# 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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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) $NaHunderline{S}O_{4}$

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) $H_{4}underline{P}_{2}O_{7}$

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) $K_{2}underline{Mn}O_{4}$

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) $Caunderline{O}_{2}$

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) $Naunderline{B}H_{4}$

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) $H_{2}underline{S}_{2}O_{7}$

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) $KAl(underline{S}O_{4})_{2}.12H_{2}O$

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) $Kunderline{I}_{3}$

(b) $H_{2}underline{S}_{4}O_{6}$

(c) $underline{Fe}_{3}O_{4}$

(d) $underline{C}H_{3}underline{C}H_{2}OH$

(e) $underline{C}H_{3}underline{C}OOH$

(a) $Kunderline{I}_{3}$

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

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

1 + 3x = 0

x = $-frac{1}{3}$

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

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

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

(b) $H_{2}underline{S}_{4}O_{6}$

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 = $+2frac{1}{2}$

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) $underline{Fe}_{3}O_{4}$

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

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

3x -8 = 0

x = $frac{8}{3}$

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.

(d) $underline{C}H_{3}underline{C}H_{2}OH$

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) $underline{C}H_{3}underline{C}OOH$

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 $CH_{3}COOH$.

3. Justify that the following reactions are redox reactions:

(a) $CuO_{(s)} ; + ; H_{2 ; (g)} ; ightarrow ; Cu_{(s)} ; + ; H_{2}O_{(g)}$

(b) $Fe_{2}O_{3 ; (s)} ; + ; 3 ; CO_{(g)} ; ightarrow ; 2 ; Fe_{(s)} ; + ; 3 ; CO_{2 ; (g)}$

(c)$4 ; BCl_{3 ; (g)} ; + ; 3 ; LiAlH_{4 ; (s)} ; ightarrow ; 2 ; B_{2}H_{6 ; (g)} ; + ; 3 ; LiCl_{(s)} ; + ; 3 ; AlCl_{3 ; (s)}$

(d) $2;K_{(s)};+;F_{2;(g)}; ightarrow;2;K;+;F_{(s)}$

(e) $4;NH_{3;(g)};+;5;O_{2;(g)}; ightarrow ; 4 ;NO_{(g)};+;6;H_{2}O_{(g)}$

(a) $CuO_{(s)} ; + ; H_{2 ; (g)} ; ightarrow ; Cu_{(s)} ; + ; H_{2}O_{(g)}$

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

Oxidation no. of $H_{2}$ is 0.

Oxidation no. of Cu is 0.

Oxidation no. of H and O in $H_{2}O$ 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 $H_{2}$. That is $H_{2}$ is oxidized to $H_{2}O$.

Therefore, the reaction is redox reaction.

(b) $Fe_{2}O_{3 ; (s)} ; + ; 3 ; CO_{(g)} ; ightarrow ; 2 ; Fe_{(s)} ; + ; 3 ; CO_{2 ; (g)}$

In the above reaction,

Oxidation no. of Fe and O in $Fe_{2}O_{3}$ 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 $CO_{2}$ is +4 and -2 respectively.

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

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

Therefore, the reaction is redox reaction.

(c) $4 ; BCl_{3 ; (g)} ; + ; 3 ; LiAlH_{4 ; (s)} ; ightarrow ; 2 ; B_{2}H_{6 ; (g)} ; + ; 3 ; LiCl_{(s)} ; + ; 3 ; AlCl_{3 ; (s)}$

the above reaction,

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

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

Oxidation no. of B and H in $B_{2}H_{6}$ 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 $AlCl_{3}$ is +3 and -1 respectively.

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

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

Therefore, the reaction is redox reaction.

(d) $2 ; K_{(s)} ; + ; F_{2 ; (g)} ; ightarrow ; 2 ; K ; + ; F_{(s)}$

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 $F_{2}$ to -1 in KF. That is $F_{2}$ is reduced to KF.

Therefore, the reaction is a redox reaction.

(e) $4 ; NH_{3 ; (g)} ; + ; 5 ; O_{2 ; (g)} ; ightarrow ; 4 ; NO_{(g)} ; + ; 6 ; H_{2}O_{(g)}$

In the above reaction,

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

Oxidation no. of $O_{2}$ is 0.

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

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

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

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

Therefore, the reaction is a redox reaction.

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

$H_{ 2 }O_{ (s) } ; + ; F_{ 2 ; (g) } ; ightarrow ; HF_{ (g) } ; + ; HOF_{ (g) }$

Justify that this reaction is a redox reaction

$H_{ 2 }O_{ (s) } ; + ; F_{ 2 ; (g) } ; ightarrow ; HF_{ (g) } ; + ; HOF_{ (g) }$

In the above reaction,

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

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

The oxidation no. of F decreased from 0 in $O_{ 2 }$ to -1 in HF.

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

5. Calculate the oxidation no. of sulphur, chromium and nitrogen in $H_{ 2 }SO_{ 5 }$, $Cr_{ 2 }O_{ 7 }^{ 2- }$ and $NO_{ 3 }^{ – }$. Suggest structure of these compounds. Count for the fallacy.

For $H_{ 2 }SO_{ 5 }$

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 $H_{ 2 }SO_{ 5 }$ 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.

For $Cr_{ 2 }O_{ 7 }^{ 2- }$

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 $Cr_{ 2 }O_{ 7 }^{ 2- }$.

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

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

For $NO_{ 3 }^{ – }$

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 $NO_{ 3 }^{ – }$.

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

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

(a) Mercury (II) chloride

$HgCl_{ 2 }$

(b) Nickel (II) sulphate

$NiSO_{ 4 }$

(c) Tin (IV) oxide

$SnO_{ 2 }$

(d) Thallium (I) sulphate

$Tl_{ 2 }SO_{ 4 }$

(e) Iron (III) sulphate

$Fe_{ 2 }(SO_{ 4 })_{ 3 }$

(f) Chromium (III) oxide

$Cr_{ 2 }O_{ 3 }$

7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

The compound where carbon has oxidation no. from -4 to +4 is as given below in the table:

 Compounds Oxidation no. of carbon $CH_{ 2 }Cl_{ 2 }$ 0 $HCequiv CH$ -1 $ClCequiv CCl$ +1 $CH_{ 3 }Cl$ -2 $CHCl_{ 3 }$, CO +2 $H_{ 3 }C-CH_{ 3 }$ -3 $Cl_{ 3 }C-CCl_{ 3 }$ +3 $CH_{ 4 }$ -4 $CCl_{ 4 }$, $CO_{ 2 }$ +4

 Compounds Oxidation no. of nitrogen $N_{ 2 }$ 0 $N_{ 2 }H_{ 2 }$ -1 $N_{ 2 }O$ +1 $N_{ 2 }H_{ 4 }$ -2 $NO$ +2 $NH_{ 3 }$ -3 $N_{ 2 }O_{ 3 }$ +3 $NO_{ 2 }$ +4 $N_{ 2 }O_{ 5 }$ +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?

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

Hence, $SO_{ 2 }$ can act as a reducing and oxidising agent.

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

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

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

9.Consider the reactions:

(a) $6 ; CO_{ 2 ; (g) } ; + ; 6 ; H_{ 2 }O_{ (l) } ; ightarrow ; C_{ 6 }H_{ 12 }O_{ 6 ; (aq) } ; + ; 6 ; O_{ 2 ; (g) }$

(b) $O_{ 3 ; (g) } ; + H_{ 2 }O_{ 2 ; (l) } ; ightarrow ; H_{ 2 }O_{ (l) } ; + ; 2 ; O_{ 2 ; (g) }$

Why it is more appropriate to write these reactions as :

(a) $6 ; CO_{ 2 ; (g) } ; + ; 12 ; H_{ 2 }O_{ (l) } ; ightarrow ; C_{ 6 }H_{ 12 }O_{ 6 ; (aq) } ; + ; 6 ; H_{ 2 }O_{ (l) } ; + ; 6 ; O_{ 2 ; (g) }$

(b) $O_{ 3 ; (g) } ; + H_{ 2 }O_{ 2 ; (l) } ; ightarrow ; H_{ 2 }O_{ (l) } ; + ; O_{ 2 ; (g) } ; + ; O_{ 2 ; (g) }$

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

(a)

Step 1 :

$H_{ 2 }O$ breaks to give $H_{ 2 }$ and $O_{ 2 }$.

$2 ; H_{ 2 }O_{ (l) } ; ightarrow ; 2 ; H_{ 2 ; (g) } ; + ; O_{ 2 ; (g) }$

Step 2 :

The $H_{ 2 }$ produced in earlier step reduces  $CO_{ 2 }$, thus produce glucose and water.

$6 ; CO_{ 2 ; (g) } ; + ; 12 ; H_{ 2 ; (g) } ; ightarrow ; C_{ 6 }H_{ 12 }O_{ 6 ; (s) } ; + ; 6 ; H_{ 2 }O_{ (l) }$

The net reaction is as given below:

[$2 ; H_{ 2 }O_{ (l) } ; ightarrow ; 2 ; H_{ 2 ; (g) } ; + ; O_{ 2 ; (g) }$] × 6

$6 ; CO_{ 2 ; (g) } ; + ; 12 ; H_{ 2 ; (g) } ; ightarrow ; C_{ 6 }H_{ 12 }O_{ 6 ; (s) } ; + ; 6 ; H_{ 2 }O_{ (l) }$

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

$6 ; CO_{ 2 ; (g) } ; + ; 12 ; H_{ 2 }O_{ (l) } ; ightarrow ; C_{ 6 }H_{ 12 }O_{ 6 ; (g) } ; + ; 6 ; H_{ 2 }O_{ (l) } ; + ; 6 ; O_{ 2 ; (g) }$

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 $H_{ 2 }O^{ 18 }$ instead of $H_{ 2 }O$.

(b)

Step 1 :

$O_{ 2 }$ is produced from each of the reactants $O_{ 3 }$ and $H_{ 2 }O_{ 2 }$. That is the reason $O_{ 2 }$ is written two times.

$O_{ 3 }$ breaks to form $O_{ 2 }$ and O.

Step 2 :

$H_{ 2 }O_{ 2 }$ reacts with O produced in the earlier step, thus produce $H_{ 2 }O$ and $O_{ 2 }$.

$O_{ 3 ; (g) } ; ightarrow ; O_{ 2 ; (g) } ; + ; O_{ (g) };$

$;H_{ 2 }O_{ 2 ; (l) } ; + ; O_{ (g) } ; ightarrow ; H_{ 2 }O_{ (l) } ; + ; O_{ 2 ; (g) }$

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