Balance the following redox reactions by ion – electron method :

(a) MnO₄ ^– (aq) + I ^– (aq) → MnO_{2 (s)} + I_2(s) (in basic medium)

(b) MnO₄ ^– (aq) + SO_{2 (g)} → Mn²⁺ (aq) + HSO₄^– (aq) (in acidic solution)

(c) H₂O_{2 (aq)} + Fe ²⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)

(d) Cr₂O₇ ^2– + SO_2(g) → Cr³⁺ _(aq) + SO₄^2– (aq) (in acidic solution)

Step 1:

The two half reactions involved in the given reaction are:

                                           -1              0

Oxidation half reaction:  l _(aq)  →  l_2(s)

                                                  +7                         +4

Reduction half reaction: Mn O^-_4(aq)   →  MnO_2(aq)

 

Step 2:

Balancing I in the oxidation half reaction, we have:

2l^-_(aq)  →  l_2(s)

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

2l^-_(aq) →  l_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.

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.

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.

MnO^-_4(aq) + 2H₂O + 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:

6l^-_(aq) →  3l_2(s) + 2e^-

2MnO^-_4(aq) + 4H₂O + 6e^-   →  2MnO_2(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) + 2MnO_2(s)  +  8OH^-_(aq)

 

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

SO_2(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)  + 5SO_2(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:

H₂O_2(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:

H₂O_2(aq)  +  2Fe²⁺_(aq)  +  2H⁺_(aq)    →   2Fe³⁺_(aq) +  2H₂O_(l)

 

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

SO_2(g)  +  2H₂O_(l)    →  SO^2-_4(aq)  +   4H⁺ _(aq)   +  2e^-

And the reduction half reaction as:

Cr₂O^2-_7(aq) + 14H⁺_(aq)  +  6e^-   →  2Cr³⁺_(aq)  + 3SO^2-_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:

Cr₂O^2-_7(aq) +  3SO_2(g)  +  2H⁺_(aq)   →  2Cr³⁺_(aq)  + 3SO^2-_4(aq) +  H₂O_(l)