# NCERT Solutions For Class 11 Chemistry Chapter 8

## NCERT Solutions Class 11 Chemistry Redox Reactions

NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions are provided here in detail. This topic is an extremely important topic of 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 Chapter 8 pdf which students can download. Practicing all the questions will be very helpful for the students as many questions are framed from the topic.

1. Assign oxidation no. to the elements underlined:

(i) NaH2PO4$NaH_2\underline{P}O_4$

(ii) NaHSO4$NaH\underline{S}O_{4}$

(iii) H4P2O7$H_{4}\underline{P}_{2}O_{7}$

(iv) K2MnO4$K_{2}\underline{Mn}O_{4}$

(v) CaO2$Ca\underline{O}_{2}$

(vi) NaBH4$Na\underline{B}H_{4}$

(vii) H2S2O7$H_{2}\underline{S}_{2}O_{7}$

(viii) KAl(SO4)2.12H2O$KAl(\underline{S}O_{4})_{2}.12H_{2}O$

(i) NaH2PO4$NaH_2\underline{P}O_4$

Let x be the oxidation no. of P.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

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$NaH\underline{S}O_{4}$

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.

(iii) H4P2O7$H_{4}\underline{P}_{2}O_{7}$

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.

(iv) K2MnO4$K_{2}\underline{Mn}O_{4}$

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.

(v) CaO2$Ca\underline{O}_{2}$

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.

(vi) NaBH4$Na\underline{B}H_{4}$

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.

(vii) H2S2O7$H_{2}\underline{S}_{2}O_{7}$

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.

(viii) KAl(SO4)2.12H2O$KAl(\underline{S}O_{4})_{2}.12H_{2}O$

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.Assign oxidation no. to the elements underlined:

(i) KI3$K\underline{I}_{3}$

(ii) H2S4O6$H_{2}\underline{S}_{4}O_{6}$

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

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

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

(i) KI3$K\underline{I}_{3}$

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 KI3$KI_{3}$.

In KI3$KI_{3}$ molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.

Therefore, in KI3$KI_{3}$ molecule, the oxidation no. of I atoms forming the molecule I2$I_{2}$ is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.

(ii) H2S4O6$H_{2}\underline{S}_{4}O_{6}$

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 = +212$+2\frac{1}{2}$

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.

(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.

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

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.

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

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 CH3COOH$CH_{3}COOH$.

3. The reactions given below are redox reactions. Justify the 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)+3CO(g)2Fe(s)+3CO2(g)$Fe_{2}O_{3 \; (s)} \; + \; 3 \; CO_{(g)} \; \rightarrow \; 2 \; Fe_{(s)} \; + \; 3 \; CO_{2 \; (g)}$

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

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

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

(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 CuO$CuO$ is +2 and -2 respectively.

Oxidation no. of H2$H_{2}$ is 0.

Oxidation no. of Cu is 0.

Oxidation no. of H and O in H2O$H_{2}O$ is +1 and -2 respectively.

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

The oxidation no. of H increased from 0 to +1 in H2$H_{2}$. That is H2$H_{2}$ is oxidized to H2O$H_{2}O$.

Therefore, the reaction is redox reaction.

(ii) Fe2O3(s)+3CO(g)2Fe(s)+3CO2(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 Fe2O3$Fe_{2}O_{3}$ 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 CO2$CO_{2}$ is +4 and -2 respectively.

The oxidation no. of Fe decreased from +3 in Fe2O3$Fe_{2}O_{3}$ to 0 in Fe. That is Fe2O3$Fe_{2}O_{3}$ is reduced to Fe.

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

Therefore, the reaction is redox reaction.

(iii)4BCl3(g)+3LiAlH4(s)2B2H6(g)+3LiCl(s)+3AlCl3(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 BCl3$BCl_{3}$ is +3 and -1 respectively.

Oxidation no. of Li, Al and H in LiAlH4$LiAlH_{4}$ is +1, +3 and -1 respectively.

Oxidation no. of B and H in B2H6$B_{2}H_{6}$ 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 AlCl3$AlCl_{3}$ is +3 and -1 respectively.

The oxidation no. of B decreased from +3 in BCl3$BCl_{3}$ to -3 in B2H6$B_{2}H_{6}$. That is BCl3$BCl_{3}$is reduced to B2H6$B_{2}H_{6}$.

The oxidation no. of H increased from -1 in LiAlH4$LiAlH_{4}$ to +1 in B2H6$B_{2}H_{6}$. That is LiAlH4$LiAlH_{4}$ is oxidized to B2H6$B_{2}H_{6}$.

Therefore, the reaction is redox reaction.

(iv) 2K(s)+F2(g)2K+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 F2$F_{2}$ to -1 in KF. That is F2$F_{2}$ is reduced to KF.

Therefore, the reaction is a redox reaction.

(v) 4NH3(g)+5O2(g)4NO(g)+6H2O(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 NH3$NH_{3}$ is -3 and +1 respectively.

Oxidation no. of O2$O_{2}$ is 0.

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

Oxidation no. of H and O in H2O$H_{2}O$ is +1 and -2 respectively.

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

The oxidation no. of O2$O_{2}$ decreased from 0 in O2$O_{ 2 }$ to -2 in NO and H2O$H_{ 2 }O$. That is O2$O_{ 2 }$ 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)$H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }$

Give reason that the above reaction is redox reaction.

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 H2O$H_{ 2 }O$ is +1 and -2 respectively.

Oxidation no. of F2$F_{ 2 }$ 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 F2$F_{ 2 }$ to +1 in HOF.

The oxidation no. of F decreased from 0 in O2$O_{ 2 }$ to -1 in HF.

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

5. Calculate the oxidation no. of chromium, sulphur and nitrogen in H2SO5$H_{ 2 }SO_{ 5 }$, Cr2O27$Cr_{ 2 }O_{ 7 }^{ 2- }$ and NO3$NO_{ 3 }^{ – }$. Give the structure for the compounds. Count for the fallacy.

(i) H2SO5$H_{ 2 }SO_{ 5 }$

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 H2SO5$H_{ 2 }SO_{ 5 }$ 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.

(ii) Cr2O27$Cr_{ 2 }O_{ 7 }^{ 2- }$

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 Cr2O27$Cr_{ 2 }O_{ 7 }^{ 2- }$.

The structure of Cr2O27$Cr_{ 2 }O_{ 7 }^{ 2- }$ is as given below.

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

(iii)  NO3$NO_{ 3 }^{ – }$

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 NO3$NO_{ 3 }^{ – }$.

The structure of NO3$NO_{ 3 }^{ – }$ is as given below.

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

(i) Mercury (II) chloride

HgCl2$HgCl_{ 2 }$

(ii) Nickel (II) sulphate

NiSO4$NiSO_{ 4 }$

(iii) Tin (IV) oxide

SnO2$SnO_{ 2 }$

(iv) Thallium (I) sulphate

Tl2SO4$Tl_{ 2 }SO_{ 4 }$

(v) Iron (III) sulphate

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

(vi) Chromium (III) oxide

Cr2O3$Cr_{ 2 }O_{ 3 }$

7.Give a list of the compounds where carbon has oxidation no. 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 CH2Cl2$CH_{ 2 }Cl_{ 2 }$ 0 HC≡CH$HC\equiv CH$ -1 ClC≡CCl$ClC\equiv CCl$ +1 CH3Cl$CH_{ 3 }Cl$ -2 CHCl3$CHCl_{ 3 }$, CO +2 H3C−CH3$H_{ 3 }C-CH_{ 3 }$ -3 Cl3C−CCl3$Cl_{ 3 }C-CCl_{ 3 }$ +3 CH4$CH_{ 4 }$ -4 CCl4$CCl_{ 4 }$, CO2$CO_{ 2 }$ +4

 Compounds Oxidation no. of nitrogen N2$N_{ 2 }$ 0 N2H2$N_{ 2 }H_{ 2 }$ -1 N2O$N_{ 2 }O$ +1 N2H4$N_{ 2 }H_{ 4 }$ -2 NO$NO$ +2 NH3$NH_{ 3 }$ -3 N2O3$N_{ 2 }O_{ 3 }$ +3 NO2$NO_{ 2 }$ +4 N2O5$N_{ 2 }O_{ 5 }$ +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?

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

Hence, SO2$SO_{ 2 }$ can act as reducing and oxidising agent.

In hydrogen peroxide (H2O2$H_{ 2 }O_{ 2 }$) 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, H2O2$H_{ 2 }O_{ 2 }$ can act as reducing and oxidising agent.

In ozone (O3$O_{ 3 }$) 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, O3$O_{ 3 }$ acts only as an oxidant.

In nitric acid (HNO3$HNO_{ 3 }$) 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, HNO3$HNO_{ 3 }$ acts only as an oxidant.

9.Consider the following reactions:

(i) 6CO2(g)+6H2O(l)C6H12O6(aq)+6O2(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)+2O2(g)$O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; 2 \; O_{ 2 \; (g) }$

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

(i) 6CO2(g)+12H2O(l)C6H12O6(aq)+6H2O(l)+6O2(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)$O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }$

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

(i)

Step 1 :

H2O$H_{ 2 }O$ breaks to give H2$H_{ 2 }$ and O2$O_{ 2 }$.

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

Step 2 :

The H2$H_{ 2 }$ produced in earlier step reduces  CO2$CO_{ 2 }$, thus produce glucose and water.

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

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

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

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

6CO2(g)+12H2O(l)C6H12O6(g)+6H2O(l)+6O2(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 H2O18$H_{ 2 }O^{ 18 }$ instead of H2O$H_{ 2 }O$.

(ii)

Step 1 :

O2$O_{ 2 }$ is produced from each of the reactants O3$O_{ 3 }$ and H2O2$H_{ 2 }O_{ 2 }$. That is the reason O2$O_{ 2 }$ is written two times.

O3$O_{ 3 }$ breaks to form O2$O_{ 2 }$ and O.

Step 2 :

H2O2$H_{ 2 }O_{ 2 }$ reacts with O produced in the earlier step, thus produce H2O$H_{ 2 }O$ and O2$O_{ 2 }$.

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 H2O182$H_{ 2 }O_{ 2 }^{ 18 }$ or O183$O_{ 3 }^{ 18 }$.

10. AgF2$AgF_{ 2 }$ is a compound which is unstable. Even if formed, the compound would act as a very strong oxidizing agent. Why?

The oxidation no. of Ag in AgF2$AgF_{ 2 }$ is +2. But, +2 is very unstable oxidation no. of Ag. Hence, when AgF2$AgF_{ 2 }$ 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, AgF2$AgF_{ 2 }$ 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.

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$P_{ 4 }$ and F2$F_{ 2 }$ are reducing and oxidizing agent respectively.

In an excess amount of P4$P_{ 4 }$ is reacted with F2$F_{ 2 }$, then PF3$PF_{ 3 }$ would be produced, where the oxidation no. of P is +3.

P4(excess)F2PF3$P_{ 4 } \; _{(excess)} \; F_{ 2 } \; \rightarrow \; PF_{ 3 }$

If P4$P_{ 4 }$ is reacted with excess of F2$F_{ 2 }$, then PF5$PF_{ 5 }$ would be produced, where the oxidation no. of P is +5.

P4+F2(excess)PF5$P_{ 4 } \; + \; F_{ 2 } \; _{(excess)} \; \rightarrow \; PF_{ 5 }$

(ii) K and O2$O_{ 2 }$ acts as a reducing agent and oxidizing agent respectively.

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

4K(excess)+O22K2O2$4 \; K \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; 2 \; K_{ 2 }O^{ -2 }$

If K reacts with an excess of O2$O_{ 2 }$, it produces K2O2$K_{ 2 }O_{ 2 }$, where the oxidation number of O is –1.

2K+O2(excess)K2O12$2 \; K \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; K_{ 2 }O_{ 2 }^{ -1 }$

(iii) C and O2$O_{ 2 }$ acts as a reducing agent and oxidizing agent respectively.

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

C(excess)+O2CO$C \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; CO$

If C is burnt in excess amount of O2$O_{ 2 }$, then CO2$CO_{ 2 }$ is produced, where the oxidation number of C is +4.

C+O2(excess)CO2$C \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; CO_{ 2 }$

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?

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

(b) KMnO4$KMnO_{ 4 }$ 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, KMnO4$KMnO_{ 4 }$ and toluene can react at a faster rate.

The redox reaction is as given below:

(ii) When concentrated H2SO4$H_{ 2 }SO_{ 4 }$ is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces H2SO4$H_{ 2 }SO_{ 4 }$ to SO2$SO_{ 2 }$ with the evolution of bromine’s red vapour.

2NaBr+2H2SO42NaHSO4+2HBr$2 \; NaBr \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HBr$

2HBr+H2SO4Br2+SO2+2H2O$2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O$

When concentrated H2SO4$H_{ 2 }SO_{ 4 }$ I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce H2SO4$H_{ 2 }SO_{ 4 }$ to SO2$SO_{ 2 }$.

2NaCl+2H2SO42NaHSO4+2HCl$2 \; NaCl \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HCl$

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)$2 \; AgBr_{ (s) } \; + \; C_{ 6 }H_{ 6 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; 2 \; HBr_{ (aq)} \; + \; C_{ 6 }H_{ 4 }O_{ 2 \; (aq) }$

(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH(aq)2Ag(s)+HCOO(aq)+4NH3(aq)+2H2O(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)+2Cu2+(aq)+5OH(aq)Cu2O(s)+HCOO(aq)+3H2O(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)+2H2O2(l)N2(g)+4H2O(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)+2H2SO4(aq)2PbSO4(aq)+2H2O(l)$Pb_{ (s) } \; + \; PbO_{ 2 \; (s) } \; + \; 2 \; H_{ 2 }SO_{ 4 \; (aq) } \; \rightarrow \; 2 \; PbSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }$

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

C6H6O2$C_{ 6 }H_{ 6 }O_{ 2 }$ => Oxidized substance

AgBr => Reduced substance

AgBr =>Oxidizing agent

C6H6O2$C_{ 6 }H_{ 6 }O_{ 2 }$ => Reducing agent

(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH(aq)2Ag(s)+HCOO(aq)+4NH3(aq)+2H2O(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)+2Cu2+(aq)+5OH(aq)Cu2O(s)+HCOO(aq)+3H2O(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)+2H2O2(l)N2(g)+4H2O(l)$N_{ 2 }H_{ 4 \; (l) } \; + \; 2 \; H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; N_{ 2 \; (g) } \; + \; 4 \; H_{ 2 }O_{ (l) }$

N2H4$N_{ 2 }H_{ 4 }$ => Oxidized substance

H2O2$H_{ 2 }O_{ 2 }$ => Reduced substance

H2O2$H_{ 2 }O_{ 2 }$ => Oxidizing agent

N2H4$N_{ 2 }H_{ 4 }$ => Reducing agent

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

PbO2$PbO_{ 2 }$ => Reduced substance

PbO2$PbO_{ 2 }$ => Oxidizing agent

Pb => Reducing agent

14. Consider the given reactions:

2S2O23(aq)+I2(s)S4O26(aq)+2I(aq)$2 \; S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; I_{ 2 \; (s) } \; \rightarrow \; S_{ 4 }O^{ 2- }_{ 6 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) }$

S2O23(aq)+2Br2(l)+5H2O(l)2SO24(aq)+4Br(aq)+10H+(aq)$S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; 2 \; Br_{ 2 \; (l) } \; + \; 5 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; Br^{ – }_{ (aq) } \; + \; 10 \; H^{ + }_{ (aq) }$

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

The average oxidation no. of S inS2O23$S_{ 2 }O_{ 3 }^{ 2- }$ is +2.

The average oxidation no. of S inS4O26$S_{ 4 }O_{ 6 }^{ 2- }$ is +2.5.

The oxidation no. of S inS2O23$S_{ 2 }O_{ 3 }^{ 2- }$ is +2.

The oxidation no. of S inSO24$SO_{ 4 }^{ 2- }$ is +6.

As Br2$Br_{ 2 }$ is a stronger oxidizing agent than I2$I_{ 2 }$, it oxidizes S of  S2O23$S_{ 2 }O_{ 3 }^{ 2- }$ to a higher oxidation no. of +6 in SO24$SO_{ 4 }^{ 2- }$.

As I2$I_{ 2 }$ is a weaker oxidizing agent so it oxidizes S of S2O23$S_{ 2 }O_{ 3 }^{ 2- }$ ion to a lower oxidation no. that is 2.5 in S4O26$S_{ 4 }O_{ 6 }^{ 2- }$ ions.

Thus, thiosulphate react differently with I2$I_{ 2 }$ and Br2$Br_{ 2 }$.

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

F2$F_{ 2 }$ can oxidize Cl$Cl^{ – }$ to Cl2$Cl_{ 2 }$, Br$Br^{ – }$ to Br2$Br_{ 2 }$, and I$I^{ – }$ to I2$I_{ 2 }$ as: