Balance the following redox reactions by ion-electron method:

(a) (aq) + I^– (aq) → MnO₂ (s) + I₂(s) (in basic medium)

(b) (aq) + SO₂ (g) → Mn²⁺ (aq) + (aq) (in acidic solution)

(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)

(d) + SO₂(g) → Cr³⁺ (aq) + (aq) (in acidic solution)

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

Oxidation half reaction:

Reduction half reaction:

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: