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Ncert Solutions for Class 11 Chemistry Chapter 8

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Adda247 provides NCERT solutions for class 11 chemistry chapter 8. The NCERT solutions for class 11 chemistry chapter 8 provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers.

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NCERT Solutions for Class 11 Chemistry Chapter – 8: Redox reactions  

Chemistry is the basic of things which we see around in our environment and is known as the “central science”. As chemistry is a mandatory subject for the students, it requires more focus from the exam perspective. The solutions provided here are equipped with all the basic details with questions which might appear in exam. Furthermore, these NCERT Solutions can also be downloaded in a PDF format.

The questions in NCERT Solution Chemistry Class 11 covered in this exercise assist students in gaining insight into the chapter, so that they can have in-depth knowledge about the topic and excel in their upcoming exams.

Redox is a type of chemical reaction in which the oxidation states of atoms are changed. Redox reactions are characterized by the actual or formal transfer of electrons between chemical species, most often with one species undergoing oxidation while another species undergoes reduction.

Download Full PDF of Class 11 Chemistry Chapter 8

An oxidation-reduction reaction is any chemical reaction in which, by obtaining or losing an electron, the oxidation number of a molecule, atom, or ion varies. An example of a redox reaction is the formation of hydrogen fluoride. To study the oxidation and reduction of reactants, we should break the reaction down.

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox reactions are provided on this page for the perusal of Class 11 Chemistry students studying under the syllabus described by CBSE. Detailed, student-friendly answers to each and exercise question provided in Chapter 8 of the NCERT Class 11 Chemistry textbook can be found here. Expert tutors have created these NCERT Solutions according to the latest CBSE syllabus and guidelines.

Furthermore, the NCERT Solutions of Class 11 Chemistry Chapter 7 can be downloaded as a PDF For free by clicking the link.

“Redox Reactions” is the eighth chapter in the NCERT Class 11 Chemistry textbook. This chapter is regarded by many as one of the most important chapters in the CBSE Class 11 Chemistry syllabus, owing to the fact that the entire field of electrochemistry deals with redox reactions. The NCERT Solutions of topics touched upon in this chapter are also a part of the JEE and NEET syllabus. The important subtopics that come under this chapter are listed below.

 

Subtopics of Class 11 Chemistry Chapter 8 Redox Reactions

  1. Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
  2. Redox Reactions in Term of electron Transfer Reactions
  • Competitive Electron Transfer Reactions
  1. Oxidation Number
  • Types of Redox Reactions
  • Balancing of Redox Reactions
  • Redox Reactions as the Basis for Titrations
  • Limitations of Concept of Oxidation Number
  1. Redox Reactions and Electrode Processes.

 

Types of Redox Reactions:

Redox reactions can be primarily classified into five different types:

 

Combination Reactions

Decomposition Reactions

Displacement Reactions

Disproportionate Reactions

 

1) Combination Reactions

In this type of oxidation and reduction reactions, that is, redox reactions, two species of any atoms or molecules combine to form a single species of the compound. This type of reaction refers to redox reactions only when both the species (A & B) or either of the species are present in their elemental form. This type of redox reaction (oxidation and reduction) reaction is just the opposite of the decomposition reaction. Type of oxidation and reduction reactions

 

Combustion Reactions

It is a subtype of combination reaction in which one of the species will always be elemental dioxygen. Refer to the examples below.

 

2) Decomposition Reactions

It is just the opposite of combination reactions. This type of reactions involves breaking down of a single compound into two or more different compounds. However, at least one product out of the two or more components must be in its elemental form.

 

3) Displacement Reactions

In this type of reactions, an atom or an ion in a compound is substituted by another element.

There are mainly two types of displacement reactions:

 

Metal-displacement reactions

Non-metal displacement reactions

  1. i) Metal-Displacement Reactions

In this type of displacement reaction, an elemental metal displaces a metallic compound present in the reaction. The metal which is a better reducing agent displaces the other metal.

 

  1. ii) Non-Metal Displacement Reactions

In this type of reactions, either a metal or a non-metal will displace another non-metal of a compound present in the reaction. Usually, the non-metal undergoing displacement is hydrogen. However, in certain cases, halogens or oxygen undergoing displacement can occur.

 

4) Disproportionation Reactions

This is a special type of oxidation and reduction reaction (redox reaction). In disproportionation reaction, the same species will undergo oxidation and reduction. This type of redox reaction can only occur if one of the elements in the reaction consists of three oxidation states.

 

Important Questions of NCERT Solutions of Chemistry Class 11 Chapter 8

 

Question: 1 Assign the oxidation number to the underlined elements in each of the following species:

  • NaH2PO4
  • NaHSO4
  • H4P2O7
  • K2MnO4
  • CaO2
  • NaBH4
  • H2S2O7
  • KAl(SO4)2.12H2O

Answer:

(a) NaH2PO4

Let assume oxidation number of P is x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have

1(+1) + 2(+1) + 1 (x) + 4(-2) = 0

= 1 + 2 + x – 8 = 0

= x – 5 = 0

= x = + 5

Hence the oxidation number of P is +5

 

(b)  NaHSO4

Let assume oxidation number of S is x.

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

1(+1) + 1(+1) + 1 (x) + 4(-2) = 0

= 1 + 1 + x – 8 = 0

= x-6 = 0

= x = +6

Hence the oxidation number of S is +6

 

(c)  H4P2O7

Let assume oxidation number of P is x.

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

4(+1) + 2(x) + 7 (-2) = 0

= 4 + 2x – 14 = 0

= 2x – 10 = 0

= 2x  = +10

= x  = +5

Hence, Oxidation number of P is +5

 

(d) K2MnO4

Let assume oxidation number of Mn is x.

Oxidation number of K = +1

Oxidation number of O = -2

Then we have:

2(+1) + 1(x) + 4 (-2) = 0

= 2 + x – 8 = 0

= x  – 6 = 0

= x  = +6

Hence, Oxidation number of Mn is +6

 

(e) CaO2

Let assume oxidation number of O is x.

Oxidation number of Ca = +2

Then we have:

1(+2) + 2(x) = 0

= 2 + 2x  = 0

= 2x = -2

= x  = -1

Hence, Oxidation number of O is -1

 

(f) NaBH4

Let assume oxidation number of B is x.

Oxidation number of Na = +1

Oxidation number of H = -1

Then we have:

1(+1) + 1(x) + 4(-1)  = 0

= 1 + x -4 = 0

= x – 3 = 0

= x  = +3

Hence, Oxidation number of B is +3.

 

(g) H2S2O7

Let assume oxidation number of S is x.

Oxidation number of O = -2

Oxidation number of H = +1

Then we have:

2(+1) + 2(x) + 7(-2)  = 0

= 2 + 2x – 14 = 0

= 2x – 12 = 0

= x  = +6

Hence, Oxidation number of S is +6.

 

(h) KAl(SO4)2.12 H2O

Let assume oxidation number of S is x.

Oxidation number of K = +1

Oxidation number of Al  = +3

Oxidation number of O  = -2

Oxidation number of H  = +1

Then we have:

1(+1) + 1 (+3) + 2(x) + 8(-2) + 24(+1) + 12 (-2)  = 0

= 1 + 3 + 2x -16 +24 -24 = 0

= 2x – 12 = 0

= 2x  = +12

= x  = +6

Hence, Oxidation number of S is +6.

 

Question: 2 What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your result?

  • KI3
  • H2S4O6
  • Fe3O4
  • CH3CH2OH
  • CH3COOH

Answer:

(a) KI3

Let assume oxidation number of l is x.

In KI3, the oxidation number (O.N.) of K is +1.

1(+1) + 3(x) = 0

= +1 +3x = 0

= 3x = -1

= x = -1/3

Hence, the average oxidation number of I is – 1/3

However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states. In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

 

(b) H2S4O6

Let assume oxidation number of S is x.

The oxidation number (O.N.) of H is +1.

The oxidation number (O.N.) of O is -2.

2(+1) + 4(x) + 6(-2) = 0

= 2 + 4x – 12 = 0

= 4x -10 = 0

= 4x  =  +10

= x  = +10/4

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

 

(c) Fe3O4

Let assume oxidation number of Fe is x.

The oxidation number (O.N.) of O is -2.

3(x) + 4(-2) = 0

= 3x  – 8 = 0

= 3x  = 8

= x  = 8/3

However, O.N. cannot be fractional.

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

 

(d) CH3CH2OH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + 2(+1) + 1(-2) + 1(+1) = 0

= x +3 + x +2 – 2  + 1 = 0

= 2x  + 4 = 0

= 2x  = -4

= x  = -2

Hence, the oxidation number of C is -2.

 

(e) CH3COOH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0

= 2x + 3 – 2 –  2  +  1 = 0

= 2x + 0 = 0

= x = 0

However, 0 is average O.N. of C.

The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.

 

Question: 3 Justify that the following reactions are redox reactions:

  • CuO(s) +  H2(g)  à  Cu(s)  +  H2O(g)
  • Fe2O3(s) +  3CO(g)  à  2Fe(s) +  3CO2(g)
  • 4BCl3(g) +  3LiAlH4(s)  à  2B2H6  +  3LiCl(s)  +  3AlCl3(s)
  • 2K(s) +  F2(g)  à  2K+F9s)
  • 4NH3(g) +  5O2(g)  à  4NO(g)  +  6H2O(g)

Answer:

(a) CuO(s) + H2(g) –> Cu(s) + H2O(g)

Let us write the oxidation number of each element involved in the given reaction as:

+2  -2        0             0             +1    -2

Cu O(s) +  H2(g)  à  Cu(s)  +  H2  O(g)

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.

 

(b) Fe2O3(s) + 3CO(g) à 2Fe(s) + 3CO2(g)

Let us write the oxidation number of each element involved in the given reaction as:

+3   -2          +2  -2            0          +4  -2

Fe2 O3(s) + 3C  O(g) → 2Fe(s) + 3C O2(g)

Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.

 

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

 

Let us write the oxidation number of each element involved in the given reaction as:

+3  -1            +1   +3   -1            -3   +1         +1  -1             +3   -1

4B Cl3(g) + 3  Li   Al   H4(s) → 2B2 H6(g) + 3Li   Cl(s)   + 3 Al  Cl3 (s)

 

In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

 

(d) 2K(s) + F2(g) → 2K+F– (s)

Let us write the oxidation number of each element involved in the given reaction as:

0               0                    +1   -1

2K(s)   +   F2(g)    →    2K+  F– (s)

 

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to – 1 in KF i.e., F2 is reduced to KF.

Hence, the above reaction is a redox reaction.

 

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

Let us write the oxidation number of each element involved in the given reaction as:

-3   +1            0              +2  -2         +1    -2

4 N  H3(g) + 5 O2(g) → 4N  O(g) + 6H2  O(g)

 

Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

 

Question: 4 Fluorine reacts with ice and results in the change:

H2O(s)  +  F2(g)  à  HF(g)  +  HOF(g)

Justify that this reaction is a redox reaction.

Answer:

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

+2     -2         0                +1  -1      +1  -2   +1

H2O + F2 à HF + HOF

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

 

Question: 5 Write the formulae for the following compounds:

  • Mercury(II) chloride
  • Nickel(II) sulphate
  • Tin(IV) oxide
  • Thallium(I) sulphate
  • Iron(III) sulphate
  • Chromium(III) oxide.

Answer:

  • Mercury(II) chloride:

HgCl2

  • Nickel(II) sulphate:

NiSO4

  • Tin(IV) oxide:

SnO2

  • Thallium(I) sulphate:

Tl2SO4

  • Iron(III) sulphate:

Fe2(SO4)3

  • Chromium(III) oxide:

Cr2O3

 

Question: 6 Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Answer:

The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table.

Substance O.N. of carbon
CH2Cl2 0
ClC=CCl +1
HC=CH -1
CHCl3, CO +2
CH3Cl -2
Cl3C-CCl3 +3
H3C-CH3 -3
CCL4, CO2 +4
CH4 -4

 

The substances where nitrogen can exhibit oxidation states from -3 to +5 are listed in the following table:

Substance O.N. of nitrogen
N2 0
N2O +1
N2H2 -1
NO +2
N2H4 -2
N2O3 +3
NH3 -3
NO2 +4
N2O5 +5

 

Question: 7 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Answer:

In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.

Therefore, SO2 can act as an oxidising as well as a reducing agent.

In hydrogen peroxide (H2O2), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H2O2 can act as an oxidising as well as a reducing agent.

In ozone (O3), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence, O3 acts only as an oxidant.

In nitric acid (HNO3), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO3 acts only as an oxidant.

 

Question: 8 The compound AgF2 is an unstable compound. However, if formed the compound acts as a very strong oxidising agent. Why?

Answer:

The oxidation state of Ag in AgF2 is +2. But, +2 is an unstable state of Ag. Therefore, whenever AgF2 is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable state +1. As a result, AgF2 acts as a very strong oxidising agent.

 

Question: 9 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Answer:

Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing ent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. Following illustrations justify this.

(i) Oxidizing agent is F2 and reducing agent is P4. When excess P4 reacts with F2, PF3 is produced in which P has +3 oxidation number.

P4( excess ) +F2 àPF3

But if fluorine is in excess, PF5 is formed in which P has oxidation number of +5.

P4 + F2( excess ) à PF5

 

(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen, K2O is formed in which oxygen has oxidation number of -2.

4K( excess ) + O2 à 2K2O

But if oxygen is in excess, then K2O2 is formed in which O has oxidation number of -1.

2K+ O2( excess ) à K2O2

 

(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.

C( excess ) +O2 àCO

When excess of oxygen is used, CO2 is formed in which C has +4 oxidation number.

C+ O2( excess ) à CO2

 

Question: 10 How do you count for the following observations?

  • Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
  • When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

Answer:

(a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.

 

(i) In a neutral medium, ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

 

(ii) KMnO4 and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO4 and toluene can react at a faster rate.

The balanced redox equation for the reaction in a neutral medium is give as below:

 

(b) When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H2SO4 to SO2 with the evolution of red vapour of bromine.

 

But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.

 

FAQs on NCERT Solutions of Chemistry Class 11 Chapter 8

  1. What are the topics and Subtopics in Class 11 Chemistry Chapter 8 Redox Reaction?

Ans. The Class 11 Chemistry Redox Reaction Chapter is an essential chapter of the class 11 syllabus. The chapter has high importance in the board exams as well as in JEE and NEET examinations. So, the students must have adequate knowledge in this chapter. The important topics and sub-topics are given below:

  1. Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
  2. Redox Reactions in Term of electron Transfer Reactions
  • Competitive Electron Transfer Reactions
  1. Oxidation Number
  • Types of Redox Reactions
  • Balancing of Redox Reactions
  • Redox Reactions as the Basis for Titrations
  • Limitations of Concept of Oxidation Number
  1. Redox Reactions and Electrode Processes.

 

  1. What can a student learn from the Chapter – 8 Chemistry Class 11 Chapter Redox Reactions?

Ans. When a student will go through Class 11 Chemistry Redox Reaction, he/she will understand what is a Redox Reaction. Redox Reaction is defined as the class of reactions where oxidation and reduction coincide. Students will also learn the meaning of oxidation, oxidizing agent, reduction and reducing agent. With the help of the knowledge of this chapter, students will be able to classify the Redox reactions into:

  • Combination Reaction
  • Decomposition Reaction
  • Displacement Reaction
  • Disproportional Reaction

Students will also learn how to balance the chemical equations using the oxidation number and half-reaction process. If you have a fair practice and understanding, you can score good marks in the examinations.

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