(a) Step 1: The two half reactions involved in the given reaction are:
Oxidation half reaction: I-1(aq) → I02(aq)
Reduction half reaction: M+7nO4-(aq) → M+4nO2(aq)
Step 2:
Balancing I in the oxidation half reaction, we have:
Now, to balance the charge, we add 2 e– to the RHS of the reaction.
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.
Now, to balance the charge, we add 4 OH– ions to the RHS of the reaction as the reaction is taking place in a basic medium.
Step 4:
In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
Step 5:
Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
Step 6:
Adding the two half reactions, we have the net balanced redox reaction as:
(b)Following the steps as in part (a), we have the oxidation half reaction as:
And the reduction half reaction as:
Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:
(c) Following the steps as in part (a), we have the oxidation half reaction as:
And the reduction half reaction as:
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
(d) Following the steps as in part (a), we have the oxidation half reaction as:
And the reduction half reaction as:
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
