Ncert Solutions Chemistry Class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iv) is the correct answer.

Redox reaction is defined as the simultaneous oxidation and reduction of reacting species. Thus, change in oxidation state will decide whether a reaction is redox or not. Thus, assigning the oxidation states as:

(i) CuO + H₂→ Cu + H₂O

 Here, oxidation of H and reduction of Cu is taking place.

(ii) Fe2O₃ + 3CO→ 2Fe + 3CO₂

Here, oxidation of C and reduction of Fe is taking place.

(iii) 2K + F₂→2KF

Here, the oxidation of K and reduction of F is taking place.

(iv) BaCl₂ + H₂SO₄→ BaSO₄ + 2HCl

This is not a redox reaction, but it is a double displacement reaction.

(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⁺