NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

Edited By Shivani Poonia | Updated on May 08, 2025 04:49 PM IST

Have you ever wondered how the process of rusting of iron or metabolism in our body takes place? Redox reactions help in understanding such important phenomena. Various industrial processes, workings of batteries, combustion of fuels, etc., are very well explained by redox reaction concepts. Chapter 7 of Class 11 Chemistry NCERT, Redox Reactions, provides detailed descriptions of various topics, such as the process of oxidation and reduction, the change in oxidation state of elements during chemical reactions, and different types of oxidation reactions, among others.

This Story also Contains
  1. Download PDF of NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions
  2. NCERT Solutions for Class 11 Chemistry Redox Reactions - Exercise Questions
  3. Class 11 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  4. Approach to Solve Questions of Chapter 7 Redox Reactions
  5. Topics of NCERT Syllabus Class 11 Chemistry Redox Reactions
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. Importance of Chapter - 7 Redox Reactions for JEE and NEET Exam
  8. NCERT Solutions for Class 11 Chemistry
  9. NCERT Solutions for Class 11 Subject Wise
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

In this Chapter, there are 30 questions in the exercise. The NCERT solutions for class 11 Chemistry Chapter 7 Redox Reactions are prepared and designed by experts. Chemistry is a scoring subject for competitive exams. These NCERT Solutions will help you in your preparation for the class 11 final examination as well as in the various competitive exams like JEE, NEET, etc.

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Download PDF of NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

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NCERT Solutions for Class 11 Chemistry Redox Reactions - Exercise Questions

Question 7.1 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 : O.N is the oxidation number

O.N of Oxygen( $O$ ) = -2 ( In the 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 7.2 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 the 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 7.3(a) Justify that the following reactions are redox reaction

$(\mathrm{a}) \mathrm{CuO} \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{Cu}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$

Answer : 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 7.3(b) Justify that the following reactions are redox reaction

$\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)}$

Answer : 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 7.3(c) Justify that the following reactions are redox reaction

$4 \mathrm{BCl}_{3(\mathrm{~g})}+3 \mathrm{LiAlH}_{4(\mathrm{~s})} \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_{6(\mathrm{~g})}+3 \mathrm{LiCl}_{(\mathrm{s})}+3 \mathrm{AlCl}_{3(\mathrm{~s})}$

Answer : 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 7.3(d) Justify that the following reactions are redox reaction

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

Answer : 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 7.3(e) Justify that the following reactions are redox reaction

$4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(g)}$

Answer : 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 7.4 Fluorine reacts with ice and results in the change

$\mathrm{H}_2 \mathrm{O}_{(s)}+\mathrm{F}_{2(g)} \rightarrow \mathrm{HF}_{(g)}+\mathrm{HOF}_{(g)}$

Justify that this reaction is a redox reaction

Answer : $F_{2}$ $HF$ $HOF$

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

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

Question 7.5 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 7.6(a) Write formulas for the following compounds:

Mercury(II) chloride

Answer : $HgCl_{2}$

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

Question 7.6(b) Write formulas for the following compounds:

Nickel(II) sulphate

Answer : $NiSO_{4}$

sulphate has $-2$ oxidation state

Question 7.6(c) Write formulas for the following compounds:

Tin(IV) oxide

Answer : $SnO_{2}$

Oxygen has $-2$ oxidation state

Question 7.6(d) Write formulas for the following compounds:

Thallium(I) sulphate

Answer : $Tl_{2}SO_{4}$

Question 7.6(e) Write formulas for the following compounds:

Iron(III) sulphate

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

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

Question 7.6(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 7.7 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-

$NH_{3}$
-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
Substance
O.N. Of N

Question 7.8 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 7.9 Consider the reactions:

$$
(\mathrm{a}) 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{aq})}+6 \mathrm{CO}_{2(\mathrm{~g})}
$$


$$
(b) \mathrm{O}_{3(g)}+\mathrm{H}_2 \mathrm{O}_{2(l)} \rightarrow \mathrm{H}_2 \mathrm{O}(l)+2 \mathrm{O}_2(\mathrm{~g})
$$

Why it is more appropriate to write these reactions as :

$(\mathrm{a}) 6 \mathrm{CO}_{2(g)}+12 \mathrm{H}_2 \mathrm{O}_{(l)} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(a q)}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}+6 \mathrm{CO}_{2(g)}$
(b) $\mathrm{O}_{3(g)}+\mathrm{H}_2 \mathrm{O}_{2(l)} \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_{2(g)}$

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 7.11 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 7.12(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 7.12(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 7.13(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 7.13(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 7.14 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 7.15 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 7.16 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 7.17(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 7.17(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 7.18(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.

$\begin{aligned} & 6 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+6 e^{-} \\ & 2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O}+6 e^{-} \rightarrow 2 \mathrm{MnO}_2+8 \mathrm{OH}^{-}\end{aligned}$

final answer-

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

Question 7.18(b) Balance the following redox reactions by ion – electron method

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

(In an 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 7.18(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 the oxidation half-reaction then add it with the reduction half-reaction

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

Question 7.18(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 7.20 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 states 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 7.21 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 the RHS side. To balance the charge add $H^{+}$ ions on RHS side and then for oxygen balance add $H_{2}O$ molecule on the 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 the 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 7.22 Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only a negative oxidation state.

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

(c) Identify the element that exhibits both positive and negative oxidation states

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

Answer :

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

(b) $Cs$ (highly electropositive)

(c) $I$ (has empty d orbitals)

(d) $Ne$ (inert gas

Question 7.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating it 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 the 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 the balanced oxidation half and the reduction half, and we get

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

Question 7.24 Refer to the periodic table given in your book and now answer the following questions:

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

(b) Select three metals that can show a disproportionation reaction

Answer :

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

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

Question 7.25 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 7.26 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) $Ag^{+}+e^{-}\rightarrow Ag(s)]*2: E^{0} = +0.80V$

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

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

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

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

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

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

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

(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 7.27(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 7.27(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 the cathode $Ag$ will be deposited.

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

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

Question 7.27(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 7.27(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 the over potential of oxygen, chloride ion gets oxidized at the anode.

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

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

Question 7.28 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 7.29 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 7.30 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 electrodes 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 (acts as an anode)

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

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

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

Class 11 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions

Question: Given below are two statements :
Statement I : Mohr's salt is composed of only three types of ions-ferrous, ammonium, and sulphate.
Statement II : If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are $\mathrm{x}_1, \mathrm{x}_2$ and $\mathrm{x}_3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by $\mathrm{x}_1+\mathrm{x}_2+2 \mathrm{x}_3$
In the light of the given statements, choose the correct answer from the options given below :

1)Both statements I and Statement II are false

2)Statement I is false but Statement II is true

3)Statement I is true but Statement II are false

4)Both statements I and Statement II are true

Answer:

Mohr's salt : $\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}$
Using Kohlrousch law

$\lambda_{\mathrm{m}}^*(\text { Mohr's salt })=\mathrm{x}_1+2 \mathrm{x}_2+2 \mathrm{x}_3$

Hence, the correct answer is option (3).


Approach to Solve Questions of Chapter 7 Redox Reactions

Sometimes, problems related to Redox Reactions seem difficult, but once we understand the basic rules and strategy, it becomes very easy to solve all the questions related to Redox reactions.

We can follow the steps given below to solve the questions based on Redox Reactions :

1). First of all we need to understand the type of question asked, basically numerical of Redox Reactions can be divided into the following categories:

  • Calculation of oxidation number
  • Balancing of Redox Reaction
  • Identifying the oxidising and reducing agent
  • Equivalent weight n-factor
  • Calculating the amount of the substance using Redox titration

2). Assign the oxidation numbers by using the oxidation number rule to find the oxidation state of each element.

3). Then we have to identify the oxidising and the reducing agent.

  • An increase in oxidation number is oxidation
  • The decrease in oxidation number is a Reduction

4). The next step to solve questions is the balancing of the Redox Reactions.

5). For redox titrations we can use the n-factor to find equivalent

These are the steps that can be followed while solving problems related to redox reactions. If students have a basic understanding of rules then redox problems can be easily solved.

Topics of NCERT Syllabus Class 11 Chemistry Redox Reactions

7.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions
7.2 Redox Reactions in Terms of Electron Transfer Reactions
7.3 Oxidation Number
7.4 Redox Reactions and Electrode Processes

What Extra Should Students Study Beyond NCERT for JEE/NEET?

ConceptJEENCERT
CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION REACTIONS
Oxidation Number
Types of Redox Reactions
Displacement Reaction
Balancing of Redox Reaction: Ion Electrode Method
Balancing of Disproportionation Redox Reaction: Ion Electrode Method
Balancing of Redox Reaction: Oxidation Number Method

Importance of Chapter - 7 Redox Reactions for JEE and NEET Exam

Redox Reaction is an important chapter from the point of view of competitive exams as the chapter is very scoring as well as it will form the basis of the chapters students are going to learn in future. Some of the topics like Redox balancing are frequently asked in JEE mains. The basics of Redox Reactions play a crucial role in Electrochemistry, Thermodynamics, Metallurgy, etc. Many JEE and NEET questions are directly asked from the concepts given in NCERT. Redox reactions act as the link between inorganic chemistry and Biology. Redox reactions develop problem-solving and analytical abilities in students.

NCERT Solutions for Class 11 Chemistry

NCERT solutions for Class 11 Chemistry Chapter 1 Some basic concepts of Chemistry
NCERT solutions for Class 11 Chemistry Chapter 2 Structure of Atom
NCERT solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

NCERT solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

NCERT solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT solutions for Class 11 Chemistry Chapter 6 Equilibrium

NCERT solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NCERT solutions for Class 11 Chemistry Chapter 8 Organic chemistry- some basic principles and techniques
NCERT solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

NCERT Solutions for Class 11 Subject Wise

Also Check,

Frequently Asked Questions (FAQs)

1. How do you balance redox reactions using the oxidation number method?

In order to balance redox reaction using oxidation number method, we have to follow the steps given below: 

  • First identify the oxidation number of all the elements given in the reaction.
  • Then determine which element undergoes oxidation and which undergoes reduction by noting the changes in oxidation number.
  • Calculate the changes in oxidation numbers, if the element undergoes oxidation the oxidation number will increase and if element undergoes reduction the oxidation number will decrease. 
  • Equate the total increase or decrease by multiplying with appropriate coefficients
  • Except H and O balance all the remaining elements 
  • Balance oxygen atom by adding H2O
  • Balance hydrogen atoms by adding Hor OH
  • Now the last step is to check whether charge and number of atoms are equal on both sides.
2. How do you identify the oxidizing and reducing agents in a reaction?

 In order to identify the oxidizing and reducing agents in the reaction, first we need to assign oxidation numbers to all the involved elements.

  • If the elements of substance show an increase in oxidation number means the elements are undergoing oxidation and are reducing agents.
  • If the elements of substance show an decrease in the oxidation number means the elements are undergoing reduction and are oxidising agents. 

 

3. What is the difference between oxidation number and valency?

Oxidation number 

Valency 

1). Imaginary charge an atom have in an ionic compound 

1). It is the combining capacity of atoms to form bonds 

2). Oxidation number can be positive, negative and zero

2). Always positive 

3). Varies for same elements in different compounds

3). Constant for an element in most of the compounds 

4). Shows loss and gain of electrons during a reaction. 

4). Valence shows the number of electrons required to complete an octet. 

5). Example : Oxidation number of Fe in FeCl3 is +3. 

5). Valency of Fe is usually 2 or 3.

4. What are the rules for assigning oxidation numbers?

 Rules of assigning oxidation number are given below: 

  • Oxidation number of atoms in its elemental form is always 0.
  • Oxidation number of a monatomic ion is equal to its charge.
  • The sum of oxidation numbers in neutral compounds is 0. 
  • In polyatomic ions, the sum of oxidation numbers is equal to the charge of ions. 
5. Why are redox reactions important in biological systems?

Redox reactions are important in the biological system because they are fundamental to metabolism and production of energy. They are used in cellular respiration, photosynthesis, detoxification, immune response and help in maintaining the cell balance in ions and molecules.

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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

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