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Class 11 Chemistry - Chapter Redox Reactions NCERT Solutions | Balance the following redox reactions by

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Redox Reactions. This page offers a step-by-step solution to the specific question from Excercise 1 , Question 18: balance the following redox reactions by ion ndas....
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)

Answer

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)

 

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