Mole Concept 2 Theory E

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CHEMISTRY

Mole Concept-2Oxidation & Reduction

Let us do a comparative study of oxidation and reduction :

Oxidation Reduction

1. Addition of Oxygen 1. Removal of Oxygen

e.g. 2Mg + O2 2MgO e.g. CuO + C Cu + CO

2. Removal of Hydrogen 2. Addition of Hydrogen

e.g. H2S + Cl

2 2HCl + S e.g. S + H

2 H

2S

3. Increase in positive charge 3. Decrease in positive charge

e.g. Fe2+ Fe3+ + e– e.g. Fe3+ + e– Fe2+

4. Increase in oxidation number 4. Decrease in oxidation number

(+2) (+4) (+7) (+2)

e.g. SnCl2 SnCl

4e.g. MnO

4– Mn2+

5. Removal of electron 5. Addition of electron

e.g. Sn2+ Sn4+ + 2e– e.g. Fe3+ + e– Fe2+

Oxidation Number

It is an imaginary or apparent charge developed over atom of an element when it goes from its

elemental free state to combined state in molecules.

It is calculated on basis of an arbitrary set of rules.

It is a relative charge in a particular bonded state.

In order to keep track of electron-shifts in chemical reactions involving formation of compounds, a

more practical method of using oxidation number has been developed.

In this method, it is always assumed that there is a complete transfer of electron from a less

electronegative atom to a more electronegative atom.

Rules governing oxidation numberThe following rules are helpful in calculating oxidation number of the elements in their different

compounds. It is to be remembered that the basis of these rule is the electronegativity of the

element .

Fluorine atom :

Fluorine is most electronegative atom (known). It always has oxidation number equal to –1 in all its

compounds

Oxygen atom :

In general and as well as in its oxides , oxygen atom has oxidation number equal to –2.

In case of (i) peroxide (e.g. H2O

2,, Na

2O

2) is –1,

(ii) super oxide (e.g. KO2) is –1/2

(iii) ozonide (e.g. KO3) is –1/3

(iv) in OF2

is + 2 & in O2F

2is +1

Hydrogen atom :

In general, H atom has oxidation number equal to +1. But in metallic hydrides ( e.g. NaH, KH), it is –1.

Halogen atom :

In general, all halogen atoms (Cl, Br , I) have oxidation number equal to –1.

But if halogen atom is attached with a more electronegative atom than halogen atom, then it will

show positive oxidation numbers.

e.g.3

5

ClOK

,3

5

OHI

,4

7

OHCI

,5

3KBrO


CHEMISTRY

Metals :

(a) Alkali metal (Li , Na, K, Rb, .......) always have oxidation number +1

(b) Alkaline earth metal (Be , Mg , Ca .......) always have oxidation number +2.

(c) Aluminium always has +3 oxidation number

Note : Metal may have negative or zero oxidation number

Oxidation number of an element in free state or in allotropic forms is always zero

e.g.0

3

0

4

0

8

0

2 O,P,S,O

Sum of the oxidation numbers of atoms of all elements in a molecule is zero.

Sum of the oxidation numbers of atoms of all elements in an ion is equal to the charge on the ion .

If the group number of an element in modern periodic table is n, then its oxidation number may vary

from

(n – 10) to (n – 18) (but it is mainly applicable for p-block elements )

e.g. N- atom belongs to 15th group in the periodic table, therefore as per rule, its oxidation number

may vary from

–3 to +5 (2

3

3

NO,HN

, 32

3

ON

,2

4

ON

,52

5

ON

)

The maximum possible oxidation number of any element in a compound is never more than the

number of electrons in valence shell.(but it is mainly applicable for p-block elements )

Calculation of average oxidation number :

Example-1 Calculate oxidation number of underlined element :

(a) Na2S

2O

3(b) Na

2S

4O

6

Solution. (a) Let oxidation number of S-atom is x. Now work accordingly with the rules given before .

(+1) × 2 + (x) × 2 + (–2) ×3 =0

x = + 2

(b) Let oxidation number of S-atom is x

(+1) × 2 + (x) × 4 + (–2) × 6 = 0

x = + 2.5

It is important to note here that Na2S

2O

3have two S-atoms and there are four S-atom in Na

2S

4O

6.

However none of the sulphur atoms in both the compounds have + 2 or + 2.5 oxidation number, it is

the average of oxidation number, which reside on each sulphur atom. Therefore, we should work to

calculate the individual oxidation number of each sulphur atom in these compounds.

Calculation of individual oxidation numberIt is important to note that to calculate individual oxidation number of the element in its compound one should

know the structure of the compound and use the following guidelines.

Formula :

Oxidation Number = Number of electrons in the valence shell – Number of electrons taken up after bonding

Guidelines : It is based on electronegativity of elements.

1. If there is a bond between similar type of atom and each atom has same type of hybridisation, then

bonded pair electrons are equally shared by each element.

Example : Calculate oxidation number of each Cl-atom in Cl2molecule

Structure :


CHEMISTRY

: Number of electrons in the valence shell = 7.

Number of electrons taken up after bonding = 7.

oxidation number = 7 – 7 = 0.

: similarly, oxidation number = 7 – 7 = 0

2. If there is a bond between different type of atoms :

e.g. A – B (if B is more electronegative than A)

Then after bonding, bonded pair of electrons are counted with B - atom .

Example : Calculate oxidation number of each atom in HCl molecule

Structure :

Note : Electron of H-atom is now counted with Cl-atom, because Cl-atom is more electronegative

than H-atom

H : Number of electrons in the valence shell = 1

Number of electrons taken up after bonding = 0

Oxidation number of H = 1 – 0 = + 1

Cl : Number of electrons in the valence shell = 7

Number of electrons taken up after bonding = 8

Oxidation number of Cl = 7– 8 = – 1

Example-2 Calculate individual oxidation number of each S-atom in Na2S

2O

3(sodium thiosulphate) with the

help of its structure .

Solution. Structure :

Note : (central S-atom) is sp3 hybridised (25% s-character) and (terminal S-atom) is sp2

hydbridised (33% s-character). Therefore, terminal sulphur atom is more electronegative

than central sulphur atom. Now, the shared pair of electrons are counted with terminal S-

atom.

, S-atom : Number of electrons in the valence shell = 6

Number of electrons left after bonding = 0

Oxidation number of central S-atom = 6 – 0 = + 6

, S-atom : Number of electrons in the valence shell = 6

Number of electrons left after bonding = 8

Oxidation number of terminal S-atom = 6 – 8 = – 2

Now, you can also calculateAverage Oxidation number of S =2

)2(6 = + 2 (as we have calculated before)


CHEMISTRY

Miscellaneous Examples :

In order to determine the exact or individual oxidation number we need to take help from

the structures of the molecules. Some special cases are discussed as follows:

The structure of CrO5is

O O

O O

Cr||O

From the structure, it is evident that in CrO5

there are two peroxide linkages and one double bond. The

contribution of each peroxide linkage is –2. Let the oxidation number of Cr is x.

x + (–2)2 + (–2) = 0 or x = 6

Oxidation number of Cr = + 6 Ans

The structure of H2SO

5is H O O S

O

OOH

From the structure, it is evident that in H2SO

5, there is one peroxide linkage, two sulphur-oxygen double

bonds and one OH group. Let the oxidation number of S = x.

(+ 1) + (– 2) + x + (–2) 2+ (–2) + 1 = 0

or x + 2 – 8 = 0 or x – 6 = 0 or x = 6

Oxidation number of S in H2SO

5is + 6 Ans.

Paradox of fractional oxidation number

Fractional oxidation number is the average of oxidation state of all atoms of element under examination and

the structural parameters reveal that the atoms of element for whom fractional oxidation state is realised a

actually present in different oxidation states. Structure of the species C3O

2, Br

3O

8and S

4O

62– reveal the

following bonding situations :

The element marked with asterisk (*) in each species is exhibiting different oxidation number from

rest of the atoms of the same element in each of the species. This reveals that in C3O

2, two carbon

atoms are present in +2 oxidation state each whereas the third one is present in zero oxidation state

and the average is + 4/3. However, the realistic picture is +2 for two terminal carbons and zero for

the middle carbon.

OC*CCO202

Structure of C3O

2

(Carbon suboxide)

Likewise in Br3O

8, eachof the two terminalbromine atoms are present in +6 oxidationstate and themiddle

bromine is present in +4 oxidation state. Once again the average, that is different from reality, is + 16/3.

In the same fashion, in the species S4O

62–, average oxidation number of S is + 2.5, whereas the

reality being +5,0,0 and +5 oxidation number respectively for respective sulphur atoms.

In general, the conclusion is that the idea of fractional oxidation state should be taken with care

and the reality is revealed by the structures only.


CHEMISTRY

Oxidising and reducing agentOxidising agent or Oxidant :

Oxidising agents are those compounds which can oxidise others and reduce itself during the chemical

reaction. Those reagents in which for an element, oxidation number decreases or which undergoes

gain of electrons in a redox reaction are termed as oxidants.

e.g. KMnO4, K

2Cr

2O

7, HNO

3, conc.H

2SO

4etc are powerful oxidising agents .

Reducing agent or Reductant :

Reducing agents are those compounds which can reduce other and oxidise itself during the chemical

reaction. Those reagents in which for an element, oxidation number increases or which undergoes

loss of electrons in a redox reaction are termed as reductants.

e.g. K , Na2S

2O

3etc are the powerful reducing agents.

Note : There are some compounds also which can work both as oxidising agent and reducing agent

e.g. H2O

2, NO

2–

HOW TO IDENTIFY WHETHER APARTICULAR SUBSTANCE IS AN OXIDISING OR AREDUCING AGENT

Redox reactionA reaction in which oxidation and reduction simultaneously take place is called a redox reaction

In all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation

number.

e.g. 10 4

2

SOFe

+5

4KMnO2

+ 8H2SO

4 342

3

SOFe5

+ 4

2

SOMn2

+ K2SO

4+ 8H

2O

Disproportionation Reaction :A redox reaction in which same element present in a particular compound in a definite oxidation state is

oxidized as well as reduced simultaneously is a disproportionation reaction.

Disproportionation reactions are a special type of redox reactions. One of the reactants in a disproportionation

reaction always contains an element that can exist in at least three oxidation states. The element in

the form of reacting substance is in the intermediate oxidation state and both higher and lower oxidation

states of that element are formed in the reaction. For example :

)aq(OH21

22

)(OH22

2

+ )g(O2

0

)s(S0

8+ 12OH¯(aq) )aq(S4

22

+ )aq(OS2

2232

+ 6H

2O ()

)g(Cl0

2+ 2OH¯(aq) )aq(ClO

1 + )aq(Cl

1 + H

2O ()

Consider the following reactions :

(a) 2KClO3

2KCl + 3O2

KClO3

plays a role of oxidant and reductant both. Here, Cl present in KClO3

is reduced and O

present in KClO3

is oxidized. Since same element is not oxidized and reduced, so it is not a

disproportionation reaction, although it looks like one.


CHEMISTRY

(b) NH4NO

2 N

2+ 2H

2O

Nitrogen in this compound has -3 and +3 oxidation number, which is not a definite value. So it is not

a disproportionation reaction. It is an example of comproportionation reaction, which is a class of

redox reaction in which an element from two different oxidation state gets converted into a single

oxidation state.

(c)5

3KClO4

7

4KClO3

+1–

KCl

It is a case of disproportionation reaction and Cl atom is disproportionating.

List of some important disproportionation reactions

1. H2O

2 H

2O + O

2

2. X2+ OH–(dil.) X¯ + XO¯ (X = Cl, Br, I)

3. X2+ OH–(conc.) X¯ + XO

3¯

F2

does not undergo disproportionation as it is the most electronegative element.

F2

+ NaOH(dil.) F– + OF2

F2

+ NaOH(conc.) F– + O2

4. (CN)2+ OH– CN– + OCN–

5. P4

+ OH– PH3

+ H2PO

2¯

6. S8+ OH– S2– + S

2O

32–

7. MnO4

2– MnO4¯ + MnO

2

8. NH2OH N

2O + NH

3

NH2OH N

2+ NH

3

9. Oxyacids of Phosphorus ( +1, +3 oxidation number)

H3PO

2 PH

3+ H

3PO

3

H3PO

3 PH

3+ H

3PO

4

10. Oxyacids of Chlorine( Halogens)( +1, +3, +5 Oxidation number)

ClO– Cl– + ClO2

–

ClO2

– Cl– + ClO3

–

ClO3

– Cl– + ClO4

–

11. HNO2

NO + HNO3

Reverse of disproportionation is called Comproportionation. In some of the disproportionation

reactions, by changing the medium (from acidic to basic or reverse), the reaction goes in backward

direction and can be taken as an example of Comproportionation reaction.

¯ + O3¯ + H+

2+ H

2O

Balancing of redox reactionsAll balanced equations must satisfy two criteria.

1. Atom balance (mass balance ) :

There should be the same number of atoms of each kind on reactant and product side.

2. Charge balance :

The sum of actual charges on both sides of the equation must be equal.

There are two methods for balancing the redox equations :

1. Oxidation - number change method

2. Ion electron method or half cell method

Since First method is not very much fruitful for the balancing of redox reactions, students are advised

to use second method (Ion electron method ) to balance the redox reactions


CHEMISTRY

Ion electron method :

By this method redox equations are balanced in two different medium.

(a) Acidic medium (b) Basic medium

Balancing in acidic medium

Students are adviced to follow the following steps to balance the redox reactions by Ion electron

method in acidic medium

Example-3 Balance the following redox reaction :

FeSO4+ KMnO

4+ H

2SO

4 Fe

2(SO

4)

3+ MnSO

4+ H

2O + K

2SO

4

Solution. Step– Assign the oxidation number to each element present in the reaction.

4

262

OSFe

+ 4

2–71

MnOK

+ 4

26

2

1

OSH

34

26

2

3

)OS(Fe

+ 4

262

OSMn

+2

2

1

OH

Step :

Now convert the reaction in Ionic form byeliminating the elements or species, which are not undergoing

either oxidation or reduction.

Fe2+ +

4

7

OMn Fe3+ + Mn2+

Step :

Now identify the oxidation / reduction occuring in the reaction

Step V : Spilt the Ionic reaction in two half, one for oxidation and other for reduction.

Fe2+ oxidationFe3+

2ductionRe4 MnMnO

Step V :

Balance the atom other than oxygen and hydrogen atom in both half reactions

Fe2+ Fe3+ MnO4

– Mn2+

Fe & Mn atoms are balanced on both side.

Step V :

Now balance O & H atom by H2O & H+ respectively by the following way : For one excess oxygen

atom, add one H2O on the other side and two H+ on the same side.

Fe2+ Fe3+ (no oxygen atom ) .................(i)

8H+ + MnO4

– Mn2+ + 4H2O ................(ii)

Step V :

Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance the

charge, add electrons to the electrically positive side.

Fe2+ oxidation Fe3+ + e– ............(1)

5e– + 8H+ + MnO4

– ductionRe Mn2+ + 4H

2O ............(2)

Step V :

The number of electrons gained and lost in each half -reaction are equalised by multiplying both the

half reactions with a suitable factor and finally the half reactions are added to give the overall balanced

reaction.

Here, we multiply equation (1) by 5 and (2) by 1 and add them :

Fe2+ Fe3+ + e– ..........(1) × 5


CHEMISTRY

OH4MnFe5MnOH8Fe5

1)2.........(OH4MnMnOH8e5

223

42

22

4

(Here, at his stage, you will get balanced redox reaction in Ionic form)

Step X :

Now convert the Ionic reaction into molecular form by adding the elements or species, which are

removed in step (2).

Now, by some manipulation, you will get :

5 FeSO4+ KMnO

4+ 4H

2SO

4

2

5Fe

2(SO

4)

3+ MnSO

4+ 4H

2O +

2

1K

2SO

4or

10FeSO4

+ 2KMnO4

+ 8H2SO

4 5Fe

2(SO

4)

3+ 2MnSO

4+ 8H

2O + K

2SO

4.

Balancing in basic medium :

In this case, except step VI, all the steps are same. We can understand it by the following example:

Example-4 Balance the following redox reaction in basic medium :

ClO– + CrO2

– + OH– Cl– + CrO4

2– + H2O

Solution. By using upto step V, we will get :

24

6Oxidation

2

3–ductionRe

1

OCrOCrClOCl

Now, students are advised to follow step VI to balance ‘O’ and ‘H’ atom.

2H+ + ClO– Cl– + H2O | 2H

2O+ CrO

2– CrO

42– + 4H+

Now, since we are balancing in basic medium, therefore add as many as OH– on both side of

equation as there are H+ ions in the equation.

2OH– + 2H+ + ClO– Cl– + H2O +2OH– 4OH– + 2H

2O + CrO

2–

CrO4

2– + 4H+ + 4OH–

Finally you will get Finally you will get

H2O + ClO– Cl– + 2OH– ...........(i) 4OH– + CrO

2– CrO

42– + 2H

2O ........... (ii)

Now see equation (i) and (ii) in which O and H atoms are balanced by OH– and H2O

Now from step VIII

2e– + H2O + ClO– Cl– + 2OH– ............. (i) ×3

4OH– + CrO2

– CrO4

2– + 2H2O + 3e– ............. (ii) ×2

–––––––––––––––––––––––––––––––––––––––––––––––––––

Adding : 3ClO– + 2CrO2

– + 2OH– 3Cl– + 2CrO4

2– + H2O

Concept of equivalentsEquivalent mass of element

Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by

mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the

equivalent weight of that element.

e.g. 2Mg + O2

2MgO

48g 32g

12g 8g

32 g of O2

reacts with 48 g of Mg

8 g of O2

=32

848= 12 g

Equivalent weight of Mg = 12


CHEMISTRY

Similarly, Zn + H2SO

4 ZnSO

4+ H

2

65.5 g 32.75

Equivalent weight of Zn =2

5.65= 32.75 g

Al +2

3Cl

2 AlCl

3

27 g2

3× 71 g

111.5 g chlorine reacts with 27 g of Al.

35.5 chlorine reacts with5.111

5.3527 = 9.0 g of Al

Equivalent weight of aluminium =3

27= 9.0

As we can see from the above examples that equivalent weight is the ratio of atomic weight and a factor

(say n-factor or valency factor) which is in above three cases is their respective valencies.

Equivalent weight (E) :

In general, Eq. wt. (E) = x

M

factorn

.wt.Mol

)f.v(factorvalency

weightMolecularorweightAtomic

Number of Equivalents = speciesthatof.wt.eq

speciesofmass

For a solution, Number of equivalents = N1V

1, where N is the normality and V is the volume in litres

Equivalent mass is a pure number which, when expressed in gram, is called gram equivalent mass.

The equivalent mass of substance may have different values under different conditions.

There in no hard and fast rule that equivalent weight will be always less than the molecular

mass.

Valency factor calculation :

For Elements :

Valency factor = valency of the element.

For Acids :

Valency factor = number of replaceable H+ ions per acid molecule

Example-5 HCl , H2SO

4H

3PO

4H

3PO

3

{see there are only two replaceable H+ions}

Solution. Valency factor 1 2 3 2

(assume 100% dissicoiation)

Eq. wt. (E) 1

M

2

M

3

M

2

M

Replaceable hydrogen atoms are those hydrogen atoms which are attached with the atoms of group

VI and group VII i.e. O,S,Se,Te, & F, Cl ,Br ,I.


CHEMISTRY

For Bases :

Valency factor = number of replacable OH– ions per base molecule.

Example-6 NaOH, KOH

Solution. v .f. 1 1

Eq. wt. 1

M

1

M

Bases may be defined as the substances in which OH group is/are directly attached with group I

elements (Li,Na, K,Rb,Cs), group II elements (Be, Mg,Ca,Ba ) or group III elements (Al, Ga,In,Tl),

transition metals, non-metallic cations like PH4

+ , NH4

+ etc.

Acid - base reaction :

In case of acid base reaction, the valence factor is the actual number of H+ or OH– replaced in the

reaction. The acid or base may contain more number of replaceble H+ or OH– than actually replaced

in reaction.

v. f. for base is the number of H+ ion from the acid replaced by each molecule of the base

Example-7 2NaOH + H2

SO4

Na2

SO4

+ 2H2O

Base Acid

Solution. Valency factor of base = 1

Here, two molecule of NaOH replaced 2H+ ion from the H2SO

4. Therefore, each molecule of NaOH

replaced only one H+ ion of acid, so v.f. = 1.

v. f. for acid is the number of OH– replaced from the base by each molecule of acid

Example-8 NaOH + H2SO

4 NaHSO

4+ H

2O

Base Acid

Solution. Valency factor of acid = 1

Here, one of molecule of H2SO

4replaced one OH– from NaOH. Therefore, valency factor for H

2SO

4

is one

Eq. wt. of H2SO

4=

1

wt.Mol

Salts :

(a) In non-reacting condition

Valency factor = Total number of positive charge or negative charge present in the compound.

Example-9 Na2CO

3, Fe

2(SO

4)3

FeSO4.7H

2O

Solution. V.f. 2 2×3 = 6 2

Eq.wt.2

M

6

M

2

M

Note : In case of hydrated salt, positive/negative charge of water molecule is not counted.

(b) In reacting condition

Example-10 Na2

CO3

+ HCl NaHCO3

+ NaCl

Base Acid


CHEMISTRY

Solution. It is an acid base reaction, therefore valency factor for Na2CO

3is one while in non-reacting condition,

it will be two.

(c) Equivalent weight of oxidising / reducing agents in a redox reaction

In case of redox change , v.f. = Total change in oxidation number per molecule .

Example-11 KMnO4

+ H2O

2 Mn2+ + O

2

Solution. Mn in KMnO4is going from +7 to +2 , so change in oxidation number per molecule of KMnO

4is 5. So

the valency factor of KMnO4is 5 and equivalent weight is

5

M.

Normality :Normality of a solution is defined as the number of equivalents of solute present in one litre (1000 mL)

solution.

Let V mL of a solution is prepared by dissolving W g of solute of equivalent weight E in water.

Number of equivalents of solute =E

W

VmL of solution containE

Wequivalents of solute

1000 mL solution will contain VE

1000W

equivalents of solute.

Normality (N) = VE

1000W

Normality (N) = Molarity x Valency factor

N × V (in mL) = M × V (in mL) × n

or

milliequivalents = millimoles × n

Example-12 Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL acidic solution.

Solution. Normality (N) =VE

1000W

Here W = 15.8 g , V = 50 mL E =factorValency

KMnOofmassmolar 4= 158/5 = 31.6

So, normality = 10 N

Example-13 Calculate the normality of a solution containing 50 mL of 5 M solution of K2Cr2O7 in acidic

medium.

Solution. Normality (N) = Molarity × valency factor

= 5 x 6 = 30 N

Law of EquivalenceThe law states that one equivalent of an element combine with one equivalent of the other. In a chemical

reaction, equivalents and milli equivalents of reactants react in equal amount to give same number of equivalents

or milli equivalents of products separately.

Accordingly

(i) aA + bB mM + nN

meq of A = meq of B = meq of M = m.eq. of N

(ii) In a compound MxN

y

meq of MxN

y= meq of M = meq of N


CHEMISTRY

Example-14 Find the number of moles of KMnO4needed to oxidise one mole Cu

2S in acidic medium.

The reaction is KMnO4

+ Cu2S Mn2+ + Cu2+ + SO

2

Solution. From law of equivalence,

equivalents of Cu2S = equivalents of KMnO

4

moles of Cu2S × v.f. = moles of kMnO

4× v.f.

1 × 8 = moles of KMnO4

× 5 moles of KMnO4

= 8/5

( v.f. of Cu2S = 2 (2 – 1) + 1 (4 – (–2))) = 8 and v.f. of KMnO

4= 1 (7 –2) = 5)

Example-15 The number of moles of oxalate ions oxidized by one mole of MnO4

– ion in acidic medium are :

(A)2

5(B)

5

2(C)

5

3(D)

3

5

Solution. Equivalents of C2O

42– = equivalents of MnO

4–

x(mole) × 2 = 1 × 5

( v.f. of C2O

42– = 2 (4 – 3) = 2 and v.f. of MnO

4– = 1 (7 – 2) = 5).

x =2

5mole of C

2O

42– ions.

Drawbacks of Equivalent concept

Since equivalent weight of a substance (for example oxidising or reducing agent) may be variable

hence it is better to use mole concept.

e.g. 5e– + 8H+ + MnO4

– Mn2+ +2H2O

Eq.wt of MnO4

– =5

MnOof.wt.Mol–

4

e.g. 3e– +2H2O + MnO

4– MnO

2+4OH–

Eq.wt of MnO4

– =3

MnOof.wt.Mol 4

Thus, the number of equivalents of MnO4

– will be different in the above two cases but

number of moles will be same.

Normality of any solution depends on reaction while molarity does not.

For example :

Consider 0.1mol KMnO4

dissolved in water to make 1L solution. Molarity of this solution is 0.1 M.

However, its normality is NOT fixed. It will depend upon the reaction in which KMnO4

participates.

e.g. if KMnO4forms Mn2+, normality = 0.1 x 5 = 0.5 N. This same sample of KMnO

4, if employed in

a reaction giving MnO2as product (Mn in +4 state), will have normality 0.1 × 3 = 0.3 N.

The concept of equivalents is handy, but it should be used with care. One must never equate

equivalents in a sequence which involves same element in more than two oxidation states. Consider

an example, KIO3reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo

solution. The reactions are :

(i) O3

– + ¯ 2

(ii) 2

+ S2O

32– S

4O

62–+ ¯

meq of hypo = meq of I2

= meq of IO3 = meq of I

meq of hypo = meq of IO3.

This is wrong. Note that I2

formed by equation (i) has v.f. = 5/3 & in equation (ii) has v.f. = 2.

v.f. of I2in both the equation are different, therefore we cannot equate milli equivalents in sequence.

In this type of case, students are advised to use mole concept.


CHEMISTRY

Example-16 How many millilitres of 0.02 M KMnO4

solution would be required to exactly titrate 25 mL of 0.2 M

Fe(NO3)

2solution in acidic medium ?

Solution. Method -1 : Mole concept method

Starting with 25 mL of 0.2 M Fe2+, we can write :

Millimoles of Fe2+ = 25 x 0.2 ........(1)

and in volume V (in milliliters) of the KMnO4,

Millimoles of MnO4¯ = V (0.02) ........(2)

The balanced reaction is :

MnO4¯ + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

This requires that at the equivalent point,

1

MnOofmoles.m –4 =

5

Feofmoles.m 2

1

)02.0(V=

5

)2.0)(25((from (1) & (2))

V = 50 mL.

Method -2 : Equivalent Method :

At the equivalence point,

milliequivalents of MnO4¯ = milliequivalents of Fe2+

M1

× vf1

× V1

= M2

× vf2

× V2

0.02 × 5 × V1

= 0.2 × 1 × 25 ( MnO4

– Mn2+ ; v.f. = 5, Fe2+ Fe3+ ; v.f. = 1)

V1

= 50 mL.

TitrationsTitration is a procedure for determining the concentration of a solution by allowing a carefully measured

volume to react with a standard solution of another substance, whose concentration is known.

Standard solution - It is a solution whose concentration is known and is taken in burette. It is also called

Titrant.

There are two type of titrants :

Primary titrants/standard - These reagents can be accurately weighed and their solutions are not

to be standardised before use.

Ex : Oxalic acid, K2Cr

2O

7, AgNO

3, CuSO

4, ferrous ammonium sulphate, hypo etc.

Secondary titrants/standard : These reagents cannot be accurately weighed and their solutions

are to be standardised before use.

Ex : NaOH, KOH, HCl, H2SO

4,

2, KMnO

4etc.

Titrate : Solution consisting of substance to be estimated, generally taken in a beaker .

Equivalence point : It is the point when number of equivalents of titrant added becomes equal to number of

equivalents of titrate.

At equivalence point :

n1V

1M

1= n

2V

2M

2

Indicator : An auxiliary substance added for physical detection of completion of titration at equivalence

point. It generally show colour change on completion of titration.

Type of Titrations :

Acid-base titrations (to be studided in Ionic equilibrium)

Redox Titrations


CHEMISTRY

Some Common Redox Titrations

Table of Redox Titrations : (Excluding Iodometric / Iodimetric titrations)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Estimation By titrating Reactions Relation*between

of with OA and RA––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

1. Fe2+ MnO4¯ Fe2+ Fe3+ + e– 5Fe2+ MnO

4¯

MnO4

– + 8H+ + 5e– Mn2+ + 4H2O Eq. wt. of Fe2+ = M/1

2. Fe2+ Cr2O

72– Fe2+ Fe3+ + e– 6Fe2+ Cr

2O

72–

Cr2O

72– + 14H+ + 6e– 2Cr3+ + 7H

2O Eq.wt. of Cr

2O

72– = M/6

3. C2O

42– MnO

4¯ C

2O

42– 2CO

2+ 2e– 5C

2O

42– 2MnO

4¯

MnO4

– + 8H+ + 5e– Mn2+ + 4H2O Eq. wt. of C

2O

42– = M/2

4. H2O

2MnO

4¯ H

2O

2 2H+ + O

2+ 2e– 5H

2O

2 2MnO

4¯

MnO4

– + 8H+ + 5e– Mn2+ + 4H2O Eq.wt. of H

2O

2= M/2

5. As2O

3MnO

4– As

2O

3+ 5H

2O 2AsO

43– + 10H+ + 4e– Eq. wt. of As

2O

3= M/4

MnO4

– + 8H+ + 5e– Mn2+ + 4H2O

6. AsO33– BrO

3– AsO

33– + H

2O AsO

43– + 2H+ + 2e– Eq. wt. of AsO

33– = M/2

BrO3

– + 6H+ + 6e– Br– + 3H2O Eq.wt. of BrO

3– = M/6

Permanganate Titrations :

KMnO4is generally used as oxidising agent in acidic medium, generally provided by dilute H

2SO

4.

KMnO4

works as self indicator persistent pink color is indication of end point.

Mainly used for estimation of Fe2+ , oxalic acid ,oxalates, H2O

2etc.

Example-17 Write the balanced reaction of titration of KMnO4

Vs oxalic acid in presence of H2SO

4.

Solution. Reaction : 2KMnO4

+ 3H2SO

4+ 5H

2C

2O

4 K

2SO

4+ 2MnSO

4+ 8H

2O + 10CO

2

Redox Changes : C2

3+ 2C4+ +2e

2

ME

422 OCH

5e + Mn7+ Mn2+

5

ME

4KMnO

Indicator : KMnO4

acts as self indicator.

Example-18 Write the balanced reaction of titration of KMnO4

Vs ferrous ammonium sulphate in presence of

H2SO

4.

Solution. Reaction : 2KMnO4+ 10[FeSO

4(NH

4)

2SO

4. 6H

2O] + 8H

2SO

4

5Fe2(SO

4)

3+ 10(NH

4)

2SO

4+ K

2SO

4+ 2MnSO

4+ 68H

2O

Redox Changes : Fe2+ Fe3+ + e

1

ME

4FeSO

Mn7+ + 5e Mn2+

5

ME

4KMnO

Indicator : KMnO4

acts as self indicator

Iodometric/Iodimetric Titrations :

Compound containing iodine are widely used in titrations.

(i) Iodide ions can be oxidised to 2

by suitable oxidising agent

2¯ (aq) 2(s) + 2e¯

(ii) Iodine (V) ions, O3¯ , will oxidise ¯ to

2

O¯ (aq) + 5¯ (aq) + 6H+ (aq) 3

2(s) + 3H

2O ()


CHEMISTRY

(iii) Thiosulphate ions, S2O

32– , can reduce iodine to iodide ions.

2S2O

(aq)

(s) S

4O

62– + 2–

colourless black colourless

Iodometric Titrations (Titration Solution is of Na2S

2O

3. 5H

2O)

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

S.No. Estimation of Reaction Relation between O.A. and R.A.––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

1. 2

2

+ 2Na2S

2O

3 2Na + Na

2S

4O

6

2 2 2Na

2S

2O

3

or 2

+ 2S2O

32– 2¯ + S

4O

62– Eq.wt. of Na

2S

2O

3= M/1

2. CuSO4

2CuSO4

+ 4K 2Cu + 2K2SO

4+

22CuSO

4

2 2 = 2Na

2S

2O

3

or 2Cu2+ + 4¯ 2Cu + 2

Eq.wt.of CuSO4

= M/1

white ppt

3. CaOCl2

CaOCl2+ H

2O Ca(OH)

2+ Cl

2CaOCl

2 Cl

2

2 2

Cl2

+ 2K 2KCl + 2

2Na2S

2O

3

Cl2

+ 2¯ 2Cl¯ + 2

Eq.wt. of CaOCl2

= M/2

4. MnO2

MnO2+ 4HCl(conc.) MnCl

2+ Cl

2+ 2H

2O MnO

2 Cl

2

2 2Na

2S

2O

3

Cl2

+ 2K 2KCl + 2

Eq.wt. of MnO2

= M/2

or MnO2

+ 4H+ + 2Cl¯ Mn2+ + 2H2O + Cl

2

Cl2

+ 2¯ 2

+ 2Cl¯

5. O3¯ O

3¯ + 5¯ + 6H+ 3

2+ 3H

2O O

3¯ 3

2 6 6Na

2S

2O

3

Eq.wt. of O3¯ = M/6

6. H2O

2H

2O

2+ 2¯ + 2H+

2+ 2H

2O H

2O

2

2 2 2Na

2S

2O

3

Eq.wt. of H2O

2= M/2

7. Cl2

Cl2

+ 2¯ 2Cl¯ + 2

Cl2

2 2 2Na

2S

2O

3

Eq.wt. of Cl2

= M/2

8. O3

O3

+ 6¯ + 6H+ 32

+ 3H2O O

3 3

2 6 6Na

2S

2O

3

Eq.wt. of O3

= M/6

9. ClO¯ ClO¯ + 2¯ + 2H+ H2O + Cl¯ +

2ClO¯

2 2 2Na

2S

2O

3

Eq.wt. of OCl– = M/2

10. Cr2O

72– Cr

2O

72– + 14H+ + 6¯ 3

2+ 2Cr3+ + 7H

2O Cr

2O

72– 3

2 6

Eq.wt. of Cr2O

72– = M/6

11. MnO4– 2MnO

4– + 10¯+ 16H+ 2MnO

4– + 5

2+ 8H

2O 2MnO

4¯ 5

2 10

Eq.wt. of MnO4¯ = M/5

12. BrO3– BrO

3– + 6¯ + 6H+ Br– + 3

2+ 3H

2O BrO

3– 3

2 6

Eq.wt. of BrO3

– = M/6

13. As(V) H2AsO

4+ 2¯+ 3H+ H

3AsO

3+ H

2O +

2H

3AsO

4

2 2

Eq.wt. of H3AsO

4= M/2

14. HNO2

2HNO2

+ 2¯ 2

+ 2NO + H2O 2HNO

2

2 2

Eq.wt. of HNO2

= M/1

15. HClO HClO + 2¯ + H+ Cl¯ + 2

+ H2O HClOI

22Na

2S

2O

3

Eq.wt. of HClO = M/2


CHEMISTRY

Iodimetric Titrations––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

S.No. Estimation of Reaction Relation between O.A. and R.A.––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

1. H2S H

2S +

2 S + 2¯ + 2H+ H

2S

2 2

(in acidic medium) Eq.wt. of H2S = M/2

2. SO32– SO

32– +

2+ H

2O SO

42– + 2¯ + 2H+ SO

32–

2 2

(in acidic medium) Eq.wt. of SO3

2– = M/2

3. Sn2+ Sn2+ + 2

Sn4+ + 2¯ Sn2+ 2 2

(in acidic medium) Eq.wt. of Sn2+ = M/2

4. As(III) (at pH 8) H2AsO

3¯ +

2+ H

2O HAsO

42– + 2¯ + 3H+ H

2AsO

3–

2 2

Eq.wt. of H2AsO

3¯ = M/2

5. N2H

4N

2H

4+ 2

2 N

2+ 4H+ + 4¯ N

2H

4= 2

2 4

Eq.wt. of N2H

4= M/4

Example-19 The sulphur content of a steel sample is determined by converting it to H2S gas, absorbing the H

2S

in 10 mL of 0.005 M I2and then back titrating the excess I

2with 0.002 M Na

2S

2O

3. If 10 mL Na

2S

2O

3

is required for the titration, how many milligrams of sulphur are contained in the sample?

Reactions :

H2S + I

2 S + 2I– + 2H+

I2+ 2S

2O

32– 2I– + S

4O

62–

Solution. Used millimoles of I2= (m.moles of I

2taken initially) –

2

usedhypoofmoles.m

= 0.005 × 10 – 0.002 ×2

10

= 0.04 = millimoles of H2S

weight of sulphur = 0.04 × 10–3 × 32 × 103 mg = 1.28 mg.

Hydrogen peroxide (H2O

2)

H2O

2can behave both like oxidising and reducing agent in both the mediums (acidic and basic).

Oxidising agent : (H2O

2H

2O)

(a) Acidic medium : 2e– + 2H+ + H2O

2 2H

2O

v.f. = 2

(b) Basic medium : 2e– + H2O

2 2OH–

v.f = 2

Reducing agent : (H2O

2 O

2)

(a) Acidic medium : H2O

2 O

2+ 2H+ + 2e–

v.f = 2

(b) Basic medium : 2OH– + H2O

2O

2+ 2H

2O + 2e–

v.f = 2

Note : Valency factor of H2O

2is always equal to 2.

Volume strength of H2O

2: Strength of H

2O

2is represented as 10V , 20 V , 30 V etc.

20V H2O

2means one litre of this sample of H

2O

2on decomposition gives 20L of O

2gas at STP.


CHEMISTRY

Decomposition of H2O

2is given as :

H2O

2 H

2O +

2

1O

2

1 mole2

1× 22.4 L O

2at STP

= 34g = 11.2 L O2at STP

To obtain 11.2 litre O2

at STP, at least 34 g H2O

2must be decomposed.

For 20 L O2, we should decompose atleast

2.11

34×20 g H

2O

2

1 L solution of H2O

2contains

2.11

34×20 g H

2O

2

1 L solution of H2O

2contains

2.11

34×

17

20equivalents of H

2O

2(

2

34

2

ME

22OH = 17)

Normality of H2O

2=

2.11

34×

17

20=

6.5

20

Normality of H2O

2(N) =

5.6

OHofstrengthVolume 22

22OHM =.f.v

N22OH

=2

N22OH

Molarity of H2O

2(M) =

11.2

OHofstrengthVolume 22

Strength (in g/L) : Denoted by S

Strength = Molarity × Mol. wt = Molarity × 34

Strength = Normality × Eq. weight = Normality × 17

Example-20 20 mL of H2O2 after acidification with dilute H2SO4 required 30 mL of12

NKMnO4 for complete

oxidation. Final the strength of H2O2 solution. [Molar mass of H2O2 = 34]

Solution. meq. of KMnO4 = meq. of H2O2

30 ×12

1= 20 × N

N =2012

30

=

8

1N

strength = N × equivalent mass =8

1× 17 = 2.12 g/L.

Hardness of water (Hard water does not give lather with soap)Temporary hardness - due to bicarbonates of Ca & Mg

Permanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we can

soften the water sample.

(a) By boiling : 2HCO3

– H2O + CO

2+ CO

32– or

By Slaked lime : Ca(HCO3)

2+ Ca(OH)

2 CaCO

3+ 2H

2O

Ca2+ + CO3

2– CaCO3

(b) By Washing Soda : CaCl2+ Na

2CO

3 CaCO

3+ 2NaCl

(c) By ion exchange resins : Na2R + Ca2+ CaR + 2Na+

(d) By adding chelating agents like (PO3

–)3etc.


CHEMISTRY

Parts Per Million (ppm)When the solute is present in very less amount, then this concentration term is used. It is defined as the

number of parts of the solute present in every 1 million parts of the solution. ppm can both be in terms of

mass or in terms of moles. If nothing has been specified, we take ppm to be in terms of mass. Hence, a 100

ppm solution means that 100 g of solute is present in every 1000000 g of solution.

ppmA

= 610massTotal

Aofmass = mass fraction × 106

Measurement of Hardness :

Hardness is measured in terms of ppm (parts per million) of CaCO3or equivalent to it.

Hardness in ppm =610

solutionofmassTotal3

CaCOofmass

Example-21 0.00012% MgSO4and 0.000111% CaCl

2is present in water. What is the measured hardness of

water and millimoles of washing soda required to purify water 1000 L water ?

Solution. Basis of calculation = 100 g hard water

MgSO4

= 0.00012g =120

00012.0mole

CaCl2

= 0.000111g =111

000111.0mole

equivalent moles of CaCO3=

111

000111.0

120

00012.0mole

mass of CaCO3

=

111

000111.0

120

00012.0× 100 = 2 × 10–4 g

Hardness (in terms of ppm of CaCO3) =

64

10100

102

= 2 ppm

CaCl2+ Na

2CO

3 CaCO

3+ 2NaCl

NaSO4+ Na

2CO

3 MgCO

3+ Na

2SO

4

Required Na2CO

3for 100g of water =

111

000111.0

120

00012.0mole

= 2 × 10–6 mole

Required Na2CO

3for 1000 litre water =

100

210

100

102 66

mole ( d = 1g/mL)

=1000

20mole = 20 m mole

Strength of Oleum :

Oleum is SO3dissolved in 100% H

2SO

4. Sometimes, oleum is reported as more than 100% by weight, say

y% (where y > 100). This means that (y 100) grams of water, when added to 100 g of given oleum sample,

will combine with all the free SO3

in the oleum to give 100% sulphuric acid.

Hence, weight % of free SO3in oleum = 80(y 100)/18


CHEMISTRY

Example-22 What volume of water is required (in mL) to prepare 1 L of 1 M solution of H2SO

4(density = 1.5g/mL)

by using 109% oleum and water only (Take density of pure water = 1 g/mL).

Solution. 1 mole H2SO

4in 1L solution = 98 g H

2SO

4in 1500 g solution = 98 g H

2SO

4in 1402 g water.

Also, in 109% oleum, 9 g H2O is required to form 109 g pure H

2SO

4& so, to prepare 98 g H

2SO

4,

water needed is 9/109 × 98 = 8.09 g.

Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL

Calculation of Available Chlorine from a sample of Bleaching Powder :

The weight of available Cl2

released from the given sample of bleaching powder on reaction with

dilute acids or CO2is called available chlorine.

224422 ClOHCaSOSOHCaOCl

2222 ClOHCaClHCl2CaOCl

222332 ClOHCOOCHCaCOOHCH2CaOCl

2322 ClCaCOCOCaOCl

Method of determination :

CaOCl2 + 2CH3COOH Ca(CH3COO)2 + H2O + Cl2(Sample of bleaching powder)

Cl2 + 2KI 2KCl + I2

I2 + 2Na2S2O3 indicatorasStarch Na2S4O6 + 2Nalv.f. = 2 v.f. = 1

End point is indicated by disappearance of blue colour.

Let M = Molarity of hypo (Na2S2O3) solution

millimoles of Cl2 produced = m.moles of I2 used by hypo

=2

VMwhere V = vol of hypo solution used in ml.

mass of Cl2 produced =2

10VM 3× 71

= 35.5 × M × V × 10–3

% of available chlorine =W

10VM5.35 3× 100

where W = amount of belaching powder taken in g.

or % of available Cl2 =W

VM3.55

Example-23 3.55 g sample of bleaching powder suspended in H2O was treated with enough acetic acid and K

solution. Iodine thus liberated required 80 mL of 0.2 M hypo for titration. Calculate the % of available

chlorine.

Solution. % of Cl2 =55.3

802.055.3 = 16%


CHEMISTRY

MISCELLANEOUS SOLVED PROBLEMS (MSPS)

1. Calculate individual oxidation number of each S-atom in Na2S

4O

6(sodium tetrathionate) with the help of its

structure .

Sol.

2. Find the average and individual oxidation number of Fe & Pb in Fe3O

4& Pb

3O

4, which are mixed oxides.

Sol. (i) Fe3O

4is mixture of FeO & Fe

2O

3in 1 : 1 ratio

so, individual oxidation number of Fe = +2 & +3

& average oxidation number =3

)3(2)2(1 = 8/3

(ii) Pb3O

4is a mixture of PbO & PbO

2in 2 : 1 molar ratio

so, individual oxidation number of Pb are +2 & +4

& average oxidation number of Pb = 3/83

)4(1)2(2

3. Balance the following equations :

(a) H2O

2+ MnO

4– Mn+2 + O

2(acidic medium)

(b) Zn + HNO3(dil) Zn(NO

3)

2+ H

2O + NH

4NO

3

(c) Cr3

+ KOH + Cl2

K2CrO

4+ KO

4+ KCl + H

2O.

(d) P2H

4 PH

3+ P

4

(e) Ca3(PO

4)

2+ SiO

2+ C CaSiO

3+ P

4+ CO

Ans. (a) 6H+ + 5H2O

2+ 2MnO

4– 2Mn+2 + 5O

2+ 8H

2O

(b) 4Zn + 10HNO3(dil) 4Zn(NO

3)

2+ 3H

2O + NH

4NO

3

(c) 2Cr3

+ 64KOH + 27Cl2

2K2CrO

4+ 6KO

4+ 54KCl + 32H

2O.

(d) 6P2H

4 8PH

3+ P

4

(e) 2Ca3(PO

4)

2+ 6SiO

2+ 10C 6CaSiO

3+ P

4+ 10CO

4. Find the valency factor for following acids

(i) CH3COOH (ii) NaH

2PO

4(iii) H

3BO

3

Ans. (i) 1 (ii) 2 (iii) 1

5. Find the valency factor for following bases :

(i) Ca (OH)2

(ii) CsOH (iii)Al(OH)3

Ans. (i) 2 (ii) 1 (iii) 3

6. Find the valence factor for following salts :

(i) K2SO

4.Al

2(SO

4)3.24H

2O

(ii) CaCO3

Ans. (i) 8 (ii) 2

7. Find the valency factor for following redox reactions :

(i)

42alkaline

2neutral

2acidic

4

MnOK

MnO

Mn

KMnO


CHEMISTRY

(ii) K2Cr

2O

7

acidic Cr3+

(iii) C2O

42 CO

2

(iv) Fe2+ Fe3+

Ans. (i) 5, 3, 1; (ii) 6 ; (iii) 2 ; (iv) 1

8. Calculate the normality of a solution obtained by mixing 50 mL of 5 M solution of K2Cr2O7 and 50 mL of

2 M K2Cr2O7 in acidic medium.

Sol. v.f. of K2Cr2O7 = 6

so Nf =21

2211

VV

VNVN

=5050

50625065

= 21 N

9. Calculate the normality of a solution containing 13.4 g of Sodium oxalate in 100 mL Sol.

Sol. Normality = litreinsolutionofvol

wt.eq/gin.wt

Here, eq. wt. of Na2C2O4 = 134/2 = 67

so N =1000/100

67/4.13= 2N

10. The number of moles of ferrous oxalate oxidised by one mole of KMnO4in acidic medium is :

(A)2

5(B)

5

2(C)

5

3(D)

3

5

Sol. Eq. of FeC2O

4= Eq. of KMnO

4

moles of FeC2O

4× 3 = moles of KMnO

4× 5

so, moles of FeC2O

4= 5/3 Ans. (D)

11. How many moles of KMnO4

are needed to oxidise a mixture of 1 mole of each FeSO4

& FeC2O

4in acidic

medium ?

(A)5

4(B)

4

5(C)

4

3(D)

3

5

Sol. Eq. of KMnO4

= Eq. of FeSO4

+ Eq. of FeC2O

4

moles of KMnO4

× 5 = moles of FeSO4

× 1 + moles of FeC2O

4× 3

moles of KMnO4

= 4/5 Ans. (A)

12. Asample of hydrazine sulphate [N2H

6SO

4] was dissolved in 100 mL water. 10 mL of this solution was treated

with excess of FeCl3Sol. Ferrous ions formed were estimated and it required 20 mL of M/50 KMnO

4solution

in acidic medium.

Fe3+ + N2H

4 N

2+ Fe2+ + H+

MnO4¯ + Fe2+ + H+ Mn2+ + Fe3+ + H

2O

(a) Write the balanced redox reactions.

(b) Estimate the amount of hydrazine sulphate in one litre of Sol.

Sol. (a) Given 4Fe3+ + N2H

4 N

2+ 4Fe2+ + 4H+

MnO4¯ +5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

(b) In 10 mL solution, eq. of N2H

6SO

4= Eq. of Fe2+ = Eq. of KMnO

4

= 20 ×50

1× 5 × 10–3 = 2 × 10–3

v.f. of N2H

6SO

4= 4

so, weight of N2H

6SO

4in 1 L solution =

104

1000102 3

× 130 = 6.5 g.


CHEMISTRY

13. Write the balanced redox reaction and calculate the equivalent weight of oxidising agent and reducing agent

for titration of K2Cr

2O

7Vs Ferrous ammonium sulphate.

Ans. The reaction : 6[FeSO4(NH

4)

2SO

4. 6H

2O] + K

2Cr

2O

7+ 7H

2SO

4

3Fe2(SO

4)

3+ Cr

2(SO

4)

3+ K

2SO

4+ 6(NH

4)

2SO

4+ 43H

2O

Redox changes :

1

ME

4FeSO ;

6

ME

722 OCrK

14. One litre of acidified KMnO4 solution containing 15.8 g KMnO4 is decolorized by passing sufficient

SO2. If SO2 is produced by FeS2, what is the amount of FeS2 required to give desired SO2 ?

Ans. 15 g.

Sol. v.f. of KMnO4

= 5 & v.f. of SO2

= 2

Now, Eq. of KMnO4

= Eq. of SO2

5/158

8.15= moles of SO

2× 2

so, moles of SO2

= 1/4

Now, applying POAC on S, we get :

2 × mole of FeS2

= 1 × moles of SO2

so, moles of FeS2

=8

1

2

1

4

1

so, weight of FeS2

=8

1× 120 = 15 g.

15. An aqueous solution containing 0.1 g KIO3

(formula weight = 214) and an excess of K was acidified with

HCl. The liberated 2consumed 45 mL of thiosulphate. The molarity of sodium thiosulphate solution is :

The reaction involved is :

O3

– + – + H+ 2

+ H2O

(A) 0.0623 M (B) 0.0313 M (C) 0.126 M (D) 0.252 M

Sol. O3

– + 5– + H+ 32

+ H2O

2Na2S

2O

3+

2 2Na + Na

2S

4O

6

Now, Moles of KO3

=214

1.0

So, Moles of 2

= 3 ×214

1.0

Now, Moles of Na2S

2O

3= 2 × 3 ×

214

1.0

M × VL

= 2 × 3 ×214

1.0 M ×

1000

45= 2 × 3 ×

214

1.0

Now, Molarity of hypo solution = 2 × 3 ×45

1000

214

1.0 = 0.0623 M Ans. (A)

16. A fresh H2O

2solution is labelled 11.2 V. This solution has the same concentration as a solution which is :

(A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these

Sol. Molarity of H2O

2=

2.11

2.11

2.11

strength.vol = 1

Now, %(w/v) = mLinsolutionof.wt

ginsoluteof.wt× 100

= Molarity × Mol. wt. of solute ×10

1

= 1 × 34 ×10

1= 3.4% Ans. (C)


CHEMISTRY

17. 100 mL each of 1 N H2O

2and 11.2 V H

2O

2solution are mixed, then the final solution is equivalent to :

(A) 3 M H2O

2solution (B) 0.5 N H

2O

2solution

(C) 25.5 g/L H2O

2solution (D) 2.55 g/L H

2O

2Sol.

Sol. Nfinal

=100100

1006.5

2.111001

VV

VNVN

21

2211

= 3/2 = 1.5 N

So, Molarity =2

5.1

.f.v

Normality = 0.75 M

Strength of solution in g/L = Molarity × Mol. wt. = 0.75 × 34 = 25.5 g/L Ans. (C)

18. Calculate the percentage of available chlorine in a sample of 3.55 g of bleaching powder which was

dissolved in 100 mL of water. 25 mL of this solution, on treatment with KI and dilute acid, required

20 mL of 0.125 N sodium thiosulphate Sol.

Ans. 10 %

Sol. CaOCl2 + H2O Ca(OH)2 + Cl2

Cl2 + 2K 2KCl + 2

2 + 2Na2S2O3 Na2S2O6 + 2Na

In 25 mL solution,

moles of Na2S2O3 =1000

20×

1

125.0= 25 × 10–4

So, moles of 2 =2

1× moles of Na2S2O3

=2

1× 25 × 10–4 = 12.5 × 10–4

So, in 100 mL solution, moles of Cl2 = 4 × 12.5 × 10–4 = 50 × 10–4

So, weight of Cl2 = 50 × 10–4 × 71 g

% of available Cl2 =55.3

711050 4

× 100 = 10%

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Mole Concept 2 Theory E