Answer :

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


The oxidation half reaction is as below :


I-(aq) I2(s)


The reduction half reaction is as below :


MnO-4(aq) MnO2(aq)


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


2I-(aq) I2(aq)


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


2I-(aq) I2(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- MnO2(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- MnO2(aq) + 4OH-


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


MnO-4(aq) + 3e- + 2H2O MnO2(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) 3I2(aq) + 6 e-


2MnO-4(aq) + 6e- + 4H2O 2MnO2(aq) + 8OH-


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


6I-(aq) + 2MnO-4(aq) + 4H2O 3I2(aq) + 2MnO2(aq) + 8OH-


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