NCERT Exemplar for Class 11 Chemistry Chapter-8 (Book Solutions)

Multiple Choice Questions (Type-I)

1. Which of the following is not an example of redox reaction? 

\[(i)CuO + H_2 \rightarrow Cu + H_2O\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{F}{{\mathbf{e}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}\text{ }+\text{ }\mathbf{3CO}\text{ }\to \text{ }\mathbf{2Fe}\text{ }+\text{ }\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}\]

\[(iii) 2K + F_2 \rightarrow 2KF \]

\[(iv) BaCl_{2} + H_{2} SO_{4} \rightarrow BaSO_{4}+2HCl \]

Ans: Redox reaction is defined as the simultaneous oxidation and reduction of reacting species. Thus, change in oxidation state will decide whether a reaction is redox or not. Thus, assigning the oxidation states as:

\[\left( \text{i} \right)\text{ CuO + }{{\text{H}}_{\text{2}}}\to \text{ Cu + }{{\text{H}}_{\text{2}}}\text{O}\]

 Here, oxidation of H and reduction of Cu is taking place.

\[\left( \text{ii} \right)\text{ F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ + 3CO}\to \text{ 2Fe + 3C}{{\text{O}}_{\text{2}}}\]

Here, oxidation of C and reduction of Fe is taking place.

\[\left( \text{iii} \right)\text{ 2K + }{{\text{F}}_{\text{2}}}\to \text{2KF}\]

Here, the oxidation of K and reduction of F is taking place.

\[\left( \text{iv} \right)\text{ BaC}{{\text{l}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{ BaS}{{\text{O}}_{\text{4}}}\text{ + 2HCl}\]

This is not a redox reaction, but it is a double displacement reaction.

Hence, option(iv) is the correct answer.

2. The more positive the value of Ee, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. Ee values: 

Fe3+/ Fe2+ = +0.77; I2 (s) / I- = +0.54; Cu2+ / Cu = +0.34; Ag+ / Ag = +0.80V

(i) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\]

(ii) \[{{\mathbf{I}}_{\mathbf{2}}}\] (s) 

(iii) \[\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}\]

(iv) \[\mathbf{A}{{\mathbf{g}}^{+}}\]

Ans: The strongest oxidising agent means it has greater tendency to oxidise other species and itself gets easily reduced. So higher the \[{{\text{E}}^{\odot }}\]values, stronger is the oxidising agent it is. Thus, Ag+ having highest positive \[{{\text{E}}^{\odot }}\] value among the given systems, is the strongest oxidising agent. Hence, option (iv) is the correct answer.

3. \[{{\mathbf{E}}^{\odot }}\]values of some redox couples are given below. On the basis of these values choose the correct option. \[{{\mathbf{E}}^{\odot }}\] values:

\[\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\text{}/\mathbf{Br}\text{}=\text{}+\text{}\mathbf{1}.\mathbf{90};\text{}\mathbf{A}{{\mathbf{g}}^{\mathbf{+}}}\text{}/\mathbf{Ag}\left(\mathbf{s}\right)\text{}=\text{}+\text{}\mathbf{0}.\mathbf{80}\text{}\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}/\mathbf{Cu}\left(\mathbf{s}\right)\text{}=\text{}+\text{}\mathbf{0}.\mathbf{34};\text{}{{\mathbf{I}}_{\mathbf{2}}}\text{}\left(\mathbf{s}\right)/{{\mathbf{I}}^{\text{ }}}=\text{ }+\text{ }\mathbf{0}.\mathbf{54}\]

(i) Cu will reduce \[\mathbf{Br}\] 

(ii) Cu will reduce Ag 

(iii) Cu will reduce \[{{\mathbf{I}}^{\text{ }}}\]

(iv) Cu will reduce \[\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\].

Ans:  Reduction potential is defined as the tendency of the specie to get reduced. More positive the value of \[{{\text{E}}^{\odot }}\text{ }\], greater is the tendency of the species to get reduced and stronger is the oxidising agent. 

On the basis of the given \[{{\text{E}}^{\odot }}\text{ }\] values, the order of getting reduced is:

\[\text{B}{{\text{r}}_{\text{2}}}\text{A}{{\text{g}}^{\text{+}}}\text{}{{\text{I}}_{^{\text{2}}}}\text{C}{{\text{u}}^{\text{2+}}}\]

Hence, Cu has the least tendency to get reduced and will itself gets oxidise and reduce other species as: \[\text{B}{{\text{r}}_{\text{2}}}\]​, \[\text{A}{{\text{g}}^{\text{+}}}\] and \[{{\text{I}}_{^{\text{2}}}}\]. Hence, option (iv) is the correct answer.

4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible. \[{{\mathbf{E}}^{\odot }}\]values:

\[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}/\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+}}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{77};\text{}{{\mathbf{I}}_{\mathbf{2}}}\text{}/{{\mathbf{I}}^{}}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{54};\text{}\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}/\mathbf{Cu}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{34};\text{}\mathbf{A}{{\mathbf{g}}^{+}}\text{}/\mathbf{Ag}\text{}=\text{}+\text{ }\mathbf{0}.\mathbf{80}\text{ }\mathbf{V}\]

(i) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\] and \[{{\mathbf{I}}^{}}\]

(ii) Ag+ and Cu 

(iii) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\] and Cu

(iv) Ag and \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\].

Ans: \[{{\text{E}}^{\odot }}\] will be negative for the pair Ag and \[\text{F}{{\text{e}}^{\text{3+}}}\]. Hence the reaction is not feasible. Hence, option(iv) is the correct answer.

5. Thiosulphate reacts differently with iodine and bromine in the reactions given below: 

\[\mathbf{2}{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}}+{{\mathbf{I}}_{\mathbf{2}}}\text{ }\to \text{ }{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}\text{ }}}^{\mathbf{2}}+\mathbf{2}{{\mathbf{I}}^{}}\text{ }\]

\[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}\text{ }}}^{\mathbf{2}}+\text{ }\mathbf{2B}{{\mathbf{r}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{5}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\to \text{ }\mathbf{2S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\text{ }+\text{ }\mathbf{2B}{{\mathbf{r}}^{}}+\mathbf{10}\text{ }{{\mathbf{H}}^{+}}\]

Which of the following statements justifies the above dual behaviour of thiosulphate? 

(i) Bromine is a stronger oxidant than iodine. 

(ii) Bromine is a weaker oxidant than iodine. 

(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions. 

(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions. 

Ans: Standard reduction potential of bromine is higher than Iodine, hence Bromine is a stronger oxidant than iodine. Therefore, option (i) is the correct answer.

6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?

(i) The oxidation number of hydrogen is always +1.

(ii) The algebraic sum of all the oxidation numbers in a compound is zero. 

(iii) An element in the free or the uncombined state bears oxidation number zero. 

(iv) In all its compounds, the oxidation number of fluorine is – 1.

Ans: In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion. Hence, option(i) is the correct answer.

7. In which of the following compounds, an element exhibits two different oxidation states. 

\[\left( \mathbf{i} \right)\text{ }\mathbf{N}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{OH}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{N}{{\mathbf{H}}_{\mathbf{4}}}\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}\]

\[\left( \mathbf{iii} \right)\text{ }{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{H}}_{\mathbf{4}}}\]

\[\left( \mathbf{iv} \right)\text{ }{{\mathbf{N}}_{\mathbf{3}}}\mathbf{H}\]

Ans: The oxidation states are given below-

In NH2OH oxidation of N is -1.

NH4NO3 exists as \[\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}\text{.N}{{\text{O}}_{\text{3}}}^{\text{-}}\text{, }\!\!~\!\!\text{ }\]

Thus, the oxidation state of N in \[\text{NH}_{4}^{+}\] is -3 while in \[\text{NO}_{3}^{-}\] is +5.

In N2H4 oxidation of N is -2

In N3H oxidation of N is \[\frac{\text{-1}}{3}\]

Hence, option (ii) is the correct answer.

8. Which of the following arrangements represent increasing oxidation number of the central atom? 

\[\left( \mathbf{i} \right)\mathbf{ Cr}{{\mathbf{O}}_{3}}^{\mathbf{ }}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\]

\[\left( \mathbf{ii} \right)\mathbf{ Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{}}\]

\[\left( \mathbf{iii} \right)\mathbf{ Cr}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{}}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\]

\[\left( \mathbf{iv} \right)\mathbf{ Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{2 }}}^{\mathbf{}}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\]

Ans: Let the oxidation number of central atom be 'y'.

Oxidation number of O = −2

\[\text{CrO}_{2}^{-}\]: y + 2 × (−2) = −1

y − 4 = −1 or y = +3

\[\text{CrO}_{4}^{2-}\] ​: y + 4 × (−2) = −2

y − 8 = −2 or y = +6

\[\text{ClO}_{3}^{-}\]: y + 3 × (−2) = −1

y − 6 = −1 or y = +5

\[\text{MnO}_{4}^{-}\]: y + 4 × (−2) = −1

y − 8 = −1 or y = +7

increasing order of oxidation number of the central atom is:

\[\text{Cr}{{\text{O}}_{\text{2}}}^{\text{--}}\mathbf{ }<\text{Cl}{{\text{O}}_{\text{3}}}^{\text{--}}\mathbf{ }<\text{Cr}{{\text{O}}_{\text{4}}}^{\text{2--}}\text{Mn}{{\text{O}}_{\text{4}}}^{\text{--}}\].

Hence, option (i) is the correct answer.

9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?

\[\left( \mathbf{i} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{1}\text{ }}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{5}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{1}}}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{5}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

Ans:  For \[\text{3}{{\text{d}}^{\text{1 }}}\text{4}{{\text{s}}^{\text{2}}}\] can exhibit highest oxidation state as 2+1 = +3,

For \[\text{3}{{\text{d}}^{\text{3}}}\text{4}{{\text{s}}^{\text{2}}}\]can exhibit highest oxidation state as 2+3 = +5

For \[\text{3}{{\text{d}}^{\text{5}}}\text{4}{{\text{s}}^{\text{1}}}\]can exhibit the highest oxidation state as 1+5 = +6

For \[\text{3}{{\text{d}}^{\text{5}}}\text{4}{{\text{s}}^{\text{2}}}\]can exhibit highest oxidation state as 2+5 = +7.

Hence, option(iv) is the correct answer.

10. Identify disproportionate reaction 

\[\left( \mathbf{i} \right)\text{ }\mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}+\mathbf{2}{{\mathbf{O}}_{\mathbf{2}}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}+\mathbf{4C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }\to \text{ }\mathbf{CC}{{\mathbf{l}}_{\mathbf{4}}}\text{ }+\text{ }\mathbf{4HCl}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{2}{{\mathbf{F}}_{\mathbf{2}}}+\mathbf{2O}{{\mathbf{H}}^{}}\text{ }\to \text{ }\mathbf{2}{{\mathbf{F}}^{}}\text{ }+\text{ }\mathbf{O}{{\mathbf{F}}_{\mathbf{2}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{2N}{{\mathbf{O}}_{\mathbf{2}}}+\mathbf{2O}{{\mathbf{H}}^{}}\text{ }\to \text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{2}\text{ }}}^{\text{ }}+\text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}^{}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-

\[\text{2N}{{\text{O}}_{\text{2}}}\text{+2O}{{\text{H}}^{\text{--}}}\to \text{N}{{\text{O}}_{\text{2 }}}^{\text{-- }}\text{+ N}{{\text{O}}_{\text{3}}}^{\text{--}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in \[\text{NO}_{3}^{-}\]​ to +5 in NO2 ​and decreases from +4 in NO to +3 in \[\text{NO}_{2}^{-}\].

Hence, option(iv) is the correct answer.

11. Which of the following elements does not show disproportionate tendency?

(i) Cl

(ii) Br 

(iii) F 

(iv) I

Ans:  As, Cl, Br, I all are having -1 to +7 oxidation state. But Oxidation state of F is fixed (-1) as it is the most electronegative element and do not loose electron. Hence, it does not show disproportionate tendency. Hence, option(iii) is the correct answer.

Multiple Choice Questions (Type-II)

In the following questions two or more options may be correct.

12. Which of the following statement(s) is/are not true about the following decomposition reaction. 

\[\mathbf{2KCl}{{\mathbf{O}}_{\mathbf{3}}}\to \text{ }\mathbf{2KCl}+\mathbf{3}{{\mathbf{O}}_{\mathbf{2}}}\] 

(i) Potassium is undergoing oxidation

(ii) Chlorine is undergoing oxidation

(iii) Oxygen is reduced 

(iv) None of the species are undergoing oxidation or reduction 

Ans: As, in the given reaction below-

\[\text{2KCl}{{\text{O}}_{\text{3}}}\to \text{2KCl+3}{{\text{O}}_{\text{2}}}\]

Potassium remains in same oxidation state and oxygen is being oxidized. Hence, options(iii) and (iv) are the correct answers.

13. Identify the correct statement (s) in relation to the following reaction: \[\mathbf{Zn}+\mathbf{2HCl}\to \text{ }\mathbf{ZnC}{{\mathbf{l}}_{\mathbf{2}}}+{{\mathbf{H}}_{\mathbf{2}}}\]

(i) Zinc is acting as an oxidant

(ii) Chlorine is acting as a reductant 

(iii) Hydrogen ion is acting as an oxidant 

(iv) Zinc is acting as a reductant 

Ans: The given equation is as-

\[\text{Zn+2HCl}\to \text{ZnC}{{\text{l}}_{\text{2}}}\text{+}{{\text{H}}_{\text{2}}}\]

In this Zn has a negative \[{{\text{E}}^{\odot }}\]value, which means it will undergo oxidation and will act as a reducing agent (reductant). Zn can produce \[{{\text{H}}_{\text{2}}}\] gas with HCl, as hydrogen has higher standard reduction potential than Zn, hydrogen will undergo reduction and will act as oxidant. Hence, option(iii) and (iv) is the correct answer.

14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds. 

\[\left( \mathbf{i} \right)\,\mathbf{3}{{\mathbf{s}}^{\mathbf{1}}}\]

\[\left( \mathbf{ii} \right)\,\mathbf{3}{{\mathbf{d}}^{\mathbf{1}\text{ }}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{iii} \right)\mathbf{3}{{\mathbf{d}}^{\mathbf{2}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{iv} \right)\mathbf{3}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{3}{{\mathbf{p}}^{\mathbf{3}}}\]

Ans:  Elements that are having only s-electrons in the valence shell do not show more than one oxidation state (shows only one oxidation state of +1).  Hence, (b), (c) having incompletely filled d-orbital's in the outermost shell show variable oxidation states.  Element with outer electronic configuration as 3d1 4S2 shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d24S2 shows variable oxidation states of +2, +3 and +4.  P-Block elements also show variable oxidation states due to involvement of d-orbital's and inert pair effect. Hence, element having 3S2 3P3 as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbital's. Hence, option(ii), (iii) and (iv) are the correct answers.

15. Identify the correct statements with reference to the given reaction-

\[{{\mathbf{P}}_{\mathbf{4}}}+\mathbf{3O}{{\mathbf{H}}^{}}+\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\to \text{ }\mathbf{P}{{\mathbf{H}}_{\mathbf{3}}}+\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{P}{{\mathbf{O}}_{\mathbf{2}}}^{}\] 

(i) Phosphorus is undergoing reduction only. 

(ii) Phosphorus is undergoing oxidation only. 

(iii) Phosphorus is undergoing oxidation as well as reduction. 

(iv) Hydrogen is undergoing neither oxidation nor reduction.

Ans:  The given reaction is as below-

\[{{\text{P}}_{\text{4}}}\text{+3O}{{\text{H}}^{\text{--}}}\text{+3}{{\text{H}}_{\text{2}}}\text{O}\to \text{P}{{\text{H}}_{\text{3}}}\text{+3}{{\text{H}}_{\text{2}}}\text{P}{{\text{O}}_{\text{2}}}^{\text{--}}\]

The above reaction is a kind of disproportionate reaction in which phosphorous is being reduced as well as oxidized whereas hydrogen remains same in +1 oxidation state. Hence, option(iii) and (iv) are the correct answers.

16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode? 

\[\left( \mathbf{i} \right)\text{ }\mathbf{Al}/\mathbf{A}{{\mathbf{l}}^{\mathbf{3}+}}~~~~~~~~~~~~~~~~{{\mathbf{E}}^{\odot }}\text{ }=\text{ }\mathbf{1}.\mathbf{66}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{Fe}/\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+}}~~~~~~~~~~~~~~~{{\text{E}}^{\odot }}\text{ }=\text{ }\text{ }\mathbf{0}.\mathbf{44}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{Cu}/\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+~}}~~~~~~~~~~~{{\mathbf{E}}^{\odot }}\text{ }=\text{ }+\text{ }\mathbf{0}.\mathbf{34}\]

\[\left( \mathbf{iv} \right)\text{ }{{\mathbf{F}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)/\mathbf{2}{{\mathbf{F}}^{}}\text{ }\left( \mathbf{aq} \right)~~~~~~~{{\mathbf{E}}^{\odot }}=\text{ }+\text{ }\mathbf{2}.\mathbf{87}\]

Ans: The ones which will act as anodes when connected to standard hydrogen electrode as they have more negative standard reduction potential as compared to standard hydrogen electrode. The one which will act as cathodes when connected to standard hydrogen electrode as they have more positive standard reduction potential as compared to standard hydrogen electrode. Hence, option(i) and (ii) are the correct answers.

Short Answer Type

17. The reaction 

\[\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)+\mathbf{2O}{{\mathbf{H}}^{}}\text{ }\left( \mathbf{aq} \right)\text{ }\to \text{ }\mathbf{Cl}{{\mathbf{O}}^{}}\left( \mathbf{aq} \right)+\text{ }\mathbf{C}{{\mathbf{l}}^{}}\left( \mathbf{aq} \right)+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\left( \mathbf{l} \right)\]

represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action. 

Ans: The given reaction is as below-

\[\text{C}{{\text{l}}_{\text{2}}}\text{ }\left( \text{g} \right)\text{+2O}{{\text{H}}^{\text{--}}}\text{ }\left( \text{aq} \right)\to \text{Cl}{{\text{O}}^{\text{--}}}\left( \text{aq} \right)\text{+ C}{{\text{l}}^{\text{--}}}\left( \text{aq} \right)\text{+}{{\text{H}}_{\text{2}}}\text{O }\left( \text{l} \right)\]

In the given reaction, oxidation number  of Cl increases from 0 (in Cl2) to +1 (in ClO-) and decreases to -1 (in Cl-). Therefore, Cl2 is both oxidized to ClO- and reduced to Cl-. Since Cl- ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), hence, Cl2 bleaches substances due to oxidizing action of hypochlorite, ClO ion.

18. \[Mn{{O}_{4}}^{2-}\] undergoes disproportionate reaction in acidic medium but \[\mathbf{Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}\] does not. Give reason. 

Ans: In \[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{2-}}\], Mn is in highest oxidation state that is +7 hence here manganese cannot undergo oxidation that is why disproportionate is not possible whereas in  \[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{2-}}\] manganese is in +6 oxidation state which can be oxidized as well as reduced.

19. PbO and PbO2 react with HCl according to following chemical equations:

2PbO + 4HCl → 2PbCl2 + 2H2O

PbO2 + 4HCl → PbCl2 + Cl2 +2H2O

Why do these compounds differ in their reactivity? 

Ans:  The given compound can differ in reactivity as-Lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO2 thus, can act as an oxidant (oxidising agent) and, therefore, can oxidise chloride ions of HCl into chlorine.

20. Nitric acid is an oxidising agent and reacts with PbO, but it does not react with PbO2 . Explain why? 

Ans: Since nitric acid itself is an oxidising agent therefore, it is unlikely that the reaction may occur between PbO2 and nitric acid. However, the acid-base reaction occurs between PbO and nitric acid because PbO is a basic oxide.

21. Write balanced chemical equation for the following reactions: 

(i) Permanganate ion (\[\mathbf{MnO}_{4}^{-}\]) reacts with sulphur dioxide gas in acidic medium to produce \[\mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}\]and hydrogen sulphate ion. (Balance by ion electron method). 

Ans: We can balance the given reaction by ion electron method as-

\[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+S}{{\text{O}}_{\text{2}}}\to \text{M}{{\text{n}}^{\text{2+}}}\text{+HS}{{\text{O}}_{\text{4}}}^{\text{-}}\left( \text{acidic medium} \right)\]

Balancing by ion electron method

\[2 \times MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4 H_{2}O \]

\[ 5 \times SO_{2} + 2H_{2}O \rightarrow HSO_{4}^{-} + 3H^{+} + 2e^{-}\]

The balanced chemical is given as-

\[2MnO_{4}^{-} + 16H^{+} + \not {10e^{-}} \rightarrow 2Mn^{2+} + 8H_{2}O \\\underline{5SO_{2}+10H_{2}O \rightarrow 5HSO_{4} ^{-} + 15H^{+} + \not {10e^{-}}}\\ \\\underline{2MnO_4^- + H^{+}+ 5SO_{2} + H_{2}O \rightarrow 2Mn^{2+} + 5HSO_{4}^{-}} \]

(ii) Reaction of liquid hydrazine (N2H2) with chlorate ion (\[ClO_3^{-}\]) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method).

Ans: We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given - 

\[3N_{2}H_{4}+4ClO_{3}^{-}\rightarrow 6NO +4Cl^{-} +6H_{2}O \]

The balanced chemical is given as-

(iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion \[(C{{l}_{2}}{{O}_{7}})\] and oxygen gas. (Balance by ion electron method).

Ans: We can balance the given reaction by ion electron method as

\[Cl_{2}O_{7} (g) + H_{2}O_{2} (aq) \rightarrow ClO_{2}^{-}+O_{2} \] (acidic medium)

Balancing bu ion electron method

2 × { Cl2O7 + 6H+ + 8e- → 2ClO2- + 3H2O

8  × { H2O2 → O2 + 2H+ + 2e- 

The balanced chemical is given as-

2Cl2O7 + 12H+ + 16e- → 4ClO2- + 6H2O

                         8H2O2 → 8O2 + 16H+ + 16e-

          2Cl2O7 + 8H2O2 → 4ClO2- + 6H2O + 8O2 + 4H+

22. Calculate the oxidation number of phosphorus in the following species. (a) \[\mathbf{HP}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2-}}\] and (b) \[\mathbf{P}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{3-}}\]

(a) \[\mathbf{HP}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2-}}\]

Ans:

Let the oxidation number of P in \[\text{HP}{{\text{O}}_{\text{3}}}^{\text{2-}}\] be x. 

Therefore, +1 + x + (-6) = -2 

x = +3

Oxidation number of phosphorus is= +3.

(b) \[\mathbf{P}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{3-}}\]

Ans: 

Let the oxidation number of P in \[\text{P}{{\text{O}}_{\text{4}}}^{\text{3-}}\] be x. 

Therefore, x + (-8) = -3

x = +5 

Oxidation number of phosphorus is= +5.

23. Calculate the oxidation number of each sulphur atom in the following compounds: 

(a) Na2S2O3

Ans: let x= oxidation number of   sulphur, and   +1 is oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 2x - 6 = 0 

x = +2. 

(b) Na2S4O6

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 4x - 12 = 0

x= +2.5 

(c) Na2SO3

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

\[\left( \mathbf{d} \right)\text{ }\mathbf{N}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}\].

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x -8=0

x = +6.

24. Balance the following equations by the oxidation number method.

\[\left( \mathbf{i} \right)\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+}}+{{\mathbf{H}}^{+}}+\mathbf{C}{{\mathbf{r}}_{\mathbf{2}}}^{\mathbf{2}}{{\mathbf{O}}_{\mathbf{7}}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{r}}^{\mathbf{3}+}}+\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\].

Ans: We can balance the given equation by oxidation number method- 

\[\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{^{\overset{\text{-2}}{\mathop{\text{2-}}}\,}}\to \overset{\text{+3}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+F}{{\text{e}}^{\text{3+}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

(a) Balance the increase and decrease in O.N.

\[6\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{^{\overset{\text{-2}}{\mathop{\text{2-}}}\,}}\to \overset{\text{+3}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+6}\overset{3+}{\mathop{\text{F}}}\,{{\text{e}}^{\text{3+}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

(b) Balancing H and O atoms by adding H+ and H2O molecules

\[6\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+14}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{2-}\to 2\overset{{}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+6}\overset{{}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{3+}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{I}{{\text{ }}_{\mathbf{2}\text{ }}}+\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}^{\text{ -}}\to \text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{I}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\].

Ans: We can balance the given equation by oxidation number method- 

Total decrease in O.N. = 1

To equilize O.N. multiply NO3-, by 10

I2 + 10 NO3- → 10NO2 + IO3-

Balancing atoms other than O and H

I2 + 10 NO3- → 10NO2 + 2 IO3-

Balancing O and H

I2 + 10 NO3- + 8H+ → 10NO2 + 2 IO3- + 4H2O

\[\left( \mathbf{iii} \right)\text{ }\mathbf{I}{{\text{ }}_{\mathbf{2}}}+\text{ }{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}}\text{ }\to \text{ }{{\mathbf{I}}^{}}\text{ }+\text{ }{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{\mathbf{2}}\].

 Ans: We can balance the given equation by oxidation number method

\[\overset{0}{\mathop{{{\text{I}}_{\text{2}}}}}\,+\overset{+2}{\mathop{{{S}_{2}}}}\,\overset{-2}{\mathop{O_{3}^{2-}}}\,\to \overset{-1}{\mathop{{{I}^{-}}}}\,+\overset{+2.5}{\mathop{{{S}_{4}}}}\,O_{6}^{2-}\]

Total increase in O.N. = 0.5 × 4 = 2

Total increase in O.N. = 1 × 2 = 2

To equilize O.N. multiply \[{{S}_{2}}O_{3}^{2-}\] and I- by 2.

\[{{I}_{2}}+2{{S}_{2}}O_{3}^{2-}\to 2{{I}^{-}}+{{S}_{4}}O_{6}^{2-}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }{{\mathbf{C}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\text{ }\to \text{ }\mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\].

Ans: We can balance the given equation by oxidation number method-

\[\overset{+4}{\mathop{Mn}}\,{{O}_{2}}+\overset{+3}{\mathop{{{C}_{2}}}}\,O_{4}^{2-}\to \overset{+2}{\mathop{M{{n}^{2+}}+}}\,\overset{+4}{\mathop{C{{O}_{2}}}}\,\]

The balanced chemical reaction is given as-

\[\text{Total increase in O}\text{.N}\text{. = 5 }\!\!\times\!\!\text{ 2= 10}\]

\[\text{To equalize O}\text{.N}\text{. multiply C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ by 2}\text{. }\]

\[\text{Mn}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + }{{\text{C}}_{\text{2}}}{{\text{O}}^{\text{2-}}}_{\text{4}}\text{ }\!\!~\!\!\text{ }\to \text{ M}{{\text{n}}^{\text{2+}}}\text{ }\!\!~\!\!\text{ + 2C}{{\text{O}}_{\text{2}}}\]

\[\text{Balance H and O by adding 2}{{\text{H}}_{\text{2}}}\text{O }\!\!~\!\!\text{ on right side, and 4}{{\text{H}}^{\text{+}}}\text{ }\!\!~\!\!\text{ on left side of equation}\text{.}\]

\[\text{Mn}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + }{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}^{\text{2-}}\text{ }\!\!~\!\!\text{ + 4}{{\text{H}}^{\text{+}}}\text{ }\!\!~\!\!\text{ }\to \text{ M}{{\text{n}}^{\text{2+}}}\text{ }\!\!~\!\!\text{ + 2C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + 2}{{\text{H}}_{\text{2}}}\text{O }\text{.}\]

25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

\[\left( \mathbf{i} \right)\text{ }\mathbf{3HCl}\left( \mathbf{aq} \right)+\mathbf{HN}{{\mathbf{O}}_{\mathbf{3}}}\left( \mathbf{aq} \right)\text{ }\to \text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)+\mathbf{NOCl}\text{ }\left( \mathbf{g} \right)+\mathbf{2}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\left( \mathbf{l}\text{ } \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[3\overset{+1}{\mathop{H}}\,\overset{-1}{\mathop{Cl}}\,(aq)+\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,\overset{-2}{\mathop{{{O}_{3}}}}\,(aq)\to \overset{0}{\mathop{C{{l}_{2}}}}\,+\overset{+3}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,\overset{-1}{\mathop{Cl(g)}}\,+2\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{O(l)}}\,\]

As, chlorine is oxidised in hydrochloric acid (behaving as reducing agent) -as its oxidation number is increases during the reaction from -1 to 0) and nitric acid is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +5 to +3). Hence, it is the redox reaction.

\[\left( \mathbf{ii} \right)\text{ }\mathbf{HgC}{{\mathbf{l}}_{\mathbf{2}}}\left( \mathbf{aq} \right)+\mathbf{2KI}\text{ }\left( \mathbf{aq} \right)\text{ }\to \text{ }\mathbf{Hg}{{\mathbf{I}}_{\mathbf{2}}}\left( \mathbf{s} \right)+\mathbf{2KCl}\left( \mathbf{aq} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{+2}{\mathop{Hg}}\,{{\overset{-1}{\mathop{Cl}}\,}_{2}}(aq)+2\overset{+1}{\mathop{K}}\,\overset{-1}{\mathop{l}}\,(aq)\to \overset{0}{\mathop{C{{l}_{2}}}}\,+\overset{+2}{\mathop{Hg}}\,\overset{-1}{\mathop{{{l}_{2}}}}\,(aq)+\overset{+1}{\mathop{K}}\,\overset{-1}{\mathop{Cl}}\,(aq)\]

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

\[\left( \mathbf{iii} \right)\text{ }\mathbf{F}{{\mathbf{e}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}\left( \mathbf{s} \right)\text{ }+\mathbf{3CO}\left( \mathbf{g} \right)\text{ }\to \mathbf{2Fe}\left( \mathbf{s} \right)+\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{\text{F}{{\text{e}}_{\text{2}}}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{3}}}}}\,\left( \text{s} \right)\text{+3}\overset{\text{+2}}{\mathop{\text{C}}}\,\overset{\text{-2}}{\mathop{\text{O}}}\,\left( \text{g} \right)\text{ }\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\overset{\text{0}}{\mathop{\text{2}}}\,\text{Fe}\left( \text{s} \right)\text{+3}\overset{\text{+4}}{\mathop{\text{C}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{2}}}}}\,\left( \text{g} \right)\]

As, carbon is oxidised in carbon monoxide (behaving as reducing agent) -as its oxidation number is increases during the reaction from +2 to +4) and ferric oxide is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +3to 0). Hence, it is the redox reaction.

\[\left( \mathbf{iv} \right)\text{ }\mathbf{PC}{{\mathbf{l}}_{\mathbf{3}}}\text{ }\left( \mathbf{l} \right)+\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\left( \mathbf{l} \right)\text{ }\to \mathbf{3HCl}\text{ }\left( \mathbf{aq} \right)+{{\mathbf{H}}_{\mathbf{3}}}\mathbf{P}{{\mathbf{O}}_{\mathbf{3}}}\left( \mathbf{aq} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{\text{P}}}\,\overset{\text{-1}}{\mathop{\text{C}{{\text{l}}_{\text{3}}}}}\,\left( \text{l} \right)\text{+3}\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{2}}}}}\,\overset{\text{-2}}{\mathop{\text{O}}}\,\left( \text{l} \right)\to \text{3}\overset{\text{+1}}{\mathop{\text{H}}}\,\overset{\text{-1}}{\mathop{\text{Cl}}}\,\left( \text{aq} \right)\text{+}\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{3}}}}}\,\overset{\text{+3}}{\mathop{\text{P}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{2}}}}}\,\left( \text{aq} \right)\]

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

\[\left( \mathbf{v} \right)\text{ }\mathbf{4N}{{\mathbf{H}}_{\mathbf{3}}}\mathbf{(l)}+\mathbf{3}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\text{ }\to \text{ }\mathbf{2}{{\mathbf{N}}_{\mathbf{2}}}\left( \mathbf{g} \right)+\mathbf{6}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\left( \mathbf{g} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{4N}}\,\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{3}}}}}\,\left( g \right)\text{+3}\overset{0}{\mathop{{{O}_{\text{2}}}}}\,\left( g \right)\to 2\overset{\text{+1}}{\mathop{{{N}_{2}}}}\,\left( g \right)\text{+6}{{\text{H}}_{\text{2}}}\overset{\text{+1}}{\mathop{\text{O}}}\,\overset{\text{-2}}{\mathop{\text{(g)}}}\,\]

As, nitrogen is oxidised in ammonia (behaving as reducing agent) -as its oxidation number is increases during the reaction from -3 to 0) and oxygen is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from 0 to -2). Hence, it is the redox reaction.

26. Balance the following ionic equations-

\[\left( \mathbf{i} \right)\text{ }\mathbf{C}{{\mathbf{r}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{7}}}{{\text{ }}^{\mathbf{2}}}\text{+ }{{\mathbf{H}}^{+}}+\text{ }{{\mathbf{I}}^{}}\to \text{ }\mathbf{C}{{\mathbf{r}}^{\mathbf{3}+}}+{{\mathbf{I}}_{\mathbf{2}}}\text{ }+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{\text{2}-}\text{+6}{{\text{I}}^{\text{-}}}\text{+14}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2C}{{\text{r}}^{\text{+3}}}\text{+3}{{\text{I}}_{\text{2}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{C}{{\mathbf{r}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{7}}}^{\mathbf{2}}+\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+\text{ }}}+{{\mathbf{H}}^{+\text{ }}}\to \text{ }\mathbf{C}{{\mathbf{r}}^{\mathbf{3}+}}+\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{\text{2-}}\text{+6F}{{\text{e}}^{\text{+2}}}\text{+14}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2C}{{\text{r}}^{\text{+3}}}\text{+6F}{{\text{e}}^{\text{+3}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[~\left( \mathbf{iii} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{4}\text{ }}}^{\text{ }}+{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}+{{\mathbf{H}}^{+}}\text{ }\to \text{ }\mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{4}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{2Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+5S}{{\text{O}}_{\text{3}}}^{\text{2-}}\text{+6}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2M}{{\text{n}}^{\text{+2}}}\text{+5S}{{\text{O}}_{\text{4}}}^{\text{2-}}\text{+3}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\text{-}}+{{\mathbf{H}}^{\mathbf{+}}}+\mathbf{B}{{\mathbf{r}}^{\mathbf{}}}\text{ }\to \mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+\text{ }\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{2Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+10B}{{\text{r}}^{\text{-}}}\text{+16}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2M}{{\text{n}}^{\text{+2}}}\text{+5B}{{\text{r}}_{\text{2}}}\text{+8}{{\text{H}}_{\text{2}}}\text{O}\]

Matching Type

27. Match Column I with Column II for the oxidation states of the central atoms. 

Ans: (i) \[\to \] (d); (ii) \[\to \] (e); (iii) \[\to \] (c); (iv) \[\to \] (a)

28. Match the items in Column I with relevant items in Column II. 

Ans:  (i) \[\to \] (e); (ii) \[\to \] (d); (iii) \[\to \] (c); (iv) \[\to \] (b); (v) \[\to \] (f).

Assertion and Reason Type

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

29. Assertion (A): Among halogens fluorine is the best oxidant. 

Reason (R): Fluorine is the most electronegative atom. 

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false. 

(iv) Both A and R are false. 

Ans: Fluorine is most electronegative element that is why it is best oxidant among halogens. Option (i) is the correct answer.

30. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent. 

Reason (R): Oxidation state of manganese changes from +2 to +7 during the reaction. 

(i) Both A and R are true and R is the correct explanation of A. 

(ii) Both A and R are true but R is not the correct explanation of A. 

(iii)A is true but R is false. 

(iv) Both A and R are false.

Ans: As permanganate ion changes to \[\text{Mn}{{\text{O}}_{\text{2}}}\]. Option(iii) is the correct answer.

31. Assertion(A): The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionate reaction. 

Reason (R): The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in O2 and -2 oxidation state in H2O.

(i) Both A and R are true and R is the correct explanation of A. 

(ii) Both A and R are true but R is not the correct explanation of A. 

(iii) A is true but R is false. 

(iv) Both A and R are false. 

Ans: Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O2 undergoing oxidation and decreases to -2 oxidation state in H2O undergoing reduction. Option (i) is the correct answer.

32. Assertion (A): Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half-cell. 

Reason (R): In the representation, \[{{\mathbf{E}}^{\odot }}_{\mathbf{F}{{\mathbf{e}}^{\mathbf{3+}}}\mathbf{/ F}{{\mathbf{e}}^{\mathbf{2+}}}}\] and \[{{\mathbf{E}}^{\odot }}{{\mathbf{ }}_{\mathbf{C}{{\mathbf{u}}^{\mathbf{+2}}}\mathbf{/Cu}}},\,{{\mathbf{E}}^{\odot }}_{\mathbf{F}{{\mathbf{e}}^{\mathbf{3+}}}\mathbf{/ F}{{\mathbf{e}}^{\mathbf{2+}}}}\], Cu2+/Cu  are redox couples. 

(i) Both A and R are true and R is the correct explanation of A. 

(ii) Both A and R are true but R is not the correct explanation of A. 

(iii) A is true but R is false.

(iv) Both A and R are false. 

Ans: A redox couple is defined as pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half reaction. Option(ii) is the correct answer.

Long Answer Type

33. Explain redox reactions on the basis of electron transfer. Give suitable examples. 

Ans: The given reaction is a redox change. 

\[\text{2Na}\left( \text{s} \right)\text{+}{{\text{H}}_{\text{2}}}\left( \text{g} \right)\to \text{2NaH}\left( \text{s} \right)\text{ }\!\!~\!\!\text{ }\]

The half reaction is:

\[\text{2Na}\left( \text{s} \right)\to \text{2N}{{\text{a}}^{\text{+}}}\left( \text{g} \right)\text{+2}{{\text{e}}^{\text{-}}}\]

The other half reaction is:

\[{{\text{H}}_{\text{2}}}\left( \text{g} \right)\text{+2}{{\text{e}}^{\text{-}}}\to \text{2}{{\text{H}}^{\text{-}}}\left( \text{g} \right)\]

This splitting of the reaction into two half-reactions automatically reveals here that sodium is oxidized, and hydrogen is reduced. Any substance which loses electron is oxidized and gains electron is reduced hence is the case of sodium and hydrogen atoms respectively. Hence, the complete reaction is a redox change.

34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for \[{{\mathbf{E}}^{\odot }}\] value). 

\[\left( \mathbf{i} \right)\text{ }\mathbf{Cu}\text{ }+\text{ }\mathbf{Z}{{\mathbf{n}}^{\mathbf{2}+}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}\text{ }+\text{ }\mathbf{Zn}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2C}{{\mathbf{l}}^{}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2B}{{\mathbf{r}}^{}}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2C}{{\mathbf{l}}^{}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2B}{{\mathbf{r}}^{}}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{Fe}\text{ }+\text{ }\mathbf{C}{{\mathbf{d}}^{\mathbf{2}+}}\text{ }\to \text{ }\mathbf{Cd}\text{ }+\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+}}\]

Ans: On the basis of standard reduction potential suggested in the reactivity series (ii) reaction can take place as Mg has more negative value of \[{{\text{E}}^{\odot }}\]cell. Hence, Mg will be oxidized by losing electron and iron will be reduced by gaining electron.

35. Why does fluorine not show disproportionate reaction? 

Ans: Disproportionate is defined as the reaction in which one compound of intermediate oxidation state converts to two compounds, one of higher and one of lower oxidation states So, to occur such type of redox reaction, the element should exist in at least three oxidation states. So that element present in the intermediate state and it can change to both higher and lower oxidation state during disproportionate reaction. Fluorine is the most electronegative element and a strong oxidizing agent and is the smallest in size of all the halogens. It does not show a positive oxidation state (shows only −1 oxidation state) and hence, does not undergo disproportionate reaction.

36. Write redox couples involved in the reactions (i) to (iv) given in question 34.

Ans: The given reactions are as given below-

\[\text{ Cu + Z}{{\text{n}}^{\text{2+}}}\to \text{ C}{{\text{u}}^{\text{2+}}}\text{ + Zn}\]

\[\text{ Mg + F}{{\text{e}}^{\text{2+}}}\to \text{ M}{{\text{g}}^{\text{2+}}}\text{ + Fe}\]

\[\text{ B}{{\text{r}}_{\text{2}}}\text{ + 2C}{{\text{l}}^{\text{--}}}\to \text{ C}{{\text{l}}_{\text{2}}}\text{ + 2B}{{\text{r}}^{\text{--}}}\]

\[\text{ Fe + C}{{\text{d}}^{\text{2+}}}\to \text{ Cd + F}{{\text{e}}^{\text{2+}}}\]

Redox couples are given as- \[\text{C}{{\text{u}}^{\text{2+}}}\text{/Cu }\!\!~\!\!\text{ }\] and \[\text{Z}{{\text{n}}^{\text{2+}}}\text{/Zn}\].

 \[\text{M}{{\text{g}}^{\text{2+}}}\text{/Mg}\] and \[\text{F}{{\text{e}}^{\text{2+}}}\text{/Fe}\].

\[\text{B}{{\text{r}}_{\text{2}}}\text{/B}{{\text{r}}^{\text{-}}}\] and \[\text{C}{{\text{l}}_{\text{2}}}\text{/C}{{\text{l}}^{\text{-}}}\].

\[\text{F}{{\text{e}}^{\text{2+}}}\text{/Fe}\] and \[\text{C}{{\text{d}}^{\text{2+}}}\text{/Cd}\].

37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine. \[\mathbf{NaCl}{{\mathbf{O}}_{\mathbf{4}}},\mathbf{NaCl}{{\mathbf{O}}_{\mathbf{3}}}\text{ },\mathbf{NaClO},\text{ }\mathbf{KCl}{{\mathbf{O}}_{\mathbf{2}}}\text{ },\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{7}}}\text{ },\mathbf{Cl}{{\mathbf{O}}_{\mathbf{3}}}\text{ },\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\mathbf{O},\text{ }\mathbf{NaCl},\text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ },\mathbf{Cl}{{\mathbf{O}}_{\mathbf{2}}}.\] which oxidation state is not present in any of the above compounds?

Ans: We can calculate the oxidation states by - ]

NaClO4 Oxidation no. of chlorine = +7 

Suppose oxidation number of chlorine is x then, 1 + x + 4 × (−2) = 0

∴ x - 7 = 0 

x = +7

We can calculate, the oxidation states, as given below-

NaClO3 Oxidation no. of chlorine = +5

​NaClO Oxidation no. of chlorine = +1

​KClO2 Oxidation no. of chlorine = +3

​Cl2O7 Oxidation no. of chlorine = +7

ClO3 Oxidation no. of chlorine = +6

Cl2O Oxidation no. of chlorine = +1 

NaCl Oxidation no. of chlorine = −1

Cl2 Oxidation no. of chlorine = 0

ClO2 Oxidation no. of chlorine = +4.

Oxidation state (+2) is not present in any of the above compounds.

38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.

Ans: We can Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant.  Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant.  Example Measurement of standard electrode potential of electrode \[E_{\text{Z}{{\text{n}}^{\text{2+}}}\text{/Zn}}^{\odot }\] using SHE as a reference electrode.

Types of Redox Reactions

The different types of Redox Reactions are:

• Decomposition Reaction

• Combination Reaction

• Displacement Reaction

• Disproportionation Reactions

Important Concepts of Class 11 Chemistry Chapter 8 Redox Reactions:

The important concepts in Class 11 Chemistry Chapter 8 are-

• Oxidation And Reduction Reactions

• Redox Reactions of Electron Transfer Reactions

• Competitive Electron Transfer Reactions

• Oxidation Numbers of elements

• Different Types Of Redox Reactions

• Balancing Of different types of Redox Reactions

• Limitations of the concept of Oxidation Numbers

• Redox Reactions and the Electrode Processes.