8.18 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) Cr2O7 2– + SO2(g) Cr3+ (aq) + SO42– (aq) (in acidic solution)


(a) Step 1: The two half-reactions involved in the given reaction are:
Oxidation                half  I-IaqI02s
reaction:                           M+7nO4-aqM+4nO2aq
Reduction half reaction:
Step 2:
Balancing I in the oxidation half reaction, we have:
2I-aqI2s
Now, to balance the charge, we add 2 e- to the RHS of the reaction.
2I-aqI2s+2e-
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.
MnO4-aq+3e-MnO2aq
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.
MnO-4aq+3e-MnO2aq+4OH-
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.
MnO4-aq+2H2O+3e-MnO2aq+4OH-
Step 5:
Equalizing the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:
6I-aq3I2s+6e-2MnO-4aq+4H2O+6e-2MnO2s+8OH-aq
Step 6:
Adding the two half-reactions, we have the net balanced redox reaction as:
6I-aq+2MnO-4aq+4H2OI3I2s+2MnO2s+8OH-aq
(b)Following the steps as in part (a), we have the oxidation half-reaction as:
SO2g+2H2OIHSO-4aq+3H+aq+2e-aq
And the reduction half-reaction as:
MnO4-aq+8H+aq+5e-Mn2+aq+4H2OI
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:
2MnO4-aq+5SO2g+2H2Og+2H2OI+H+aq2Mn2+aq+5HSO4-aq
(c) Following the steps as in part (a), we have the oxidation half reaction as:
Feaq2+Fe3+aq+e-
And the reduction half reaction as:
H2O2aq+2H+aq+2e-2H2OI
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half-reaction, we have the net balanced redox reaction as:
H2O2aq+2Fe2+aq+2H+aq2Fe3+aq+2H2OI
(d) Following the steps as in part (a), we have the oxidation half reaction as:
SO2g+2H2OlSO42-aq+4H+aq+2e-
And the reduction half reaction as:
Cr2O72-aq+14H+aq+6e-2Cr3+aq+7H2Ol
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
Cr2O72-aq+3SO2g+2H+aq2Cr3+aq+3SO42-aq+H2Ol