01_Stoichiometry & Redox Reaction_final

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Stoichiometry & Redox Reaction

Stoichiometry & Redox Reaction

Stoichiometry & Redox Reaction

Stoichiometry & Redox Reaction

Amedeo Avogadro

Born 09 August 1776

Died 09 July 1856

Nationality Italy

Field Chemistry

Known for Avogadro’s number

Lorenzo Romano Amedeo Carlo Avogadro, (1776 - 1856) was an Italian savant chemist, most noted for his contributions to the theory of molarity and molecular weight. As a tribute to him, the number of atoms of one mole of a substance, 6.02 × 1023 _{is known as Avogadro's number.}

In honour of Avogadro's contributions to the theory of molarity and molecular weights, the number of molecules in one mole was renamed Avogadro's number, NA. It is approximately 6.0221415 × 1023_.

Loschmidt first calculated the value of Avogadro's number, now called Avogadro's constant, which is still sometimes referred to as the Loschmidt number in German-language countries (Loschmidt constant now has another meaning). Avogadro's number is commonly used to compute the results of chemical reactions. It allows chemists to determine the exact amounts of substances produced in a given reaction.

Clausius, by his kinetic theory on gases, was able to give another confirmation of Avogadro's law. Not long after, in his researches regarding dilute solutions (and the consequent discovery of analogies between the behaviour of solutions and gases), J. H. van 't Hoff added his final consensus for the triumph of the Italian scientist, who since then has been considered the founder of the atomic-molecular theory.

SECTION-I

(For Students of Class XI)

1.1 The Concept of Atoms And Molecules

The Atoms

The smallest particle of an element that takes part in chemical reaction.

Dalton’s Atomic Theory

The main postulates of this atomic theory are

(i) Matter is discrete (i.e., discontinuous) and is made up of atoms. An atom is the smallest (chemically) indivisible particle of an element, which can take part in a chemical reaction.

(ii) Atoms of different elements have different properties and different weights.

(iii) Atoms cannot be created or destroyed. So a chemical reaction is nothing but a rearrangement of atoms and the same number of atoms must be present before and after the reaction.

(iv) A compound is formed by the union of atoms of one element with atoms of another in a fixed ratio of small whole numbers (1 : 1, 1 : 2, 2 : 3 etc).

All the postulates of Dalton’s atomic theory have been proved to be incorrect.

An atom is divisible in the sense that it has a structure. Sub-atomic particle are known as electron, proton and neutron.

The existence of isotopes for most elements show that atoms of the same element need not have the same mass. The atomic weight of an element is, in fact, a mean of the atomic masses of the different isotopes of the element.

Atomic Weight

An atom is so minute that it cannot be detected even with the most powerful microscope, let alone placed on a balance pan and weighed. So there is no question of determining the absolute weight of an atom. So chemists decided to determine the relative masses of atoms (i.e., how many times one atom of an element is heavier than another). Hydrogen atom was first selected as standard. i.e.

Atomic weight of an element = Weight of 1 atom of the element

Weight of 1 atom of hydrogen

When we state that the atomic weight of chlorine is 35.5, we mean that an atom of chlorine is 35.5 times heavier than an atom of hydrogen. It was later felt that the standard for reference for atomic weight may be oxygen, the advantage being that the atomic weights of most other elements became close to whole numbers.

Atomic weight of an element = Weight of 1 atom of the element

1

weight of 1 atom of oxygen 16×

The modern reference standard for atomic weights is carbon isotope of mass number 12. Atomic weight of an element = Weight of 1 atom of the element

1

weigth of 1 atom of carbon 12

12× −

On this basis, atomic weight of oxygen 16 was changed to 15.9994.

Nowadays atomic weight is called relative atomic mass and denoted by amu (atomic mass unit). The standard for atomic mass is C12_.

(i) Atomic weight is not a weight but a number.

(ii) Atomic weight is not absolute but relative to the weight of the standard reference element (C12_).

(iii) Gram atomic weight is atomic weight expressed in grams, but it has a special significance with reference to a mole.

Dulong and Petit measured the specific heats of a number of metals and found that the product of the specific heat and the atomic weight is a constant, having an approximate value of 6.4.

Specific heat (cal/g-deg) × atomic weight (g/g–atom) ≃ 6.4 (cal/deg.g.atom).

This correlation has been used to ‘correct’ the atomic weights of some elements in the periodic table. Dulong and Petit’s law is applicable only to metals.

The Molecule

Avogadro (1811) suggested that the fundamental chemical unit is not an atom but a molecule, which may be a cluster of atoms held together in some manner causing them to exist as a unit. The term molecule means the smallest particle of an element or a compound that can exist free and retain all its properties.

Consider a molecule of sulphur dioxide. It has been established that it contains one atom of sulphur and two atoms of oxygen. This molecule can split up into atoms of sulphur and oxygen. So the smallest particle of sulphur dioxide that can exist free and retain all its properties is the molecule of sulphur dioxide.

The term molecule is also applied to describe the smallest particle of an element which can exist free. Thus a hydrogen molecule is proved to contain two atoms; when it is split up into atoms, there will be observed a change in properties. The number of atoms of an element in a molecule of the element is called its atomicity.

Molecular Mass

Molecular mass of a substance is an additive property and can be calculated by adding the atomic masses of all the atoms of different elements present in one molecule.

Methods of determining Molecular Masses

(a) Vapour density method

V. D. = 1

2

Mass of volatile substance

Volume of vapour of its chloride at STP × 22400

Molecular mass = 2 × V.D. (b) Diffusion method

According to Graham’s law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

1 2 r r = 2 1 M M

(c) Victor Meyer method

This method applies to volatile organic liquids. Suppose vapour of an organic liquid having mass W g occupies a volume of V mL at STP. Then its molecular mass is,

Mol. mass = W

V × 22,400.

Avogadro Number (NA). The number of carbon atoms present in one gram-atom

(1 mole atom) of C–12 isotope is called Avogadro’s number. One gram-atom (12 grams) of C–12 contains 6.02 × 1023 atoms. Thus the numerical value of Avogadro’s number (NA) is

6.02 × 1023 _{per mol. It should be noted that}

1 a.m.u.= 1

12 th of mass of a C

12 _atom

Avogadro Hypothesis

It states that equal volumes of gases at the same temperature and pressure contain equal number of molecules. It means that 1 ml of hydrogen, oxygen, ammonia, or a mixture of gases taken at the same temperature and pressure contains the same number of molecules.

Application

(a) To prove that simple elementary gas molecules are diatomic

Consider the experimental result,

1 volume of hydrogen + 1 volume of chlorine → 2 volumes of hydrogen chloride at the same temperature and pressure

1 volume contains ‘n’ molecules.

Then n molecules of hydrogen + n molecules of chlorine → 2n molecules of hydrogen chloride.

Cancelling the common ‘n’, we have 1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chloride.

A molecule of hydrogen chloride should contain at least 1 atom of hydrogen and 1 atom of chlorine. Two molecules of hydrogen chloride should contain at least 2 atoms of hydrogen and 2 atoms of chlorine and these should have come from 1 molecule of hydrogen and 1 molecule of chlorine respectively. Thus Avogadro’s hypothesis enables us to establish that hydrogen and chlorine molecules must contain at least 2 atoms (Measurement of the ratio of specific heats of these gases at constant pressure and at constant volume, CP/CV = 1.4, establishes that these gases are diatomic).

(b) [To establish the relationship between molecular weight and vapour density of a gas. The absolute density of gas is the weight of 1 litre (dm3_{) of the gas at S.T.P. [Standard}

Temperature (0°C and Pressure (1 atmosphere)].

The relative density or vapour density of gas = Density of the gas

Density of hydrogen

Vapour density of a gas = Weight of 1 litre of gas at STP

Weight of 1 litre of hydrogen at STP

= Weight of a certain volume of the gas

Weight of the same volume of hydrogen at the same temperature and pressure

So the vapour density of a gas is defined as the ratio of the weight of a certain volume of the gas to the weight of the same volume of hydrogen at the same temperature and pressure.

∴ Vapour density (V.D.) of a gas

= Weight of 'V ' litres of the gas

Weight of 'V ' litres of hydrogen at the same temperature and pressure

Let ‘V’ litres of the gas contains ‘n’ molecules.

V.D. of a gas = Weight of 'n ' molecules of the gas Weight of 1 molecule of the gas

Weight of 'n ' molecules of hydrogen=Weight of 1 molecule of hydrogen

V.D. of a gas = Weight of 1 molecule of the gas 1 Weight of 1 molecule of the gas

Weight of 2 atoms of hydrogen = ×2 Weight of 1 atom of hydrogen

V.D. of a gas = 1

2 × Molecular weight of the gas

∴ Molecular weight of the gas = 2 × Vapour density of the gas.

(c) The volume occupied by a gram molecular weight of any gas is called a molar volume and it is 22.4 L at STP.

1.2 Mole Concept

A mole of any substance is related to : (a) mass of a substance

(b) number of particles

Mole-Particle Relationship

A mole is a collection of 6.023 × 1023 _{particles, ions, atoms etc.}

i.e. (i) 6.023 × 1023 _{atoms of Na constitute one mole atom of Na.}

(ii) 6.023 × 1023 _{molecules of oxygen constitutes 1 mole of oxygen molecules.}

(iii) 6.023 × 1023 _{electrons constitute one mole of electrons.}

Mole-Weight Relationship

One mole of every substance weighs equal to the gram atomic weight of the substance or to the gram molecular weight of the substance.

e.g. (i) 1 mole of sodium weighs 23 g of Na. (ii) 1 mole of CaCO3 weighs 100 g.

Mole-Volume Relationship

One mole of every gas occupies 22.4 lit. of volume at STP. i.e. 1 mole of O2 occupies 22400 ml of volume at STP.

1 mole of He occupies 22400 ml of volume at STP.

Illustra tions

Illustration 1

Determine the number of moles of H2 in 0.224 litre of hydrogen gas at STP (assuming ideal gas behaviour).

Solution

22.4 litres of H2 at STP contain 1 mole of H2. 0.224 litre of H2 at STP has = 0.01 mole

Illustration 2

What will be the number of moles in 13.5 g of SO2Cl2 ?

Solution

Molecular mass of SO2Cl2 = 32 + 32 + 71 = 135 g mol–1 No. of moles = Mass in gram

gram mol. wt. = 13.5

135 = 0.1 mole

Illustration 3

What will be the number of moles of oxygen in one litre of air containing 21% oxygen by volume under standard conditions.

Solution

1000 ml of air at STP contains = 21

100 × 1000 = 210 ml O2 ∴ No. of moles = Vol. in litres under STP conditions

22.4 litre

= 210

22400 = 0.009 mole.

Practice Exercise

1. Calculate the total number of electrons present in 1.6 g of methane. Molecular wt. of methane = 16 a.m.u.

2. 0.24 gm of a volatile substance upon vaporisation gives 45 ml vapours at STP. What will be the vapour density of the substance ?

Answers

1. 6.023 × 1023 _{2. 59.}

1.3 Concept of Limiting Reagent

In reactions involving more than one reactant, one has to identify first, of all the reactant, which is completely consumed (limiting reagent), one can identify the limiting reagent as follow

2 2 3

N 3H 2NH

Initial mole 5 12

+ →

−

If N2 is the limiting reactant moles of NH3 produced = 10. If H2 is the limiting reactant moles of NH3 produced = 3 12 8

2× = . The reactant producing the least number of mole of the product is

the limiting reactant, hence H2 is the limiting reactant.

The limiting reactant can also be ascertained by knowing the initial number of equivalents or milli equivalents of each reactant. The reactant with least number of equivalents or milli equivalents is the limiting reactant. The equivalent methods to identify the limiting reactant used not require balancing of chemical equation.

1.4 Gravimetric Analysis

Calculation involving in mass – mass relationship

In general, the following steps are adopted in making necessary calculations 1. Write down balanced molecular equation for the chemical change

2. Write down the no of moles below the formula of each of the reactant and product

3. Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and product.

4. By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors are determined.

Calculation involving percent yield

In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the percentage yield can be calculated as

% yield = Actual yield 100 Theoretical yield×

Weight-Volume Relationship

2

2Mg(s)+O (g) → 2MgO(s)

In the above reaction, one can find out the volume of O2 at STP required to react with 10 gm of Mg. The moles of Mg is 10

24. The moles of O2 required would be 1/2 the moles of Mg. Therefore

moles of 2

1 10 O

2 24

= × . Since 1 mole of gas (ideal) occupies 22.4 L at STP, therefore 1 10

2×24 moles of

O2 would occupy, 1 10

2×24 × 22.4 L = 4.67 L.

Volume-volume Relationship

Let us consider the reaction H2(g) + ½ O2(g) → H2O(l). We are given 10 L of H2 at a given temperature and pressure. How many litres of O2 would react with hydrogen at the same temperature and pressure?

From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T,

2 2 2 2

H H

O O

V n

V =n . Since the molar ratio is 2 : 1 (H2 : O2), ∴ the volume ratio would also be 2 : 1.

Therefore the volume of O2 required would be 5L. On the other hand if we need to calculate the volume of O2 at a different T and P, then

2 2 2 2 2 2 2 2 H H H H O O O O P V =n R T , P V =n RT , and dividing we get 2 2 2 2 2 2 2 2 H H H H O O O O P V n T P V =n T , ∴ 2 2 2 2 2 2 2 2 H H O O O O H H P V n T V P n T = × .

1.5 Concentration Units

There are several methods to express the concentration of solution. We are going to describe some very important terms used to represent the concentration of solution.

(a) Percent by weight (b) Percent by volume (c) Percent by weight/volume

(d) Mole fraction (e) Molarity (f) Molality (g) Normality (h) ppm Percent by Weight (w/w)

It is the weight of the solute as a per cent of the total weight of the solution. That is, %by weight of solute weight of solute 100

weight of solution

= ×

For example, if a solution of HCI contains 36 % HCI by weight, it has 36 g of HCI in 100 g of solution.

Percent by Volume (v/v)

% by volume = volume of solute 100 volume of solution×

For example, it we have 35% C2H5OH solution by volume means 25 ml C2H5OH is present per 100 ml of the solution.

Percent by weight/volume (w/v)

% = weight of solution 100 volume of solution×

i.e.30% HCl solution, means 30 g of HCl is present per 100 ml of solution

Mole Fraction (X)

A simple solution is made of two substances; one is the solute and the other is solvent. Mole fraction, X, of solute is defined as the ratio of the number of moles of solute and the total number of moles of solute and solvent. Thus,

(

)

solute

moles of solute X

moles of solute moles of solvent

=

+

If n represents moles of solute and N number of moles of solvent, solute n X n N = +

Notice that mole fraction of solvent would be solvent N X n N = +

Mole fraction is unitless and

(

)

Molarity (M)

In current practice, concentration is most often expressed as molarity. Molarity is defined as the number of moles of solute per litre of solution.

Molarity moles of solute volume in litres = or = = × ′ w / M w 1000 M v / 1000 M V

w = Mass of solute in grams

M = molecular weight of solute in gm/mol. V = value of solution in ml.

It’s unit is mol/L.

Molality (m)

Molality of a solution is defined as the number of moles of solute per kilogram of solvent:

Molality moles of solute

mass of solvent in ki log rams

= or w w 1000 M M W _MW 1000 × = = where

w = mass of solute in grams M = molecular wt of solute W = mass of solvent in gram.

Normality (N)

Normality of a solution is defined as number of equivalents of solute per litre of the solution:

Normality =Number of gram equivalents of solute Volume of solution in litres or

× = = w w 1000 E N V _VE 1000 w = mass of solute in gram

V = volume of solution in ml E = equivalent wt of solute

ppm (parts per million)

It is the mass of solute in grams present per 106 _{grams of solution.} 6 weight of solute 10 ppm weight of solution   = ×   

and also volume of particular gas 6

10 ppm

volume of gaseous mixture

  ×    

Illustra tions

A solution of NaCl 0.5% by wt. If the density of the solution is 0.997 g/ml, calculate the molality, molarity, normality and mole fraction of the solute.

Solution

Number of moles of NaCI mass of NaCI

Molecular mass of NaCI

0.5

0.0085 58.5

= =

By definition,

Molality No. of moles of solute 1000 Mass of solvent in grams

×

= =0.00854 1000 0.086m

99.5

× ₌

Volume of the solution Mass of solution in grams Density in gm per m

=

l

= 100 =100.3 0.997

Now Molarity Nimber of moles of soloute Volume of solution in litres

=

= 0.00854 1000× =0.08514M 100.3

and Normality Number of gram equivalents of solute volume of solution in litres

= × = × 0.5 1000 100.3 58.5 = 0.0852

To calculate mole fraction of the solute No. of moles of water in 99.5g = 99.5

18 = 5.5277 Moles of NaCl = 0.5 = × −3 8.547 10 58.5 2 NaCl NaCl NaCl H O n X n n = + = 3 3 3 8.547 10 1.5433 10 M 5.5277 8.547 10 − − − × _{= ×} + × = − = 2 H O NaCl X 1 X 0.9984

SECTION-II

(For Students of Class XII)

1.6 Law of Equivalents

Before heading for the law of equivalents, let us first discuss certain definitions.

Molarity (M)

It is defined as the number of moles of solute present in one litre of solution. Molarity (M) = number of moles of solute

Volume of solution (in litres)

Let the weight of solute be w g, molar mass of solute be M1g./mol and the volume of solution be V litre. Number of moles of solute =

1

weight of solute w Atomic or molar mass of solute=M

∴ 1 w 1 M M V(in litres) = ×

∴ Number of moles of solute =

1

w

M V

M = × (in litres)

Normality (N)

It is defined as the number of moles of equivalents of a solute present in one litre of solution. Equivalent is also the term used for amount of substance like mole with the different that one equivalent of a substance in different reactions may be different as well as the one equivalent of each substance is also different.

Normality (N) = number of equivalents of solute

Volume of solution (in litres)

Let the weight of solute be w g, equivalent mass of solute be E g/eqv. And the volume of solution be V litre.

Number of equivalents of solute = weight of solute w

Equivalent mass of solute= E

∴ N w 1

E V(in litres)

= ×

∴ Number of equivalents of solute = N × V (in litre)

Equivalent Mass

Equivalent mass = Atomic or molecular mass M1

'n ' factor = n

∴ Number of equivalents of solute =

1 1

w w w n

E M / n M

×

∴ Number of equivalents of solute = n × number of moles of solute Also,

1 1

w 1 w 1

N n

M / n v (in litre) M V (in litre)

= × = × ×

N = M × n

∴ Normality of solution = n × molarity of solution.

Illustra tions

Illustration 4

1.60 g of a metal were dissolved in HNO3 to prepare its nitrate. The nitrate was strongly heated

to give 2 g oxide. Calculate equivalent weight of metal.

Solution 3 HNO

3 n 2 n

M →M(NO ) →∆ M O

where n is valency of metal

∴ neq. of metal = neq. of nitrate = neq. of metal oxide = neq. of oxygen

metal metal w E = oxygen oxygen w E 1.60 E = 2 1.60 8 − E = 32 Illustration 5

1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate the equivalent weight of metal. M(NO3)n→ M2(SO4)n

∴ neq. of M(NO3)n = neq. of M2(SO4)n

M 1 E +62 = 0.86 E 48+ 1 E 62+ = 0.86 E 48+ ∴ E = 38 Illustration 6

1.5276 g of CdCl2 was found to contain 0.9367 g of Cd. Calculate atomic weight of Cd.

Weight of Cd = 0.9367 g

∴ Weight of Cl = 1.5276 – 0.9367 = 0.5909 g For CdCl2; neq. of cadmium = neq. of chlorine

0.9367 E =

0.5909 35.5

∴ E = 56.275

∴ Atomic weight = Eq. wt. × Valency = 56.275 × 2 = 112.55.

Practice Exercise

3. 1.8 g of an element displaces 2.04 g copper from CuSO4 solution. If equivalent weight of Cu =

31.7, what is equivalent weight of iron ?

4. 2 g of a metal in H2SO4 gives 4.51 g of the metal sulphate. The specific heat of metal is 0.057

cal/g. Calculate the valency and atomic weight of metal.

5. A hydrated sulphate of metal contained 8.1% metal and 43.2% SO4–2 by weight. The specific

heat of metal is 0.24 cal/g. What is hydrated sulphate ?

Answers

3. 27.97 4. 114.72 5. M2(SO4)3.18H2O

Dilution Effect

When a solution is diluted, the moles and equivalents of solute donot change but molarity and normality changes while on taking out a small volume of solution from a larger volume, the molarity and normality of solution do not change but moles and equivalents change proportionately.

In stoichiometry, the biggest problem is that for solving a problem we need to know a balanced chemical reaction. Since the number of chemical reactions are too many, it is not possible to remember all those chemical reactions. So, there is need to develop an approach which does not require the use of balanced chemical reaction. This approach makes use of a law called law of equivalence. The law of equivalence provide us the molar ratio of reactants and products without knowing the complete balanced reaction, which is as good as having a balanced chemical reaction. The molar ratio of reactants and products can be known by knowing the n-factor of relevant species.

According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant.

Let us suppose we have a reaction, A + B → C + D. In this reaction, the number of moles of electrons lost by 1 mole of A are x and the number of mole of electrons gained by 1 mole of B are y. Since, the number of mole of electrons lost and gained are not same, the molar ratio in which A & B react cannot be 1 : 1. Thus, if we take y moles of A, then the total moles of electrons lost by y moles of A would be (x × y). Similarly, if x moles of B are taken, then the total mole of electrons gained by x moles of B would be (y × x). Thus, the number of electrons lost by A and number of electrons gained by B becomes equal. For reactant A, its n-factor is x and the number of moles used are y. So,

The equivalents of A reacting = moles of A reacting × n-factor of A = y × x. Similarly, for reactant B, its n-factor is y and the number of moles used are x. So,

The equivalents of B reacting = moles of B reacting × n-factor of B = x × y

Thus, the equivalents of A reacting would be equal to the equivalents of B reacting. Thus, the balancing coefficients of the reactant would be as

(n factor y) (n factor x )

yA xB C D

− =

− = + → +

The n-factor of A & B are in the ratio of x : y, and their molar ratio is y : x. Thus, molar ratio is inverse of the n-factor ratio.

In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be b : a.

To get the equivalents of a substance, its n-factor is to be known. Let the weight of the substance used in the reaction be w g.

Then, equivalents of substance reacted would be W

E or 1 1

w w

n

M / n=M × (where E and

M1 are the equivalent mass and molar mass of the substance) Thus, in order to calculate the equivalents of a substance, knowledge of n-factor is a must.

Illustra tions

Illustration 7

Calculate normality and molarity of the following? (a) 0.74 g of Ca(OH)2 in 5 ml of solution.

(b) 3.65 g of HCl in 200 ml of solution.

(c) 1/10 moles of H2SO4 in 500 ml of solution. Solution

(a) Eq. of Ca(OH)2 = 0.74

74 / 2 w Eq. E  ₌     

Volume of solution = 5/1000 litre

∴ N = 0.74 1000 2

74 5

× × ×

N= 4 ∴ M N 4

Total charge on anion or cation 2

= = =2

(b) Eq. of HCl = 3.65

36.5

and Volume of solution = 200 / 1000 litre

∴ N = 3.65 1000 0.5 36.5 200

× ₌

and M N 0.5 0.5 Valency 1 = = = (c) Eq. of H2SO4 = 1 10 × 2 ( Meq. = N × V in ml) = 0.2 Molar H2SO4 = 0.4 Normal H2SO4. Illustration 8

What volume of water is required to make 0.20 N solution from 1600 ml of 0.2050 N solution.

Solution

Meq. of conc. solution = 1600 × 0.2050 = 328 Let after dilution volume becomes V ml Meq. of dil. solution = 0.20 × V

∴ 328 = 0.20 × V

∴ V = 1640 ml

Thus volume of water used to prepare 1640 ml of 0.20N solution = 1640 – 1600 = 40ml

Illustration 9

What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3 .H2O to 100 ml of 0.1 N H2SO4 ? Solution neq. of = 0.62 1000 62 × = 0.01 neq. of H2SO4= 100 0.1 1000× = 0.01 2 3 2 4 2 4 2 2 eq. Na CO H SO Na SO H O CO n added 0.01 0.01 0 0 0 0 0 0.01 0.01 0.01 Meq. left + → + + ↑ ∴ Na SO_{2 4} 0.01 N 1000 100 = × =0.1

Solution becomes neutral since both acid and base are used up and Na2SO4 does not show hydrolysis.

Illustration 10

How much AgCl will be formed by adding 200 ml of 5N HCl to a solution containing 1.7 g AgNO3 ?

Solution

3 3

eq.

2

eq.

AgNO HCl AgCl HNO 1.7 n initially 200 5 0 0 170 10 1 eq 0 0 n after reaction 0 1 0.01 .01 .01 − + → + × = = − ∴ neq. of AgCl formed = 0.01

∴ Amount of AgCl formed = 0.01 × 143.5 = 1.435 g

∴ wAgCl = 1.435 g.

Illustration 11

Find the weight of H2SO4 in 1200 ml of a solution of 0.2N strength. Solution

neq. of H2SO4 = 0.2 1200

1000

×

= 0.24 ( Meq. = N × V in ml) amount of H2SO4 = neq. × neq. wt. of H2SO4

= 0.24 × 49 = 11.76 g w = 11.76g

Practice Exercise

6. The density of 3M solution of Na2S2O3 is 1.25g ml–1. Calculate (a) the % by weights of Na2 S2 O3

(b) mole fraction of Na2 S2 O3 (c) the molalities of Na+ and 2

2 3

S O− ions.

7. A sample of H2SO4 (density 1.787g ml–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2M H2SO4?

8. What weight of AgCl will be precipitated when a solution containing 4.77 g NaCl is added to a solution of 4.77 g of AgNO3 ?

9. How much BaCl2 would be needed to make 250 ml of a solution having same concentration of Cl– as the one containing 3.78 g of NaCl per 100 ml ?

Answers

1.7 ‘n’ Factors

In Non Redox Change

(i) n-factor for element: Valency of the element

(ii) For Acids: Acids will be treated as species which furnish H+ _{ions when dissolved in a} solvent. The n factor of an acid is the no. of acidic H+ _{ions that a molecule of the acid would} give when dissolved in a solvent (Basicity).

e.g. for HCl (n = 1), HNO3 (n = 1), H2SO4 (n = 2), H3PO4 (n = 3) and H3PO3 (n = 2)

(iii) For Bases: Bases will be treated as species which furnish OH– _{ions when dissolved in a} solvent. The n factor of a base is the no. of OH– _{ions that a molecule of the base would give} when dissolved in a solvent (Acidity).

e.g. NaOH (n = 1), Ba(OH)2 (n = 2), al(OH)3 (n = 3), etc.

(iv) For Salts: A salts reacting such that no atom of the salt undergoes any change in oxidation state. e.g. 2AgNO3 + MgCl2 → Mg(NO3)2 + 2AgCl

In this reaction it can be seen that the oxidation state of Ag, N, O, Mg and Cl remains the same even in the product. The n factor for such a salt is the total chare on cation or anion.

Illustra tions

Illustration 12

Calculate the n-factor in the following chemical changes.

(i) H 2 4 KMnO →+ Mn + (ii) H O2 4 4 KMnO →Mn + (iii) OH 6 4 KMnO −→Mn + (iv) H 3 2 2 7 K Cr O →+ Cr + (v) 2 2 4 2 C O − →CO (vi) FeSO4 ¾→ Fe2O3 (vii) Fe2O3 ¾→ FeSO4 Solution

(i) In this reaction, KMnO4 which is an oxidizing agent, itself gets reduced to Mn2+ under acidic conditions.

n = |1 × (+ 7) – 1 × (+ 2)| = 5

(ii) In this reaction, KMnO4 gets reduced to Mn4+ under neutral or slightly (weakly basic conditions.

n = |1 × (+ 7) – 1 × (+ 4)| = 3

(iii) Here KMnO4 gets reduced to Mn6+ under basic conditions. n = |1 × (+ 7) – 1 × (+ 6) = 1

(iv) Here K2Cr2O7 which acts as an oxidizing agent gets reduced to Cr+3 under acidic conditions. (It does not react under basic conditions).

n = +1 × (+7) – 1 × (+ 6)| = 1

(v) C2O42– (oxalate ion) gets oxidized to CO2 when it is reacted with an oxidizing agent. N = |2 × (+ 3) – 2 × ( 4) = –2

(vi) Ferrous ions get oxidized to ferric ions n = |1 × (+ 2) –1 × (+ 3) = 1

(vii) Here ferric ions are getting reduced to ferrous ions N = |2 × (+ 3) – 2 × (+ 2)| = 2

Illustration 13

Predict the molar ratio in which the following two substances would react if they are assumed to be salts Ba3(PO4)2 and AlCl3

Solution

N factor of Ba3 (PO4) = 3 × (+2) = 6 = n1 While n factor of AlCl3 = 1 × (+ 3) = 3 = n2

1 2 n 6 n =3 If 1 2 n x n = y Molar ratio = y

x (inverse of equivalent ratio)

∴ molar ratio in which Ba3 (PO4) and AlCl3 will react = 3 : 6 = 1 : 2. Illustration 14

Find the n-factor for the reactants in each of the following cases

(i) I– → _{I2 (ii) I2} → _I–

(iii) S2O32– → S4O62– (iv) IO3– → Cu2+ + SO2 Solution

(i) n factor = 1 × |0 – (–1)| = 1 (ii) n factor = 2 × |–1 – 0| = 2

(iii) n factor = 2 × |2.5 – 2| = 1 (iv) n factor = 1 × |1 – 5| = 4

Illustration 15

Calculate the n-factor in the following redox change

3

2 4 2

Solution

Here Fe2+ _{is getting oxidized to Fe}3+ _{and C}3+ _{is getting oxidized to C}4+_{. The n factor w.r.t. Fe is +1} and w.r.t. C2O42– is 2 (as calculated earlier). Therefore total n factor is 3.

In Redox Change

For oxidizing agent or reducing agent n-factor is the change in oxidation number per mole of the substance.

(i) If we have a salt which react in a fashion that atoms of one of the element are getting oxidized and the atoms of another element are getting reduced and no other element on the reactant side is getting oxidized or reduced, than the n–factor of such a salt can be calculated either by taking the total number of moles of electrons lost or total number of mole of electrons gained by one mole of the salt.

For example, decomposition reaction of KClO3 is represented as

5 2 1 0 2 3 KClO KCl O + − − → +

In this reaction, O2– _{is getting oxidized to O2 and Cl}+5 _{is getting reduced to Cl}–1_{. In each case,} 6 mole of electrons are exchanged whether we consider oxidation or reduction.

n-factor of KClO3 considering oxidation = |3 (–2) – 3 (0)| = 6 or n–factor of KClO3 considering reduction = |1 × (+5) –1 × (–1)| = 6.

(ii) Disproportionation reactions in which moles of compound getting oxidsed and reduced are not same i.e. moles of oxidizing agent and reducing agent are not same.

For example,

6Br2 + 12 OH¯ → 10 Br¯ + 2BrO₃− + 6H2O

In this reaction, the mole of electrons lost by the oxidation of some of the moles of Br3 are same as the number of mole of electrons gained by the reduction of rest of the moles of Br3. Of the 6 moles of Br2 used, one mole is getting oxidized, loosing 10 electrons (as reducing agent) and 5 moles of Br2 are getting reduced and accepts 10 moles of electrons (as oxidizing agent).

Br₂

5Br₂ + 10 e¯ _{10 Br¯} 2Br+5 _{+ 10e¯}

Br₂ + 5Br₂ 10 Br¯ + 2Br+5

Thus, n-factor of Br2 acting as oxidizing agent is 2 and that Br2 acting as reducing agent has n-factor 10.

Or when the reaction is written as 5

2

where, Br2 is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compound in such reactions, first find the total number of mole of electrons exchanged (lost or gained) using the balanced equation and divide it with the number of mole of Br2 involved in the reaction to get the number of mole of electrons exchanged by one mole of Br2.

In the overall reaction, the number of mole of electrons exchanged (lost or gained) is 10 and the moles of Br2 used in the reaction are 6. Thus, each mole of Br2 has exchanged 10/6 or 5/3 mole of electrons. Therefore, the n-factor of Br2 when the reaction is written without segregating oxidizing and reducing agent is 5/3.

5 2 (n 5) (n 1) (n 5 / 3) 6Br 10Br− 2Br+ = = = → +

1.8 Application of The Law of Equivalence

Simple titration

In this, we can find the concentration of a substance with the help of the concentration of another substance which can react with it.

For example: Let there be a solution of a substance A of unknown concentration. We are given another substance B whose concentration is known (N1). We take a certain known volume of A in a flask (V2) and then we add B to A slowly till all the A is consumed by B. (This can be known with the help of indicators). Let us assume that the volume of B consumed is V1. According to the law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents of B.

∴ N1V1 = N2V2, where N2 is the concentraton of a. From this we can calculate the value of N2.

Acid Base titration

In this type of titration, the concentration of an acid in a solution is estimated by adding solution of standard base and vice versa. The equivalence point is detected by adding a few drops of a suitable indicator to the solution whose concentration is to be estimated. An acid-base indicator gives different colours with acids and acid-bases. The choice of indicator in a particular titration depends on the pH-range of the indicator and the pH change near the equivalence point. For example.

(i) Strong Acid-Strong Base Titration

In the titration of HCl Vs NaOH, the equivalence point lies in the pH-range of 4–10. Thus, methyl red (pH-range 4.2 to 6.3), methyl orange (pH-range 3.1 to 4.4) and phenolphthalein (pH-range 8.3 – 10) are suitable indicators of such titrations.

(ii) Weak Acid–Strong Base Titration

In the titration of CH3COOH and NaOH, the equivalence point lies between 7.5 and 10. Thus, phenolphthalein is the suitable indicator.

(iii) Weak Base-Strong Acid Titration

In the titration of NH4OH and HCl, the equivalence point lies in the pH range of 4 to 6.5. Thus, methyl orange and methyl red are suitable indicators.

(iv) Weak Acid-Weak Base Titration

In the titration of CH3COOH and NH4OH, the equivalence point lies between 6.5 and 7.5 and the pH change is not sharp at the equivalence point. Thus, no simple indicator can be employed to detect the equivalence point.

Iodometric and Iodimetric titration

The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titration’s.

2

I +2e− →2I (reduction)−

2

2I− →I +2e (oxidation)−

These are divided into two types

(i) Iodometric titration: In iodometric titrations, an oxidizing agent in allowed to react in

neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine. KI + oxidizing agent → I2

Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate Halogen, dichroamtes, cupric ion, peroxides, etc can be estimated by this method.

I2 + 2Na2S2O3 → 2NaI + Na2S4O6 2CuSO4 + 4KI → Cu2I2 + 2K2SO4 + I2

K2Cr2O7 + 6KI + 7H2SO4 → Cr2(SO4)3 + 4K2SO4 + 7H2O + 3I2

(ii) Iodimetric titration: These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI solution.

KI + I2 → KI3 (Potassium triiodide)

This solution is first standardizes before use with the standard solution of I2 substance such as sulphite, thiosulphate, arsenite etc. are estimated.

The iodimetric and iodometric titrations, starch solution is used as indicator. Starch solution gives blue or violet colour with iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.

Back titration

Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gms and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert. We are provided with two solutions A and B, where the concentration of B is known (N1) and that of A is not known. This type of titration will work only if the following conditions are satisfied A, B and C should be such compounds that A and B can react with each other, A and C should not react with B.

Now we take a certain volume of A in a flask (the A taken should be such that the gm equivalents of A taken should be ≥ gm equivalents of C in the sample. (This can be done by taking A in excess). Now we perform a simple titration using B. Let us assume that the volume of B used is V1. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is complete, the solution is being titrated with B. Let us assume that the volume of B used up is V2.

Gram equivalents of B used in the first titration = N1V1

∴ gm. Equivalents fo A inititally = N1V1

gm. Equivalents of B used in the second titration is N1V1

∴ gm. Equivalents of A left in excess after reacting with C = N1V2 gm. Equivalents of A that reacted with C = N1V1 – N1V2

gm. Equivalents of pure C = N1V1 – N1V2.

If the n factor of C is x, then the moles of pure C = N V1 1 N V1 2

x − ∴ the weight of C = N V1 1 N V1 2 x − _× Molecular weight of C. ∴ percentage of C = N V1 1 N V1 2 Molecular wt. of C 100 x w − _{× ×} . Double titration

The method involves two indicator (indicators are substance that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities weighing w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample, say, HCl along with its concentration (M1).

We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with HCl.

NaOH is a strong base while Na2CO3 is a weak base. So it is safe to assume that NaOH reacts with HCl first, completely and only then does Na2CO3 react.

Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps. Na2CO3 + HCl → NaHCO3 + NaCl and then,

NaHCO3 + HCl → NaCl + CO2 + H2O

As can be seen, when we go on adding more and more of HCl the pH of the solution keeps on falling. When Na2CO3 is converted to NaHCO3, completely, the solution is weakly basic due to the presence of NaHCO3 (which is a weaker base as compared to Na2CO3). It this instant phenolphthalein changes colour since it requires this weakly basic solution to change its colour. Therefore, remember that phenolphthalein changes colour only when the weakly basic NaHCO3 is present. As we keep adding HCl, the pH again falls and when also the NaHCO3 reacts to form NaCl, CO2 and H2O the solution becomes weakly acidic due to the presence of the weak acid H2CO3(CO2 + H2O). At this instance methyl orange changes colour since it requires this weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when H2CO3 is present.

Now, let us assume that the volume of HCl used up for the first and the second reaction.

i.e., NaOH + HCl → NaCl + H2O and Na2CO3 + HCl → NaHCO3 + NaCl be V1 (this is the volume of HCl from the beginning of the titration up to the point when phenolphthalein changes colour). Let the volume of HCl required for the last reaction, i.e., NaHCO3 + HCl → NaCl + CO2 + H2O be V2 (this is the volume of HCl from the point where phenolphthalein had changed colour upto the point when methyl orange changes (colour). Then,

moles of HCl used for reacting with NaHCO3 = moles of NaHCO3 reacted = M1V2 moles of NaHCO3 produced by the Na2CO3 = M1V2

moles of Na2CO3 that gave M1V2 moles of NaHCO3 = M1V2 Mass of Na2CO3 = M1V2 × 106

% Na2CO3 = M V1 2 106 100

w

× _×

moles of HCl used for the first two reactions = M1V1 moles of Na2CO3 = M1V2

moles of HCl used for reacting with Na2CO3 = M1V2

moles of HCl used for reacting with only NaOH = M1V1 – M1V2

∴ moles of NaOH = M1V1 – M1V2 Mass of NaOH = (M1V1 – M1V2) × 40 % NaOH = (M V1 1 M V ) 401 2 ₁₀₀

w

Illustra tions

Illustration 16

50 mL of a solution, containing 1 g each of Na2CO3, NaHCO3 and NaOH, was titrated with 1(N) HCl. What will be titre reading if

(a) only phenolphthalein is used as indicator

(b) only methyl orange is used as indicator from the very beginning? (c) methyl orange is added after the first end point with phenolphthalein?

Solution

(a) The titration reactions in this case are

Na2CO3 + HCl → NaHCO3 + NaCl and NaOH + HCl → NaCl + H2O Thus, we have

m.e. of Na2CO3 + m.e. of NaOH = m.e. of v1 mL (say) of 1(N) HCl

1 1

1000 1000

106× +40× = 1 × v1 ; ∴ v1 = 34.4 mL.

(b) The reactions in this case are,

Na2CO3 + HCl → NaHCO3 + NaCl 3 2 2 (produced) NaHCO +HCl→NaCl+H O CO+ 3 2 2 (originally present) NaHCO +HCl→NaCl H O CO+ +

and NaOH + HCl → NaCl + H2O Thus, we have,

2 3 3 3

(produced) (originally present)

m.e. of Na CO +m.e. of NaHCO +m.e. of NaHCO +m.e. of NaOH

= m.e. of v2 mL (say) of N HCl 2 1 1 1 1 1000 1000 1000 1000 1 v 106× +106× +84× +40× = × ∴ v2 = 55.8 mL.

(c) The reaction in this case are

NaHCO3 (produced) + HCl → NaCl + H2O + CO2 and NaHCO3 (originally present) + HCl → NaCl + H2O + CO2 Thus we have,

2 3 3 3

(produced) (originally present)

m.e. of NaH CO +m.e. of NaHCO =m.e. of v mL (say) of 1(N) HCl

3 1 1 1000 1000 1 v 106× +84× = × ∴ v3 = 21.3 mL.

Practice Exercise

10. A solution contained Na2CO3 and NaHCO3. 25 mL of this solution required 5 mL of 0.1 N HCl for titration with phenolphthalein as indicator. The titration was repeated with the same volume of the solution but with methyl orange. 12.5 mL of 0.1 N HCl was required this time. Calculate the amount of Na2CO3 and NaHCO3 in the solution.

Answers

10. .053 g, .021 g

1.9 Volume of Strength of H

O

x volume of H2O2 means x litre of O2 is liberated by 1 litre of H2O2 on decomposition at STP

2 2 2 2

68 gm 22.4 lit

2H O → 2H O+ O

22.4 lit (at STP) of O2 is given by 68 gm of H2O2 x litre of O2 is released from

68 x 22.4 gm of H2O2 = 17x 5.6 ∴ x lit of O2 is given by = 68 x 17x 22.4= 5.6 (strength) Strength, S = 17x 5.6 Normality = S 17x x E=5.6 34 / 2× =5.6 [x = N × 5.6] Molarity = 1 2 Normality = x 11.2.

1.10 Strength of Oleum

Oleum means mixture of conc. H2SO4 and SO3

Thus, Oleum = x H2SO4 + y SO3

(x H2SO4 + y SO3) + y H2O → (x + y) H2SO4

Strength of oleum is expressed in terms of percentage which is equal to weight of H2SO4

obtained when water is added to 100 gms of oleum. e.g. 109% oleum means

100 gm oleum + water → 109 gm H2SO4

Moles of H2O added = 9

18 = Moles of SO3 present in oleum sample.

Mass of SO3 in oleum = 9 80

18× = 40 g

Thus, oleum sample contained 40% SO3 and 60% H2SO4.

1.11 Oxidation and Reduction

Oxidation is

(i) the gain of oxygen

(ii) the loss of hydrogen

(iii) the loss of electrons (de-electronation)

(iv) the increase of O.N. Reduction is

(i) the loss of oxygen

(ii) the gain of hydrogen

(iii) the gain of electron

(iv) the decrease in O.N.

2

O.N.Fe0 Fe2 2e

+ − = → + +

• Fe loses electrons

• There is increase in O.N.

• Hence Fe is said to be oxidized

• It is a source of electron hence it can act as a reducing agent (R.A.)

• Any species that can be oxidized is a reducing agent (R.A.)

• 2 O.N. 2 0 Cu + 2e Cu + + → • Cu2+ _{gains electron}

• There is decrease in O.N.

• Hence Cu2+ _{is said be reduced.}

• Since it can gain electrons, hence it can act as an oxidizing agent (O.A.)

• Any species that can be reduced is an oxidizing agent (O.A.), or oxidant. Above examples represent half reactions.

A complete reaction showing oxidation and reduction together is called a Redox reaction.

R O

Now that you are clear on what is oxidation – reduction, we are now in the right position to know how to balance a redox (oxidation – reduction) reaction. This is important because many of the problems of stoichiometry would be based on such redox eractions.

Oxidation State/Oxidation Number

There are several chemical reactions in which oxidation – reduction takes place. Oxidation refers to a reaction in which electrons are removed from an atom and reduction refers to a reaction in which electrons are added to an atom. To describe these changes, the concept of oxidation state becomes necessary. For ionic species, the charge on each ion is said to be the oxidation state for that atom. For example in NaCl, Na exists as Na+ _{and Cl exists as Cl}−_. Therefore the oxidation state of Na in NaCl is +1 and that of Cl− is –1. But in covalent molecules, the charge on an atom would be so small that sometimes it becomes impossible to calculate the exact charge on each atom of a molecule. Therefore, the Oxidation State (O.S.) or Oxidation Number (O.N.) is defined as the charge, an atom would have in a molecule if all the bonds associated with this atom in the molecule are considered to be completely ionic. For example in H2O there are two O–H bonds. If we assume both the O–H bonds to be completely ionic, then each H would possess a charge of + 1, while O possess a charge of –2. This is because oxygen is more electronegative than hydrogen. On the other hand, in H2O2 there are two OH bonds and one O–H bond (H–O–O–H). Considering each O–H bond to be ionic both the oxygen atoms acquire a charge of –1 and both the H, +1. This is because O – O bond can not be assumed to be ionic as both the atoms have the same electronegativity.

To calculate the oxidation state of an element in a molecule you need not always know the structure of the molecule. There are certain set of rules used to assign oxidation states in polyatomic molecules.

(i) The O.S. of all elements (in any allotropic form) is zero.

(ii) O.S. of oxygen is –2 in all its compounds except peroxides like H2O2 and Na2O2 where it is –1 and superoxides like KO2 where it is –1/2. In OF2 and O2F2, oxygeneous +2, +1 respectively.

(iii) O.S. of hydrogen is +1 in all of its compounds except those with the metals where it is –1.

(iv) O.S. of all alkali metals is +1 and alkaline earth metals is +2 in all their compounds.

(v) O.S. of all the halogens is –1 in all their compounds except where they are combined with an element of higher electronegativity. Since fluorine is the most electronegative of all elements, its O.S. is always –1.

Illustra tions

Illustration 17

Calculate the O.S. of the atoms in the following species:

(i) In ClO¯ (ii) NO2¯ (iii) NO3¯ (iv) CCl4 (v) K2CrO4 (vi) KMnO4 Solution

(i) In ClO¯, the net charge on the species is –1 and therefore the sum of the oxidation states of Cl and O must be equal to –1. Oxygen will have an O.S. of –2 and if the O.S. of Cl is assumed to be ‘x’ then x – 2 should be equal to –1.

∴ x is + 1

(ii) NO2¯: (2 × –2) + x = –1 (where ‘x’ is O.S. of N)

∴ x = + 3

(iii) NO3¯: : x + (3 × –2) = –1 (where ‘x’ is O.s. of N)

∴ x + 5

(iv) In CCl4, Cl has an O.S. of –1

∴ x + 4 × –1 = 0

∴ x = + 4 (where ‘x’ is O.S. of C)

(v) K2CrO4: K has O.S. of + 1 and O has O.S. of –2 and let Cr has O.S. ‘x’ then, 2 × + 1 + x + 4 × –2 = 0

∴ x = + 6

(vi) KMnO4 : + 1 + x + (4 × –2) = 0

∴ x = + 7 (where x is O.S. of Mn).

Balancing Redox Equations

Some examples of redox reactions are

(a) R O Sn2+ _{+ 2Hg}2+ Hg₂2+ _{+ Sn}4+ (b) R O MnO₄¯ _{+ 5Fe}2+ _{+ 8H}+ 5Fe3+ _{+ Mn}2+ _{+ 4H} 2O

(c) R O Cr₂O₇2+ _{+ 6Fe}2+ _{+ 14H}+ _6Fe3+ _{+ 2Cr}3+ _{+ 7H} 2O (d) _R O

3Cu + 2NO₃¯ + 8H+ _{2NO + 3Cu}2+ _{+ 4H} 2O

If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side.

R

3Cl₂ + 6OH¯ 5Cl¯ + ClO₃¯ _{+ 3H} 2O

0 –1 +5

O.N. =

In this we find the Cl2 has been oxidized as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are

R O

2Cu+ Disproportionates

Cu + Cu2+

R

2HCHO + OH¯ CH₃OH + HCOO¯

O R O 4ClO₃¯ 3ClO₄¯ + Cl¯ R O 3MnO₄2- _{+ 2H} 2O MnO2 + 2MnO4 - _{+ 4OH}

-How to Balance a Redox Reaction Ion Electron Method

(i) Divide the complete equation into two half reactions, one representing oxidation and the other reduction.

(ii) Balance the atoms in each half reaction separately according to the following steps (a) First of all balance the atoms other than H and O.

(b) In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H+ _{ions to the other side deficient in hydrogen atoms.} On the other hand, in alkaline medium (OH¯), for every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH¯ to the other side. In case hydrogen is still unbalanced, then balance by adding one OH¯, for every excess of H atom on the same side and one H2O on the other side.

(c) Equalize the charge on both sides by adding suitable number of electrons to the side deficient in negative charge.

(iii) Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction.

(iv) Add the two balanced half equations and cancel any term common to both sides.

Illustra tions

Step 4 No other atom (except H and O) is unbalanced therefore no need for this step. Step 5 Balancing O atom. This is made by using H2O and H+ io ns. Add desired molecules

of H2O on the side deficient in O atom and double H+ on opposite side. Therefore

3 2 2 4 4

NO− +H S+H O →NH+ +HSO− +2H+

Step 6 Balancing charge by H+

3 2 2 4 4

NO−+H S+H O+3H+ →NH+ +HSO− +2H+

∴ Balanced equation is

3 2 2 4 4

(b) Step 1 Fe → Fe2+ _{+ 2e} Step 2 N22– + 2e → 2N3–

Step 3 Fe + N2H4 → Fe(OH)2 + 2NH3

Step 4 No other atom (except H and O) is unbalanced and therefore no need for this step.

Step 5 Fe N H2 4 4OH Fe(OH)2 2NH3 2H O2

−

+ + → + +

Step 6 Balance charge by H+

∴ Fe + N2H4 + 4OH¯ + 4 H+ → Fe(OH)2 + 2NH3 + 2H2O Finally Fe + N2H4 + 2H2O → Fe(OH)2 + 2NH3

Practice Exercise

11. Balance the redox equation, HNO3 + H2S → NO + S 12. Balance the following redox equation,

FeC2O4 + KMnO4 + H2SO4 → Fe2(SO4)3 + CO2 + MnSO4 + K2SO4.

Answers

11. 2HNO3 + 3H2S → 3S + 2NO + 4H2O

12. 10FeC2O4 + 6KMnO4 + 24H2SO4 → 5Fe2(SO4)3 + 6MnSO4 + 3K2SO4 + 20CO2 + 24H2O

Oxidation Number Method

This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are

(i) Write the equation (if it is not complete, then complete it) representing the chemical changes.

(ii) By knowing oxidation number of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.

(iii) Add required electrons on the right hand side of oxidation reaction and on the left hand side of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.

(iv) Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.

(v) Transfer the coefficient of the oxidizing and reducing agents and their products to the main equation. By inspection, arrive at the co-efficient of the species not undergoing oxidation or reduction.

Illustra tions

Illustration 19

Balance the equation K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2 Solution 6 1 3 0 2 2 7 3 2 2 K Cr O HCl KCl CrCl H O Cl + − + + → + + +

Thus here cr of K2Cr2O7 is reduced to CrCl3*(+6 → + 3) and Cl of HCl is oxidized to Cl2 (–1 → 0). In short. Oxidation: 0 2 2Cl− →Cl ↑2 …(a) Reduction: 6 3 2 Cr+ → 2Cr+ ↓6 …(b) Step (iii) 1 0 6 3 2 2 2Cl− → Cl +2e ; Cr− + +6e− →2Cr +

Step (iv) Multiply equation (a) by 6 and (b) by 2

1 0 2 6 3 2 6 1 3 0 2 2 12Cl 6Cl 12e 2Cr 12e 4Cr 2Cr 12Cl 4Cr 6Cl − − + − + + − → + + → + → + or 6 3 0 2 2 Cr+ +6Cl− →2Cr+ +3Cl Step (v) K2Cr2O7 + 6HCl → 2CrCl3 + 3Cl2

Step (vi) Making provision of KCl and H2O in the product: Since the reactant has 7 oxygen atoms in the product 7H2O must be present. For accounting 14 hydrogen atoms of water in the product, the reactants must have 12 HCl molecules (the only H containing species). For accounting the 2K atoms and 14 – 12 = 2 additional Cl atoms in the reactant, the product must have 2KCl. Hence the balanced equation is K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

Practice Exercise

13. Balance the following oxidation-reduction equationm KMnO4 + KCl + H2SO4 → MnSO4 + K2SO4 + H2O + Cl2 14. Balance the following redox equation

K2Cr2O7 + HCl → KCl + CrCl3 + Cl2 + H2O

Answers

13. 2KMnO4 + 10KCl + 8H2SO4 → 2MnSO4 + 6K2SO4 + 8H2O + 5Cl2

14. K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 3Cl2 + 7H2O

Miscellaneous

Miscellaneous

Miscellaneous

Miscellaneous Problems

Objective Type

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8 g of sulphur is burnt to form SO2, which is oxidized by Cl2 water. The solution is then treated with BaCl2

solution. The amount of BaSO4 precipitated would be

(a) 1 mole (b) 0.5 mole (c) 0.24 mole (d) 0.25 mole

Solution 2 2 2 O Cl water 2 BaCl 2 4 4 S→SO →SO−→BaSO +2Cl− Moles of sulphur = 8 1 32 =4 = Moles of SO2

∴ Moles of BaSO4 precipiatated = 1

4 (applying principle of atom conservation)

∴ ∴ ∴

∴ Ans. (d)

*

The normally of 0.3 M H3PO3 when it undergoes the following reaction, H PO_{3 3}+2OH− →HPO₃2−+2H O₂

would be

(a) 0.6 N (b) 0.15 N (c) 0.9 N (d) 0.1 N

Solution

Molar ratio of reactants being 1 : 2, so the n-factor ratio has to be 2 : 1, as n-factor of OH ¯ is 1.

∴ n-factor of H3PO3 in the given reaction is 2.

∴ Normally of H3PO3 = 0.3 × 2 = 0.6 N ∴ ∴ ∴ ∴ Ans. (a)

*

The moles of ammonium sulphate needed to react with one mole of MnO2 in acidic medium in a reaction

giving MnSO4 and (NH4)2S2O8 is

(a) 2 (b) 3 (c) 1 (d) 1/3 Solution 2 4 2 2 4 4 4 2 2 8 2 MnO (NH ) 2H SO MnSO (NH ) S O 2H O n 2 n 1 n 1 + + → + + = = =

Law of equivalence is not applicable

∴ ∴ ∴

*

The number of moles of 2 2 7

Cr O− needed to oxidize 0.136 equivalents of N H2 5

+ by the reaction 2 3 2 5 2 7 2 2 N H++Cr O− →N +Cr++H O is (a) 0.136 (b) 0.272 (c) 0.816 (d) 0.0227 Solution ‘n’ factor of 2 2 7

Cr O − ion in the given reaction is 6. 6 2 3 2 7 Cr O Cr + − +  _→     

Equivalents of Cr2O72– needed to oxide N H_{2 5}+ = Equivalents of N H_{2 5}+ =0.136

∴ Moles of 2 2 7 Cr O − needed to oxides N H2 5 + 0.136 6 = = 0.0227 ∴ ∴ ∴ ∴ Ans. (d)

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A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate

calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of

CaO is 1.62 g. The % by mass of NaCl in the original mixture is

(a) 15.2% (b) 67.9% (c) 21.8% (d) 32.1%

Solution

2 2 3 3

CaCl +Na CO →CaCO +2NaCl

3 2

CaCO →∆ CaO+CO

Moles of CaO 1.62 56

= = Moles of CaCO3 = Moles of CaCl2 [from equation (i) and (ii)]

Mass of CaCl2 = 1.62 111 3.21 g 56 × = % of CaCl2 = 3.21 100 32.1% 10 × = and % of NaCl = 100 – 32.1 = 67.9 % ∴ ∴ ∴ ∴ Ans. (d)

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Equal volumes of 1 M each of KMnO4 and K2Cr2O7 are used to oxidize Fe(II) solution in acidic medium. The

amount of Fe oxidized would be

(a) more with KMnO4 (b) equal with both oxidizing agents

(c) more with K2Cr2O7 (d) cannot be determined

The ‘n’ factor of KMnO4 in its reaction with Fe(II) would be 5 while that of K2Cr2O7 would be 6. So, for the

same number of moles, K2Cr2O7 will have greater equivalents than KMnO4. So, more equivalents of Fe(II)

would be oxidized by K2Cr2O7 than by KMnO4.

∴ ∴ ∴

∴ Ans. (c)

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It takes 2.56 × 10–3 _{equivalents of KOH to neutralize 0.1254 g H2XO4. The number of neutrons in X would be}

(a) 16 (b) 8 (c) 18 (d) 32

Solution

Let ‘a’ be the atomic mass of X. Moles of H2XO4 = 0.1254

(a+66)

‘n’ factor of H2XO4 = 2 [as H2XO4 is a dibasic acid]

Equivalents of KOH = Equivalents of H2XO4

∴ 2.56 × 10–3 ₌ 0.1254 ₂

(a+66)×

a = 31.96 g/mol ≈ 32 g/mol

∴ X is sulphur and sulphur has 16 protons and 16 neutrons.

∴ ∴ ∴

∴ Ans. (a)

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10 ml of a solution of H2O2 requires 25 ml of 0.1 N KMnO4 for complete reaction. The volume strength of

H2O2 would be

(a) 1.4 (b) 1.2 (c) 0.25 (d) 2.5

Solution

In the reaction between KMnO4 and H2O2, KMnO4 acts as oxidizing agent, changing to Mn2+ and H2O2

behaves as reducing agent oxidizing to O2.

2

4 2 2 2

(n 5) ( n 2)

KMnO H O Mn + O

= + = → +

Equivalents of KMnO4 = Equivalents of H2O2.

25 × 10–3× _{0.1 = 10} × ₁₀–3× _N N = 0.25 ∴ Volume strength of H2O2 = 5.6 × 0.25 = 1.4 ∴ ∴ ∴ ∴ Ans. (a)

Subjective Type

P and Q are two elements which forms P2Q3 and PQ2. If 0.15 mole of P2Q3 weights 15.9 g and 0.15 mole of

PQ2 weights 9.3 g, what are atomic weights of P and Q?

Solution

Let atomic weight of P and Q a and b respectively

∴ Molecular weight of P2Q3 = 2a + 3b

and Molecular weight of PQ2 = a + 2b

Now given that 0.15 mole of P2Q3 weight 15.9 g

(2a + 3b) = 15.9 0.15 wt. mole mol. wt   ∴ =     Similarly, (a + 2b) = 9.3 0.15

Solving these two equations b = 18

a = 26

Example 2

Potassium selenate is isomorphous with potassium sulphate and contains 45.52% selenium by weight. Calculate the atomic weight of selenium. Also report the equivalent weight of potassium selenate.

Solution

Potassium seleante is isomorphous to K2SO4 and thus its molecular formula of K2SeO4.

Now molecular weight of K2SeO4 = (39 × 2 + a + 4 × 16)

where a is atomic weight of Se (142 + a)g K2SeO4 has Se = ag

∴ 100 g K2SeO4 has Se = a 100 142 a × + ∵ % of Se = 45.52 ∴ a 100 142 a × + = 45.52 ∴ a = 118.168 = 118.2

Also equivalent of K2SeO4 = Mol. wt. 2 39 118.2 64

2 2

× + +

Example 3

A sample of H2SO4 (density 1.787 g mL–1) is labeled as 86% by weight. What is molarity of acid? What

volume of acid has to be used to make 1 litre of 0.2 H2SO4?

Solution

H2SO4 is 86% be weight

∴ Weight of H2SO4 = 86 g

Weight of solution = 100 g

∴ Volume of solution = 100 mL 100 litre

1.787 =1.787 1000× ∴ H SO_{2 4} 86 M 15.68 100 98 1.787 1000 = = × ×

Let V mL of this H2SO4 are used to prepare 1 litre of 0.2 M H2SO4

∴ mM of H2SO4 conc. = mH of H2SO4 dilute

V × 15.68 = 1000 × 0.2 ∴ V = 12.75 mL.

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The molecular mass of an organic acid was determined by the study of its barium salt 4.209 g of salt was quantitatively converted to free acid by the reaction with 21.64 ml of 0.477 M H2SO4. The barium salt was

found to have two mole of water of hydration per Ba2+ _{ion and the acid is mono basic. What is molecular}

weight of anhydrous acid?

Solution

Meq. of barium salt = meq. of acid

4.290 1000

M / 2× = 21.64 × 0.4777 × 2

Molecular weight of salt = 415.61

Molecular weight of anion = 415.61 137.36 121.31 2

− ₌

∴ Molecular weight of acid = 121.31.

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25 mL of a solution of Na2CO3 having a specific gravity of 1.25 g mL–1 required 32.9 mL of a solution of HCl

containing 109.5 gm of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H2SO4

that will be completely neutralized by 125 g of Na2CO3 solution.

Solution HCl 109.5 N 3 36.5 1 = = ×

Since Na2CO3 is completely neutralized by HCl