Question

Balance the following redox reactions by ion-electron method: 

(a) MnO₄^- (aq) + I– (aq) → MnO_{2 (s)} + I_2(s) (in basic medium) 

(b) MnO₄^- _(aq) + SO_{2 (g)} → Mn²⁺ _(aq) + HSO^4-_(aq) (in acidic solution) 

(c) H₂O_{2 (aq)} + Fe²⁺ _(aq) → Fe³⁺ _(aq) + H₂O _(l) (in acidic solution) 

(d) Cr₂O₇^2- + SO_2(g) → Cr³⁺ (aq) +SO₄^2- (aq) (in acidic solution) 

Answer 1

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

Oxidation half reaction:      I^-1_(aq)  →  I⁰_2(aq)

Reduction half reaction:     M⁺⁷nO₄^-_(aq)  →  M⁺⁴nO_2(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: