NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Edited By Sumit Saini | Updated on Sep 10, 2022 05:28 PM IST

NCERT exemplar Class 11 Chemistry solutions chapter 8 is the quintessential workbook for students who have an inclination towards expanding their practical understanding regarding a theoretical concept. NCERT exemplar Class 11 Chemistry chapter 8 solutions commit towards focussing upon the practical implication of the principles and rules of a topic by forming them in challenging question format for the interested students. This chapter of NCERT Class 11 Chemistry Solutions introduces the students to redox reactions and guides them around their different types and how to deal with them by understanding their specific rules and properties.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: MCQ (Type 1)

Question:1

Which of the following is not an example of redox reaction?
(i)CuO + H_{2}\rightarrow Cu + H_2O
(ii) Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_{2}
(iii) 2K + F_{2} \rightarrow 2KF
(iv) BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + 2HCl

Answer:

The answer is the option (iv) BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + 2HCl
Explanation: The correct answer is the option (iv) because only its oxidation no. changes while that of all elements remains unchanged.

Question:3

E^{}\Theta values of some redox couples are given below. On the basis of these values choose the correct option.
E^{}\Theta values: Br_{2}/Br^{-}=+1.90; Ag^{+}/Ag(s)=+0.80
Cu^{2+}/Cu(s)=+0.34; I_{2}(s)/I^{-}=+0.54

(i) Cu will reduce Br^{-}
(ii) Cu will reduce Ag
(iii) Cu will reduce I^{-}
(iv) Cu will reduce Br_{2}
Answer:

The answer is the option (iv) Cu will reduce Br2
Explanation: Compared to Cu, Br2 is a better reducing agent.

Question:4

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
E^{}\Theta values: Fe^{3+}/Fe^{2+} =0.77;I_{2}/I^{-}=+0.54;
Cu^{2+}/Cu=0.34;Ag^{+}/Ag=+0.80V


(i) Fe^{3+} and \;I^{-}
(ii) Ag^{+} and \;Cu
(iii) Fe^{3+} and \;Cu
(iv) Ag \;and \;Fe^{3+}
Answer:

The answer is the option (iv) Ag \;and \;Fe^{3+}
Explanation: To make the reaction feasible, E^{0}cell need to be positive for the pair Ag and Fe^{3+}, but it is negative. Hence the reaction is not feasible.

Question:5

Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2S_{2}O^{2-}_{3}+I_{2}\rightarrow S_{4}O_{6}^{2-} +2I^{-}
S_{2}O^{2-}_{3}+2Br_{2} +5H_{2}O\rightarrow 2SO_{4}^{2-} +2Br^{-}+10H^{+}
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.
Answer:

Ans. The answer is the option (i) Bromine is a stronger oxidant than iodine.
Explanation: To be a better oxidant, the standard reduction potential of the particular element should be higher than the other, but, here, the standard reduction potential of bromine is higher than iodine. Hence, bromine is a stronger oxidant than iodin

Question: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.
Answer:

The answer is the option (i) The oxidation no. of hydrogen is always +1.
Explanation: In ionic hydrides, hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.

Question:7

In which of the following compounds, an element exhibits two different oxidation states.
(i) NH_{2}OH
(ii) NH_{4}NO_{3}
(iii) N_{2}H_{4}
(iv) N_{3}H

Answer:

The answer is the option (ii) NH_{4}NO_{3}
Explanation: Here, Nitrogen exists in two different oxidation states, i.e., +5 and -3.

Question:8

Which of the following arrangements represent increasing oxidation number of the central atom?
(i) CrO_{2}^{-}, ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-}
(ii) ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-},CrO_{2}^{-}
(iii) CrO_{2}^{-}, ClO_{3}^{-},MnO_{4}^{-},CrO_{4}^{2-}
(iv) CrO_{4}^{2-}, MnO_{4}^{-},CrO_{2}^{-}, ClO_{3}^{-}

Answer:

The answer is the option (i) CrO_{2}^{-}, ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-}
Explanation: the above arrangement represents increasing oxidation no. of the central atom because here the oxidation no. of the central element increases as +3, +5, +6, +7, respectively.

Question: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?
(i) 3d^1 4s^2
(ii) 3d^3 4s^2
(iii) 3d^5 4s^1
(iv) 3d^5 4s^2
Answer:

The answer is the option (iv) 3d^5 4s^2
Explanation: the above arrangement will exhibit the largest oxidation number since total no. of electrons present in d and s subshell = 7.

Question:10

Identify disproportionation reaction
(i) CH_4 + 2O_{2} \rightarrow CO_{2} + 2H_{2}O
(ii) CH_{4} + 4Cl_{2}\rightarrow CCl_{4} + 4HCl
(iii) 2F_{2} + 2OH^{-}\rightarrow 2F^{-} + OF_{2} + H_{2}O
(iv) 2NO_{2} + 2OH^{-} \rightarrow NO_{2}^{-} + NO_{3}^{-} + H_2O
Answer:

The answer is the option (iv) 2NO_{2} + 2OH^{-} \rightarrow NO_{2}^{-} + NO_{3}^{-} + H_2O
Explanation: Oxidation no. of nitrogen decreases by 1 from NO_{2} to NO_{2}^{-} and increase by +1 from NO_{2} to NO_{3}^{-}

Question:11

Which of the following elements does not show disproportionation tendency?
(i) Cl
(ii) Br
(iii) F
(iv) I
Answer:

The answer is the option (iii) F
Explanation: Fluorine is the most electronegative element; therefore, it does not show a disproportionation tendency.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: MCQ (Type 2)

Question:12

Which of the following statement(s) is/are not true about the following decomposition reaction.
2KClO_{3} \rightarrow 2KCl + 3O_{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
Answer:

The answer is the option (i) and (iv)
Explanation: The statements above are not true because it is evident in the reaction that, potassium remains in the same oxidation state, and oxygen is being oxidised

Question:13

Identify the correct statement (s) in relation to the following reaction:
Zn + 2HCl \rightarrow ZnCl_{2} + H_{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
Answer:

The answer is the option (iii) and (iv)
Explanation: The above statement is correct because you according to the reaction zinc is being oxidised and hydrogen is reduced.

Question: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.
(i) 3s^{1}
(ii) 3d^{1}4s^{2}
(iii) 3d^{2}4s^{2}
(iv) 3s^{2} 3p^{3}
Answer:

The answer is the option (iii) and (iv)
Explanation: Atoms having electronic configurations such as in Option (iii) and (iv) will exhibit more than one oxidation state in the compound because in option (iii) electron can be removed from 4s as well as 3d. Similarly, in option (iv) electron can be removed from 3p and 3s both.

Question:15

Identify the correct statements with reference to the given reaction
P_{4} + 3OH^{-} + 3H_{2}O \rightarrow PH_{3} + 3H_{2}PO_{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.
Answer:

The answer is the option (iii) and (iv)
Explanation: This is a kind of disproportionation reaction in which phosphorous is being reduced as well as oxidised, as given in opt (iii); whereas hydrogen remains same in +1 oxidation state as in opt (iv). Therefore, option (iii) and (iv) are correct.

Question:16

Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(i)Al/Al^{3+} , E^{\Theta }=-1.66
(ii)Fe/Fe^{2+} , E^{\Theta }=-0.44
(iii)Cu/Cu^{2+} , E^{\Theta }=+0.34
(iv)F_{2}(g)/2F^{-}(aq) , E^{\Theta }=+2.87

Answer:

The answer is the option (i) and (ii)
Explanation: They will act as anodes because both of the electrodes has a negative value of standard reduction potential

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: Short Answer Type

Question:17

The reaction
Cl_{2} (g) + 2OH^{-} (aq) \rightarrow ClO^{-} (aq) + Cl^{-} (aq) + H_{2}O(l)
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

Answer:

The hypochlorite ion (ClO^{-}) formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds; therefore, in the above reaction, the hypochlorite ion acts as a bleach

Question:18

MnO_{4}^{2-}undergoes disproportionation reaction in acidic medium but MnO_{4}^{-} does not. Give reason.
Answer:

In MnO_{4}^{-}, Mn has the highest oxidation state (+7); therefore, it cannot undergo oxidation which is required for disproportionation here. Hence MnO_{4}^{-} doesn’t undergo disproportionation reaction. Whereas in MnO_{4}^{2-} manganese is in +6 (i.e., less than that of in MnO_{4}^{-}) oxidation state which can be oxidised as well as reduced, hence it undergoes disproportionation.

Question:19

PbO and PbO2 react with HCl according to following chemical equations :
2PbO + 4HCl \rightarrow 2PbCl_{2} + 2H_{2}O
PbO_{2} + 4HCl \rightarrow PbCl_{2} + Cl_{2} + 2H_{2}O
Why do these compounds differ in their reactivity?

Answer:

Lead is present in +4 oxidation state and the stable oxidation state of lead in PbO is +2. PbO_{2} thus it can act as an oxidising agent. Therefore, it can oxidise Cl^{-} ions of HCl into Chlorine.

Question:20

Nitric acid is an oxidising agent and reacts with PbO but it does not react with PbO_{2}. Explain why?
Answer:

We know that HNO3 is and oxidising agent and PbO_2 is a basic oxide. Therefore, it is unlikely that any reaction will occur between them, however, the acid base reaction occurs between these two compounds, as shown below:
2PbO + 4HNO_{3} \rightarrow 2Pb(NO_{3})_{2} + 2H_{2}O

Question:22

Calculate the oxidation number of phosphorus in the following species.
(a) HPO_{3}^{2-} and
(b) PO_{4}^{3-}

Answer:

(a) HPO_{3}^{2-}

Let us consider the oxidation number of phosphorus to be x.

+1 + x + (-2) \times 3 = -2
+1 + x - 6 = -2
X -5 = -2
X = -2 +5
X = +3
Thus, O.S. of phosphorous is +3.

(b) PO_{4}^{3-}
x + (-2) \times 4 = -3
x-8 = -3
x = -3 +8
x = +5
Thus, O.S. of phosphorous in this ion is +5.

Question:23

Calculate the oxidation number of each sulphur atom in the following compounds:
(a) Na_{2}S_{2}O_{3}
(b) Na_{2}S_{4}O_{6}
(c) Na_{2}SO_{3}
(c) Na_{2}SO_{4}

Answer:

(a) Consider oxidation number of sulphur atom x.

2 \times (+1) + 2x + (-2) \times 3 = 0
2 + 2x -6 = 0
2x -4 = 0
2x = 4
x = +2

(b) Na_{2}S_{4}O_{6}
2 \times (+1) + 2x + (-2) \times 6 = 0
2 + 2x -12 = 0
2x -10 = 0
X = +5

(c) Na_{2}SO_{3}
2 \times (+1) + x + (-2) X = 0
2 + x - 6 = 0
x = +4

(d) Na_{2}SO_{4}
2 \times (+1) + x + (-2) \times 4 = 0
2 + x - 8 = 0
x = +6

Question:24

Balance the following equations by the oxidation number method.
(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O
(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}
(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}
(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}

Answer:

(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O
2+ 6+ 2- +3 +3
Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O
(a) Balance the inceases and decreases in O.N.
2+ 6+ 2- +3 +3
6Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+H_{2}O
(b) Balancing H and O atoms by adding H^{+} and H_{2}O molecules
6Fe^{2+}+14H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_{2}O
(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}

I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}

O.N. increases by +5
Now, we know that the total increase in O.N. = 5\times2= 10 and the reduction in O.N. = 1
In order to equalize O.N., we will multiply NO_{3}^{-} with 10, which will give us,
I_{2} + 10NO_{3} \rightarrow 10NO_{2} + IO_{3}^{-}
Now, we will be balancing the atoms other than O and H, to give us
I_{2} + 10NO_{3} \rightarrow 10NO_{2} + 2IO_{3}^{-}
Now, on balancing O and H we will get,
I_{2} + 10NO_{3} + 8H^{+}\rightarrow 10NO_{2} + 2IO_{3}^{-}+4H_{2}O
(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}
0 +2 -1 +2.5
I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}
Total increase in O.N. =0.5 \times 4 = 2
Total decrease in O.N. =1 \times 2 = 2
To equalise O.N. multiply S_{2}O_{3}^{2-} and I^{-}by 2.
I_{2}+S_{2}O_{3}^{2-}\rightarrow 2I^{-}+S_{4}O_{6}^{2-}
(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}
+4 +3 +2 +4
MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}
Total increase in O.N. =1 \times 2 = 2
Total decrease in O.N. =2
Now, in order to equalize O.N., we will multiply CO_{2} with 2 to give us,
MnO_2 + C_{2}O^{2-}_{4} \rightarrow Mn^{2+} + 2CO_{2}
Now, we will balance H and O by adding 2H2O on the right side, and 4H+ on the left side of equation.
MnO_2 + C_{2}O^{2-}_{4} +4H^{+}\rightarrow Mn^{2+} + 2CO_{2}+2H_{2}O

Question:25

Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
(i)3HCl(aq) +HNO_{3} (aq)\rightarrow Cl_{2}(g)+NOCl(g)+2H_{2}O(l)
(ii)HgCl_{2}(aq) +2KI (aq)\rightarrow HgI_{2}(s)+2KCl(aq)
(iii)Fe_{2}O_{3}(s)+3CO(g)\overset{\Delta }{\rightarrow} 2Fe(s)+3CO_{2}(g)
(iv)PCl_{3}(l)+3H_{2}O(l)\rightarrow 3HCl(aq)+H_{3}PO_{3}(aq)
(v)4NH_{3}+3O_{2}(g)\rightarrow 2N_{2}(g)+6H_{2}O(g)

Answer:

HCl(aq) +HNO_{3} (aq)\rightarrow Cl_{2}(g)+NOCl(g)+2H_{2}O(l)

In the above reaction, it can be seen that Cl oxidizes from -1 to 0 and Nitrogen reduces from +5 to +3. Thus, Cl oxidizes to act as a reducing agent and the nitric acid works as the oxidizing agent.

Therefore, in the above reaction, the reducing agent is HCl and the oxidizing agent is HNO_{3}
(ii)HgCl_{2}(aq) +2KI (aq)\rightarrow HgI_{2}(s)+2KCl(aq)
The above mentioned reaction is the perfect example of a displacement reaction. As a result, none of the components undergo the process of reduction or oxidation. This is because the oxidation states of Hg, Cl, K, & I are unchanged in both the products and reactants.
(iii)Fe_{2}O_{3}(s)+3CO(g)\overset{\Delta }{\rightarrow} 2Fe(s)+3CO_{2}(g)


In the above reaction, the oxidising agent is Fe_{2}O_{3}and the reducing agent is CO.
(iv)PCl_{3}(l)+3H_{2}O(l)\rightarrow 3HCl(aq)+H_{3}PO_{3}(aq)

In the above reaction, oxidising agent is O_2and the reducing agent is NH_3.

Question:26

Balance the following ionic equations

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O
(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O

(iii) MnO^{-}_{4} + SO^{2-}_{3} + H^{+}\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O
(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O


Answer:

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O
As step (i), we will be generating the unbalanced skeleton i.e.,
(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Oxidation is 2I^{-}\rightarrow I_{2}+2e^{-}and Reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

Oxidation is 2I^{-}\rightarrow I_{2}+2e^{-} and Reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

The oxidation is 2I^{-}\rightarrow I_{2}+2e^{-} and reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Hence, oxidation is 2I^{-}\rightarrow I_{2}+2e^{-}and

reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction. We will now multiply the oxidation reaction by 3, which will give us,

The oxidation as 6I^{-}\rightarrow 3I_{2}+6e^{-} and
the reduction as Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

6I^{-}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}+6e^{-}

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
6I^{-}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}

Now, we know that the charge on the LHS is +6 which is same as the RHS of the equation, thereby enduring that the reaction is balanced.

(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O

As step (i), we will be generating the unbalanced skeleton i.e.,

Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O


As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Thus, oxidation is Fe^{2+}\rightarrow Fe^{3+}+e^{-} and the reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

Thus, oxidation is Fe^{2+}\rightarrow Fe^{3+}+e^{-} and the reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

Thus, oxidation Fe^{2+}\rightarrow Fe^{3+}+e^{-}is and
reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, oxidation isFe^{2+}\rightarrow Fe^{3+}+e^{-} and
reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is6Fe^{2+}\rightarrow 6Fe^{3+}+6e^{-} and
reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

6Fe^{2+}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 6Fe^{3+}+6e^{-}+2Cr^{3+}+7H_{2}O
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_{2}O
We will now verify whether all the charges are balanced.

So, the LHS = 6\times +2 + 1\times -2 + 14 = 24 and the RHS = 6\times +3 + 2\times +3 + 7\times 0 = 24

Thus the sum total is same on both the sides, therefore the solve reaction is right.


(iii) MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O

As step (i), we will be generating the unbalanced skeleton i.e.,

MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O

Here in Mn undergoes reduction and S undergoes oxidation.

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Thus, oxidation is SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.

Thus, oxidation is SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-} and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

Thus, oxidation is SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+} and
reduction is MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}


As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, oxidation is SO^{2-}_{3}+H_{2}O \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+}and
reduction is MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O


Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is 5SO^{2-}_{3}+5H_{2}O \rightarrow 5SO^{2-}_{4}+10e^{-}+10H^{+}and
reduction is 2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O


Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

2MnO_{4}^{-}+10e^{-}+16H^{+}+5SO_{3}^{2-} +5H_{2}O\rightarrow 2Mn^{2+}+8H_{2}O+5SO_{4}^{2-}+10e^{-}+10H^{+}

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

2MnO_{4}^{-}+6H^{+}+5SO_{3}^{2-} \rightarrow 2Mn^{2+}+3H_{2}O+5SO_{4}^{2-}

We will now verify whether all the charges are balanced.

So, LHS = 2\times-1+6+5\times-2= -6and the RHS = 2\times+2+3\times 0+5\times-2 = -6
Equal charges on both sides imply towards a balanced equation.


(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn_{2+} + Br_{2} + H_{2}O

As step (i), we will be generating the unbalanced skeleton i.e.,


(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.


Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-} and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.
As we can see that the atoms are balanced, therefore the reaction is going to be the same as the previous step.

Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-} and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.


Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-}and
reduction is MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}


As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, the oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-} and
reduction is MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is 10Br^{-}\rightarrow 5Br_{2}+10e^{-}and
reduction is 2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O


Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

2MnO_{4}^{-}+5Br^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}+10e^{+}

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

2MnO_{4}^{-}+10Br^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}

Thus, the charge on LHS and RHS is equal.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: Matching Type

Question:27

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

COLUMN I

COLUMN II

(i) Cr_{2}O_{7}^{2-}

(a)+3

(ii) MnO_{4}^{-}

(b)+4

(iii)VO_{3}^{-}

(c)+5

(iv)FeF_{6}^{3-}

(d)+6


(e)+7

Answer:

(i) → (d); (ii) → (e); (iii) → (c); (iv) → (a)

Question:28

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

COLUMN I

COLUMN II

(i) Ions having positive charge

(a) +7

(ii) The sum of oxidation number of all atoms in a netural molecules

(b) -1

(iii) Oxidation number of hydrogen ion (H^{+})

(c)+1

(iv) Oxidation number of fluorine in NaF

(d) 0

(v) Ions having negative charge

(e)Cation


(f) Anion

Answer:

(i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: Assertion and Reason Type

Question:29

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.
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.
Answer:

The answer is the option (ii) Both A and R are true but R is not the correct explanation of A.
Explanation: Fluorine is considered as the best oxidant among halogens because it is the most electronegative element. There is no other element that can oxidize fluorine. Its electronegativity is 3.98, which is the highest in the periodic table. This results in good oxidant properties.

Question:30

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.
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.
Answer:

The answer is the option (iii) A is true, but R is false.Explanation: Due to the change of permanganate ion changes to MnO_{2}

As per this equation, the maximum oxidation state that Mn is capable of reaching as per its electronic configuration is +7. Therefore, being an oxidising agent, it cannot undergo oxidation and hence undergoes reduction. On the other hand, Iodine undergoes oxidation from -1 to 0.

Question:31

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.
Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O_{2} and –2 oxidation state in H_{2}O.

(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.
Answer:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: 2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2�}(g)
Here the oxygen of peroxide is present in -1 state, and it is converted to zero oxidation state in O_2 undergoing oxidation and decreases to -2 oxidation state in H_{2}Oundergoing reduction.

Question:32

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.
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 E^{\Theta }_{Fe^{3+}/Fe^{2+}} and E^{\Theta }_{Cu^{2+}/Cu}, Fe^{3+}/Fe^{2+} and Cu^{2+}/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.
Answer:

The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: A redox couple is defined as a pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half-reaction.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: Long Answer Type

Question:33

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

When a redox reaction takes place, the species which loses the electrons is regarded as undergoing the oxidation reaction. It therefore acts as an oxidizing agent or oxidant. For the species that accepts electrons is regarded as undergoing reduction and therefore it behaves as a reductant.
The above mentioned transfer is largely based on the relative electronegativity of difference between two interacting species. The higher electronegative element attracts more electrons while on the other hand the electropositive element loses electrons.
Splitting of the reaction into two halves here proves that sodium is oxidised, and hydrogen is reduced. Here, the sodium atom is oxidised, and the hydrogen atom is reduced; therefore, it is a redox reaction.

2Na(s) + H_{2}(g) \rightarrow 2NaH(s) is a redox change.
2Na(s) \rightarrow 2Na+(g) + 2e^{-} and the other half-reaction is:
H_{2}(g) + 2e^{-} \rightarrow 2H^{-}(g)

Question:34

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

(i) Cu + Zn^{2+}\rightarrow Cu^{2+} + Zn
(ii) Mg + Fe^{2+}\rightarrow Mg^{2+} + Fe
(iii) Br_{2} + 2Cl^{-}\rightarrow Cl_{2} + 2Br^{-}
(iv) Fe + Cd^{2+}\rightarrow Cd + Fe^{2+}
Answer:

If a reaction might take place or not is dependent on the net cell EMF of the cell. The equation for the same is
E^{^{\circ}}_{cell}=E^{^{\circ}}_{cathode}-E^{^{\circ}}_{anode}
In option (ii) it is clearly visible that, the reaction can take place as Mg has a more negative value of E^{^{\circ}}_{cell}. Hence, Mg is oxidised by losing electron and iron is reduced by gaining an electron.
Mg + Fe^{2+}\rightarrow Mg^{2+} + Fe
We can say that Fe undergoes reduction and Mg undergoes oxidation,

E^{^{\circ}}_{cathode}=-0.44V
E^{^{\circ}}_{anode}=-2.36V
E^{^{\circ}}_{cell}=-0.44- \left (-2.36 \right )V
E^{^{\circ}}_{cell}=+1.92V

Question:35

Why does fluorine not show disproportionation reaction?
Answer:

We already know that; fluorine has the highest electronegativity in the entire periodic table with EN value of 3.98. Therefore, it is the most electronegative element in its period as well as in the group. It shows only -1 oxidation state and is smallest in size of all the halogens. We know that fluorine has the greatest reduction potential (E^{^{\circ}}cell =2.87)therefore, it is not able to undergo oxidation by itself. Hence, fluorine shows a disproportionation reaction.

Question:37

Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
NaClO_{4}, NaClO_{3}, NaClO, KClO_{2}, Cl_{2}O_{7}, ClO_{3}, Cl_{2}O, NaCl, Cl_{2}, ClO_{2}.
Which oxidation state is not present in any of the above compounds?

Answer:

For all the chlorine atoms, we will assume the oxidation state to be x, and for atoms Na, K = +1, O = -2. Now, we know that, the sum of oxidation numbers of all the atoms in a compound is the same as the charge on that compound for an ion or zero.

Hence,

NaClO_{4}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +7
NaClO_{3}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +5


In the Cl_{2}O_{7} structure there are two atoms of chlorine with varying types of O ? O, Cl ? Cl. The structure is explained as follows: -


Cl_{2}O_{7}\rightarrow 2\times x+7\times -2 = 0\rightarrow x = +7
ClO_{3}\rightarrow x+2\times -2 = 0\rightarrow x = +6
Cl_{2}O\rightarrow 2 \times x -2 = 0\rightarrow x = +1
NaCl\rightarrow +1+ x = 0\rightarrow x = -1
Cl_{2}\rightarrow As chlorine is in its naturalthe oxidation number is zero.
ClO_{2}\rightarrow x +2\times -2= 0\rightarrow x = +4
Therefore the ascending order is,
NaCl (-1), Cl_{2}(0), Cl_{2}O(+1), KClO_{2}(+3), ClO_{2}(+4), NaClO_{3}(+5), ClO_{3}(+6), Cl_{2}O_{7}=NaClO_{4}(+7).

Question:38

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

In order to determine the strength of a reductant/oxidant, we can use the titration method; it can be performed with the help of a redox-sensitive indicator.
The usage indicators in redox titration are as shown below:

In one scenario, the reagent itself is intensely coloured, e.g., permanganate ion, MnO_{4}^{-}. In here, MnO_{4}^{-} acts as a self-indicator. In this case, the visible endpoint is attained after the last of the reductant(Fe^{2+} or C_{2}O_{4}^{2-})is oxidised. Now, the first lasting tinge of pink colour appears as MnO_{4}^{-} concentration as low as 10^{-5}mol (10^{-6} mol L^{-1}). This make sure of a minimal ‘overshoot’ in colour beyond the equivalence point, which is the point from where the reductant and the oxidant are equal in terms of their mole stoichiometry.

The NCERT exemplar Class 11 Chemistry solutions chapter 8 pdf download is a best-suited format offered to students who want an all-time access to the topics’ questions and answers and want to go back to them as per their requirement.
Also, check - NCERT Solutions for Class 11

Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Class 11 Chemistry NCERT Exemplar Solutions Chapter 8 include the following topics:

1. Classical Idea of Redox Reactions – Oxidation And Reduction Reactions

2. Redox Reactions In Terms Of Electron Transfer Reactions

2.1 Competitive Electron Transfer Reactions

3. Oxidation Number

3.1 Types Of Redox Reactions

3.2 Balancing Of Redox Reactions

3.3 Redox Reactions As The Basis For Titrations

3.4 Limitations Of Concept Of Oxidation Number

3.5 Redox Reactions And Electrode Processes.

What The Students Will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions?

  • After studying Class 11 Chemistry NCERT exemplar solutions chapter 8, students will have been enlightened upon the concept of the occurrence of a specific type of chemical reaction named as redox reactions.

  • They will have the complete knowledge as to the reasons and factors that influence the occurrence of them and how we can manage these factors.

  • After getting a hold on the theoretical aspect of redox reactions, students will learn to evaluate them after gaining the expertise to balance their equations so that they can draw a conclusion about the reaction.

  • NCERT Exemplar Class 11 Chemistry solutions chapter 8 also imparts knowledge about the practical applications of redox reactions, namely, titrations.

  • NCERT exemplar solutions for Class 11 Chemistry chapter 8 makes the students familiar with the concept of oxidation and oxidation number after which they engage in learning about the concluding topic of the chapter that is redox reactions and electrode processes.

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

  • Students develop an understanding of the concept of redox reactions and the conditions under which they can occur along with their limitations. They are introduced with the concept of oxidation number and how to calculate it.
  • After learning how to calculate oxidation numbers, students can work on several problems where they have to calculate the oxidation number of the missing element in the given compound. Most problems in this chapter are based on this concept.
  • Class 11 Chemistry NCERT Exemplar Solutions Chapter 8 also engages in the terms oxidising and reducing agents and teaches the reader how to identify the two in a compound or a chemical equation by highlighting both of their individual properties.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Check Solutions of Textbook Chapters

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Can NCERT exemplar class 11 Chemistry solutions chapter 8 be used as a reference module by the students for respective chapters?

Yes. Since this includes all the basic elements of a reference book, i.e. questions and accompanying answers which strengthen the concepts of students in the respective topics, it can be used as a guide/ reference.

2. Can this be used by students for last moment revision by students?

Yes. The module contains questions including all the important topics of this chapter and thus can be used for an efficient, quick last-minute revision.

3. Is NCERT exemplar solutions for class 11 Chemistry chapter 8 a good option for students?

Yes, the solutions mentioned are designed by experts in a specific manner to make it easy and efficient for the students to study and comprehend and take the best route to solve a question.

4. What if the students want to save the contents for their future reference or usage?

Students can refer to the information provided anytime by making use of the NCERT exemplar class 11 Chemistry solutions chapter 8 pdf download.

Articles

Upcoming School Exams

Application Date:07 October,2024 - 22 November,2024

Application Date:07 October,2024 - 22 November,2024

Application Correction Date:08 October,2024 - 27 November,2024

View All School Exams
Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top