Question 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) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
Step 1:
The two half reactions involved in the given reaction are:
-1 0
Oxidation half reaction: l (aq) → l2(s)
+7 +4
Reduction half reaction: Mn O-4(aq) → MnO2(aq)
Step 2:
Balancing I in the oxidation half reaction, we have:
2l-(aq) → l2(s)
Now, to balance the charge, we add 2 e- to the RHS of the reaction.
2l-(aq) → l2(s) + 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.
MnO-4(aq) + 3e- →MnO2(aq)
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-4(aq) + 3e- →MnO2(aq) + 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.
MnO-4(aq) + 2H2O + 3e- →MnO2(aq) + 4OH-
Step 5:
Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6l-(aq) → 3l2(s) + 2e-
2MnO-4(aq) + 4H2O + 6e- → 2MnO2(s) + 8OH-(aq)
Step 6:
Adding the two half reactions, we have the net balanced redox reaction as:
6l-(aq) + 2MnO-4(aq) + 4H2O(l) → 3l2(s) + 2MnO2(s) + 8OH-(aq)
(b) Following the steps as in part (a), we have the oxidation half reaction as:
SO2(g) + 2H2O(l) → HSO-4(aq) + 3H+(aq) + 2e-(aq)
And the reduction half reaction as:
MnO-4(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
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:
2MnO-4(aq) + 5SO2(g) + 2H2O(l) + H+(aq) → Mn2+(aq) + HSO-4(aq)
(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe2+(aq) → Fe3+(aq) + e-
And the reduction half reaction as:
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)
(d) Following the steps as in part (a), we have the oxidation half reaction as:
SO2(g) + 2H2O(l) → SO2-4(aq) + 4H+ (aq) + 2e-
And the reduction half reaction as:
Cr2O2-7(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
Cr2O2-7(aq) + 3SO2(g) + 2H+(aq) → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)
