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Balance the following redox reactions by ion-electron method: 

(a) MnO4(aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium) 

(b) MnO4- (aq) + SO2 (g) → Mn2+ (aq) + HSO4-(aq) (in acidic solution) 

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

(d) Cr2O72- + SO2(g) → Cr3+ (aq) +SO42- (aq) (in acidic solution) 

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(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: 

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