NCERT Solutions For Class 11 Chemistry Chapter 8

NCERT Solutions Class 11 Chemistry Redox Reactions

NCERT Solutions For Class 11 Chemistry Chapter 8 is provided here. This topic is an extremely important topic in chemistry and is important for class 11, Class 12 and for competitive exams like JEE and NEET. Students must have a good knowledge of the topic in order to excel in the examination. We at BYJU'S provide the NCERT Solutions For Class 11 Chemistry Redox Reactions pdf which students can download. Practicing all the questions will be very helpful for the students as many questions are framed from the topic.

NCERT Solutions for Class 11 Chemistry Chapter 8 includes the topic - 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: Combination reactions, Decomposition reactions, Displacement reactions, Disproportionation 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 Class 11 Chemistry Redox Reactions PDF is provided here for better understanding and clarification of the chapter.

In chemistry there is transformation of matter from one form to another through various chemical reactions. One among them is redox reaction. Where there is oxidation there is always reduction. The process in which the change in chemical substance occurs due to the addition of oxygen is called oxidation. Example of oxidation: Oxidation of ethanol to ethanal. The process in which the change in chemical substance occurs due to the addition of hydrogen or removal of oxygen is called reduction. Example of reduction: The reduction process is used in many ways - Reducing ores to obtain metals. Both these reactions occur in the presence of either oxidising agent or reducing agent. Oxidizing agents give oxygen. Reducing agents remove oxygen. This is about the oxidation and reduction process.

1. Assign oxidation no. to the elements underlined:

(i) NaH2PO4

(ii) NaHSO4

(iii) H4P2O7

(iv) K2MnO4

(v) CaO2

(vi) NaBH4

(vii) H2S2O7

(viii) KAl(SO4)2.12H2O

 

Answer:

(i) NaH2PO4

Let x be the oxidation no. of P.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

1.1a

Then,

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

1 + 2 + x -8 = 0

x = +5

Therefore, oxidation no. of P is +5.

(ii) NaHSO4

1.2a

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.

(iii) H4P2O7

1.3a

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.

(iv) K2MnO4

1.4a

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.

(v) CaO2

1.5a

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.

(vi) NaBH4

1.6a

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.

(vii) H2S2O7

1.7a

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.

(viii) KAl(SO4)2.12H2O

1.8a

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.Assign oxidation no. to the elements underlined:

(i) KI3

(ii) H2S4O6

(iii) Fe3O4

(iv) CH3CH2OH

(v) CH3COOH

 

Answer:

(i) KI3

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

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

1 + 3x = 0

x = 13

Oxidation no. cannot be fractional. Hence, consider the structure of KI3.

In KI3 molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.

2.1a

Therefore, in KI3 molecule, the oxidation no. of I atoms forming the molecule I2 is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.

(ii) H2S4O6

2.2a

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 = +212

Oxidation no. cannot be fractional. Therefore, S would be present with different oxidation state in molecule.

2.21a

The oxidation no. of two out of the four S atoms is +5 while that of other two atoms is 0.

(iii) Fe3O4

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

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

3x -8 = 0

x = 83

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.

2.3a

(iv) CH3CH2OH

2.4a

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.

(v) CH3COOH

2.5a

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 CH3COOH.

2.51a

 

 

3. The reactions given below are redox reactions. Justify the reactions.

(i) CuO(s)+H2(g)Cu(s)+H2O(g)

(ii) Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)

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

(iv) 2K(s)+F2(g)2K+F(s)

(v) 4NH3(g)+5O2(g)4NO(g)+6H2O(g)

 

Answer:

(i) CuO(s)+H2(g)Cu(s)+H2O(g)

Oxidation no. of Cu and O in CuO is +2 and -2 respectively.

Oxidation no. of H2 is 0.

Oxidation no. of Cu is 0.

Oxidation no. of H and O in H2O is +1 and -2 respectively.

The oxidation no. of Cu decreased from +2 in CuO to 0 in Cu. That is CuO is reduced to Cu.

The oxidation no. of H increased from 0 to +1 in H2. That is H2 is oxidized to H2O.

Therefore, the reaction is redox reaction.

(ii) Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)

In the above reaction,

Oxidation no. of Fe and O in Fe2O3 is +3 and -2 respectively.

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 CO2 is +4 and -2 respectively.

The oxidation no. of Fe decreased from +3 in Fe2O3 to 0 in Fe. That is Fe2O3 is reduced to Fe.

The oxidation no. of C increased from 0 to +2 in CO to +4 in CO2. That is CO is oxidized to CO2.

Therefore, the reaction is redox reaction.

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

the above reaction,

Oxidation no. of B and Cl in BCl3 is +3 and -1 respectively.

Oxidation no. of Li, Al and H in LiAlH4 is +1, +3 and -1 respectively.

Oxidation no. of B and H in B2H6 is -3 and +1 respectively.

Oxidation no. of Li and Cl in LiCl is +1 and -1 respectively.

Oxidation no. of Al and Cl in AlCl3 is +3 and -1 respectively.

The oxidation no. of B decreased from +3 in BCl3 to -3 in B2H6. That is BCl3is reduced to B2H6.

The oxidation no. of H increased from -1 in LiAlH4 to +1 in B2H6. That is LiAlH4 is oxidized to B2H6.

Therefore, the reaction is redox reaction.

(iv) 2K(s)+F2(g)2K+F(s)

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 F2 to -1 in KF. That is F2 is reduced to KF.

Therefore, the reaction is a redox reaction.

(v) 4NH3(g)+5O2(g)4NO(g)+6H2O(g)

In the above reaction,

Oxidation no. of N and H in NH3 is -3 and +1 respectively.

Oxidation no. of O2 is 0.

Oxidation no. of N and O in NO is +2 and -2 respectively.

Oxidation no. of H and O in H2O is +1 and -2 respectively.

The oxidation no. of N increased from -3 in NH3 to +2 in NO.

The oxidation no. of O2 decreased from 0 in O2 to -2 in NO and H2O. That is O2 is reduced.

Therefore, the reaction is a redox reaction.

 

 

4. Give the reaction of fluorine when reacts with ice:

H2O(s)+F2(g)HF(g)+HOF(g)

Give reason that the above reaction is redox reaction.

 

Answer:

H2O(s)+F2(g)HF(g)+HOF(g)

In the above reaction,

Oxidation no. of H and O in H2O is +1 and -2 respectively.

Oxidation no. of F2 is 0.

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 F2 to +1 in HOF.

The oxidation no. of F decreased from 0 in O2 to -1 in HF.

Therefore, F is both reduced as well as oxidized. So, it is redox reaction.

 

 

5. Calculate the oxidation no. of chromium, sulphur and nitrogen in H2SO5, Cr2O27 and NO3. Give the structure for the compounds. Count for the fallacy.

 

Answer:

(i) H2SO5

Let x be the oxidation no. of S.

Oxidation no. of O= -2

Oxidation no. of H = +1

5.1a

 

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 H2SO5 is as given below:

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.

 

(ii) Cr2O27

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 Cr2O27.

The structure of Cr2O27 is as given below.

Each of the two Cr atoms has the oxidation no. of +6.

5.2a

(iii)  NO3

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 NO3.

The structure of NO3 is as given below.

5.3a

Nitrogen atom has the oxidation no. of +5.

 

6.Give the formula for the given compounds:

(i) Mercury

(II) chloride

(ii) Nickel (II) sulphate

(iii) Tin (IV) oxide

(iv) Thallium (I) sulphate

(v) Iron (III) sulphate

(vi) Chromium (III) oxide

 

Answer:

(i) Mercury (II) chloride

HgCl2

 

(ii) Nickel (II) sulphate

NiSO4

 

(iii) Tin (IV) oxide

SnO2

 

(iv) Thallium (I) sulphate

Tl2SO4

 

(v) Iron (III) sulphate

Fe2(SO4)3

 

(vi) Chromium (III) oxide

Cr2O3

 

7.Give a list of the compounds where carbon has oxidation no. 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
CH2Cl2 0
HCCH -1
ClCCCl +1
CH3Cl -2
CHCl3, CO +2
H3CCH3 -3
Cl3CCCl3 +3
CH4 -4
CCl4, CO2 +4

 

Compounds Oxidation no. of nitrogen
N2 0
N2H2 -1
N2O +1
N2H4 -2
NO +2
NH3 -3
N2O3 +3
NO2 +4
N2O5 +5
  1. Hydrogen peroxide and sulphur dioxide can act as oxidizing and reducing agents in the reactions, nitric acid and ozone act only as oxidants. Why?

 

Answer:

In sulphur dioxide (SO2) the oxidation no. of S is +4 and the range of oxidation no. of sulphur is from +6 to -2.

Hence, SO2 can act as reducing and oxidising agent.

 

In hydrogen peroxide (H2O2) the oxidation no. of O is -1 and the range of the oxidation no. of oxygen is from 0 to -2. Oxygen can sometimes attain the oxidation no. +1 and +2.

Therefore, H2O2 can act as reducing and oxidising agent.

 

In ozone (O3) the oxidation no. of O is 0 and the range of the oxidation no. of oxygen is from 0 to –2. Hence, the oxidation no. of oxygen only decreases in this case.

Therefore, O3 acts only as an oxidant.

 

In nitric acid (HNO3) the oxidation no. of nitrogen is +5 and the range of the oxidation no. that nitrogen can have is from +5 to -3. Hence, the oxidation no. of nitrogen can only decrease in this case.

Therefore, HNO3 acts only as an oxidant.

9.Consider the following reactions:

(i) 6CO2(g)+6H2O(l)C6H12O6(aq)+6O2(g)

(ii) O3(g)+H2O2(l)H2O(l)+2O2(g)

It is suitable to write the above equations as given below. Why?

(i) 6CO2(g)+12H2O(l)C6H12O6(aq)+6H2O(l)+6O2(g)

(ii) O3(g)+H2O2(l)H2O(l)+O2(g)+O2(g)

Give a technique to find the path of above (i) and (ii) redox reactions.

Answer:

(i)

Step 1 :

H2O breaks to give H2 and O2.

2H2O(l)2H2(g)+O2(g)

Step 2 :

The H2 produced in earlier step reduces  CO2, thus produce glucose and water.

6CO2(g)+12H2(g)C6H12O6(s)+6H2O(l)

The net reaction is as given below:

2H2O(l)2H2(g)+O2(g) × 6

6CO2(g)+12H2(g)C6H12O6(s)+6H2O(l)

————————————————————————————————————————–

6CO2(g)+12H2O(l)C6H12O6(g)+6H2O(l)+6O2(g)

 

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 H2O18 instead of H2O.

 

(ii)

Step 1 :

O2 is produced from each of the reactants O3 and H2O2. That is the reason O2 is written two times.

 

O3 breaks to form O2 and O.

Step 2 :

H2O2 reacts with O produced in the earlier step, thus produce H2O and O2.

 

O3(g)O2(g)+O(g)

 

H2O2(l)+O(g)H2O(l)+O2(g)

———————————————————————————————-

H2O2(l)+O3(g)H2O(l)+O2(g)+O2(g)

 

The path can be found with the help of H2O182 or O183.

 

 

10. AgF2 is a compound which is unstable. Even if formed, the compound would act as a very strong oxidizing agent. Why?

 

Answer:

 

The oxidation no. of Ag in AgF2 is +2. But, +2 is very unstable oxidation no. of Ag. Hence, when AgF2 is formed, silver accepts an electron and forms Ag+. This decreases the oxidation no. of Ag from +2 to +1. +1 state is more stable. Therefore, AgF2 acts as a very strong oxidizing agent.

 

11. 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.

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) P4 and F2 are reducing and oxidizing agent respectively.

In an excess amount of P4 is reacted with F2, then PF3 would be produced, where the oxidation no. of P is +3.

P4(excess)F2PF3

If P4 is reacted with excess of F2, then PF5 would be produced, where the oxidation no. of P is +5.

P4+F2(excess)PF5

(ii) K and O2 acts as a reducing agent and oxidizing agent respectively.

If an excess of K reacts with O2, it produces K2O. Here, the oxidation number of O is -2.

4K(excess)+O22K2O2

If K reacts with an excess of O2, it produces K2O2, where the oxidation number of O is –1.

2K+O2(excess)K2O12

(iii) C and O2 acts as a reducing agent and oxidizing agent respectively.

If an excess amount of C is reacted with insufficient amount of O2, then it produces CO, where the oxidation number of C is +2.

C(excess)+O2CO

If C is burnt in excess amount of O2, then CO2 is produced, where the oxidation number of C is +4.

C+O2(excess)CO2

12.How do you count for the following observations?

(i) Acidic potassium permanganate and alkaline 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.

(ii) 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:

(i) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.

(a) In a neutral medium, OH ions are produced in the reaction. Due to that, the cost of adding an acid or a base can be reduced.

(b) KMnO4 and alcohol are homogeneous to each other as they are polar. Alcohol and toluene are homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium compared to heterogeneous medium. Therefore, in alcohol, KMnO4 and toluene can react at a faster rate.

The redox reaction is as given below:

12.1a

(ii) When concentrated H2SO4 is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces H2SO4 to SO2 with the evolution of bromine’s red vapour.

2NaBr+2H2SO42NaHSO4+2HBr

2HBr+H2SO4Br2+SO2+2H2O

When concentrated H2SO4 I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce H2SO4 to SO2.

2NaCl+2H2SO42NaHSO4+2HCl

13. Find the oxidizing agent, reducing agent, the substance oxidized and reduced for the given reactions:

(i) 2AgBr(s)+C6H6O2(aq)2Ag(s)+2HBr(aq)+C6H4O2(aq)

(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH(aq)2Ag(s)+HCOO(aq)+4NH3(aq)+2H2O(l)

(iii) HCHO(l)+2Cu2+(aq)+5OH(aq)Cu2O(s)+HCOO(aq)+3H2O(l)

(iv) N2H4(l)+2H2O2(l)N2(g)+4H2O(l)

(v) Pb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(aq)+2H2O(l)

Answer:

(i) 2AgBr(s)+C6H6O2(aq)2Ag(s)+2HBr(aq)+C6H4O2(aq)

C6H6O2 => Oxidized substance

AgBr => Reduced substance

AgBr =>Oxidizing agent

C6H6O2 => Reducing agent

(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH(aq)2Ag(s)+HCOO(aq)+4NH3(aq)+2H2O(l)

HCHO => Oxidized substance

[Ag(NH3)2]+ => Reduced substance

[Ag(NH3)2]+ => Oxidizing agent

HCHO=> Reducing agent

(iii) HCHO(l)+2Cu2+(aq)+5OH(aq)Cu2O(s)+HCOO(aq)+3H2O(l)

HCHO => Oxidized substance

Cu2+ => Reduced substance

Cu2+ => Oxidizing agent

HCHO => Reducing agent

(iv) N2H4(l)+2H2O2(l)N2(g)+4H2O(l)

N2H4 => Oxidized substance

H2O2 => Reduced substance

H2O2 => Oxidizing agent

N2H4 => Reducing agent

(v) Pb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(aq)+2H2O(l)

Pb=> Oxidized substance

PbO2 => Reduced substance

PbO2 => Oxidizing agent

Pb => Reducing agent

14. Consider the given reactions:

2S2O23(aq)+I2(s)S4O26(aq)+2I(aq)

S2O23(aq)+2Br2(l)+5H2O(l)2SO24(aq)+4Br(aq)+10H+(aq)

Thiosulphate, the reductant, react differently with bromine and iodine. Why?

Answer:

The average oxidation no. of S inS2O23 is +2.

The average oxidation no. of S inS4O26 is +2.5.

The oxidation no. of S inS2O23 is +2.

The oxidation no. of S inSO24 is +6.

As Br2 is a stronger oxidizing agent than I2, it oxidizes S of  S2O23 to a higher oxidation no. of +6 in SO24.

As I2 is a weaker oxidizing agent so it oxidizes S of S2O23 ion to a lower oxidation no. that is 2.5 in S4O26 ions.

Thus, thiosulphate react differently with I2 and Br2.

15.Among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.Justift this statement with the help of examples

Answer:

F2 can oxidize Cl to Cl2, Br to Br2, and I</