NCERT Solutions for Class 11 Chemistry Chapter 8 Redox ...

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

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

following species

Answer :

solution-

O.N is the oxidation number

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

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

O.N of sodium ( ) = +1

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O.N of aluminium ( ) = +3

O.N of potassium ( )= +1

O.N of calcium ( ) = +2

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

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

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

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

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

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

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

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

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

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

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

Question 8.2 What are the oxidation number of the underlined elements in each of the

following and how do you rationalize your results

Answer :

Solution-

O.N of potassium ( )= +1

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

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

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1*1 + 3*x = 0

x = (-1/3)

average O. N. Of is . But it is wrong because O.N cannot be fractional. So lets try

with structure of

O.N of = -1 (because a coordinate bond is formed between and ion. Hence O. N

of three atoms are 0,0 and -1, O.N 0 in moleculeand -1 in ion.)

Assume O.N of S is x

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

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]

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If you calculate the oxidation number of Fe in it would be 8/3 and however, O.N

cannot be in fractional.

Here one iron atom has

+2 O.N and the other two are of +3 O.N.

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.

suppose the oxidation number of Carbon is x.

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

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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-

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

Answer :

Solution-

Let us write the O.N of each element

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

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Answer :

Solution-


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

Answer :

Solution-


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 to +1 in . Hence it is a redox reaction.

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

Answer :

Solution-

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We know that oxidation = loosing of by atom

and reduction = gaining of by another atom

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

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

Answer :

Solution-

here oxidation reaction

and 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

Justify that this reaction is a redox reaction

Answer :

Solution-

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Oxidation state of

Here 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 , and . Suggest the structure of these compounds. Count for

the fallacy.

Answer :

(i) let the oxidation number of sulphur be x

So,

There is a fallacy Sulphur cannot have +8 oxidation state because it has maximum +6

oxidation number, not more than that.The structure of is shown as follows:

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(ii)

let the oxidation number of chromium be x

now

There is no fallacy here

(iii)

let assume oxidation number of N is x

Now,

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here is no fallacy about the O.N of N in

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

Mercury(II) chloride

Answer :

Answer-

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

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

Nickel(II) sulphate

Answer :

Answer-

sulphate has oxidation state

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

Tin(IV) oxide

Answer :

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Oxygen has oxidation state

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

Thallium(I) sulphate

Answer :

Answer-

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

Iron(III) sulphate

Answer :

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

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

Chromium(III) oxide

Answer :

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

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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 :

Answer-

the substance of Carbon -

Substance O.N. Of C

-4

-3

-2

-1

0

+1

+2

+3 Aak

ash I

nstitu

te

Substance O.N. Of C

+4

substance for Nitrogen-

-3

-2

-1

0

+1

+2

+3

+4

+5

Substance O.N. Of N

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 :

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• sulphur dioxide 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 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 , 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 , 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:

Why it is more appropriate to write these reactions as :

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

reactions.

Answer :

Answer-

(a) In the photosynthesis process-

step 1- the liberation of and -->

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step-2 The produced in above reduces the into glucose and

water

So, the final net reaction is

It is more appropriate to write the reaction as above because water molecule

also produced in photosynthesis reaction.

The path of reaction can be investigated by using the radioactive instead

of

(b)

(the final net reaction)

Dioxygen is produced from both steps, one from the decomposition of ozone ( ) and

other is from the reaction of hydrogen peroxide with(O)

• The path of the reaction can be investigated by using .

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 :

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Answer-

These can be understood by the following examples-

• is reducing agent and is an oxidizing agent

[O.N of phosphorus +5] Higher O.S of P

[O.N of phosphorus +3] Lower O.S of P

• is an is a reducing agent and oxidising agent

(O is in excess) [O.N of C +4]

(C is in excess) [O.N of C +2]

• is a reducing agent and is an oxidizing agent

(K is in excess) [O.N of O -2] (lower O.S.)

(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 :

Answer(a)-

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Alcohol and 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-

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 :

Answer(b)-

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

HCl is a weak reducing agent and it cannot reduce to thats why we get

colourless pungent smelling gas HCl.

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

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When conc. sulphuric acid is added to an inorganic mixture containing bromide initially it

produces and it is a strong reducing agent so it reduces to 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

Answer :

Answer(a)-

Substance reduced/oxidizing agent-

Substance oxidized/reducing agent-

Question 8.13(b) Identify the substance oxidised reduced, oxidising agent and reducing


Answer :

Answer(b)-

Substance reduced/oxidising agent-

Substance oxidised/reducing agent-

Question 8.13(d) Identify the substance oxidised reduced, oxidising agent and reducing


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Answer :

Answer-



Question 8.13(e) Identify the substance oxidised reduced, oxidising agent and reducing


Answer :

Answer-



Question 8.14 Consider the reactions :

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

Answer :

Answer-

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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 :

Answer-

part(i)

Fluorine can oxidize other halogen ions. On the other hand cannot

oxidize

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

part(ii)

& are able to reduce but are unable to reduce

sulphuric acid.

So here we can say that & are better reductant than .

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Again can only able to reduce but cannot.

Hence among hydrohalic compound hydroiodic acid is the best reductant.

Question 8.16 Why does the following reaction occur?

What conclusion about the compound (of which is a part) can be

drawn from the reaction ?

Answer :

Answer-

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

and oxidation state of F changes from -1 to 0

is reduced by accepting an electron.

It is a strong oxidizing agent than F

Question 8.17(a) Consider the reactions:

(a)

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What inference do you draw about the behaviour of and from these reactions

?

Answer :

In the first reaction, we can see that oxidizes the phosphorus from (+1 +5) also

in second, we clearly see that oxidize the phosphorus from (+1 +5).

Both are oxidizing agents.

Question 8.17(b) Consider the reactions:

No change observed

Answer :

Here, by looking at the reaction, we conclude that oxidises and in the

second reaction not able to oxidise. So we can say that is stronger oxidizing

agent than .

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

(In basic medium)

Answer :

reduction half reaction

(+7 to +4)

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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

oxidation half

balance it

equalising the no. of electrons by multiplying the oxidation half by 3 and reduction half

by 2 and then add it.

final answer-

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

(In Acidic medium)

Answer -

oxidation half reaction

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reduction half reaction

Balancing the reaction

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

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

(c)

in acidic medium

Answer :

In acidic medium

oxidation half reaction-

reduction half reaction-

Balancing the reaction

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

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

in acidic medium

Answer :

Half-reaction

oxidation half

reduction half

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

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

Answer :

Answer-

Carbon shows different oxidation state according to the compound formula.

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

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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 ion is unstable in solution and undergoes disproportionation

to give , and ion. Write a balanced ionic equation for the reaction

Answer :

Answer-

The base equation

write oxidation half with their oxidation state

Balance the charge on by adding 1 on RHS side. To balance charge

add ions on RHS side and then for oxygen balance add molecule on LHS

side.

reduction half

balancing the reduction half by adding 1 on LHS side

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Add both balanced reduction half and oxidation half

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

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

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

oxidation state.

Answer :

Answer-

(a) (because of highly electronegative in nature)

(b) (highly electropositive)

(c) (has emplty d orbitals)

(d) (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 :

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Answer-

Base equation- --------------(have to remember)

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

oxidation half =

balancing - Oxygen is balanced by adding water molecule, Hydrogen is balanced

by ion and for charge add (electron)

Reduction hallf =

Balancing - to balance charge add an electron

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

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 :

Answer

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

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

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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 = 10/17 = 0.588

No. of moles of oxygen = 20/32= 0.625

Balanced Reaction

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

0.588 moles of ammonia = moles of . But we have only

0.625 moles of .

It means oxygen is a limiting reagent and the maximum weight of nitric oxide can

be produced by 0.635 moles of

So, 5 moles of produced 4 moles of C.

therefore 0.625 moles of = moles of .

from Eq. 1

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mass of = number of moles * molecular weight

=

= 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) and

(b) and Cu(s)

(c) (aq) and Cu(s)

(d) Ag(s) and (aq)

(e) and .

Answer :

If for the overall reaction is positive feasible

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negative not feasible

(a)

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

(b)

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

(c)

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

(d)

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

(e)

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

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

(i) An aqueous solution of with silver electrodes

Answer :

Answer-

(i) dissociate into and

@ Cathode - (reduction potential of silver is higher

than )

@ Anode - ( oxidation potential of silver is higher than water

molecule.So silver electrode oxidized )

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

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An aqueous solution with platinum electrodes

Answer :

Answer-

(ii) since platinum electrode cannot easily oxidize. So at the anode will

oxidize and liberate oxygen and at cathode will be deposited.

At cathode-

At anode-

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

(iii) A dilute solution of with platinum electrodes

Answer :

Answer-

given sulphuric acid is dilute.

ionize into

At cathode

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

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

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An aqueous solution of CuCl2 with platinum electrodes

Answer :

Answer-

(iv) In aqueous solution ionise into and

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-

@ cathode-

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.

Question 8.29 Given the standard electrode potentials

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arrange these metals in their increasing order of reducing power.

Answer :

Answer-

A negative electrode potential means redox couple is a stronger reducing agent.

So as per data the increasing order of the following is-

Question 8.30 Depict the galvanic cell in which the reaction

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

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(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-

At Cathode

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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox ...

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