NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions

NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions/

NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions

Multiple Choice Questions
Single Correct Answer Type

Q1. Which of the following is not an example of redox reaction? 
(a) CuO + H₂ → Cu + H₂0                    
(b) Fe₂0₂ + 3CO → 2Fe + 3C0₂
(c).2K + F₂ →2KF                                   
(d) BaCl₂ + H₂S0₄ →BaS0₄ + 2FIC1
Sol: (d) BaCl₂ + H₂S0₄ —> BaS0₄ + 2HC1 is not a redox reaction. It is an example of double displacement reactions.

Q2. The more positive the value of E°, 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 oxidizing agent. E° values: Fe³⁺/Fe²⁺ = +0.77; I₂(g)/I = +0.54;
Cu²⁺/Cu = +0.34; Ag⁺/Ag = +0.80 V
(a) Fe³⁺
(b) I₂(g)                    
(c) Cu²⁺                    
(d) Ag⁺
Sol: (d) Since Ag⁺/Ag has highest positive value of E°, therefore, Ag⁺ is the strongest oxidizing agent with highest tendency to get reduced

Q3. E° values of some redox complexes are given below. On the basis of these values choose the correct option.
E° values: Br₂/Br^– = +1.90; Ag⁺/Ag(s) = +0.80 Cu²⁺/Cu(s) = +0.34; I₂(s)/I^– = +0.54 V
(a) Cu will reduce Br^–
(b) Cu will reduce Ag
(c) Cu will reduce I^–                               
(d) Cu will reduce Br₂

Sol: (d) Copper will reduce Br₂, if the E° of the redox reaction, 2Cu + Br₂
CuBr₂ is +ve.

Since E° of this reaction is +ve, therefore, Cu can reduce Br₂ and hence option (d) is correct.

Q4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
E° values: Fe³⁺/Fe²⁺ = +0.77; I₂/I^– = +0.54;
Cu²⁺/Cu = +0.34; Ag⁺/Ag = +0.80 V
(a) Fe³⁺ and I
(b) Ag⁺ and Cu
(c) Fe³⁺ and Cu
(d) Ag and Fe³⁺
Sol: (d)

Q5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2S₂0₃^2_ + I₂→S₄0₆^2- + 2I^–
S₂0₃^2- + 2Br₂ + 5H₂0 →2S0₄^2- + 4Br^– + 10H⁺
Which of the following statements justifies the above dual behaviour of thiosulphate?
(a) Bromine is a stronger oxidant than iodine.
(b) Bromine is a weaker oxidant than iodine.
(c) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(d) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

Bromine being stronger oxidizing agent than I₂, it oxidises S of S₂O^2-₃ to SO₄^2- whereas I₂ oxidises it only into S₄O₆^2- ion.

Q6. 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?
(a) The oxidation number of hydrogen is always +1.
(b) The algebraic sum of all the oxidation numbers in a compound is zero.
(c) An element in the free or the uncombined state bears oxidation number zero.
(d) In all its compounds, the oxidation number of fluorine is -1.
Sol: (a) Oxidation number of hydrogen is -1 in metal hydrides like NaH.

Q7. In which of the following compounds, an element exhibits two different oxidation states.
(a) NH₂OH
(b) NH₄NO₃
(c) N₂H₄
(d) N₃H

Sol: (b) NH₄NO₃ is an ionic compound consisting of NH₄⁺ and NO₃^– ion.
The oxidation number of N in two species is different as shown below:
In NH;,
Let the oxidation state of N in NH₄⁺ be x.
x + 4x(+l) = +l
x = -3
In NO₃^–
Let the oxidation state of N in NO₃^–  be y,
y + 3 x (-2) = -1
y = +5

Only in arrangement (a), the O.N. of central atom increases from left to right. Therefore, option (a) is correct.

Q9. 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?
(a) 3d¹4s²               
(b) 3d² 4s²               
(c) 3d⁵4s¹                 
(d) 3d⁵4s²
Sol: (d) Highest O.N. of any transition element = (n – 1)d electrons +ns electrons. Therefore, larger the number of electrons in the 3d orbitals, higher is the maximum O.N.
(a) 3d¹4s²= 3; ‘
(b) 3d² 4s² = 3 + 2 = 5;
(c) 3d⁵4s¹=5 + 1=6
(d) 3d⁵4s² = 5+2 = 7

Q10. Identify disproportionation reaction
(a) CH₄ + 20₂ → C0₂ + 2H₂0
(b) CH₄ + 4C1₂ → CC1₄ + 4HCl
(c) 2F₂ + 20H^–→2F^– + OF₂ + H₂0
(d) 2N0₂ + 20H^– → N0₂ + NO^–₃ + H₂0
Sol: (d) Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,

Thus, in reaction (d), N is both oxidized as well as reduced since the O.N. of N increases from +4 in N0₂ to +5 in N0^–₃ and decreases from +4 in N0₂ to +3 in N0^–₂.

Q11. Which of the following elements does not show disproportionation tendency?
(a) Cl
(b) Br  
(c) F  
(d) I
Sol: (c) Being the most electronegative element, F can only be reduced and hence it always shows an oxidation number of-1. Further, due to the absence of d-orbitals, it cannot be oxidized and hence it does not show +ve oxidation numbers. In other words, F cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option (c) is correct.

More than One Correct Answer Type

Q12. Which of the following statement(s) is/are not true about the following decomposition reaction?
2KCIO3 →2KC1 + 30₂
(a) Potassium is undergoing oxidation.
(b) Chlorine is undergoing oxidation.
(c) Oxygen is reduced.
(d) None of the species are undergoing oxidation or reduction.
Sol: (a, b, c, d) Writing the oxidation number of each element above its symbol,

(a) The O.N. of K does not change, K. undergoes neither reduction nor oxidation. Thus, option (a) is not correct.
(b) The O.N. of chlorine decreases from +5 in KCl0₃ to -l in KCl, hence, Cl undergoes reduction.
(c) Since, O.N. of oxygen increases from -2 in KC10₃ to 0 in 0₂, oxygen is oxidized.
(d) This statement is not correct because Cl is undergoing reduction and O is undergoing oxidation.

Q13. Identify the correct statement(s) in relation to the following reaction:
Zn + 2HCl → ZnCl₂ + H₂
(a) Zinc is acting as an oxidant.
(b) Chlorine is acting as a reductant.
(c) Hydrogen ion is acting as an oxidant.
(d) Zinc is acting as a reductant.

(a) The O.N. of Zn increases from 0 to +2 (in ZnCl₂) and therefore, Zn acts as a reductant and not as an oxidant. Hence, option (a) is not correct.
(b) The O.N. of Cl does not change and therefore, it neither acts as a reductant nor an oxidant. Hence, option (b) is not correct.
(c) The O.N. of H decreases from +1 in H⁺ to 0 in H₂. Therefore, H⁺ acts an oxidant. This option is correct.
(d) Zinc acts as reductant because its O.N. changes from 0 to +2. This option is correct.

Q14. 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/their compounds.
(a) 3s¹
(b) 3d^l4s²                 
(c)  3d²4s²
(d) 3s²3p³

Sol:(b, c, d) Elements which have only s-electron in the valence shell do not show more than one oxidation state. Thus, element with 3s¹ as outer electronic configuration shows only one oxidation state of +1.
Transition elements, i.e., elements (b, c) having incompletely filled orbitals in the penultimate shell show variable oxidation states. Thus, 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 a number of reasons such as involvement of J-orbitals and inert pair effect. For example, element (d) with 3s² 3p³ as (i.e., P) as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbitals.

Q15. Identify the correct statements with reference to the given reaction.
P₄ + 30H^– + 3H₂0→ PH₃ + 3H₂ P0^–₂
(a) Phosphorus is undergoing reduction only.
(b) Phosphorus is undergoing oxidation only.
(c) Phosphorus is undergoing oxidation as well as reduction.
(d) Hydrogen is undergoing neither oxidation nor reduction
Sol:

Because O.N. of P increases from 0 (P₄) to +1 (H₂P0^–₂) and decreases from 0 (P₄) to -3 (PH₃), therefore, P has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because O.N. of H remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.

Q16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(a)   A1^3-/A1;  E°= -1.66 V
(b)    Fe²⁺ /Fe;  E°= -0.44 V
(c) Cu²⁺/ Cu E°=34 V
(d) F₂(g)/2F^–(aq) E°= 2.87 V
Sol:(a, b) The electrodes having negative electrode potentials are stronger reducing agents than H2 gas and therefore, will act as anodes.

Short Answer Type Questions
Q17. The reaction Cl₂(g) + 20H^–(aq)→ Cl0^–(aq) + Cl^–(aq) + H₂0(l) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidizing action.
Sol:

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

Q18. Mn0^2-₄ undergoes disproportionation reaction in acidic medium but Mn0^–₄ does not. Give reason.

Q19. PbO and Pb0₂ react with HC1 according to following chemical equations:
2PbO + 4HCl → 2PbCl₂ + 2H₂0
Pb0₂ + 4HC1 → PbCl₂ + Cl₂ + 2H₂0
Why do these compounds differ in their  reactivity?

In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acid-base reaction, because PbO is a basic oxide which reacts with HCl acid.
The reaction (ii) is a redox reaction in which Pb0₂   gets reduced and acts as an oxidizing agent.

Q20. Nitric acid is an oxidizing agent and reacts with PbO but it does not react with Pb0₂. Explain why?
Sol: Nitric acid is an oxidizing agent and reacts with PbO to give a simple acid- base reaction without any change in oxidation state. In Pb0₂, Pb is in +4 oxidation state and cannot be oxidized further, hence no reaction takes place between Pb0₂ and HN0₃.

Q21. Write balanced chemical equations for the following reactions:
(i) Permanganate ion (Mn0^–₄) reacts with sulphur dioxide gas in acidic medium to produce Mn²⁺ and hydrogensulphate ion. (Balance by ion- electron method)
(ii) Reaction of liquid hydrazine (N₂H₄) with chlorate ion (C10^–₃) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)
(iii) Dichlorine heptaoxide (C1₂0₇) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (C10^–₂) and oxygen gas. (Balance by ion-electron method)
Sol: (i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺→5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

Q22. Calculate the oxidation number of phosphorus in the following species.
(a) HPO₃^2- and (b) P0₄^3-
Sol: (a) Let the oxidation number of P inHPO₃^2- be x. So,
+1 + x + (—6) = —2 x = +3
(b) Let the oxidation number of P in P0³₄^– be x. So, x + (-8) = -3
x = +5

Q23. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) Na₂S₂0₃               
(b) Na₂S₄0₆             
(c) Na₂S0₃
(d) Na₂S0₄
Sol: (a) Na₂S₂0₃

Thus, the oxidation states of two S atoms in Na₂S₂0₃ are -2 and +6.
(b) Na₂S₄0₆

Q24. Balance the following equations by the oxidation number method.
(i) Fe²⁺ + H⁺ + Cr₂0₇^2- →Cr³⁺ + Fe³⁺ + H₂0
(ii) I₂ + N0^–₃→ N0₂ +I0₃
(iii) I₂ + S₂0₃^2- →I^– + S₄0₆^2- ‘
(iv) MnO, + C₂0₄^2-→ Mn²⁺ + CO₂

Total increase in O.N. = 5×2= 10
Total decrease in O.N. = 1
To equalize O.N. multiply NO₃^–, by 10
I₂ + 10no₃^–→ 10no₂ + IO₃^–
Balancing atoms other than O and H
I₂ + 10no₃^–→ 10NO₂ + 2 IO₃^–
Balancing O and H
I₂ + lO no₃^– + 8H⁺→ 10NO₂ + 2 IO₃^– + 4H₂0

To equalize O.N. multiply C0₂ by 2. –
Mn0₂ + C₂0^2-₄ →Mn²⁺ + 2CO₂
Balance H and O by adding 2H₂0 on right side, and 4H⁺ on left side of equation.
MnO₂ + C₂0^2-₄ + 4H⁺ →Mn²⁺ + 2CO₂ + 2H₂0

Q25. Identify the redox reactions out of the following reactions and identify the oxidizing and reducing agents in them

Sol: (i) Writing the O.N. of on each atom,

Here, O.N. of Cl increases from -1 (in HCl) to 0 (in Cl₂). Therefore, Cl^– is oxidized and hence HCl acts as a reducing agent.
The O.N. of N decreases from +5 (in HN0₃) to +3 (in NOC1) and therefore, HN0₃ acts as an oxidizing agent.
Thus, reaction (i) is a redox reaction.

Here O.N. of none of the atoms undergo a change and therefore, this is not a redox reaction.

Here, O.N. of Fe decreases from +3 (in Fe₂0₃) to 0 (in Fe) and therefore, Fe₂0₃ acts as an oxidizing agent.
O.N. of C increases from +2 (in CO) to +4 (in C0₂) and therefore, CO acts as a reducing agent. Thus, this is a redox reaction.

Here O.N. of none of the atoms undergo a change and therefore, it is not a redox reaction.

Here. O.N. of N increases from -3 (in NH₃) to 0 in (N₂) and therefore, NH₃ acts as a reducing agent. O.N. of O decreases from 0 (in 0₂) to -2 (in H,0) and therefore, 0₂ acts as an oxidizing agent.
Thus, reaction (v) is a redox reaction.

Dividing the equation into two half reactions:
Oxidation half reaction: I^–→ I₂
Reduction half reaction: Cr₂0₇^2-→ Cr³⁺
Balancing oxidation and reduction half reactions separately as:

(ii) Step-1: Separate the equation into two half reactions.
The oxidation number of various atoms are shown below:

(i) Balance all atoms other than H and O. This step is not needed, because, it is already balanced.
(ii) The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as:
Fe²⁺→Fe³⁺ + e^–
(iii) Charge is already balanced.
(iv) No need to add H or O.
The balanced half equation is:
Fe²⁺→ Fe³⁺ + e^–         …(i)
Consider the second half equation
Cr₂0₇^2-→ Cr³⁺

(i) Balance the atoms other than H and O.

Cr₂0₇^2-→2Cr³⁺
(ii) The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there are two Cr atoms, add 6e^– on the left.
Cr₂0₇^2- + 6e ^–→ 2Cr³⁺
(iii) Since the reaction takes place in acidic medium add 14H⁺ on the left to equate the net charge on both sides.
Cr₂0₇^2-+6e^–+14H⁺→2Cr³⁺

(iv) To balance FI atoms, add 7H₂O molecules on the right.

Cr₂0₇^2- + 6e^– + 14H⁺→ 2Cr³⁺ + 7H₂0          .. .(ii)

This is the balanced half equation.

Step-3: Now add up the two half equations. Multiply eq. (i) by 6 so that electrons are balanced.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions help you. If you have any query regarding NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions, drop a comment below and we will get back to you at the earliest.