Question

Balance the following redox reactions by ion-electron method:

1. \[\ce{MnO-_4 (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)}\]
2. \[\ce{MnO-_4 (aq) + SO2 (g) → Mn^{2+} (aq) + HSO-_4  (aq) (in acidic solution)}\]
3. \[\ce{H2O2 (aq) + Fe^{2+} (aq) → Fe^{3+} (aq) + H2O (l) (in acidic solution)}\]
4. \[\ce{Cr_2O^{2-}_7 + SO2(g) → Cr^{3+} (aq) + SO^{2-}_4 (aq) (in acidic solution)}\]

Answer 1

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

Oxidation half reaction \[\ce{^{-1}I_{(aq)} -> ^0I_{2(s)}}\]

Reduction half reaction: 

\[\ce{^{+7}MnO-_{4(aq)} -> ^{+4}MnO_{2(aq)}}\]

Step 2:

Balancing I in the oxidation half-reaction, we have:

\[\ce{2I-_{(aq)} -> I_{2(s)}}\]

Now, to balance the charge, we add 2 e^– to the RHS of the reaction

\[\ce{2I-_{(aq)}  -> I_{2(s)} + 2e-}\]

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.

\[\ce{MnO-_{4(aq)} + 3e- -> MnO_{2(aq)}}\]

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.

\[\ce{MnO-_{4(aq)} + 3e- -> MnO_{2(aq)} + 4OH-}\]

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.

\[\ce{MnO-_{4(aq)} + 2H_2O + 3e- -> MnO_{2(aq)} + 4OH-}\]

Step 5:

Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

\[\ce{6I-_{(aq)} -> 3I_{2(s)} + 6e-}\]

\[\ce{2MnO-_{4(aq)} +  4H_2O + 6e- -> 2MnO_{2(s)} + 8OH-_{(aq)}}\]

Step 6:

Adding the two half reactions, we have the net balanced redox reaction as:

\[\ce{6I-_{(aq)}  +  2MnO-_{4(aq)} + 4H_2O_{(l)} ->3I_{2(s)} + 2MnO_{2(s)} + 8OH-_{(aq)}}\]

b) Following the steps as in part (a), we have the oxidation half reaction as:

\[\ce{SO_{2(g)} + 2H_2O_{(l)} -> HSO-_{4(aq)} + 3H+_{(aq)} + 2e-_{(aq)}}\]

And the reduction half reaction as:

\[\ce{MnO-_{4(aq)}  + 8H+_{(aq)} + 5e-  -> Mn^{(2+)}_{(aq)} + 4H_2O_{(l)}}\]

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:

\[\ce{2MnO-_{4(aq)} + 5SO_{2(g)} + 2H_2O_{(l)} + H+_{(aq)} -> 2Mn^{2+}_{(aq)}  + 5HSO-_{4(aq)}}\] 

(c) Following the steps as in part (a), we have the oxidation half reaction as:

\[\ce{Fe^{2+}_{(aq)} -> Fe^{3+}_{(aq)} + e-}\]

And the reduction half reaction as:

\[\ce{H_2O_{2(aq)} + 2H+_{(aq)} + 2e- -> 2H_2O_{(l)}}\]

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

\[\ce{H_2O_{2(aq)} + 2Fe^{2+}_{(aq)} + 2H^+_{(aq)} -> 2Fe^{3+}_{(aq)} + 2H_2O_{(l)}}\]

(d) Following the steps as in part (a), we have the oxidation half reaction as:

\[\ce{SO_{2(g)} + 2H_2O_{(l)} -> SO^{2-}_{4(aq)} + 4H+_{(aq)} + 2e-}\]

And the reduction half reaction as:

\[\ce{Cr_2O^{2-}_{7(aq)} + 14H+_{(aq)} + 6e- -> 2Cr^{3+}_{(aq)} + 7H_2O_{(l)}}\]

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

\[\ce{Cr2O^{2-}_7_{(aq)} + 3SO_{2(g)} + 2H+_{(aq)} -> 2Cr^{3+}_{(aq)} + 3SO^{2-}_4_{(aq)} + H_2O_{(l)}}\]