NCERT Solutions for Class 11 Chemistry: Chapter 8 (Redox Reactions)

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 prescribed by CBSE. Detailed, student-friendly answers to each and every intext and exercise question provided in chapter 8 of the NCERT class 11 chemistry textbook can be found here. Furthermore, these NCERT solutions for class 11 chemistry chapter 8 can also be downloaded for free in a PDF format by clicking the download button provided above.

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NCERT Solutions for Chemistry – Class 11, 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 CBSE class 11 chemistry syllabus owing to the fact that the entire field of electrochemistry deals with redox reactions. The topics touched upon in this chapter are also a part of the JEE & NEET syllabi. 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 Terms Of Electron Transfer Reactions
    • Competitive Electron Transfer Reactions
  3. Oxidation Number
    • Types Of Redox Reactions
    • Balancing Of Redox Reactions
    • Redox Reactions As The Basis For Titrations
    • Limitations Of Concept Of Oxidation Number
  4. 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:

(i) NaH2PO4NaH_2\underline{P}O_4

(ii) NaHSO4NaH\underline{S}O_{4}

(iii) H4P2O7H_{4}\underline{P}_{2}O_{7}

(iv) K2MnO4K_{2}\underline{Mn}O_{4}

(v) CaO2Ca\underline{O}_{2}

(vi) NaBH4Na\underline{B}H_{4}

(vii) H2S2O7H_{2}\underline{S}_{2}O_{7}

(viii) KAl(SO4)2.12H2OKAl(\underline{S}O_{4})_{2}.12H_{2}O

 

Answer:

(i) NaH2PO4NaH_2\underline{P}O_4

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) NaHSO4NaH\underline{S}O_{4}

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) H4P2O7H_{4}\underline{P}_{2}O_{7}

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) K2MnO4K_{2}\underline{Mn}O_{4}

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) CaO2Ca\underline{O}_{2}

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) NaBH4Na\underline{B}H_{4}

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) H2S2O7H_{2}\underline{S}_{2}O_{7}

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)2.12H2OKAl(\underline{S}O_{4})_{2}.12H_{2}O

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.What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

(i) KI3K\underline{I}_{3}

(ii) H2S4O6H_{2}\underline{S}_{4}O_{6}

(iii) Fe3O4\underline{Fe}_{3}O_{4}

(iv) CH3CH2OH\underline{C}H_{3}\underline{C}H_{2}OH

(v) CH3COOH\underline{C}H_{3}\underline{C}OOH

 

Answer:

(i) KI3K\underline{I}_{3}

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

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

1 + 3x = 0

x = 13-\frac{1}{3}

Oxidation no. cannot be fractional. Hence, consider the structure of KI3KI_{3}.

In KI3KI_{3} molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.

2.1a

Therefore, in KI3KI_{3} molecule, the oxidation no. of I atoms forming the molecule I2I_{2} is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.

(ii) H2S4O6H_{2}\underline{S}_{4}O_{6}

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+2\frac{1}{2}

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\underline{Fe}_{3}O_{4}

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

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

3x -8 = 0

x = 83\frac{8}{3}

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\underline{C}H_{3}\underline{C}H_{2}OH

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\underline{C}H_{3}\underline{C}OOH

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 CH3COOHCH_{3}COOH.

2.51a

 

 

3. Justify that the following reactions are redox reactions:

(i) CuO(s)  +  H2  (g)    Cu(s)  +  H2O(g)CuO_{(s)} \; + \; H_{2 \; (g)} \; \rightarrow \; Cu_{(s)} \; + \; H_{2}O_{(g)}

(ii) Fe2O3  (s)  +  3  CO(g)    2  Fe(s)  +  3  CO2  (g)Fe_{2}O_{3 \; (s)} \; + \; 3 \; CO_{(g)} \; \rightarrow \; 2 \; Fe_{(s)} \; + \; 3 \; CO_{2 \; (g)}

(iii)4  BCl3  (g)  +  3  LiAlH4  (s)    2  B2H6  (g)  +  3  LiCl(s)  +  3  AlCl3  (s)4 \; BCl_{3 \; (g)} \; + \; 3 \; LiAlH_{4 \; (s)} \; \rightarrow \; 2 \; B_{2}H_{6 \; (g)} \; + \; 3 \; LiCl_{(s)} \; + \; 3 \; AlCl_{3 \; (s)}

(iv) 2  K(s)  +  F2  (g)    2  K  +  F(s)2\;K_{(s)}\;+\;F_{2\;(g)}\;\rightarrow\;2\;K\;+\;F_{(s)}

(v) 4  NH3  (g)  +  5  O2  (g)    4   NO(g)  +  6  H2O(g)4\;NH_{3\;(g)}\;+\;5\;O_{2\;(g)}\;\rightarrow \; 4  \;NO_{(g)}\;+\;6\;H_{2}O_{(g)}

 

Answer:

(i) CuO(s)  +  H2  (g)    Cu(s)  +  H2O(g)CuO_{(s)} \; + \; H_{2 \; (g)} \; \rightarrow \; Cu_{(s)} \; + \; H_{2}O_{(g)}

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

Oxidation no. of H2H_{2} is 0.

Oxidation no. of Cu is 0.

Oxidation no. of H and O in H2OH_{2}O is +1 and -2 respectively.

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

The oxidation no. of H increased from 0 to +1 in H2H_{2}. That is H2H_{2} is oxidized to H2OH_{2}O.

Therefore, the reaction is redox reaction.

(ii) Fe2O3  (s)  +  3  CO(g)    2  Fe(s)  +  3  CO2  (g)Fe_{2}O_{3 \; (s)} \; + \; 3 \; CO_{(g)} \; \rightarrow \; 2 \; Fe_{(s)} \; + \; 3 \; CO_{2 \; (g)}

In the above reaction,

Oxidation no. of Fe and O in Fe2O3Fe_{2}O_{3} 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 CO2CO_{2} is +4 and -2 respectively.

The oxidation no. of Fe decreased from +3 in Fe2O3Fe_{2}O_{3} to 0 in Fe. That is Fe2O3Fe_{2}O_{3} is reduced to Fe.

The oxidation no. of C increased from 0 to +2 in CO to +4 in CO2CO_{ 2 }. That is CO is oxidized to CO2CO_{ 2 }.

Therefore, the reaction is redox reaction.

(iii)4  BCl3  (g)  +  3  LiAlH4  (s)    2  B2H6  (g)  +  3  LiCl(s)  +  3  AlCl3  (s)4 \; BCl_{3 \; (g)} \; + \; 3 \; LiAlH_{4 \; (s)} \; \rightarrow \; 2 \; B_{2}H_{6 \; (g)} \; + \; 3 \; LiCl_{(s)} \; + \; 3 \; AlCl_{3 \; (s)}

the above reaction,

Oxidation no. of B and Cl in BCl3BCl_{3} is +3 and -1 respectively.

Oxidation no. of Li, Al and H in LiAlH4LiAlH_{4} is +1, +3 and -1 respectively.

Oxidation no. of B and H in B2H6B_{2}H_{6} 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 AlCl3AlCl_{3} is +3 and -1 respectively.

The oxidation no. of B decreased from +3 in BCl3BCl_{3} to -3 in B2H6B_{2}H_{6}. That is BCl3BCl_{3}is reduced to B2H6B_{2}H_{6}.

The oxidation no. of H increased from -1 in LiAlH4LiAlH_{4} to +1 in B2H6B_{2}H_{6}. That is LiAlH4LiAlH_{4} is oxidized to B2H6B_{2}H_{6}.

Therefore, the reaction is redox reaction.

(iv) 2  K(s)  +  F2  (g)    2  K  +  F(s)2 \; K_{(s)} \; + \; F_{2 \; (g)} \; \rightarrow \; 2 \; K \; + \; 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 F2F_{2} to -1 in KF. That is F2F_{2} is reduced to KF.

Therefore, the reaction is a redox reaction.

(v) 4  NH3  (g)  +  5  O2  (g)    4  NO(g)  +  6  H2O(g)4 \; NH_{3 \; (g)} \; + \; 5 \; O_{2 \; (g)} \; \rightarrow \; 4 \; NO_{(g)} \; + \; 6 \; H_{2}O_{(g)}

In the above reaction,

Oxidation no. of N and H in NH3NH_{3} is -3 and +1 respectively.

Oxidation no. of O2O_{2} is 0.

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

Oxidation no. of H and O in H2OH_{2}O is +1 and -2 respectively.

The oxidation no. of N increased from -3 in NH3NH_{ 3 } to +2 in NO.

The oxidation no. of O2O_{2} decreased from 0 in O2O_{ 2 } to -2 in NO and H2OH_{ 2 }O. That is O2O_{ 2 } is reduced.

Therefore, the reaction is a redox reaction.

 

 

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

H2O(s)  +  F2  (g)    HF(g)  +  HOF(g)H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }

Justify that this reaction is a redox reaction

 

Answer:

H2O(s)  +  F2  (g)    HF(g)  +  HOF(g)H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }

In the above reaction,

Oxidation no. of H and O in H2OH_{ 2 }O is +1 and -2 respectively.

Oxidation no. of F2F_{ 2 } 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 F2F_{ 2 } to +1 in HOF.

The oxidation no. of F decreased from 0 in O2O_{ 2 } 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 H2SO5H_{ 2 }SO_{ 5 }, Cr2O72Cr_{ 2 }O_{ 7 }^{ 2- } and NO3NO_{ 3 }^{ – }. Give the structure for the compounds. Count for the fallacy.

 

Answer:

(i) H2SO5H_{ 2 }SO_{ 5 }

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 H2SO5H_{ 2 }SO_{ 5 } 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) Cr2O72Cr_{ 2 }O_{ 7 }^{ 2- }

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 Cr2O72Cr_{ 2 }O_{ 7 }^{ 2- }.

The structure of Cr2O72Cr_{ 2 }O_{ 7 }^{ 2- } is as given below.

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

5.2a

(iii)  NO3NO_{ 3 }^{ – }

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 NO3NO_{ 3 }^{ – }.

The structure of NO3NO_{ 3 }^{ – } is as given below.

5.3a

Nitrogen atom has the oxidation no. of +5.

 

6. Write formulas for the following 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

HgCl2HgCl_{ 2 }

 

(ii) Nickel (II) sulphate

NiSO4NiSO_{ 4 }

 

(iii) Tin (IV) oxide

SnO2SnO_{ 2 }

 

(iv) Thallium (I) sulphate

Tl2SO4Tl_{ 2 }SO_{ 4 }

 

(v) Iron (III) sulphate

Fe2(SO4)3Fe_{ 2 }(SO_{ 4 })_{ 3 }

 

(vi) Chromium (III) oxide

Cr2O3Cr_{ 2 }O_{ 3 }

 

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
CH2Cl2CH_{ 2 }Cl_{ 2 } 0
HCCHHC\equiv CH -1
ClCCClClC\equiv CCl +1
CH3ClCH_{ 3 }Cl -2
CHCl3CHCl_{ 3 }, CO +2
H3CCH3H_{ 3 }C-CH_{ 3 } -3
Cl3CCCl3Cl_{ 3 }C-CCl_{ 3 } +3
CH4CH_{ 4 } -4
CCl4CCl_{ 4 }, CO2CO_{ 2 } +4

 

Compounds Oxidation no. of nitrogen
N2N_{ 2 } 0
N2H2N_{ 2 }H_{ 2 } -1
N2ON_{ 2 }O +1
N2H4N_{ 2 }H_{ 4 } -2
NONO +2
NH3NH_{ 3 } -3
N2O3N_{ 2 }O_{ 3 } +3
NO2NO_{ 2 } +4
N2O5N_{ 2 }O_{ 5 } +5
  1.  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 (SO2SO_{ 2 }) the oxidation no. of S is +4 and the range of oxidation no. of sulphur is from +6 to -2.

Hence, SO2SO_{ 2 } can act as reducing and oxidising agent.

 

In hydrogen peroxide (H2O2H_{ 2 }O_{ 2 }) 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, H2O2H_{ 2 }O_{ 2 } can act as reducing and oxidising agent.

 

In ozone (O3O_{ 3 }) 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, O3O_{ 3 } acts only as an oxidant.

 

In nitric acid (HNO3HNO_{ 3 }) 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, HNO3HNO_{ 3 } acts only as an oxidant.

9.Consider the reactions:

(i) 6  CO2  (g)  +  6  H2O(l)    C6H12O6  (aq)  +  6  O2  (g)6 \; CO_{ 2 \; (g) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (aq) } \; + \; 6 \; O_{ 2 \; (g) }

(ii) O3  (g)  +H2O2  (l)    H2O(l)  +  2  O2  (g)O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; 2 \; O_{ 2 \; (g) }

Why it is more appropriate to write these reactions as :

(i) 6  CO2  (g)  +  12  H2O(l)    C6H12O6  (aq)  +  6  H2O(l)  +  6  O2  (g)6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (aq) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; + \; 6 \; O_{ 2 \; (g) }

(ii) O3  (g)  +H2O2  (l)    H2O(l)  +  O2  (g)  +  O2  (g)O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions

(i)

Step 1 :

H2OH_{ 2 }O breaks to give H2H_{ 2 } and O2O_{ 2 }.

2  H2O(l)    2  H2  (g)  +  O2  (g)2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; H_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }

Step 2 :

The H2H_{ 2 } produced in earlier step reduces  CO2CO_{ 2 }, thus produce glucose and water.

6  CO2  (g)  +  12  H2  (g)    C6H12O6  (s)  +  6  H2O(l)6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 \; (g) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (s) } \; + \; 6 \; H_{ 2 }O_{ (l) }

The net reaction is as given below:

[2  H2O(l)    2  H2  (g)  +  O2  (g)2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; H_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }] × 6

6  CO2  (g)  +  12  H2  (g)    C6H12O6  (s)  +  6  H2O(l)6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 \; (g) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (s) } \; + \; 6 \; H_{ 2 }O_{ (l) }

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

6  CO2  (g)  +  12  H2O(l)    C6H12O6  (g)  +  6  H2O(l)  +  6  O2  (g)6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (g) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; + \; 6 \; O_{ 2 \; (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 H2O18H_{ 2 }O^{ 18 } instead of H2OH_{ 2 }O.

 

(ii)

Step 1 :

O2O_{ 2 } is produced from each of the reactants O3O_{ 3 } and H2O2H_{ 2 }O_{ 2 }. That is the reason O2O_{ 2 } is written two times.

 

O3O_{ 3 } breaks to form O2O_{ 2 } and O.

Step 2 :

H2O2H_{ 2 }O_{ 2 } reacts with O produced in the earlier step, thus produce H2OH_{ 2 }O and O2O_{ 2 }.

 

O3  (g)    O2  (g)  +  O(g)  O_{ 3 \; (g) } \; \rightarrow \; O_{ 2 \; (g) } \; + \; O_{ (g) }\;

 

  H2O2  (l)  +  O(g)    H2O(l)  +  O2  (g)\;H_{ 2 }O_{ 2 \; (l) } \; + \; O_{ (g) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) }

———————————————————————————————-

H2O2  (l)  +  O3  (g)    H2O(l)  +  O2  (g)  +  O2  (g)H_{ 2 }O_{ 2 \; (l) } \; + \; O_{ 3 \; (g) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }

 

The path can be found with the help of H2O218H_{ 2 }O_{ 2 }^{ 18 } or O318O_{ 3 }^{ 18 }.

 

 

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

 

Answer:

 

The oxidation no. of Ag in AgF2AgF_{ 2 } is +2. But, +2 is very unstable oxidation no. of Ag. Hence, when AgF2AgF_{ 2 } is formed, silver accepts an electron and forms Ag+Ag^{ + }. This decreases the oxidation no. of Ag from +2 to +1. +1 state is more stable. Therefore, AgF2AgF_{ 2 } acts as a very strong oxidizing agent.

 

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) P4P_{ 4 } and F2F_{ 2 } are reducing and oxidizing agent respectively.

In an excess amount of P4P_{ 4 } is reacted with F2F_{ 2 }, then PF3PF_{ 3 } would be produced, where the oxidation no. of P is +3.

P4  (excess)  F2    PF3P_{ 4 } \; _{(excess)} \; F_{ 2 } \; \rightarrow \; PF_{ 3 }

If P4P_{ 4 } is reacted with excess of F2F_{ 2 }, then PF5PF_{ 5 } would be produced, where the oxidation no. of P is +5.

P4  +  F2  (excess)    PF5P_{ 4 } \; + \; F_{ 2 } \; _{(excess)} \; \rightarrow \; PF_{ 5 }

(ii) K and O2O_{ 2 } acts as a reducing agent and oxidizing agent respectively.

If an excess of K reacts with O2O_{ 2 }, it produces K2OK_{ 2 }O. Here, the oxidation number of O is -2.

4  K  (excess)  +  O2    2  K2O24 \; K \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; 2 \; K_{ 2 }O^{ -2 }

If K reacts with an excess of O2O_{ 2 }, it produces K2O2K_{ 2 }O_{ 2 }, where the oxidation number of O is –1.

2  K  +  O2  (excess)    K2O212 \; K \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; K_{ 2 }O_{ 2 }^{ -1 }

(iii) C and O2O_{ 2 } acts as a reducing agent and oxidizing agent respectively.

If an excess amount of C is reacted with insufficient amount of O2O_{ 2 }, then it produces CO, where the oxidation number of C is +2.

C  (excess)  +  O2    COC \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; CO

If C is burnt in excess amount of O2O_{ 2 }, then CO2CO_{ 2 } is produced, where the oxidation number of C is +4.

C  +  O2  (excess)    CO2C \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; CO_{ 2 }

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, OHOH^{ – } ions are produced in the reaction. Due to that, the cost of adding an acid or a base can be reduced.

(b) KMnO4KMnO_{ 4 } 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, KMnO4KMnO_{ 4 } and toluene can react at a faster rate.

The redox reaction is as given below:

12.1a

(ii) When concentrated H2SO4H_{ 2 }SO_{ 4 } is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces H2SO4H_{ 2 }SO_{ 4 } to SO2SO_{ 2 } with the evolution of bromine’s red vapour.

2  NaBr  +  2  H2SO4    2  NaHSO4  +  2  HBr2 \; NaBr \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HBr

2  HBr  +  H2SO4    Br2  +  SO2  +  2  H2O2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O

When concentrated H2SO4H_{ 2 }SO_{ 4 } I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce H2SO4H_{ 2 }SO_{ 4 } to SO2SO_{ 2 }.

2  NaCl  +  2  H2SO4    2  NaHSO4  +  2  HCl2 \; NaCl \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HCl

13. Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:

(i) 2  AgBr(s)  +  C6H6O2  (aq)    2  Ag(s)  +  2  HBr(aq)  +  C6H4O2  (aq)2 \; AgBr_{ (s) } \; + \; C_{ 6 }H_{ 6 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; 2 \; HBr_{ (aq)} \; + \; C_{ 6 }H_{ 4 }O_{ 2 \; (aq) }

(ii) HCHO(l)  +  2  [Ag(NH3)2](aq)+  +  3  OH(aq)    2  Ag(s)  +  HCOO(aq)  +  4  NH3  (aq)  +  2  H2O(l)HCHO_{ (l) } \; + \; 2 \; [Ag(NH_{ 3 })_{ 2 }]^{ + }_{ (aq) } \; + \; 3 \; OH^{ – }_{ (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 4 \; NH_{ 3 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }

(iii) HCHO(l)  +  2  Cu(aq)2+  +  5  OH(aq)    Cu2O(s)  +  HCOO(aq)  +  3  H2O(l)HCHO_{ (l) } \; + \; 2 \; Cu^{ 2+ }_{ (aq) } \; + \; 5 \; OH^{ – }_{ (aq) } \; \rightarrow \; Cu_{ 2 }O_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 3 \; H_{ 2 }O_{ (l) }

(iv) N2H4  (l)  +  2  H2O2  (l)    N2  (g)  +  4  H2O(l)N_{ 2 }H_{ 4 \; (l) } \; + \; 2 \; H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; N_{ 2 \; (g) } \; + \; 4 \; H_{ 2 }O_{ (l) }

(v) Pb(s)  +  PbO2  (s)  +  2  H2SO4  (aq)    2  PbSO4  (aq)  +  2  H2O(l)Pb_{ (s) } \; + \; PbO_{ 2 \; (s) } \; + \; 2 \; H_{ 2 }SO_{ 4 \; (aq) } \; \rightarrow \; 2 \; PbSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }

Answer:

(i) 2  AgBr(s)  +  C6H6O2  (aq)    2  Ag(s)  +  2  HBr(aq)  +  C6H4O2  (aq)2 \; AgBr_{ (s) } \; + \; C_{ 6 }H_{ 6 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; 2 \; HBr_{ (aq)} \; + \; C_{ 6 }H_{ 4 }O_{ 2 \; (aq) }

C6H6O2C_{ 6 }H_{ 6 }O_{ 2 } => Oxidized substance

AgBr => Reduced substance

AgBr =>Oxidizing agent

C6H6O2C_{ 6 }H_{ 6 }O_{ 2 } => Reducing agent

(ii) HCHO(l)  +  2  [Ag(NH3)2](aq)+  +  3  OH(aq)    2  Ag(s)  +  HCOO(aq)  +  4  NH3  (aq)  +  2  H2O(l)HCHO_{ (l) } \; + \; 2 \; [Ag(NH_{ 3 })_{ 2 }]^{ + }_{ (aq) } \; + \; 3 \; OH^{ – }_{ (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 4 \; NH_{ 3 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }

HCHO => Oxidized substance

[Ag(NH3)2]+[Ag(NH_{ 3 })_{ 2 }]^{ + } => Reduced substance

[Ag(NH3)2]+[Ag(NH_{ 3 })_{ 2 }]^{ + } => Oxidizing agent

HCHO=> Reducing agent

(iii) HCHO(l)  +  2  Cu(aq)2+  +  5  OH(aq)    Cu2O(s)  +  HCOO(aq)  +  3  H2O(l)HCHO_{ (l) } \; + \; 2 \; Cu^{ 2+ }_{ (aq) } \; + \; 5 \; OH^{ – }_{ (aq) } \; \rightarrow \; Cu_{ 2 }O_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 3 \; H_{ 2 }O_{ (l) }

HCHO => Oxidized substance

Cu2+Cu^{ 2+ } => Reduced substance

Cu2+Cu^{ 2+ } => Oxidizing agent

HCHO => Reducing agent

(iv) N2H4  (l)  +  2  H2O2  (l)    N2  (g)  +  4  H2O(l)N_{ 2 }H_{ 4 \; (l) } \; + \; 2 \; H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; N_{ 2 \; (g) } \; + \; 4 \; H_{ 2 }O_{ (l) }

N2H4N_{ 2 }H_{ 4 } => Oxidized substance

H2O2H_{ 2 }O_{ 2 } => Reduced substance

H2O2H_{ 2 }O_{ 2 } => Oxidizing agent

N2H4N_{ 2 }H_{ 4 } => Reducing agent

(v) Pb(s)  +  PbO2  (s)  +  2  H2SO4  (aq)    2  PbSO4  (aq)  +  2  H2O(l)Pb_{ (s) } \; + \; PbO_{ 2 \; (s) } \; + \; 2 \; H_{ 2 }SO_{ 4 \; (aq) } \; \rightarrow \; 2 \; PbSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }

Pb=> Oxidized substance

PbO2PbO_{ 2 } => Reduced substance

PbO2PbO_{ 2 } => Oxidizing agent

Pb => Reducing agent

14. Consider the reactions :

2  S2O3  (aq)2  +  I2  (s)    S4O6  (aq)2  +  2  I(aq)2 \; S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; I_{ 2 \; (s) } \; \rightarrow \; S_{ 4 }O^{ 2- }_{ 6 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) }

S2O3  (aq)2  +  2  Br2  (l)  +  5  H2O(l)    2  SO4  (aq)2  +  4  Br(aq)  +  10  H(aq)+S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; 2 \; Br_{ 2 \; (l) } \; + \; 5 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; Br^{ – }_{ (aq) } \; + \; 10 \; H^{ + }_{ (aq) }

Why does the same reductant, thiosulphate react differently with iodine and bromine ?

Answer:

The average oxidation no. of S inS2O32S_{ 2 }O_{ 3 }^{ 2- } is +2.

The average oxidation no. of S inS4O62S_{ 4 }O_{ 6 }^{ 2- } is +2.5.

The oxidation no. of S inS2O32S_{ 2 }O_{ 3 }^{ 2- } is +2.

The oxidation no. of S inSO42SO_{ 4 }^{ 2- } is +6.

As Br2Br_{ 2 } is a stronger oxidizing agent than I2I_{ 2 }, it oxidizes S of  S2O32S_{ 2 }O_{ 3 }^{ 2- } to a higher oxidation no. of +6 in SO42SO_{ 4 }^{ 2- }.

As I2I_{ 2 } is a weaker oxidizing agent so it oxidizes S of S2O32S_{ 2 }O_{ 3 }^{ 2- } ion to a lower oxidation no. that is 2.5 in S4O62S_{ 4 }O_{ 6 }^{ 2- } ions.

Thus, thiosulphate react differently with I2I_{ 2 } and Br2Br_{ 2 }.

15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer:

F2F_{ 2 } can oxidize ClCl^{ – } to Cl2Cl_{ 2 }, BrBr^{ – } to Br2Br_{ 2 }, and II^{ – } to I2I_{ 2 } as:

F2  (aq)  +  2  Cl(s)    2  F(aq)  +  Cl2  (g)F_{ 2 \; (aq) } \; + \; 2 \; Cl^{ – }_{ (s) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; Cl_{ 2 \; (g) }

F2  (aq)  +  2  Br(aq)    2  F(aq)  +  Br2  (l)F_{ 2 \; (aq) } \; + \; 2 \; Br^{ – }_{ (aq) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; Br_{ 2 \; (l) }

F2  (aq)  +  2  I(aq)    2  F(aq)  +  I2  (s)F_{ 2 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; I_{ 2 \; (s) }

But, Cl2Cl_{ 2 }, Br2Br_{ 2 }, and I2I_{ 2 } cannot oxidize FF^{ – } to F2F_{ 2 }. The oxidizing power of halogens increases in the order as given below:

I2I_{ 2 }< Br2Br_{ 2 }< Cl2Cl_{ 2 }<F2F_{ 2 }

Therefore, fluorine is the best oxidant among halogens.

HIHI and HBrHBr can reduce H2SO4H_{ 2 }SO_{ 4 } to SO2SO_{ 2 }, but HClHCl and HFHF cannot. Hence, HIHI and HBrHBr are stronger reductants compared to HClHCl and HFHF.

2  HI  +  H2SO4    I2  +  SO2  +  2  H2O2 \; HI \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; I_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O

2  HBr  +  H2SO4    Br2  +  SO2  +  2  H2O2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O