NCERT Solutions for Class 11 Chemistry Chapter 8 – Free PDF Download
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 latest term – I syllabus of 202122 prescribed by CBSE. Detailed, studentfriendly answers to each and every intext and exercise question provided in Chapter 8 of the NCERT Class 11 Chemistry textbook can be found here. Subject experts have created these NCERT Solutions for Class 11 Chemistry according to the latest termwise CBSE Syllabus and its guidelines.
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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions
“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 term – I CBSE Class 11 Chemistry Syllabus 202122, 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 syllabi. The important subtopics that come under this chapter are listed below.
Subtopics of Class 11 Chemistry Chapter 8 Redox Reactions
 Classical Idea Of Redox Reactions – Oxidation And Reduction Reactions

Redox Reactions In Terms Of Electron Transfer Reactions
 Competitive Electron Transfer Reactions

Oxidation Number
 Types Of Redox Reactions
 Balancing Of Redox Reactions
 Redox Reactions As The Basis For Titrations
 Limitations Of Concept Of Oxidation Number
 Redox Reactions And Electrode Processes.
NCERT Solutions for Class 11 Chemistry Chapter 8
1. Assign oxidation number to the underlined elements in each of the following species:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Answer:
(a)
Let x be the oxidation no. of P.
Oxidation no. of Na = +1
Oxidation no. of H = +1
Oxidation no. of O = 2
Then,
1(+1) + 2(+1) + 1(x) + 4(2) = 0
1 + 2 + x 8 = 0
x = +5
Therefore, oxidation no. of P is +5.
(b)
Let x be the oxidation no. of S.
Oxidation no. of Na = +1
Oxidation no. of H = +1
Oxidation no. of O = 2
Then,
1(+1) + 1(+1) + 1(x) + 4(2) = 0
1 + 1 + x 8 = 0
x = +6
Therefore, oxidation no. of S is +6.
(c)
Let x be the oxidation no. of P.
Oxidation no. of H = +1
Oxidation no. of O = 2
Then,
4(+1) + 2(x) + 7(2) = 0
4 + 2x – 14 = 0
2x = +10
x = +5
Therefore, oxidation no. of P is +5.
(d)
Let x be the oxidation no. of Mn.
Oxidation no. of K = +1
Oxidation no. of O = 2
Then,
2(+1) + x + 4(2) = 0
2 + x – 8 = 0
x = +6
Therefore, oxidation no. of Mn is +6.
(e)
Let x be the oxidation no. of O.
Oxidation no. of Ca = +2
Then,
(+2) + 2(x) = 0
2 + 2x = 0
2x = 2
x = 1
Therefore, oxidation no. of O is 1.
(f)
Let x be the oxidation no. of B.
Oxidation no. of Na = +1
Oxidation no. of H = 1
Then,
1(+1) + 1(x) + 4(1) = 0
1 + x 4 = 0
x = +3
Therefore, oxidation no. of B is +3.
(g)
Let x be the oxidation no. of S.
Oxidation no. of H = +1
Oxidation no. of O = 2
Then,
2(+1) + 2(x) + 7(2) = 0
2 + 2x – 14 = 0
2x = +12
x = +6
Therefore, oxidation no. of S is +6.
(h)
Let x be the oxidation no. of S.
Oxidation no. of K = +1
Oxidation no. of Al = +3
Oxidation no. of O= 2
Oxidation no. of H = +1
Then,
1(+1) + 1(+3) + 2(x) + 8(2) + 24(+1) + 12(2) = 0
1 + 3 + 2x – 16 + 24 – 24 = 0
2x = +12
x = +6
Therefore, oxidation no. of S is +6.
OR
Ignore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.
1(+1) + 1(+3) + 2(x) + 8(2) = 0
1 + 3 + 2x 16 = 0
2x = 12
x = +6
Therefore, oxidation no. of S is +6.
2.What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(a)
(b)
(c)
(d)
(e)
Answer:
(a)
Let x be the oxidation no. of I.
Oxidation no. of K = +1
Then,
1(+1) + 3(x) = 0
1 + 3x = 0
x =
Oxidation no. cannot be fractional. Hence, consider the structure of
In
Therefore, in
(b)
Let x be the oxidation no. of S.
Oxidation no. of H = +1
Oxidation no. of O = 2
Then,
2(+1) + 4(x) + 6(2) = 0
2 + 4x 12 = 0
4x = 10
x =
Oxidation no. cannot be fractional. Therefore, S would be present with different oxidation state in molecule.
The oxidation no. of two out of the four S atoms is +5 while that of other two atoms is 0.
(c)
Let x be the oxidation no. of Fe.
Oxidation no. of O = 2
Then,
3(x) + 4(2) = 0
3x 8 = 0
x =
Oxidation no. cannot be fractional.
One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.
(d)
Let x be the oxidation no. of C.
Oxidation no. of O= 2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 1(2) = 0
2x + 4 2 = 0
x = 2
Therefore, oxidation no. of C is 2.
(e)
Let x be the oxidation no. of C.
Oxidation no. of O= 2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 2(2) = 0
2x + 4 4 = 0
x = 0
Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in
3. Justify that the following reactions are redox reactions:
(a)
(b)
(c)
(d)
(e)
Answer:
(a)
Oxidation no. of Cu and O in
Oxidation no. of
Oxidation no. of Cu is 0.
Oxidation no. of H and O in
The oxidation no. of Cu decreased from +2 in
The oxidation no. of H increased from 0 to +1 in
Therefore, the reaction is redox reaction.
(b)
In the above reaction,
Oxidation no. of Fe and O in
Oxidation no. of C and O in CO is +2 and 2 respectively.
Oxidation no. of Fe is 0.
Oxidation no. of C and O in
The oxidation no. of Fe decreased from +3 in
The oxidation no. of C increased from 0 to +2 in CO to +4 in
Therefore, the reaction is redox reaction.
(c)
the above reaction,
Oxidation no. of B and Cl in
Oxidation no. of Li, Al and H in
Oxidation no. of B and H in
Oxidation no. of Li and Cl in LiCl is +1 and 1 respectively.
Oxidation no. of Al and Cl in
The oxidation no. of B decreased from +3 in
The oxidation no. of H increased from 1 in
Therefore, the reaction is redox reaction.
(d)
In the above reaction,
Oxidation no. of K is 0.
Oxidation no. of F is 0.
Oxidation no. of K and F in KF is +1 and 1 respectively.
The oxidation no. of K increased from 0 in K to +1 in KF. That is K is oxidized to KF.
The oxidation no. of F decreased from 0 in
Therefore, the reaction is a redox reaction.
(e)
In the above reaction,
Oxidation no. of N and H in
Oxidation no. of
Oxidation no. of N and O in NO is +2 and 2 respectively.
Oxidation no. of H and O in
The oxidation no. of N increased from 3 in
The oxidation no. of
Therefore, the reaction is a redox reaction.
4. Fluorine reacts with ice and results in the change:
Justify that this reaction is a redox reaction
Answer:
In the above reaction,
Oxidation no. of H and O in
Oxidation no. of
Oxidation no. of H and F in HF is +1 and 1 respectively.
Oxidation no. of H, O and F in HOF is +1, 2 and +1 respectively.
The oxidation no. of F increased from 0 in
The oxidation no. of F decreased from 0 in
Therefore, F is both reduced as well as oxidized. So, it is redox reaction.
5. Calculate the oxidation no. of sulphur, chromium and nitrogen in
Answer:
For
Let x be the oxidation no. of S.
Oxidation no. of O= 2
Oxidation no. of H = +1
Then,
2(+1) + 1(x) + 5(2) = 0
2 + x – 10 = 0
x = +8
But the oxidation no. of S cannot be +8 as S has 6 valence electrons. Therefore, the oxidation no. of S cannot be more than +6.
The structure of
Now,
2(+1) + 1(x) + 3(2) + 2(1) = 0
2 + x – 6 – 2 = 0
x = +6
Therefore, the oxidation no. of S is +6.
For
Let x be the oxidation no. of Cr.
Oxidation no. of O= 2
Then,
2(x) + 7(2) = 2
2x 14 = 2
x = +6
There is no fallacy about the oxidation no. of Cr in
The structure of
Each of the two Cr atoms has the oxidation no. of +6.
For
Let x be the oxidation no. of N.
Oxidation no. of O= 2
Then,
1(x) + 3(2) = 1
x – 6 = 1
x = +5
There is no fallacy about the oxidation no. of N in
The structure of
Nitrogen atom has the oxidation no. of +5.
6. Write formulas for the following compounds:
(a) Mercury (II) chloride (b) Nickel (II) sulphate
(c) Tin (IV) oxide (d) Thallium (I) sulphate
(e) Iron (III) sulphate (f) Chromium (III) oxide
Answer:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) sulphate
(e) Iron (III) sulphate
(f) Chromium (III) oxide
7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Answer:
The compound where carbon has oxidation no. from 4 to +4 is as given below in the table:
Compounds  Oxidation no. of carbon 
0  
1  
+1  
2  
+2  
3  
+3  
4  
+4 
Compounds  Oxidation no. of nitrogen 
0  
1  
+1  
2  
+2  
3  
+3  
+4  
+5 
8. 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 (
Hence,
In hydrogen peroxide (
Therefore,
In ozone (
Therefore,
In nitric acid (
Therefore,
9.Consider the reactions:
(a)
(b)
Why it is more appropriate to write these reactions as :
(a)
(b)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions
Answer:
(a)
Step 1 :
Step 2 :
The
The net reaction is as given below:
[————————————————————————————————————————–
This is the suitable way to write the reaction as the reaction also produce water molecules in the photosynthesis process.
The path can be found with the help of radioactive
(b)
Step 1 :
Step 2 :
———————————————————————————————
The path can be found with the help of
10. The compound AgF_{2} is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?
Answer:
The oxidation no. of Ag in
11. 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.
Justify the above statement with three examples.
Answer:
When there is a reaction between reducing agent and oxidizing agent, a compound is formed which has lower oxidation number if the reducing agent is in excess and a compound is formed which has higher oxidation number if the oxidizing agent is in excess.
(i)
In an excess amount of
If
(ii) K and
If an excess of K reacts with
If K reacts with an excess of
(iii) C and
If an excess amount of C is reacted with insufficient amount of
If C is burnt in excess amount of
12. How do you count for the following observations?
(a) 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.
(b) 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) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.
(i) In a neutral medium,
(ii)
The redox reaction is as given below:
(b) When concentrated
When concentrated
13. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
(a)
(b)
(c)
(d)
(e)
Answer:
(a)
AgBr => Reduced substance
AgBr =>Oxidizing agent
(b)
HCHO => Oxidized substance
HCHO=> Reducing agent
(c)
HCHO => Oxidized substance
HCHO => Reducing agent
(d)
(e)
Pb=> Oxidized substance
Pb => Reducing agent
14. Consider the reactions :