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NCERT Exemplar · Class 11 · Chemistry

Chapter 8: Redox Reactions

Practice each question, then open the answer or worked solution.

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Multiple Choice Questions (Type-II)

5 questions
Q12Written answer

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 

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Worked solution

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.


Q13Written answer

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 

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Worked solution

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.


Q14Written answer

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. 

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

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

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

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

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Worked solution

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.


Q15Written answer

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.

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Worked solution

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.


Q16Written answer

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

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

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

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

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

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Worked solution

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.