# Balance the following redox reactions by ion-electron method:

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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: +1 vote