactions, we have the net balanced redox reaction as:-

6I^-_(aq) + 2MnO^-_4(aq) + 4H₂O → 3I_2(aq) + 2MnO_2(aq) + 8OH^-