NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Edited By Sumit Saini | Updated on Sep 10, 2022 - 5:28 p.m. 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:2

The more positive the value of $E^{}\Theta$ , 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.
$E^{}\Theta$ values: $Fe^{3+}/Fe^{2+}= +0.77; I_{2}(s)/I^{-}=+0.54;$
$Cu^{2+}/Cu= +0.34; Ag^{+}/Ag=+0.80V$
$(i) Fe^{3+}$
$(ii) I_{2} (s)$
$(iii) Cu^{2+}$
$(iv) Ag^{+}$

Answer:

The answer is the option $(iv) Ag^{+}$
Explanation: Reduction potential of $Ag^{+}/Ag$ is the highest.

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 PbO₂react 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:21

Write balanced chemical equation for the following reactions:
(i) Permanganate ion $(MnO_{4}^{-})$ reacts with sulphur dioxide gas in acidic medium to produce $Mn^{2+}$ and hydrogensulphate ion.
(Balance by ion electron method)
(ii) Reaction of liquid hydrazine $(N_{2}H_{4})$ with chlorate ion $(ClO_{3}^{-})$ in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(iii) Dichlorine heptaoxide $(Cl_{2}O_{7})$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $(ClO_{2}^{-})$ and oxygen gas.
(Balance by ion electron method)
Answer:

$(i) 2MnO_{4}^{-} + 5SO_{2}+ 2H_{2}O + H^{+} \rightarrow 2Mn^{+2}+5HSO_{4}^{-}$

$(ii) N_{2}H_{4} + ClO_{3}^{-} \rightarrow Cl^{-} + N_{2}O + 2H_{2}O$

$(iii) Cl_{2}O_{7} + 4H_{2}O_{2} \rightarrow 2ClO_{2}^{-} + 3H_{2}O + 4O_{2}+ 2H^{+}$

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 $O_2$ and 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 is $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$ 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 is $6Fe^{2+}\rightarrow 6Fe^{3+}+6e^{-}$ 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 = $2\times-1+6+5\times-2= -6$ and 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 $H_{2}O$ undergoing 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:36

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

$(i) Cu/Cu^{2+}\; and\; Zn^{2+} / Zn$
$(ii) Mg/Mg^{+2} \;and \;Fe^{2+}/ Fe$
$(iii) Br_{2}/Br^{-} \;and\;Cl^{-}/ Cl_{2}$
$(iv) Fe/Fe^{2+}\; and\; Cd^{2+}/ Cd$

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

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 9 Hydrogen

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

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.

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

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

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

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

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

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8: Matching Type

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

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

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

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

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

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

Check Solutions of Textbook Chapters

NCERT Exemplar Class 11 Solutions

Frequently Asked Question (FAQs) - NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Question: What if the students want to save the contents for their future reference or usage?

Question: Is NCERT exemplar solutions for class 11 Chemistry chapter 8 a good option for students?

Question: Can this be used by students for last moment revision by students?