Redox Reactions

(i) We can write the given reaction along with their oxidation numbers as-

As, chlorine is oxidised in hydrochloric acid (behaving as reducing agent) -as its oxidation number is increases during the reaction from -1 to 0) and nitric acid is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +5 to +3). Hence, it is the redox reaction.

(ii) We can write the given reaction along with their oxidation numbers as-

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

(iii) We can write the given reaction along with their oxidation numbers as-

As, carbon is oxidised in carbon monoxide (behaving as reducing agent) -as its oxidation number is increases during the reaction from +2 to +4) and ferric oxide is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +3to 0). Hence, it is the redox reaction.

(iv) We can write the given reaction along with their oxidation numbers as-

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

(v) We can write the given reaction along with their oxidation numbers as-

As, nitrogen is oxidised in ammonia (behaving as reducing agent) -as its oxidation number is increases during the reaction from -3 to 0) and oxygen is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from 0 to -2). Hence, it is the redox reaction.

This is a Short answer type question as classified in NCERT Exemplar

(i) We can balance the given equation by oxidation number method- 

(a)Balance the increase and decrease in O.N.

  (b) Balancing H and O atoms by adding H⁺ and H₂O molecules

 

(ii) We can balance the given equation by oxidation number method- 

Total decrease in O.N. = 1

To equilize O.N. multiply NO₃^-, by 10

I₂ + 10 NO₃^-? 10NO₂ + IO₃^-

Balancing atoms other than O and H

I₂ + 10 NO₃^-? 10NO₂ + 2 IO₃^-

Balancing O and H

I₂ + 10 NO₃^- + 8H⁺? 10NO₂ + 2 IO₃^- + 4H₂O

 

(iii) We can balance the given equation by oxidation number method

Total increase in O.N. = 0.5 * 4 = 2

Total increase in O.N. = 1 * 2 = 2

to equilize O.N. multiply S₂O₃^2– and I^- by 2.

 

(iv) We can balance the given equation by oxidation number method-

The balanced chemical reaction is given as-

Total increase in O.N. = 5 * 2= 10

To equalize O.N. multiply CO₂ by 2. 

MnO₂   + C₂O₄^2-   ?  Mn₂⁺   + 2CO₂

Balance H and O by adding 2H₂O   on right side, and 4H+  on left side of equation.

MnO₂   + C₂O₄^2-   + 4H⁺   ?  Mn₂⁺   + 2CO₂   + 2H₂O .

 

(a) let x= oxidation number of   sulphur, and   +1 is oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 2x - 6 = 0 

x = +2. 

(b) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(c) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(d)  let x= oxidation number of sulphur, and  +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 8=0

x = +6.

We can balance the given reaction by ion electron method as

Cl₂O₇ (g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

This is a Short answer type question as classified in NCERT Exemplar

Balancing bu ion electron method

2 * { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  * { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺

This is a Short answer type question as classified in NCERT Exemplar

(i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺?5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

(ii) We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given

3N₂H₄ + 4ClO₃^- → 6NO + 4Cl^- + 6H₂O

The balanced chemical is given as-

(iii) We can balance the given reaction by ion electron method as

Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

Balancing bu ion electron method

2 * { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  * { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺