NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction

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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

Edited By Sumit Saini | Updated on Aug 22, 2022 08:49 AM IST

NCERT solutions for class 11 chemistry chapter 8 Redox Reactions- NCERT Solutions for class 11 chapter 8 Redox Reactions discusses the concept of reduction and oxidation in detail and various insights about the redox reaction. Also, it discusses oxidation number, balancing redox reactions, types of a redox reaction, loss and gain of electrons by the elements in the reaction, change in oxidation number and applications of redox reactions.

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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions - Exercise Questions

Question 8.1 Assign oxidation number to the underlined elements in each of the following species

$(a) NaH_{2}\bar{P}O_{4}$

$(b)NaH\bar{S}O_{4}$

$(c) H_{4}\bar{P_{2}}O_{7}$

$(d) K_{2}\bar{Mn}O_{4}$

$(e) Ca\bar{O_{2}}$

$(f) Na\bar{B}H_{4}$

$(g) H_{2}\bar{S_{2}}O_{7}$

$(h) KAl(\bar{S}O_{4})_{2}.12H_{2}O$

Answer :

solution-

O.N is the oxidation number

O.N of Oxygen( $O$ ) = -2 ( In case of peroxide and superoxide it wil be different ON)

O.N of hydrogen( $H$ )= +1 (In case of metalic hydride, -1)

O.N of sodium ( $Na$ ) = +1

O.N of aluminium ( $Al$ ) = +3

O.N of potassium ( $K$ )= +1

O.N of calcium ( $Ca$ ) = +2

In neutral compounds the sum of O.N of all the atoms is zero.

(a) Let the O.N of P be x

$\therefore\:\:\:1\ast 1+2\ast 1+x+4\ast (-2) = 0 \Rightarrow x = +5$

(b) Let the O.N of S be x

$\therefore \:\:1\ast 1 + 1\ast 1+x +4\ast (-2) = 0\Rightarrow x = +6$

(c) Let the O.N of P be x

$\therefore \:\:4*1 +2*x +7*(-2) = 0 \Rightarrow x = +5$

(d) Let the O.N of Mn be x

$\therefore \:\:2*1 + x + 4*(-2) = 0\Rightarrow x = +6$

(e) Let the O.N of O be x

Ca is an alkaline earth metal so its O.N. is +2

$\therefore \:\:1\ast 2 + 2\ast x = 0\Rightarrow x = -1$

(f) Let the O.N of B be x

Note that in this H exists as hydride ion $H^{-}$ so its O.N. is -1

$\therefore \:\:1*1+x+4*(-1) = 0 \Rightarrow x = +3$

(g) Let the O.N of S be x

$\therefore \:\: 2*1 +2*x+7*(-2) =0 \Rightarrow x = +6$

(h) Let the O.N of S be x

$\therefore \:\:1*1+1*3+[x+(-2)*4]*2 +12*[1*2+(-2)] = 0\Rightarrow x = +6$

Question 8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results

$(a) K\underline{I}_{3}$

$(b) H_{2}\underline{S_{4}}O_{6}$

$(c) \underline{Fe}_{3}O_{4}$

$(d) \underline{C}H_{3}\underline{C}H_{2}OH$

$(e) \underline{C}H_{3}\underline{C}OOH$

Answer :

Solution-

O.N of potassium ( $K$ )= +1

O.N of hydrogen( $H$ )= +1 (In case of metalic hydride, -1)

O.N of Oxygen( $O$ ) = -2 ( In case of peroxide and superoxide it wil be different ON)

$(a) K\underline{I}_{3}$

1*1 + 3*x = 0

x = (-1/3)

average O. N. Of $I$ is $-\frac{1}{3}$ . But it is wrong because O.N cannot be fractional. So lets try with structure of $KI_{3}$

$K^+(I-I\leftarrow I)^{-1}$

O.N of $I$ = -1 (because a coordinate bond is formed between $I_{2}$ and $I^{-}$ ion. Hence O. N of three $I$ atoms are 0,0 and -1, O.N 0 in $I_{2}$ moleculeand -1 in $I^{-}$ ion.)

$(b) H_{2}\underline{S_{4}}O_{6}$

Assume O.N of S is x

$2*1 + 4*x + 6*(-2) = 0 \Rightarrow x=2.5$

Fractional O.N is not possible so try with structure -

1594293130665

The two S atom present in the middle has zero (0) O.N and other two have (+5) O.N [ -2 from two O atom and -1 from OH]

$(c) \underline{Fe}_{3}O_{4}$

If you calculate the oxidation number of Fe in $Fe_{3}O_{4}$ it would be 8/3 and however, O.N cannot be in fractional.

$Fe_{3}O_{4} \:\:\: is \:an\:equimolar \:mixture\: of \:(FeO)\: and \:(Fe_{2}O_{3})$ Here one iron atom has +2 O.N and the other two are of +3 O.N.

$(d) \bar{C}H_{3}\bar{C}H_{2}OH$

let assume carbon has x oxidation Number

So,[ x + 1(3) +x +1(2) +(-2)+1 = 0]

2x = -4

x=-2

In this molecule two carbon atoms present in different environments. Hence, they cannot have the same O.N.Thus, C exhibits the O. S of -3 and -1.

1594293648509

$(e) \bar{C}H_{3}\bar{C}OOH$

suppose the oxidation number of Carbon is x.

If we calculate the O.N of x we get x=0

However, 0 is average O.N. of C atoms. In this molecule two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O.S of +3 and –3 in CH3COOH This can be more understood by structure-

1594293781269

Here we can see that at right C, +3 O.N (-1 from OH and -2from O atom) and in left C, -3 O.N(contribution from H atom only)

Question 8.3(a) Justify that the following reactions are redox reaction

$(a) CuO_{(s)}+H_{2}_{(g)}\rightarrow Cu_{(s)}+ H_{2}O_{(g)}$

Answer :

Solution-

Let us write the O.N of each element

$\overset{\:\:2+\:\:-2}{CuO}\:+\:\overset{0}{H_{2}}\rightarrow \overset{0}{Cu}\:+ \:\overset{+1\:-2}{H_{2}O}$

Here, the O.N of Cu decreases from +2 to 0 i.e., CuO is reduced to Cu. Also, the O.N of H increases from 0 to +1 i.e., H2 is oxidized to H2O. Hence it is a redox reaction.

Question 8.3(b) Justify that the following reactions are redox reaction

$Fe_{2}O_{3}_{(s)}+3CO \rightarrow 2Fe_{(s)}+3CO_{2}_{(g)}$

Answer :

Solution-

Let us write the O.N of each element

$\overset{\:\:+3\:\:-2}{Fe_{2}O_{3}}\:+ 3\overset{+2\:-2}{CO}\rightarrow 2\overset{0}{Fe}\:\:+$ $3\overset{+4\:\:-2}{CO_{2}}$

Here, the O.N of Fe decreases from +3 to 0. Also, the O.N of C increases from +2 to +4 . Hence it is a redox reaction.

Question 8.3(c) Justify that the following reactions are redox reaction

$4BCl_{3}_{(g)}+3LiAlH_{4}_{(s)}\rightarrow 2B_{2}H_{6}_{(g)}+3LiCl_{(s)}+3AlCl_{3}_{(s)}$

Answer :

Solution-

Let us write the O.N of each element

$4\overset{+3\:\:-1}{BCl_{3}}\:\:+\:3\overset{+1\:\:+3\:\:-1}{LiAlH_{4}}\:\rightarrow\:2\overset{-3\:\:+1}{B_{2}H_{6}}\:\:+\:\:3\overset{+1\:-1}{LiCl}\:\:+\:3\overset{+3\:-1}{AlCl_{3}}$

Here, the O.N of Fe decreases from +3 in BCl3 to –3 in B2H6. And, the O.N of H increases from –1 in $LiAlH_{4}$ to +1 in $B_{2}H_{6}$ . Hence it is a redox reaction.

Question 8.3(d) Justify that the following reactions are redox reaction

$2K_{(s)}+F_{2}_{(g)}\rightarrow 2K^{+}F^{-}_{(s)}$

Answer :

Solution-

We know that oxidation = loosing of $e^{-}$ by atom

and reduction = gaining of $e^{-}$ by another atom

here $K$ lose its electron and $F$ accept it, Hence it is a redox reaction

Question 8.3(e) Justify that the following reactions are redox reaction

$4NH_{3}_{(g)}+5O_{2}_{(g)}\rightarrow4NO_{(g)}+6H_{2}O_{(g)}$

Answer :

Solution-

here $N (-3)\rightarrow N(+2)$ oxidation reaction

and $O(0)\rightarrow O(-2)$ reduction reaction (oxidation state of oxygen is zero at molecular state )

hence it's a redox reaction

Question 8.4 Fluorine reacts with ice and results in the change

$H_{2}O_{(s)}+F_{2}_{(g)}\rightarrow HF_{(g)}+HOF_{(g)}$

Justify that this reaction is a redox reaction

Answer :

Solution-

$F_{2}$ $HF$ $HOF$

Oxidation state of $F$ $\rightarrow$ $0$ $-1$ $+1$

Here $F$ is oxidized and reduced as well. So, it is a redox reaction.

Question 8.5 Calculate the oxidation number of sulphur, chromium, and nitrogen in $H_{2}SO_{5}$ , $Cr_{2}O_{7}^{2-}$ and $NO_{3}^{-}$ . Suggest the structure of these compounds. Count for the fallacy.

Answer :

(i) $H_{2}SO_{5}$ let the oxidation number of sulphur be x

So,

$\\2*1+x+5(-2)= 0\\ x= +8$

There is a fallacy Sulphur cannot have +8 oxidation state because it has maximum +6 oxidation number, not more than that.The structure of $H_{2}SO_{5}$ is shown as follows:

1594293301896

$2(H)+1(S)+3(O)+2(O \:\:in \:peroxy\: linkage)$

$\Rightarrow 2(+1) + 1(x) + 3(-2) + 2(-1) = 0$

$\Rightarrow x = +6( Answer))$

(ii) $Cr_{2}O_{7}^{2-}$

let the oxidation number of chromium be x

now

$\\2x+7*(-2)=-2\\ x= (-2+14)/2\\ x= +6$

There is no fallacy here

1594293400354

(iii) $NO_{3}^{-}$

let assume oxidation number of N is x

Now, $\\x+(-2)*3= -1\\(x-6)=-1\\x= +5$

1594293355171

here is no fallacy about the O.N of N in $NO_{3}^{-}$

Question 8.6(a) Write formulas for the following compounds:

Mercury(II) chloride

Answer :

Answer-

$HgCl_{2}$

in this formula, we can see that mercury has $+2$ oxidation state

Question 8.6(b) Write formulas for the following compounds:

Nickel(II) sulphate

Answer :

Answer-

$NiSO_{4}$

sulphate has $-2$ oxidation state

Question 8.6(c) Write formulas for the following compounds:

Tin(IV) oxide

Answer :

$SnO_{2}$

Oxygen has $-2$ oxidation state

Question 8.6(d) Write formulas for the following compounds:

Thallium(I) sulphate

Answer :

Answer-

$Tl_{2}SO_{4}$

Question 8.6(e) Write formulas for the following compounds:

Iron(III) sulphate

Answer :

Answer- Formula of the compounds: Iron(III) sulphate is

$Fe_{2}(SO_{4})_{3}$

Question 8.6(f) Write formulas for the following compounds:

Chromium(III) oxide

Answer :

Answer- Formula of the Chromium(III) oxide compounds is

$Cr_{2}O_{3}$

Question 8.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5

Answer :

the substance of Carbon -

Substance	O.N. Of C
$CH_{4}$	-4
$C_{2}H_{6}$	-3
$C_{2}H_{4}\: or\: CH_{3}Cl$	-2
$C_{2}H_{2}$	-1
$CH_{2}Cl_{2}$	0
$C_{6}Cl_{6}$	+1
$CHCl_{3}$	+2
$(COOH)_{2}$	+3
$CO_{2} \: or\: CCl_{4}$	+4

substance for Nitrogen-

Substance	O.N. Of N
$NH_{3}$	-3
$N_{2}H_{4}$	-2
$N_{2}H_{2}$	-1
$N_{2}$	0
$N_{2}O$	+1
$NO$	+2
$N_{2}O_{3}$	+3
$N_{2}O_{4}$	+4
$N_{2}O_{5}$	+5

Question 8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Answer :

sulphur dioxide $(SO_{2})$ here oxidation state of sulphur is +4 and the range of oxidation number of S is from -2 to +6 . It means it can accept an electron and lose as well, therefore, it can behave as oxidant and reductant both.
In case of hydrogen peroxide $(H_{2}O_{2})$ oxidation state is -1 and the oxidation state of O can vary from 0 to -2 . So it shows both oxidizing and reducing properties.
For $HNO_{3}$ , Nitrogen has +5 oxidation state and it varies from +5 to -3. So it only accepts electrons. The oxidation state of N only decreases. Hence it acts as only oxidants.
And in case of $O_{3}$ , the oxidation state of O is zero(0) and the range of oxidtion number of O is 0 to -2. It only decreases in this case also so therefore it acts as only oxidants.

Question 8.9 Consider the reactions:
$(a) 6CO_{2}_{(g)}+6H_{2}O_{(l)}\rightarrow C_{6}H_{12}O_{6}_{(aq)}+6CO_{2}_{(g)}$

$(b) O_{3}_{(g)}+H_{2}O_{2}_{(l)}\rightarrow H_{2}O(l) +2O_{2}(g)$

Why it is more appropriate to write these reactions as :

$(a) 6CO_{2}_{(g)}+12H_{2}O_{(l)}\rightarrow C_{6}H_{12}O_{6}_{(aq)}+6H_{2}O_{(l)}+6CO_{2}_{(g)}$

$(b) O_{3}_{(g)}+H_{2}O_{2}_{(l)}\rightarrow H_{2}O(l) +O_{2}(g)+O_{2}_{(g)}$

Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer :

(a) In the photosynthesis process-

step 1- the liberation of $O_{2}$ and $H_{2}$ --> $2H_{2}O\rightarrow O_{2}+2H_{2}$

step-2 The $H_{2}$ produced in above reduces the $CO_{2}$ into glucose $(C_{6}H_{12}O_{6})$ and water $(H_{2}O)$

$6CO_{2}+12H_{2}\rightarrow C_{6}H_{12}O_{6}+6H_{2}O$

So, the final net reaction is

$2H_{2}O\rightarrow CO_{2}+2H_{2}]*6\\+6CO_{2}+12H_{2}\rightarrow C_{6}H_{12}O_{6}+6H_{2}O\\------------------\\6CO_{2}+12H_{2}O\rightarrow C_{6}H_{12}O_{6}+6H_{2}O$

It is more appropriate to write the reaction as above because water $(H_{2}O)$ molecule also produced in photosynthesis reaction.

The path of reaction can be investigated by using the radioactive $H_{2}O^{18}$ instead of $(H_{2}O)$

(b)

$\\O_{3}(g)\rightarrow O_{2}(g)+O(g)\\ H_{2}O_{2}(l)+O(g)\rightarrow H_{2}O(l)+O_{2}(g)\\ ---------------\\ H_{2}O_{2}(l)+O_{3}(g)\rightarrow H_{2}O(l)+O_{2}(g)+O_{2}(g)$ (the final net reaction)

Dioxygen is produced from both steps, one from the decomposition of ozone ( $O_{3}$ ) and other is from the reaction of hydrogen peroxide with(O)

The path of the reaction can be investigated by using $O^{18}_{3}/ H_{2}O^{18}$ .

Question 8.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations?

Answer :

These can be understood by the following examples-

$P_{4}$ is reducing agent and $Cl_{2}$ is an oxidizing agent

$P_{4}+10Cl_{2}(Excess)\rightarrow PCl_{5}$ [O.N of phosphorus +5] $\Rightarrow$ Higher O.S of P

$P_{4}(Excess)+6Cl_{2}\rightarrow 4PCl_{3}$ [O.N of phosphorus +3] $\Rightarrow$ Lower O.S of P

is an $O_{2}$ is a reducing agent and $C$ oxidising agent

$C+O_{2}\rightarrow CO_{2}$ (O is in excess) [O.N of C +4]

$C+O_{2}\rightarrow CO$ (C is in excess) [O.N of C +2]

$K$ is a reducing agent and $O_{2}$ is an oxidizing agent

$K+O_{2}\rightarrow K_{2}O$ (K is in excess) [O.N of O -2] (lower O.S.)

$K+O_{2}\rightarrow K_{2}O_{2}$ (O is in excess) [O.N of O -1] (lower O.S.)

Question 8.12(a) How do you count for the following observations?

Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.

Answer :

Alcohol and $KMnO_{4}$ both are polar in nature and alcohol is homogenous to toluene because both are organic compounds. So the reaction is faster in the homogenous medium rather than heterogeneous medium. And hence all the compounds react at a faster rate.

Chemical equation-

$C_{6}H_{6}CH_{3}+MnO_{4}^{-}(alc.)\rightarrow C_{6}H_{6}COO^{-}+MnO_{2}+H_{2}O(l)+OH^{-}(aq)$

Question 8.12(b) How do you count for the following observations?

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

Answer :

Concentrated sulphuric acid is added to an inorganic mixture containing chloride-

$2NaCl\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HCl$

HCl is a weak reducing agent and it cannot reduce $H_{2}SO_{4}$ to $SO_{2}$ thats why we get colourless pungent smelling gas HCl.

Concentrated sulphuric acid is added to an inorganic mixture containing bromide-

$2NaBr\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HBr$

$2HBr + H_{2}SO_{4}\:\rightarrow \:Br_{2} ( Red \:Vapour)\:+\:SO_{2}\:+\:2H_{2}O$

When conc. sulphuric acid is added to an inorganic mixture containing bromide initially it produces $HBr$ and it is a strong reducing agent so it reduces $H_{2}SO_{4}$ to $SO_{2}$ with evolution of is a red vapor of bromine.

Question 8.13(a) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(a)2AgBr(s)+C_{6}H_{6}O_{2}(aq)\rightarrow 2Ag(s)+2HBr(aq)+C_{6}H_{4}O_{2}(aq)$

Answer :

Substance reduced/oxidizing agent- $AgBr$

Substance oxidized/reducing agent- $C_{6}H_{6}O_{2}]$

Question 8.13(b) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(b) HCHO(l)+2[Ag(NH_{3})]^+(aq) +3OH^-(aq)\rightarrow 2Ag(s)+HCOO^- (aq)+4NH_{3}(aq) +2H_{2}O(l)$

Answer :

Substance reduced/oxidising agent- $[Ag(NH_{3})_{2}]^+$

Substance oxidised/reducing agent- $HCHO$

Question 8.13(d) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(d) N_{2}H_{4}(l)+2H_{2}O_{2}(l)\rightarrow N_{2}(g)+4H_{2}O(l)$

Answer :

Substance oxidized/reducing agent- $N_{2}H_{4}$

Substance reduced/oxidizing agent- $H_{2}O_{2}$

Question 8.13(e) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(e)Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)\rightarrow 2PbSO_{4}(s)+2H_{2}O(l)$

Answer :

Substance oxidized/reducing agent- $Pb$

Substance reduced/oxidizing agent- $PbO_{2}$

Question 8.14 Consider the reactions :
$2S_{2}O_{3}^{2-}(aq)+I_{2}(s)\rightarrow S_{4}O_{6}^{2-}(aq)+2I^{-}(aq)$

$2S_{2}O_{3}^{2-}(aq)+2Br_{2}(l)+5H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+4Br^{-}(aq)+10H^{+}(aq)$

Why does the same reductant, thiosulphate react differently with iodine and bromine?

Answer :

$F_{2}>Cl_{2}>Br_{2}>I_{2}$ oxidizing power order

Bromine is a stronger oxidizing agent than iodine. So in the case of bromine (avg. oxidation number of sulphur is changed from +2 to +6)

and in case of iodine it (+2 to +2.5). So that's why thiosulphate reacts differently with bromine and iodine.

Question 8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer :

part(i)

Fluorine can oxidize other halogen ions. On the other hand $Br_{2},I_{2},Cl_{2}$ cannot oxidize $2F^{-} \rightarrow F_{2}$

$F_{2}+2I^{-}\rightarrow 2F^{-}+I_{2} \\F_{2}+2Br^{-}\rightarrow 2F^{-}+Br_{2} \\F_{2}+2Cl^{-}\rightarrow 2F^{-}+Cl_{2}$

And hence we say that fluorine is the better oxidant among halogen.

part(ii)

$HI$ & $HBr$ are able to reduce $H_{2}SO_{4}\rightarrow SO_{2}$ but $HCl,HF$ are unable to reduce sulphuric acid.

So here we can say that $HI$ & $HBr$ are better reductant than $HCl,HF$ .

Again $I^{-}$ can only able to reduce $Cu^{2+}\rightarrow Cu^{+}$ but $Br^{-}$ cannot.

$4I^{-}+2Cu^{2+}\rightarrow Cu_{2}I_{2}+I_{2}$

Hence among hydrohalic compound hydroiodic acid is the best reductant.

Question 8.16 Why does the following reaction occur?

$XeO_{6}^{4-}(aq)+2F^{-}(aq)+6H^{+}(aq)\rightarrow XeO_{3}(g)+F_{2}(g)+3H_{2}O(l)$

What conclusion about the compound $Na_{4}XeO_{6}$ (of which $XeO_{6}^{4-}$ is a part) can be drawn from the reaction ?

Answer :

we conclude that the oxidation state of Xenon changes from +8 to +6

$XeO_{6}^{4-} (+8)\rightarrow XeO_{3}(+6)$

and oxidation state of F changes from -1 to 0

$F^{-}(-1)\rightarrow F_{2}(0)$

$Na_{4}XeO_{6}$ is reduced by accepting an electron.

It is a strong oxidizing agent than F

Question 8.17(a) Consider the reactions:

(a) $H_{3}PO_{2}(aq)+4AgNO_{3}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+4Ag(s)+4HNO_{3}(aq)$

$(b)H_{3}PO_{2}(aq)+2CuSO_{4}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+2Cu(s)+4H_{2}SO_{4}(aq)$

What inference do you draw about the behaviour of $Ag^{+}$ and $Cu^{+}$ from these reactions ?

Answer :

In the first reaction, we can see that $Ag^{+}$ oxidizes the phosphorus from (+1 $\rightarrow$ +5) also in second, we clearly see that $Cu^{+}$ oxidize the phosphorus from (+1 $\rightarrow$ +5).

Both are oxidizing agents.

Question 8.17(b) Consider the reactions:

$(c)C_{6}H_{5}CHO(l)+2[Ag(NH_{3})_{2}]^{+}(aq)+3OH^{-}(aq)\rightarrow C_{6}H_{5}COO^{-}(aq)+2Ag^{+}(s)+4NH_{3}(aq)+2H_{2}O(l)$

$(d)C_{6}H_{5}CHO(l)+2Cu^{2+}(aq)+5OH^{-}(aq)\rightarrow$ No change observed

Answer :

Here, by looking at the reaction, we conclude that $Ag^{+}$ oxidises $C_{6}H_{5}CHO$ and in the second reaction $Cu^{+}$ not able to oxidise. So we can say that $Ag^{+}$ is stronger oxidizing agent than $Cu^{+}$ .

Question 8.18(a) Balance the following redox reactions by ion – electron method

$MnO_{4}^{-}(aq)+ I^{-}(aq)\rightarrow MnO_{2}(s)+ I_{2}(s)$ (In basic medium)

Answer :

reduction half reaction

$MnO_{4}^{-}\rightarrow MnO_{2}$ (+7 to +4)

Add 3 electron on LHS side and after that to balance charge add OH ions. And to balance O atom add water molecule on whichever side it needed

balance it

$MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2} +4OH^{-}$

oxidation half

$I\rightarrow I_{2}$

balance it

$2I^{-}\rightarrow I_{2}+2e^{-}$

equalising the no. of electrons by multiplying the oxidation half by 3 and reduction half by 2 and then add it.

$6I^{-}\rightarrow 3I_{2}+ 6e^{-}\\&\ 2MnO_{4}^{-}+4H_{2}O+6e^{-}\rightarrow 2MnO_{2}+8OH^{-}$

final answer-

$2MnO_{4}^{-}+4H_{2}O+6I^{-}\rightarrow 2MnO_{2}+3I_{2}+8OH^{-}$

Question 8.18(b) Balance the following redox reactions by ion – electron method

$MnO_{4}^{-}(aq)+SO_{2}(g) \rightarrow Mn^{2+}(aq)+HSO_{4}^{-}$

(In Acidic medium)

Answer -

oxidation half reaction

$SO_{2}+2H_{2}O\rightarrow HSO_{4}^{-}+3H^{+}+2e^{-}$

reduction half reaction

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

Balancing the reaction

multiply the oxidation half by 5 and reduction half by 2 and then add these two reactions

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

Question 8.18(c) Balance the following redox reactions by ion – electron method

(c) $H_{2}O_{2}(aq)+Fe^{3+}(aq)\rightarrow Fe^{3+}(aq)+H_{2}O(l)$

in acidic medium

Answer :

In acidic medium

oxidation half reaction-

$Fe^{2+}\rightarrow Fe^{3+}+e^{-}$

reduction half reaction-

$H_{2}O_{2}+2H^{+}+2e^{-}\rightarrow 2H_{2}O$

Balancing the reaction

multiply by 2 on oxidation half-reaction then add it with reduction half reaction

$H_{2}O_{2}+2Fe^{2+}+2H^{+}\rightarrow 2Fe^{3+}+2H_{2}O$

Question 8.18(d) Balance the following redox reactions by ion – electron method

$Cr_{2}O_{7}^{2-}+SO_{2}(g) \rightarrow Cr^{3+}(aq)+SO_{4}^{2-}(aq)$

in acidic medium

Answer :

Half-reaction

oxidation half $SO_{2}+2H_{2}O\rightarrow SO_{4}^{2-}+4H^{+}+2e^{-}$

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

balancing them by multiplying oxidation half by 3 and adding the reaction

$Cr_{2}O_{7}^{2-}+3SO_{2}+2H^{+}\rightarrow 2Cr^{3+}+3SO_{4}^{2-}+H_{2}O$

Question 8.20 What sorts of information can you draw from the following reaction?

$(CN)_{2}(g)+2OH^{-}(aq)\rightarrow CN^{-}(aq) +CNO^{-}(aq)+H_{2}O(l)$

Answer :

Carbon shows different oxidation state according to the compound formula.

here we can clearly say that Carbon is in its +3 oxidation state.

$\\(CN)_{2}(+3)\rightarrow CN^{-}(+2)\\ (CN)_{2}(+3)\rightarrow CNO^{-}(+4)$

The oxidation state of carbon is increased(oxidized) and decreased(reduced) as well in the product side. So it is a redox reaction and more specifically we can say it disproportion redox reaction.

Question 8.21 The $Mn^{3+}$ ion is unstable in solution and undergoes disproportionation to give $Mn^{2+}$ , $MnO_{2}$ and $H^{+}$ ion. Write a balanced ionic equation for the reaction

Answer :

The base equation

$Mn^{3+}\rightarrow Mn^{2+}+MnO_{2}+H^{+}$

write oxidation half with their oxidation state

$Mn^{3+}(+3)\rightarrow MnO_{2}(+4)$

Balance the charge on $Mn$ by adding 1 $e^{-}$ on RHS side. To balance charge add $H^{+}$ ions on RHS side and then for oxygen balance add $H_{2}O$ molecule on LHS side.

$Mn^{3+}+2H_{2}O\rightarrow MnO_{2}+1e^{-}+4H^{+}$

reduction half

$Mn^{3+}(+3)\rightarrow Mn^{2+}(+2)$

balancing the reduction half by adding 1 $e^{-}$ on LHS side

$Mn^{3+}+1e^{-}\rightarrow Mn^{2+}$

Add both balanced reduction half and oxidation half

$2Mn^{3+}+2H_{2}O\rightarrow Mn^{2+}+MnO_{2}+4H^{+}$

Question 8.22 Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer :

(a) $F$ (because of highly electronegative in nature)

(b) $Cs$ (highly electropositive)

(d) $Ne$ (enert gas

Question 8.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water

Answer :

Base equation- $SO_{2}+Cl_{2}\rightarrow Cl^{-}+SO_{4}^{2-}$ --------------(have to remember)

Now we have to balance the oxidation half and reduction half.

oxidation half = $SO_{2}\rightarrow SO_{4}^{2-}$

balancing - Oxygen is balanced by adding water molecule, Hydrogen is balanced by $H^{+}$ ion and for charge add $e^{-}$ (electron) $SO_{2} +2H_{2}O\rightarrow SO_{4}^{2-}+4H^{+}+2e^{-}$

Reduction hallf = $Cl_{2}\rightarrow Cl^{-}$

Balancing - to balance charge add an electron $Cl_{2}+2e^{-}\rightarrow 2Cl^{-}$

Now add both balanced oxidation half and reduction half, we get

$Cl_{2}+SO_{2}+2H_{2}O\rightarrow 2Cl^{-}+SO_{4}^{2-}+4H^{+}$

Question 8.24 Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non-metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction

Answer :

(a) Phosphorus, sulphur and chlorine can show disproportionation reaction.

(b) Manganese, copper, indium and gallium can show disproportionation reaction.

Question 8.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer :

Answer-

we have,

number of moles(n) = given mass/ molecular mass ------------------------------(eq.1)

No. of moles of ammonia $(NH_{3})$ = 10/17 = 0.588

No. of moles of oxygen $(O_{2})$ = 20/32= 0.625

Balanced Reaction $4NH_{3}+5O_{2}\rightarrow 4NO + 6H_{2}O$

Here we see that 4 moles of ammonia required 5 moles of oxygen. So

0.588 moles of ammonia = $(5/4)*0.588 = 0.735$ moles of $(O_{2})$ . But we have only 0.625 moles of $(O_{2})$ .

It means oxygen is a limiting reagent and the maximum weight of nitric oxide $(NO)$ can be produced by 0.635 moles of $(O_{2})$

So, 5 moles of $(O_{2})$ produced 4 moles of C.

therefore 0.625 moles of $(O_{2})$ = $(4/5)*0.625 = 0.5$ moles of $(NO)$ .

from Eq. 1

mass of $(NO)$ = number of moles $(NO)$ * molecular weight $(NO)$

= $0.5* 30$

= 15 g

Alternate Method

directly consider the molecular weight

(17*4) g of NH3 required (5*32) g of O to produce (30*4) g of NO

So, 10g of NH3 required= (5*32/17*4)*10 = 23.5g of O. But we have only 20g (means O is limiting reagent) whatever the max. NO produce is from 20g of O.

and we know that 5*32g of O produce 30*4 g of NO

So, 20g of O produce =(30*4/5*32)*20 g of NO = 15g of NO

Question 8.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible

(a) $Fe^{3+}(aq)$ and $I^{-}(aq)$
(b) $Ag^{+}(aq)$ and Cu(s)
(c) $Fe^{3+}$ (aq) and Cu(s)
(d) Ag(s) and $Fe^{3+}$ (aq)
(e) $Br_{2}(aq)$ and $Fe^{2+}(aq)$ .

Answer :

If $E^{0}$ for the overall reaction is positive $\rightarrow$ feasible

negative $\rightarrow$ not feasible

(a) $[Fe^{3+}+ e^{-}\rightarrow Fe^{2+} ]*2: E^{0} = 0.77V$

$2I^{-}\rightarrow I_{2}+2e^{-}: E^{0} = -0.54V$

---------------------------------------------------------------------------------------

$2Fe^{3+}+2I^{-}\rightarrow Fe^{2+}+I_{2}: E^{0} = +0.23V$

(b) $Ag^{+}+e^{-}\rightarrow Ag(s)]*2: E^{0} = +0.80V$

$Cu\rightarrow Cu^{2+}+2e^{-}:E^{0} =-0.34V$

---------------------------------------------------------------------------------------

$2Ag^{+}+Cu\rightarrow 2Ag+Cu^{2+}: E^{0}=+0.46V$

$2Br^{-}\rightarrow Br+2e^{- }:E^{0}=-1.09V$

-------------------------------------------------------------------------

$2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{3+}+Br_{2}:E^{0}=-0.32$

(d) $Ag\rightarrow Ag^{+}+e^{-}:E^{0}=-0.80V$

$Fe^{3+}+e^{-}\rightarrow Fe^{2+}:E^{0}==0.77V$

------------------------------------------------------------------------------

$Ag+Fe^{3+}\rightarrow Ag^{+}+Fe^{2+}:E^{0}=-0.03V$

(e) $Br_{2}+2e^{-}\rightarrow 2Br^{-}:E^{0}=+1.09V$

$Fe^{2+}\rightarrow Fe^{3+}+e^{-}]*2:E^{0}=-0.77V$

-------------------------------------------------------------------------------

$Br_{2}+2Fe^{2+}\rightarrow 2Br^{-}+2Fe^{3+}:E^{0}=+0.32V$

Question 8.27(i) Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $AgNO_{3}$ with silver electrodes

Answer :

Answer-

(i) $AgNO_{3}$ dissociate into $Ag^{+}$ and $NO_{3}^{-}$

@ Cathode - $Ag^{+}(aq)+e^{-}\rightarrow Ag(s)$ (reduction potential of silver is higher than $H_{2}O$ ) $E^{0} = 0.80V$

@ Anode - $Ag(s)\rightarrow Ag^{+}(aq)+e^{-}$ ( oxidation potential of silver is higher than water molecule.So silver electrode oxidized ) $E^{0} = -0.83V$

Question 8.27(ii) Predict the products of electrolysis in each of the following:

An aqueous solution $AgNO_{3}$ with platinum electrodes

Answer :

Answer-

(ii) since platinum $(Pt)$ electrode cannot easily oxidize. So at the anode $H_{2}O$ will oxidize and liberate oxygen and at cathode $Ag$ will be deposited.

At cathode- $Ag^{+}+e^{+}\rightarrow Ag$

At anode- $H_{2}O\rightarrow O_{2}+4H^{+} +4e^{-}$

Question 8.27(iii) Predict the products of electrolysis in each of the following:

(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes

Answer :

Answer-

given sulphuric acid is dilute.

ionize into $H_{2}SO_{4}\rightarrow 2H^{+}+ SO_{4}^{2-}$

At cathode $H^{+}+e^{-}\rightarrow 1/2H_{2}(g)$

At anode, There will be -(liberation of oxygen gas)

$H_{2}O\rightarrow 1/2O_{2}+4H^{+}+4e^{-}$

Question 8.27(iv) Predict the products of electrolysis in each of the following:

An aqueous solution of CuCl2 with platinum electrodes

Answer :

Answer-

(iv) In aqueous solution $CuCl_{2}$ ionise into $Cu^{2+}$ and $2Cl^{-}$

At the cathode , the copper ion will be deposited because it has a higher reduction potential than the water molecule

At the anode , the lower electrode potential value will be preferred but due to overpotential of oxygen, chloride ion gets oxidized at the anode.

@ Anode- $2Cl^{-}\rightarrow Cl_{2}+2e^{-}$

@ cathode- $Cu^{2+}+2e^{-}\rightarrow Cu$

Question 8.28 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn

Answer :

Answer-

In order to displace a metal from its metal salt is done only when the other metal has higher electrode potential.

$Mg(-2.36),Al(-1.66),Zn(-0.76),Fe(-0.44),Cu(+0.34)$

Question 8.29 Given the standard electrode potentials

$K^{+}/K = -2.93 V$ $Ag^{+}/Ag = 0.80V$

$Hg^{2+}/Hg= 0.79V$

$Mg^{2+}/Mg = -2.37V$ $Cr^{3+}/Cr = -0.74V$

arrange these metals in their increasing order of reducing power.

Answer :

Answer-

A negative electrode potential $(E^{0})$ means redox couple is a stronger reducing agent. So as per data the increasing order of the following is-

$Ag<Hg<Cr<Mg<K$

Question 8.30 Depict the galvanic cell in which the reaction

$Zn(s)+2Ag^{+}\rightarrow Zn^{2+}(aq)+2Ag(s)$
takes place, Further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Answer :

Answers-

(i) $Zn$ electrode is negatively charged because it loses electrons (act as an anode)

(ii) electron flow from negatively charged electrode to a positively charged electrode (anode to cathode) and the flow of current is just reversed. So current flow through silver cathode to the zinc anode.

(iii) At Anode- $Zn\rightarrow Zn^{2+}+2e^{-}$

At Cathode $Ag^{+}+e^{-}\rightarrow Ag$

More About Redox Reactions Class 11 NCERT Chemistry Chapter

Redox reactions are the class of reactions in which reduction and oxidation occur simultaneously. In NCERT book for class 11 chemistry chapter 8 Redox Reactions you will get to know three-tier conceptualization of redox reactions which are classical, electronic, and oxidation number. Two methods, oxidation number, and half reaction method are used to balance chemical equations for redox reactions. Oxidation number method is based on the change in the oxidation number of oxidizing agent and the reducing agent and in half-reaction method, redox reaction split into two half reactions-one involving reduction and the other involving oxidation.

After completing NCERT solutions for class 11 chemistry chapter 8 Redox Reactions you will be able to identify redox reaction; define terms like reduction, oxidation, reducing agent and oxidising agent; explain mechanism of redox reaction by electron transfer process; identify reductant and oxidation by using the concept of oxidation number; balance chemical equations by using two methods-oxidation number method and half reaction method.

Important Points of Class 11 Chemistry Chapter 8 Redox Reactions-

1. Redox reactions are the class of reactions in which reduction and oxidation occur simultaneously.

2. The oxidation number represents the total number of electrons gained or lost by an atom.

3. Oxidation is the process which involves the addition of oxygen or removal of hydrogen.

4. The reduction is the process which involves removal of oxygen or addition of hydrogen.

Topics of NCERT Syllabus Class 11 Chemistry Chapter 8 Redox Reactions

8.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions

8.2 Redox Reactions in Terms of Electron Transfer Reactions

8.3 Oxidation Number

8.4 Redox Reactions and Electrode Processes

NCERT Solutions for Class 11 Chemistry

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-13	Hydrocarbons
Chapter-14	Environmental Chemistry

NCERT Solutions for Class 11 Subject Wise

NCERT solutions for Class 11 Biology

NCERT Solutions for Class 11 Maths

NCERT solutions for Class 11 Chemistry

NCERT solutions for Class 11 Physics

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Revision will be a lot much easier such that you always remember the concepts and get very good marks in your class.
Homework will be easy now, all you need to do is check the detailed CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reactions and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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