Chemical reactions and equations Class 10 science PDF

CHEMICAL REACTIONS
AND EQUATIONS
1.1 INTRODUCTION
Any change with the formation of new substance called chemical
reaction. There are many type of chemical reaction e.g. displacement
reaction, combination reaction etc.Achemical equation represents a
chemical reaction. In a chemical reaction balancing is very important.
It can be done by various methods. Oxidation and reduction reaction
are types of chemical reaction. These reactions can be balanced by
ion electron method and oxidation number method. Corrosion and
rancidity are the effects of oxidation reactions.
1.2 CHEMICAL REACTION
A chemical reaction is a process which transforms one or more
substances into new substances. During chemical reactions, new
substances with new properties are formed . The substances which
take part in chemical reactions are called reactants and the
substances which are formed as a result of chemical reactions are
called products. For example, in the reaction between sodium
hydroxide and hydrochloric acid to give sodium chloride and water.
NaOH + HCl aCl + H2
O
Sodium Hydroxide Hydrochloric acid Sodium Chloride Water
Reactants Products
The chemical reactions involved the breaking of bonds between the
atoms of the reacting substances and making of new bonds between
atoms of products.
The following observation helps us to determine whether a chemical
reaction has taken place.
• change in state • change in colour
• evolution of a gas • change in temperature
What is a Chemical Equation?
Symbolic representation of chemical reaction in terms of sym-
bols and formulas of reactants and the products which will
give idea about true chemical change.
1.1 Introduction
1.2 Chemical reactions
1.3 Chemical equations
1.4 Different types of
chemical reacions
1.5 Oxidation and
reduction reaction
1.6 Oxidising and reducing
agents
1.8 Effects of oxidation
reactions in everyday life
1.9 Rancidity
“IIT-JEE Foundation”
*1.5.1 Modern Concept of
Oxidation & Reduction
*1.5.2 Redox reactions
*1.6.3 Valency, Oxidation state
and Oxidation Number
*1.6.4 Oxidation and
Reduction in terms of
oxidation number
*1.6.5 Oxidation number or
Oxidation state
*1.6.6 Distinction between
valency & Oxidation
states/Oxidation
number
*1.6.7 Oxidation state/
Oxidation number in
complex molecules
*1.6.8 Fractional values of
oxidation numbers.
*1.7 Balancing of redox
reaction.
VAVA CLASSES/CHEM/10TH
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1.3 CHEMICAL EQUATION
When a magnesium ribbon is burnt in oxygen, it gets converted to magnesium oxide.
The word equation for the above reaction would be –
Magnesium + Oxygen  Magnesium oxide
(Reactants) (Product)
Substances which react among themselves to bring about the chemical change are called reactants and
the substances which are produced as a result of chemical change are called products.
Writing a Chemical Equation : Chemical equations can be made more concise and useful if we use
chemical formulae instead of words.
2Mg + O2
 2MgO
Various steps which must be kept in mind while writing a chemical equation are :
(i) The reactants are always written on the left hand side while the products are always written on the
right hand side.
(ii) The formulae of various reactants are separated by + sign.Similarly, the formulae of the products are
also separated by + sign.
(iii) An arrow () sign is put between the reactants and the products.
Using these three steps, the reaction between zinc and sulphuric acid to produce zinc sulphate and hydro-
gen can be written in the form of a chemical equation as:
Zn H2
SO4
ZnSO4
H2
Means reacts means means
with gives alongwith
produce
Essentials of Chemical Equations :
Essentials of chemical equations are:
(i) It must represent a true chemical change.
(ii) All the reactants and the products of the chemical reaction must be in the form of their representative
chemical formulae or symbols.
(iii) The total number of atoms of all elements must be same on both sides.
(iv) It must be molecular.
For example, Pt + 4HCl  PtCl4
+ 2H2
, though a balanced equation algebraically, is not a true chemical
reaction and hence, not a chemical equation since platinum does not react with hydrochloric acid to give
hydrogen. KClO3
 KCl + 3O, is a true chemical reaction and balanced too, yet oxygen is not shown its
molecular form, hence, it is yet an incomplete chemical equation .
Balanced and Unbalanced Chemical Equations : A chemical equation in which number of atoms of
different elements in the reactants and the products are the same is known as a balanced chemical
equation. For example, let us consider the chemical equation
2KClO3
 2KCl + 3O2
Significance or Important Information Conveyed by a Chemical Equation : A balanced chemical
equation conveys a lot of other informations, in addition to its telling about the feasibility of a reaction.
Some important information conveyed by a chemical equation is given below:
1. An equation, tells us the names of the reactants and the products. For example, chemical equation.
CaCO3
+ 2HCl  CaCl2
+ CO2
+ H2
O
2. It conveys to us the relative number of molecules of the reactants and the products.
Zn + H2
SO4
 ZnSO4
+ H2

3. It conveys to us the relative number of moles of reactants and the products.
4. Relative masses of reactants and the products become known to us.
5. It also tells about the relative volumes of reactants and the products, if they happen to be gaseous.
In reaction (i) above, when 100.0 g of CaCO3
reacts with 73.0 g of HCl, 111 g of CaCl2
, 22.4 litres of CO2
gas at N.T.P. and 18 g of H2
O (or 22.4 litres of steam at N.T.P., if the reaction occurs above 100°C) are
produced.
Limitations of Chemical Equation : Inspite of giving so much useful information, a chemical equation
fails to convey much about some other important aspects, unless, extra notation is written, in the balanced
equation.
1. It does not tell anything about the physical state, nature, etc., of the substance that take part in the
reaction.
2. In cases where instead of pure solid compounds, solution are used, it does not tell whether these
solutions are concentrated or dilute.
3. The time required for completion of the reaction or speed of the reaction is also not conveyed.
4. Conditions, such as temperature, pressure and catalyst (if any) required are not given in the equation.
5. It does not convey any information about the heat or energy evolved or absorbed during the chemical
change.
Balanced Chemical Equations: The total mass of the elements present in the products of a chemical
reaction has to be equal to the total mass of the elements present in the reactants.
Zinc + Sulphuric acid  Zinc sulphate + Hydrogen
The above word-equation may be represented by the following chemical equation –
Zn + H2
SO4
 ZnSO4
+ H2
Step I : To balance a chemical equaiton, first draw boxes around each formula. Do not change anything
inside the boxes while balancing the equation.
Fe + H2
O  Fe3
O4
+ H2
Step II : List the number of atoms of different elements present in the unbalanced equation.
Step III: It is often convenient to start balancing with the compound that contains the maximum number
of atoms. It may be a reactant or a product.
To equalise the number of atoms, it must be remembered that we cannot alter the formulae of the com-
pounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coeffi-
cient ‘4’ as 4 H2
O and not H2
O4
. Now the partly balanced equation becomes
Fe + 4H2
O  Fe3
O4
+ H2
Step IV : Fe and H atoms are still not balanced. Pick any of these elements to proceed further. Let us
balance hydrogen atoms in the partly balanced equaiton.
To equalise the number of H atoms, make the number of molecules of hydrogen as four on the RHS.
The equation would be –
Fe + 4H2
O  Fe3
O4
+ 4H2
Step V : Examine the above equation and pick up the third element which is not balanced. You find that
only one element is left to be balanced, that is, iron.
To equalise Fe, we take three atoms of Fe on the LHS.
3Fe + 4H2
O  Fe3
O4
+ 4H2
Step VI : Finally, to check the correctness of the balanced equation, we count atoms of each element on
both sides of the equation.
3Fe + 4H2
O  Fe3
O4
+ 4H2
The numbers of atoms of elements on both sides of equation are equal. This equation is now balanced.

This method of balancing chemical equaitons is called hit-and -trial method as we make trials to balance
the equation by using the smallest whole number coefficient.
Step VII : Writing Symbols of Physical States :
To make a chemical equation more informative,
(i) The physical states of the reactants and products are mentioned along with their chemical formulae.
(ii) The gaseous, liquid, aqueous and solid states of reactants and products are represented by the nota-
tions (g), (l), (aq) and (s), respectively.
(iii) The word aqueous (aq) is written if the reactant or product is present as a solution in water.
The balanced eq. becomes
3Fe (s) + 4H2
O (g)  Fe3
O4
(s) + 4H2
(g)
Note that the symbol (g) is used with H2
O to indicate that in this reaction water is used in the form of
steam.
(iv) Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc. , for the reaction are
indicated above and /or below the arrow in the equation. For example,
CO (g) + 2H2
(g) 

 
340atm CH3
OH(I)
6CO2
(aq) + 6H2
O (I) l
Chlorophyl
Sunlight


 
 C6
H12
O6
(aq) + 6O2
(aq)
Illustration 1
Balance the given equation
(i) CH4 + O2 — CO2 + H2O (not balanced)
(ii) Fe + O2 — Fe2O3 (not balanced)
Solution
(i) CH4 + 2O2 — CO2 + 2H2O (balanced)
(ii) 4Fe + 3O2 — 2Fe2O3 (balanced)
1.4 DIFFERENT TYPES OF CHEMICAL REACTIONS
1.4.1 Combination Reactions
Those chemical reactions which involve the combination of two or more substances to form a single new
substance are called combination reactions. Combination reactions may involve either (i) combination
of two elements or (ii) combination of an element and a compound to form a new compound or (iii)
combination of two compounds.
Let us now discuss all these types of combination reactions one by one.
(a) Combination reactions involving two elements: Some examples of combination reactions in-
volving two elements are:
(i) Carbon (charcoal) burns in air to form carbon dioxide
C(s) + O2
(g)  CO2
(g)
Carbon Oxygen Carbon dioxide
(Charcoal)
(ii) Hydrogen burns in oxygen to form water
2H2
(g) + O2
(g) 


 

spark
Electric 2H2
O(l)
Hydrogen Oxygen Water

(b) Combination reactions involving an element and a compound: Some examples of combination
reactions involving an element and a compound are:
(i) Nitric oxide combines with oxygen at room temperature to form brown fumes of nitrogen dioxide
(NO2
)
2NO + O2
 2NO2
Nitric oxide Oxygen Nitrogen dioxide
(c) Combination reactions involving two compounds: Some examples of such reactions are:
(i) Quick lime (CaO) reacts with water to form calcium hydroxide (slaked lime)
CaO(s) + H2
O(l)  Ca(OH)2
(aq)
Calciumoxide Water Calciumhydroxide
(Quicklime) (Slakedlime solution)
or (Lime water)
Formationof Slaked Lime (CalciumHydroxide)
A solution of slaked lime as produced by the above reaction is used for white washing purpose. Calcium
hydroxide reacts slowly with carbon dioxide in the air to form a thin layer; of calcium carbonate on the
walls. The formation of calcium carbonate takes place in two to three days after white washing which
gives a shine to the walls. It is also interesting to note that the chemical formula for marble is also CaCO3
Ca(OH)2
(aq) + CO2
(g)  CaCO3
(s) + H2
O(l)
Calciumhydroxide (From air) Calcium carbonate
(Shine on walls)
Exothermic and Endothermic Reactions
Exothermic Reactions: The chemical reactions in which formation of products, is accompanied by
evolution of heat are known as exothermic reactions.
Some examples of exothermic reactions are:
(i) Burning of coal :
C(s) + O2
(g)  CO2
(g) + Heat
(ii) Burning of natural gas :
CH4
(g) + O2
(g)  CO2
(g) + 2H2
O(l) + Heat
(iii) Formation of slaked lime from quick lime :
CaO(s) + H2
O(l)  Ca(OH)2
(aq) + Heat
Quicklime Slaked lime
(iv) Respiration. Do you know that respiration is also an exothermic process ? This energy is
generally supplied by food we eat. Bread, potatoes and rice etc. which we eat all contains carbohydrates.
C6
Hl2
O6
(aq) + 6O2
(aq) 6CO2
(aq) + 6H2
O(l) + Energy
The reaction is known by a special name respiration.
(v) Decomposition of vegetable matter on a compost heap is also an example of exothermic reaction.
Endothermic Reactions: The chemical reactions in which formation of products is accompanied by the
absorption of heat are known as endothermic reactions.
Some examples of endothermic reactions are:
(i) N2
(g) + O2
(g) 2NO(g) – Heat
(ii) H2
(g) + I2
(g) 2HI(g) – Heat

(iii) C(s) + 2S(g)  CS2
(l) – Heat
(iv) C(s) + H2
O(g)  CO(g) + H2
(g) – Heat
1.4.2 Decomposition Reactions
(i) Those chemical reactions in which a compound breaks down to produce two or more simpler sub-
stances are known as decomposition reactions.
(ii) These reactions take place when the energy is supplied in the form of heat, light or electricity.
(iii) It may be noted that decomposition reactions are just the reverse of combination reactions.
(a) Thermal Decomposition Reactions: Chemical reactions in which the decomposition is achieved
by supplying heat energy are called thermal decomposition reactions.
(i) Decomposition of ferrous sulphate. Ferrous sulphate on heating decomposes as given below:
2FeSO4
(s) 
 
Heat Fe2
O3
(s) + SO2
(g) + SO3
(g)
Ferrous sulphate Ferric oxide Sulphurdioxide Sulphurtrioxide
(i) Decomposition of lead nitrate
2Pb(NO3
)2
Heat
ion
Decomposit 



 
 2PbO + 4NO2
+ O2
Lead nitrate Lead oxide Nitrogen dioxide Oxygen
(ii) Decomposition of calcium carbonate
CaCO3
(s)
Heat
ion
Decomposit 



 
 CaO(s) + CO2
(s)
Calcium carbonate Calcium oxide Carbon dioxide
Calcium oxide obtained in this process is called lime or quick lime, it has many uses. The most important
is in the manufacture of cement.
(b) Electrolytic Decomposition Reactions: Chemical reactions in which decomposition is achieved
by passing electric current are called electrolytic decomposition reactions. This process of electrolytic
decomposition of a substance is also known as electrolysis.
Some examples of electrolytic decomposition reactions are:
(i) Electrolytic decomposition of water  When electric current is passed through acidulated water, it
decomposes to give hydrogen gas and oxygen gas.
2H2
O(l) 



 
 current
Electric 2H2
(g) + O2
(g)
Water Hydrogen Oxygen
(c) Decomposition in the presence of sun light: Some compounds decompose when placed in sun
light.
2AgCl (s) 

 
Sunlight 2Ag(s) + Cl2
(g)
(Grey) (Chlorine)
If we take silver bromide in place of silver chloride, it is also converted into grey coloured silver.
2AgBr 

 
Sunlight
2Ag + Br2
Silver bromide Silver metal Bromine
(Yellow) (Grey)
Decomposition reactions are generally endothermic in nature.

1.4.3 Advantages of Decomposition Reactions
Decomposition reactions find a large number of its advantages in industry and our daily life. Some impor-
tant advantages of decomposition reactions are:
(i) Extraction of metals. Metals like sodium, potassium, aluminium, calcium, magnesium, etc. can be ex-
tracted by the electrolytic decomposition of their molten salts. When fused (molten) metal chloride or
oxide is decomposed by passing electricity, the metal is produced at the cathode (negative electrode).
(ii) Isolation of some non-metals. Some non-metals like hydrogen, oxygen and chlorine etc. can be ob-
tained on large scale by the electrolytic decomposition of their respective compounds. For example,
hydrogen and oxygen can be obtained by the electrolysis of acidulated water.
(iii) Digestion of food in our body. For example, starch which we take in the form of food like rice, wheat,
potatoes, etc. decomposes in our bodies to produce simple sugars like glucose. Similarly, proteins con-
sumed by us in different forms decompose to form amino acids.
1.4.4 Displacement Reactions
Those reactions in which a more active element displaces or removes another less active element from a
compound are called displacement reactions.
(i) Displacement of copper by iron: When a piece of iron metal is dipped in a solution of copper sulphate,
deep blue colour of copper sulphate starts fading and starts converting into green colour. This is due to
displacement of copper from copper sulphate solution by more reactive iron metal which results in the
formation of green coloured ferrous sulphate solution along with the deposition of reddish brown copper
metal on the surface of iron metal.
Fe(s) + CuSO4
(aq)  FeSO4
(aq) + Cu (s)
Iron Copper sulphate Iron (II) sulphate Copper
(Blue) (Green)
(ii) Displacement of copper by zinc: If we dip a strip of zinc metal in copper sulphate solution, zinc
displaces copper from copper sulphate forming zinc sulphate and copper metal
Zn(s) + CuSO 4
(aq)  ZnSO4
(aq) + Cu (s)
Zinc Copper sulphate Zinc sulphate Copper
solution (Blue) (Colourless)
As zinc sulphate solution is colourless, therefore, the blue colour of copper sulphate solution goes on
fading with the passage of time. At the same time, a reddish brown deposit of copper metal is formed on
the zinc strip.
One more example of displacement of copper from its salt solution is given below :
Pb(s) + CuCl2
(aq)  PbCl2
(aq) + Cu(s)
Lead Copper chloride Lead chloride
Let us now consider a reaction in which more reactive copper displaces less reactive metal from its salt
solution.
(iii) Displacement of less reactive non metal by more reactive metal
Like metals a more reactive non-metal also displaces a less reactive non-metal from its compound.
For example, more reactive chlorine gas when bubbled through a colourless solution of potassium
iodide, displaces iodine from it as violet vapours.
2KI(aq) + Cl2
(g) 2KCl (aq) + I2
Potassiumiodide Chlorine Potassium chloride Iodine
(Colourless) (Violet)

1.4.5 Double Displacement Reactions
The reaction in which two different atoms or groups of atoms are displaced by other atoms or groups of
atoms or in which two compounds react by an exchange or displacement of ions to form new compounds
are called double displacement reactions.
Ex. Na2
SO4
(aq) + BaCl2
(aq)  2NaCl(ag) + BaSO4
(s)
Sodiumsulphate Bariumchloride Sodiumchloride Barium sulphate
(White ppt.)
In this reaction, –
2
4
SO ions displace Cl–
ions and Cl–
ions displace –
2
4
SO ions. Since the reactions involves
the displacement of two chemical species, therefore, it is known as double displacement reaction.
These reactions usually occur in ionic compounds.
The double decomposition reactions can be further classified in two types :
(a) Precipitation reactions (b) Neutralisation reactions
Let us discuss these double decomposition reactions, individually.
(a) Precipitation reaction: Those reactions in which two clear and transparent solutions on mixing
result in the formation of an insoluble product are known as precipitation reactions and the insoluble
product is known as precipitate.
An example of precipitation reactions is given below:
Ex. Na2
S(aq) + (CH3
COO)2
Pb(aq)  PbS(s)  + 2CH3
COONa(aq)
Sodiumsulphide Lead acetate Lead sulphide Sodium acetate
(Blackprecipitate)
(b) Neutralisation Reactions :
Reactions in which an acid and a base react with each other to produce salt and water are known as
neutralisation reactions.
When an aqueous solution of hydrochloric acid is mixed with an aqueous solution of sodium hydroxide in
equivalent amounts, a reaction takes place to form sodium chloride and water
Ex. HCl (aq) + NaOH(aq)  NaCl(aq) + H2
O(l)
Hydrochloric acid Sodiumhydroxide Sodiumchloride Water
(Acid) (Base) (Salt) (Water)
Such a reaction is termed as a neutralisation reaction. The hydrogen (H+
) ions which were responsible
for the acidic properties of HCl have reacted with hydroxyl (OH–
) ions which were responsible for the
basic properties of NaOH to produce neutral water. The Na+
and Cl–
ions have undergone no chemical
change and appear in the form of crystalline sodium chloride on evaporation.
Since HCl, NaOH and NaCl are all soluble strong electrolytes, therefore, the above equation can also be
written in an ionic form as under:
H+
(aq) + Cl–
(aq) + Na+
(aq) + OH–
(aq)  Na+
(ag) + Cl–
(aq) + H2
O(l)
Cancelling the common ions on both sides, the net ionic equation is:
H+
(aq) + OH–
(aq)  H2
O(l)
neutralisation reaction is infact a combination of H+
ions of the acid and OH–
ions of the base to produce
H2
O.
Application of neutralization reaction in daily life:
(i) The most important use of neutralization reaction is in the form of antacids. The substances which
remove excess acid in our stomach are called antacids.
(ii) In other words, antacids are simple bases that neutralize digestive acids. Their ability to neutralize
acids is due to the hydroxide, carbonate or bicarbonate they contain.

Ex. (i) Digel available in the market is a mixture of Mg(OH)2
and CaCO3
.
(ii) Milk of magnesia - [Mg(OH)2
], Maalox [a mixture of Mg(OH)2
and Al (OH)3
], Amphojel [Al(OH)3
]
etc.)
Illustration 2
Identify the type of chemical reaction
(a) 2KNO3 — 2KNO2 + O2
(b) N2 + 3H2 — 2NH3
(c) CuSO4 + Fe — FeSO4 + Cu
(d) AgNO3 (aq) + NaCl (aq) — AgCl (s) + NaNO3 (aq)
Solution
(a) Decomposition (b) Combination
(c) Displacement (d) Double displacement reaction
Illustration 3
What are condition required for decomposition reaction.
Solution
Heat, light and electricity
1.5 OXIDATION AND REDUCTION REACTIONS
Oxidation : (i) The addition of oxygen to a substance is called oxidation.
(ii) The removal of hydrogen from a substance is also called oxidation.
Reduction : (i) The addition of hydrogen to a substance is called reduction.
(ii) The removal of oxygen from a substance is also called reduction.
The process of reduction is just the opposite of oxidation. Moreover, oxidation and reduction occur
together.
Oxidising agent :
(i) The substance which gives oxygen for oxidation is called an oxidising agent.
(ii) The substance which removes hydrogen is also called an oxidising agent.
Reducing agent :
(i) The substance which gives hydrogen for reduction is called a reducing agent.
(ii) The substance which removes oxygen is also called a reducing agent.
The oxidation and reduction reactions are also called redox reactions (In the name ‘redox’, the
term ‘red’ stands for ‘reduction’ and ‘ox’ stands for oxidation). We will now give some examples of
oxidation and reduction reactions.
Example 1. When zinc oxide is heated with carbon, then zinc metal and carbon monoxide are formed :
ZnO + C 
 
Heat Zn + CO
Zinc oxide Carbon Zinc Carbon monoxide
In this reaction,
(i) zinc oxide (ZnO) is losing oxygen, so it is being reduced to zinc (Zn).
(ii) carbon (C) is gaining oxygen, so it is being oxidised to carbon monoxide (CO). In this reaction, zinc
oxide is the oxidising agent whereas carbon is the reducing agent. Carbon is used in the form of coke for
the extraction of zinc metal.
Example 2. When manganese dioxide reacts with hydrochloric acid, then manganese dichloride, chlorine
and water are formed :

MnO2
+ 4HCl  MnCl2
+ Cl2
+ 2H2
O
Manganese Hydrochloric Manganese Chlorine Water
dioxide acid dichloride
In this reaction, MnO2
is losing oxygen to form MnCl2
, so manganese dioxide (MnO2
) is being reduced to
manganese dichloride (MnCl2
). On the other hand, HC1 is losing hydrogen to form Cl2
, so hydrochloric
acid (HC1) is being oxidised to chlorine (Cl2
). In this reaction, manganese dioxide (MnO2
) is the oxidising
agent whereas hydrochloric acid (HCl) is the reducing agent.
*1.5.1 Modern concept of oxidation and reduction
Electronic concept: This concept is applicable when reactants are ionic in nature.
As per this concept oxidation is defined as a process in which an atom, ion or molecule looses one or
more eletrons.
Mg — Mg+2
+ 2
Na — Na+
+
Cu — Cu2+
+ 2
Fe2+
— Fe3+
+
H2
O2
— O2
+ 2H+
+ 2
Reduction is defined as a process in which atom, ion or molecule gains one or more electrons.
Cl + — Cl–
Zn2+
+ 2 — Zn
Sn4+
+ 2 — Sn2+
Cr2
–
2
7
O + 14H+
+ 6 — 2Cr3+
+ 7H2
O
H2
O2
+ 2H+
+ 2 — 2H2
O
*1.5.2 Redox Reaction
The chemical reactions involving loss of electrons by one species (oxidation) and simultaneous gain of
electron (reduction) gives oxidation–reduction system or called redox reaction.
Redox reactions involve two half reactions, one corresponding to oxidation and the other reduction.
Example:
Zn + Cu2+
— Zn+2
+ Cu
Zn — Zn2+
+ 2 (oxidation half reaction)
Cu2+
+ 2 — Cu (Reduction half reaction)
1.6 OXIDISING AND REDUCING AGENTS
1.6.1 Oxidising agent
A substance i.e. atom, ion or molecule that oxidises other one and in return gets reduced itself by gain of
electron is called Oxidising agent.
Examples of Oxidising agent
(i) Non-metals such as halogens and oxygen, ozone are good oxidising agents.Among halogens fluorine is
the strongest oxidising agent.
(ii) Oxides of elements such as CaO, MgO, CuO, P4
O10
, Na2
O etc.

(iii)Acidified K2
Cr2
O7
Cr2
–
2
7
O + 14H+
+ 6 — 2Cr+3
+ 7H2
O
(iv) H2
O2
H2
O2
+ 2H+
+ 2 — 2H2
O
(v) Potassium permangnate
(a) In acidic medium
Mn –
4
O + 8H+
+ 5 — Mn2+
+ 4H2
O
i.e. Mn+7
+ 5 — Mn+2
(b) In alkaline medium
Mn –
4
O + — Mn –2
4
O
i.e. Mn+7
+ — Mn+6
(c) In neutral medium
Mn –
4
O + 2H2
O + 3 — MnO2
+ 4OH–
i.e. Mn+7
+ 3 — Mn+4
1.6.2 Reducing agent
A substance i.e., atom, ion or molecule which reduces the other one and gets oxidised it self by loss
of electrons is called a reducing agent
Example:
(i) All the metals are strong reducing agents. For e.g. , Na, K, Zn, Al, V, Cr, Fe etc.
(ii) Metallic hydrides such as, CaH2
, NaH, LiH etc.
(iii) Strongest reducing power is shown by Lithium in its solution state.
(iv) Compounds such as FeSO4
, HI, HCl, HBr, H2
S, SnCl2
, O3
and H2
O2
etc. also show strong reducing
power.
(v) In the reaction of stannous chloride and mercuric chloride, stannous chloride acts as a reducing agents
as it oxidation state is increased from +2 to +4
Sn+2
— Sn+4
+ 2e–
(Oxidation)
2Hg+2
+ 2e–
— 2Hg+1
(Reduction)
The overall reaction can be written as
Sn+2
+ 2Hg+2
— 2Hg+1
+ Sn+4
(vi) Oxalic acid









COOH
|
COOH
— C2
–2
4
O + 2 
H
C2
–2
4
O — 2CO2
+ 2e–
The substance which act as both oxidising and reducing agents are O3
, H2
O2
, H2
SO3
, HNO2
, NaNO2
,
SO2
, Na2
S2
O3
etc.

* 1.6.3 Valency, Oxidation State and Oxidation Number
Valency of an element is defined as number indicating its combining capacity. For example,
(i) It represents the number of hydrogen atoms which can combine with a given atom.
(ii) it also represents the number of single bonds which an atom can form.
(iii) It is also defined as a number of electrons its atom is able to lend, borrow or share.
(iv) In any case valency is a pure number and has no plus or minus sign associated with it.
In ionic compounds the hereby oxidation state of an element is the same as the charge on the ion formed
from an atom of the element. For example, in potassium bromide potassium is said to be in the + 1
oxidation state and bromine in – 1 oxidation state. It ionizes as
KBr = K+
+ Br–
(v) Oxidation state of aluminium inAl2
O3
is +3 and the total oxidation number of two aluminium atoms is
+6.
(vi) Thus oxidation state of an element is its oxidation number per atom.
There may actually be a
Difference between the magnitude of valency and the oxidation number.
For example, consider the following compounds of carbon:
CH4
CH3
Cl CH2
Cl2
Methane Methyl chloride Methylene chloride
CHCl3 CCl4
Chloroform Carbon tetrachloride
In each case one atom of carbon shares a total of 4 pairs of electrons with other atoms. Carbon atom is,
therefore, tetravalent in each case.
Oxidation number for carbon in CH4, CH3Cl, CH2Cl2, CHCl3 and CCl4 is –4, –2, 0, +2 and +4 respectively.
Difference between oxidation number and valency
Thus while valency of carbon remains constant (=4) in each one of the five compounds, its oxidation
number varies from –4 to +4.
* 1.6.4 Oxidation and Reduction in terms of Oxidation Number
The term oxidation refers to any chemical change involving increase in oxidation number whereas
the term reduction applies to any chemical change involving decrease in oxidation number
Consider the following chemical changes:
(i) 2H2
+ O2
— 2H2
O
Here in oxidation number of hydrogen changes from 0 (in H2
) to + 1 (in H2
O). It is, therefore, a case of
oxidation of hydrogen.
(ii) Sugar (C12
H22
O11
) burns to give CO2
and water. In this oxidation number of carbon increases from 0
(in C12
H22
O11
) to +4 in CO2
. The sugar is, therefore, said to have undergone oxidation.
(iii) When oxygen reacts with hydrogen to give water [example (i)] the oxidation number of oxygen
decreases from 0 (in O2
) to –2 (in H2
O). It is, therefore, a case of reduction of oxygen.
In the same reaction, oxidation number of hydrogen increases, and that of oxygen decreases, i.e., hydrogen
undergoes oxidation while oxygen undergoes reduction. Thus oxidation and reduction occur together.
An oxidising agent is a substance which brings about oxidation. It contains an atom which undergoes
a decrease in oxidation number. It can also be defined as a substance which picks up electrons and
thus brings about de-electronation.

ESC/Chemistry/Class-X CH-1: CHEMICALREACTIONS AND EQUATIONS
*1.6.5 Oxidation Number or Oxidation State
Definition of Oxidation Number
The oxidation number is defined as a positive or negative number that represents a charge that an atom
appears to have in a given species when the bonding electrons are counted as per the certain prescribed
set of rules.
Rules for assigning oxidation number
Oxidation number for atoms & ions can be assigned using the following set of rules.
Rule 1
The oxidation number of an atom in an element in its free uncombined state is zero, regardless of whether
the element exists as monoatomic or polyatomic molecule. For example, each fluorine atom in F2
, each
phosphorus atom in P4
& the silver atom in Ag, is assigned as oxidation number of zero.
Rule 2
The oxidation number of a monoatomic ion is same as the charge on the ion. For example, the oxidation
number of calcium ion is +2, in sulphur S–2
ion has oxidation number of –2.
Rule 3
Oxidation numbers conventionally assigned to atoms in their chemical compounds are as follows:
(a) Oxygen = –2 (except in peroxides where it is –1). For example, the oxidation state of oxygen in SO2
,
KClO3
and KMnO4
is –2. In Na2
O2
& H2
O2
, the oxidation number of oxygen is –1. In the very rare
instance when oxygen is bound to an element that is more electronegative than itself, such as in OF2
,
Oxygen exhibits an oxidation number of +2.
(b) Hydrogen = +1 (except in metallic hydrides where it is –1). For example the oxidation number of
hydrogen atom in H2
O, H2
O2
NH3
, CH3
COOH is +1. In LiH, it is –1.
(c) Group IA elements (alkali metals) = +1
(d) Group IIA elements (alkaline earth elements) = +2
(e) Halogen atoms in binary ionic compounds (halides) = –1. The halogen atom in Na+
, KBr, CsI has an
oxidation number of –1.
Rule 4
The algebric sum of the positive & negative oxidation numbers in a compound is zero.
The oxidation number of a specified atom in a compound can therefore be determined as illustrated
below.
Oxidation number of Mn in KMnO4
Let oxidation number of Mn be x
Oxidation number of oxygen is –2
& Oxidation number of K is +1
 +1 + x + 4 (–2) = 0
 x = + 7
Therefore, the oxidation number of Mn in KMnO4 is +7
Rule 5
The algebraic sum of the positive and negative oxidation states or numbers of the atoms in a polyatomic
ion is equal to charge on the ion.
Ex. (i) Let us find out the oxidation number of chromium in Cr2
2
7
O
Let the oxidation number of chromium be x and oxygen as –2,
Sum of oxidation numbers = 2 x + 7 (–2) = –2

2 x – 14 = –2
x = +6
Thus the oxidation number of chromium in Cr2
2
7
O
ion is +6.
(ii) Oxidation number of S in H2SO4 is as,
Let oxidation number of S be x.
Sum of oxidation numbers of various atoms in H2SO4 = 2x (+1) + x + 4 x (–2)
= 2 + x – 8 = x – 6
This sum must be zero (rule b). Hence
x – 6 = 0
when x = 6
or oxidation number of S in H2SO4 = +6
Putting oxidation number of S in Na2S2O3
2x – 4 = 0, we have
x = +2
 Oxidation number of S in Na2S2O3 = +2
(iii) Oxidation state of Mn in Mn2O7 is as,
Let the oxidation state of manganese be x.
Sum of oxidation numbers of various atoms in Mn2O7 is as,
2(x) + 7 (–2) = 0
 2 x = + 14
x = +7
Rule 6
The oxidation numbers of atoms in covalent compounds can be derived by assigning the electrons of each
bond to the more electromagnetic atom of the bonded atoms.
(i) Oxidation state of S in per monosulphuric acid (H2SO5).
Let the oxidation number S be x, oxygen –2, hydrogen +1 and the oxidation state of oxygen in peroxylinkage
is – 1.
Sum of oxidation number of various atoms is,
+ 2 + x + 3 (–2) – 2 = 0
+ 2 + x – 6 – 2 = 0 H
—
O
—
—O
O
S
O
—
O
—
H


x = + 6
*1.6.6 Distinction between valency and oxidation state / oxidation number
(i) Valency is always a whole number, on the other hand the oxidation number of the element may be a
whole number or fractional.
(ii) Valency of the element is never zero except of noble gases but the oxidation number of the element
may be zero.
(iii)Valency is the combining power of an element with no plus or minus sign. On the other hand oxidation
number is the charge present on the atom of the element while being in combination. It may have plus or
minus sign.
The oxidation number changes with the following cases.
Oxidation  Oxidation number increases
Reduction  Oxidation number decreases

Oxidizing agent  Oxidation number decreases
Reducing agent  Oxidation number increases
*1.6.7 Oxidation State / Oxidation number in Complex molecules
(i) Carbon in Glucose (C6
H12
O6
)
Let the oxidation number of carbon be x, hydrogen +1 and oxygen –2
Sum of oxidation number of various atoms in C6
H12
O6
is
6x + 12 (+1) + 6 (–2) = 0
6x + 12 – 12 = 0
6x = 0
x = 0
In glucose the oxidation state of carbon is zero.
(ii) Ni in [Ni(CO)4
]
The oxidation state of CO is zero, hence the oxidation state of nickel will also zero.
(iii) Sulphur in (CH3
)2
SO i.e., dimethyl sulphoxide.
Let the oxidation state of S be x, oxygen –2, and each methyl group is +1
Sum of oxidation numbers of various atoms in (CH3
)2
SO is
+2 + x – 2 = 0
x = 0
Thus the oxidation state of sulphur in (CH3
)2
SO is zero.
(iv) Boron in Li BH4
In metal hydrides, the oxidation state of hydrogen is –1 and the metal Lithium is +1
Sum of oxidation number of various atoms in LiBH4
is,
+1 + x – 4 = 0
x = +3
Thus the oxidation state of B in LiBH4
is +3
*1.6.8 Fractional values of Oxidation numbers are possible with the following
components
(i) Hydrazoic acid N3
H
Let the oxidation no. of nitrogen be x and hydrogen +1
3x + 1 = 0
3x = – 1
x = 
3
1
x = – 0.333
The oxidation number of N in N3
H is –1/3
(ii) Na2
S4
O6
Let the oxidation no. of sulphur be x, sodium +1 and oxygen –2,
2 + 4x + 6(–2) = 0
2 + 4x – 12 = 0

4x = + 10
x = 10/4= 2.5
The oxidation number of S in Na2
S4
O6
is 2½
Illustration 4
Find out the oxidation number / oxidation state of
(a) S in H2S2O7 (b) S in Na2S2O3 (c) Cr in Cr(CO)6
(d) Fe in Fe2(CO)9 (e) Fe in Fe3O4 (f) Mn in MnO4
–
Solution
(a) Let the oxidaiton no. of S be x,
H – 1 and O – 2
2 (+1) + 2x + 7 (–2) = 0
2 + 2x – 14 = 0
2x = +12
x = +6
(b) Let the oxidation number of S be x,
Na +1 and O –2
2 + 2x – 6 = 0
2x = +4
x = +2
(c) Let oxidation number of Cr be x, and CO = zero.
x + 6 (0) = 0
x = 0
(d) Let oxidation number of Fe be x, and CO = zero.
2x + 9 (0) = 0
2x = 0
x = 0
(e) Let oxidation number of Fe be x, and O = –2.
3x + 4 (–2) = 0
3x – 8 = 0
x =
3
8

(f) Let the oxidation number of Mn be x and O = –2.
x + 4 (–2) = –1
x = +7
*1.7 BALANCING OF REDOX REACTIONS
The redox reactions can be balanced by following methods
(i) Oxidation number method
(ii) Ion-electron method
1.7.1 Balancing by oxidation number method
The various steps involved in balancing a redox equation by oxidation number method are:

(i) Write the skeleton equation.
(ii) Indicate the oxidation numbers of all the atoms involved in the equation above their symbols.
(iii) Identify the elements which undergo change in oxidation number.
(iv) Calculate the increase and decrease in oxidation number per atom with respect to the reactants. If
more than one atom is involved, then multiply with the number of the atoms undergoing the change to
calculate the total change in oxidation number.
(v) Equate the increase and decrease in oxidation number on the reactant side by multiplying the formulae
of the oxidising and reducing agents suitably.
(vi) Balance the equation with respect to all the atoms except hydrogen and oxygen.
(vii) Finally balance hydrogen and oxygen atoms also.
(viii) In the reactions taking place in the acidic medium, balance the O atoms by adding required number
of H2
O molecules to the side deficient in O atoms. Then balance the H atoms by adding H+
to the side
deficient in H atoms.
(ix) In the basic medium, first balance the number of negative charges by adding required number of OH–
ions to the side deficient in the magnitude of the charges. Then add H2
O molecules on the other side in
order to balance the OH–
ions added.
Let us try to balance a few chemical equations by oxidation number method.
Illustration 5
Balance the following chemical equations by the oxidation number method
CuO + NH3
— Cu + N2
+ H2
O
Solution
The balancing is done in the following steps:
1. Write the O.N. of each atom in the skeleton equation
2
Cu
 2
–
O +
3
N
 1
3
H

—
0
Cu +
0
2
N +
1
2
H
 2
O

2. Identify the atoms which undergo change in O.N.
2
Cu

O +
3
N

H3 —
0
Cu +
0
2
N + H2O
3. Calculate the increase and decrease in O.N. w.r.t. reactant atoms
4. Equate the increase and decrease in O.N. on the reactant side.
3CuO + 2NH3 — Cu + N2 + H2O
5. Balance the number of Cu and N atoms on both sides of the equation.
3CuO + 2NH3 — 3Cu + N2 + H2O
6. Now balance H and O atoms by hit and trial method
3CuO + 2NH3 — 3Cu + N2 + 3H2O
(i) Cu + 
3
NO — NO2 + Cu2+
(Acidic medium)

1. Write the O.N. of each atom in the skeleton equation.
0
Cu + (
5
N
 2
3
O

)–
—
4
N
 2
2
O

+ (
2
Cu

)2+
0
Cu + (
5
N

O3)–
—
4
N

O2 + (
2
Cu

)2+
3. Calculate the increase and decrease in O.N. w.r.t. to reactant atoms
4. Equate the increase and decrease in O.N. on the reactant side.
Cu + 2 
3
NO — NO2 + Cu2+
5. Balance the number of Cu and N atoms on both sides of the equation.
Cu + 2 
3
NO — 2NO2 + Cu2+
6. As the reaction is carried in the acidic medium balance the number of O atoms by adding two H2O
molecules on the product side.
Cu + 2 
3
NO — 2NO2 + Cu2+
+ 2H2O
7. To balance the number of H atoms, add 4H+
on the reactant side
Cu + 2 
3
NO + 4H+
— 2NO2 + Cu2+
+ 2H2O.
The final equation is balanced w.r. to charge also.
[Cr(OH)4]–
+ H2O2 — Cr 
2
4
O + H2O (Basic Medium)
1. Write the O.N. of each atom in the skeleton equation:
1
2
1
2
1
2
4 O
H
(OH)
cr












2
2 1 1 2 6 2 1 2
4
4 2 2
Cr(OH) H O CrO H O
 
       
   
  
 
   
1
2
2O
H 
[
3
Cr

(OH)4]–
+ H2
1
2
O

— (
6
Cr

O4)2–
+ H2
2
O

3. Calculate the increase and decrease in O.N. w.r. to reactant atoms
[Cr(OH) ] + H O — (CrO ) + H O
4 2 2 4 2
– 2–

Decrease in O.N. = 1 × 2 = 2
Increase in O.N. = 3
+3 –1 +6 –2

4. Equate the increase and decrease in O.N. in the reactant side.
2[Cr(OH)4]–
+ 3H2O2 — (CrO4)2–
+ H2O
5. Balance the number of Cr atoms in the equation.
2[Cr(OH)4]–
+ 3H2O2 — 2(CrO4)2–
+ H2O
6. In order to balance the number of oxygen atoms, add five H2O molecules on the product side
2[Cr(OH)4]–
+ 3H2O2 — 2(CrO4)2–
+ 6H2O
7. As the reaction is carried in the basic medium, in order to balance the number of negative charges add
two OH–
ions on the reactant sides and two H2O molecules on the product side.
2[Cr(OH)4]–
+ 3H2O2 + 2OH–
— 2(CrO4)2–
+ 6H2O + 2H2O
2[Cr(OH)4]–
+ 3H2O2 + 2OH–
— 2(CrO4)2–
+ 8H2O
Illustration 6
C6H6 + O2  CO2 + H2O
Solution
1
–
6
C
1
6
H

+
0
2
O —
4
C
 2
–
2
O +
1
2
H
 2
O

1
–
6
C H6 +
0
2
O —
4
C

O2 + H2
2
2
O

3. Calculate the total increase and decrease in O.N. w.r. to reactant atoms.
C H + O — CO + H O
6 6 2 2 2

Decrease in O.N. = 2 × 2 = 4
Increase in O.N. = 5 × 6 = 30
–1 +4 –2
4. Equate the increase and decrease in O.N. on the reactant side after taking out a common fator of 2.
2C6H6 + 15O2 — CO2 + H2O
5. Balance the number of C and O atoms on both sides of the equation.
2C6H6 + 15O2 — 12CO2 + 6H2O
The H atoms are already balanced in the above equation
Illustration 7
SnO2 + C — Sn + CO
Solution
4
Sn
 2
2
O
 +
0
C —
0
Sn +
2
C
 2
O

4
Sn

O2 +
0
C —
0
Sn +
2
C

O

3. Calculate the total increase and decrease in O.N. w.r. to reactant atoms.
4. Equate the increase and decrease in O.N. on the reactant side after taking out a common fator of 2.
SnO2 + 2C — Sn + CO
5. Balance the number of Sn and C atoms on both sides of the equation.
SnO2 + 2C — Sn + 2CO
The O atoms are already balanced in the above equation
1.7.2 Balancing by Ion-electron method (or half reaction method)
Balancing can also be done by another method known as ion-electron method. It is based on the
principle that the electrons lost during oxidation half reaction in a particular redox reaction is
equal to the electrons gained in the reduction half reaction. The method is, therefore, called half
reaction method. The balancing is completed in the following steps:
(i) Write the redox reaction in ionic form.
(ii) Find out species which are getting oxidised and also which are getting reduced.
(iii) Split the whole equation into two half reactions i.e. oxidation half reaction and reduction half reaction.
(iv) While balancing each half reaction add electrons for the number of atoms of each element.
(v) In the acidic medium, and neutral medium add water molecules to the side dificient in O and H+
to the
side deficient in hydrogen.
(vi) In the basic medium, for each excess of oxygen, add one water molecule to the same side and two
OH+
ions to the other side. If hydrogen is still unbalanced, add on OH-
ion for each excess hydrogen on
the same side and one water molecule to the other side.
(vii) Multiply one or both half reactions by suitable number so that the number of e
s become equal in
both the equation.
(viii) Add the two balanced half reactions and cancel any term common to both sides.
Illustration 8
Balance the following chemical equation by ion-electron method.
Cr2

2
7
O + Fe2+
+ H+
— Cr3+
+ Fe3+
+ H2O
Solution
Step I. Separation of the equation in two half reactions
(i) Write the O.N. of all the atoms involved in the skeleton equation
(
6
2
Cr
 2
7
O

)2–
+ ( 2
Fe

)2+
+ ( 1
H

)+
— (
3
Cr

)3+
+ ( 3
Fe

)3+
+
1
2
H
 2
O

(ii) Identify the atoms which undergo change in O.N.
(
6
2
Cr

O7)2–
+ ( 2
Fe

)2+
+ ( 1
H

)+
— ( 3
Cr

)3+
+ ( 3
Fe

)3+
+ H2O
(iii) Find out the species involved in the oxidation and reduction half reactions.

Thus, the two half reactions are:
Oxidation half reaction: Fe2+
— Fe3+
Reduction half reaction: (Cr2O7)2–
— Cr3+
Step II. Balancing of oxidation half reaction:
The oxidation half reaction is: Fe2+
— Fe3+
(i) As the increase in O.N. as a result of oxidation is 1, add one e–
on the product side to balance change in
O.N.
Fe2+
— Fe3+
+ e–
(ii) The charge is already balanced, and thus the equation is alo balanced
Fe2+
— Fe3+
+ e–
.....(i)
Step III. Balancing of reduction half reaction
The reduction half reaction is: (
6
2
Cr

O7)2–
— (
3
Cr

)3+
(i) The decrease in O.N. per Cr atom is 3 and the total decrease in O.N. for two Cr atoms is 6. Therefore,
add 6e–
on the reactant side
(Cr2O7)2–
+ 6e–
— Cr3+
(ii) Balance Cr atoms on both sides of the equation
(Cr2O7)2–
+ 6e–
— 2Cr3+
(iii) In order to balance O atoms add seven H2O molecules on the product side and then to balance H atoms
add 14 H+
on the reactant side.
(Cr2O7)2–
+ 6e–
+ 14H+
— 2Cr3+
+ 7H2O .....(ii)
Step IV. Adding the two half reactions:
In order two equate the electrons, multiply the equation (i) by 6 and then add to equation (ii) in order to get
the final equation.
 
6
]
e
Fe
Fe
O
7H
2Cr
6Fe
14H
O
Cr
6Fe
O
7H
2Cr
6e
14H
O
Cr
3
2
2
3
3
2
7
2
2
2
3
2
7
2























Illustration 9
Peramanganate (VII) ion, Mn 
4
O in basic solution oxidises iodide ion, I–
to produce molecular
iodine (I2) and manganese (IV) oxide (MnO2). Write a balanced ionic equation to represent this
redox reaction.
Solution
Step 1. First we write the skeletal ionic equation which is
Mn 
4
O (aq) + I–
(aq) — MnO2 (s) + I2(s)
Step 2. The two half-reactions are:
Oxidation half:
1
(aq)
I

 —
0
2
I (s)

Reduction half:
7
Mn


4
O (aq) —
4
Mn

O2 (s)
Step 3. To balance the I atoms in the oxidation half reaction, we rewrite it as:
2I–
(aq) — I2 (s) + 2e
Step 4. To balance the O atoms in the reduction half reaction, we add two water molecules on the right:
Mn 
4
O (aq) — MnO2 (s) + 2H2O (l)
To balance the H atoms, we add four H+
ions on the left:
Mn 
4
O (aq) + 4H+
(aq) — MnO2 (s) + 2H2O (l)
As the reaction takes place in a basic solution, therefore, for four H–
ions, we add four OH–
ions to both
sides of the equation:
Mn 
4
O (aq) + 4H+
(aq) + 4OH–
(aq) — MnO2 (s) + 2H2O (l) + 4OH–
(aq)
Replacing the H+
and OH–
ions with water, the resultant equation is:
Mn 
4
O (aq) + 2H2O (l) — MnO2 (s) + 4OH–
(aq)
Step 5. In this step we balance the charges of the two half-reactions in the manner depicted as:
2I–
(aq) — I2 (s) + 2e–
Mn 
4
O (aq) + 2H2O (l) 3e–
— MnO2 (s) + 4OH–
(aq)
Step 6. Add two half-reactions to obtain the net reactions after cancelling electrons on both sides.
6I–
(aq) + 2Mn 
4
O (aq) + 4H2O (l) — 3I2 (s)+ 2MnO2 (s) + 8OH–
(aq)
Step 7.Afinal vertification shows that the equation is balanced in respect of the number of atoms and charge on
both sides.
1.8 EFFECTS OF OXIDATION REACTIONS IN EVERYDAY LIFE
Oxidation has damaging effect on metals as well as on food. The damaging effect of oxidation on metals
is studied as corrosion and that on food is studied as rancidity. Thus, there are two common effects of
oxidation reactions which we observe in daily life. These are :
1. Corrosion of metals , and
2. Rancidity of food.
Corrosion : Corrosion is the process in which metals are eaten up gradually by the action of air, moisture
or a chemical (such as an acid) on their surface. Corrosion is caused mainly by the oxidation of metals by
the oxygen of air. Rusting of iron metal is the most common form of corrosion. When an iron object is left
in damp air for a considerable time, it gets covered with a red-brown flaky substance called ‘rust’. This
is called rusting of iron.
Rusting involves unwanted oxidation of iron metal which occurs in nature on its own.
Rust is a soft and porous substance which gradually falls off from the surface of an iron object, and then
the iron below starts rusting. Thus, rusting of iron (or corrosion of iron) is a continuous process which, if
not prevented in time, eats up the whole iron object. Corrosion weakens the iron and steel objects
and structures such as railings, car bodies, bridges and ships, etc., and cuts short their life. A lot
of money has to be spent every year to prevent the corrosion of iron and steel objects, and to replace the
damaged iron and steel structures.The black coating on silver and the green coating on copper are other
examples of corrosion.

1.9 RANCIDITY
When the fats and oils present in food materials get oxidised by the oxygen (of air), their oxidation
products have unpleasant smell and taste. The condition produced by aerial oxidation of fats and
oils in foods marked by unpleasant smell and taste is called rancidity. Rancidity spoils the food
materials prepared fats and oils which have been kept for a considerable time and makes them unfit for
eating.
(i) Rancidity can be prevented by adding anti-oxidants to foods containing fats and oils. Anti-oxi-
dant is a substance (or chemical) which prevents oxidation. The two common anti-oxidants used in foods
to prevent the development of rancidity are BHA (Butylated Hydroxy-Anisole) and BHT (Butyated
Hydroxy-Toluene).
(ii) Rancidity can be prevented by packaging fat and oil containing foods in nitrogen gas. When the
packed food is surrounded by an unreactive gas nitrogen, there is no oxygen (of air) to cause its oxidation
and make it rancid. The manufacturers of potato chips (and other similar food products) fill the plastic
bags containing chips with nitrogen gas to prevent the chips from being oxidised and turn rancid.
(iii) Rancidity can be retarded by keeping food in a refrigerator. The refrigerator has a low temperature
inside it. When the food is kept in a refrigerator, the oxidation of fats and oils in it is slowed down due to
low temperature. So, the development of rancidity due to oxidation is retarded.
(iv) Rancidity can be retarded by storing food in air-tight containers. When food is stored in air-tight
containers, then there is little exposure to oxygen of air. Due to reduced exposure to oxygen, the oxidation
of fats and oils present in food is slowed down and hence the development of rancidity is retarded.
(v) Rancidity can be retarded by storing foods away from light. In the absence of light, the oxidation of
fats and oils present in food is slowed down and hence the development of rancidity is retarded.
Solved Examples
Example 1
Find oxidation state of Chromium.
CrO5
Solution
x + 4(–1) + (–2) = 0
x = +6
Example 2
Find oxidation state of Nitrogen
NH4NO3
Solution

4
NH –
3
NO
x + 4(+1) = +1 x + 3(–2) = –1
x = –3 x – 6 = –1
x = +5
Example 3
Find oxidation state of Chlorine
CaOCl2

Solution
Ca2+
OCl–
Cl–
OCl–
Cl–
(–2) + x = –1
x = –1 x = +1
Example 4
Find oxidation of Bromine
Solution
The O.N. of two bromine atoms at terminal positions is +6 while that of the third atom is +4. The average
oxidation number of bromine atom is 




 

3
4
6
6
= 





3
16
.
Example 5
Point out the oxidising and reducing agents in the following reactions:
SO2 + 2HNO3 — H2SO4 + 2NO2
Solution
(i)
HNO3 is an oxidising agent decrease in O.N. because N atom undergoes decrease in O.N.
SO2 is a reducing agent, because S atom undergoes increase in O.N.
(ii)
SO2 is an oxidising agent because S atom undergoes decrease in O.N.
H2S is a reducing agent, because S atom undergoes increase in O.N.
Example 6
Balance the equation in acidic medium: (by ion electron method)
When chloride ion is oxidised to Cl2 by Mn 
4
O
Solution
(a) The skeleton equation is:
Cl–
+ 4
MnO
— Cl2 + Mn2+
Step 1. Separation of the equation in two half reactions:
(i) Write the O.N. of the atoms involved in the skeleton equation
1
(Cl)


+
7 2
4
MnO

 
 
 
 
—
0
2
Cl +
2
2
(Mn)


(ii) Identify the atoms which undergo change in O.N.
1
(Cl)


+
7 2
4
MnO

 
 
 
 
—
0
2
Cl +
2
2
(Mn)


(iii) Find out the species involved in the oxidation are reduction half reactions

Thus, the two half reactions are:
Oxidation half reaction: Cl–
— Cl2
Reduction half reaction: Mn 
4
O — Cl2 + Mn2+
Step 2. Balancing of oxidation half reaction:
The oxidation half reaction is: 2Cl–
— Cl2
(i) As the increase in O.N. per atom as a result of oxidation is 1, therefore, add two e–
on the product side
to balance change in O.N.
2Cl–
— Cl2 + 2e–
Thus, oxidation half reaction is balanced.
Step 3. Balancing of reduction half reaction:
The reduction half reaction is:
7 2
4
MnO

 
 
 
 
—
2
(Mn)

(i) The decrease in O.N. of Mn is 5. Therefore, add 5e–
on the reactant side to balance change in O.N.
Mn 
4
O + 5e–
— Mn2+
(ii) In order to balance the no. of O atoms, add four H2O molecules on the product side and then to
balnace H atoms add 8H+
on the reactant side.
Mn 
4
O + 8H+
+ 5e–
— Mn2+
+ 4H2O
Step 4. Adding the two half reactions:
In order to equate the electrons, multiply equation (i) by 5 and equation (ii) by 2 and then add the two
equations to get the final equation.
5
]
2e
Cl
—
2Cl
O
8H
2Mn
5Cl
—
16H
2MnO
10Cl
2
O]
4H
Mn
—
5e
8H
MnO
2
–
2
2
2
–
4
–
2
2
–
4



















Example 7
Balance the equation in acidic medium: (by ion electron method)
Chlorate ion (Cl 
3
O ) oxidises Mn2+
to MnO2 (s)
Solution
The skeleton equation:
3
ClO
+ Mn2+
— MnO2 + Cl–
Step 1. Separation of the equation in two half reactions:
(i) Write the O.N. of all the atoms involved in the equation:
(
2
3
5
O
Cl


)–
+ (
2
Mn

)2+
—
2
2
4
O
Mn


+ (
1
–
Cl )–
(ii) Identify in atoms which undergo change in O.N.
( 3
5
O
Cl

)–
+ (
2
Mn

)2+
— 2
5
O
Mn

+ (
1
–
Cl )–
(iii) Find out the species involved in the oxidation and reduction half reactions
The two half reactions are:
Oxidation half reaction: Mn2+
— MnO2
Reduction half reaction: –
3
ClO — Cl–

Step 2. Balancing of oxidation half reaction
The oxidation half reaction is: Mn2+
— MnO2
(i) As the increase in O.N. is 2, therefore, add two e–
on the product side to balance change in O.N.
Mn2+
— MnO2 + 2e–
(ii) In order to balance oxygen atoms, add two molecules of H2O on the reactant side. Then add four H+
ions on the product side to balance the hydrogen atoms
Mn2+
2H2O — MnO2 + 4H+
+ 2e–
.....(i)
Thus, oxidation half reaction is balnaced.
Step 3. Balalncing of reduction half reaction:
The reduction half reaction is: –
3
ClO — Cl–
(i) As the decrease in O.N. is 6, add six e–
on the reactant side to balance change in O.N.
–
3
ClO + 6e–
— Cl–
(ii) In order to balance oxygen atoms, add three molecules of H2O on the product side. Then add 6H+
iions
on the reactant side to balance the hydrogen atoms
–
3
ClO +6H+
+ 6e–
— Cl–
+ 3H2O .....(ii)
The reduction half reaction is balanced.
Step 4. Adding the two half reactions:
In order to equate the electrons, multiply equation (i) by 3 and add to equation (ii)
3
]
2
4H
MnO
O
2H
Mn
Cl
6H
3MnO
O
3H
ClO
3Mn
O
3H
Cl
6
6H
ClO
2
2
2
2
2
3
2
2
3

























e
e
Example 8
Balance the equation by oxidation number method.
P + HNO3 — HPO3 + NO + H2O (Neutral medium)
Solution
Step1. Write the O.N. of each atom in the skeleton equation:
0
P +
2
3
5
1
O
N
H



—
2
3
5
1
O
P
H



+
2
2
O
N


+
2
1
2 O
H


Step 2. Identify the atoms which undergo change in O.N.
0
P + 3
5
O
N
H

— 3
5
O
P
H

+ O
N
2

+ H2O
Step 3. Calculate the increase and decrease in O.N. W.V. to the reactant atoms.
Step 4. Equate the increase and decrease in O.N. on the reactant side.
3P + 5HNO3 — 3HPO3 + 5NO + H2O
The O and H atoms are already balnced in the equation balanced.

EXERCISE-I
Each question carry 1 mark:
1. Out of ‘Word equation’and ‘symbol equation’which is more informative?
2. State wheather the following information is true or false:
‘chemical equations can be balanced easily by altering the formulae of the substances involved’.
3. Aluminium burns in chlorine to form aluminium chloride. Write a balanced equation for the reaction.
4. What do the following symbols signify in a chemical equation.
(i) An arrow pointing downwards ()
(ii)An arrow pointing upwards ()
5. What does the symbol ‘aq’ represent in a chemical equation?
6. When a burning magnesium ribbon is placed in a jar of carbon dioxide, it continue burning. Write the
chemical equation involved in the reaction.
7. Give one example of decomposition which is useful in photography?
8. Can oxidation occur without reduction?
9. What is the main formula of rusts?
10. In the electrolysis of water, in which ratio are the gases hydrogen and oxygen collected?
11. Why do gold and silver not corrode in moist air?
12. What will happen to the beaker in which exothermic reaction is carried?
13. Why is double displacement reaction so named?
14. Predict whether the following displacement reaction is correct.
Cu (s) + FeSO4
(aq)  CuSO4
(aq) + Fe (s)
15. What is wrong with the following equation?
Ca + O  CaO
Correct it and write it in balanced form.
16. Following equation is balanced :
CaOH + HCl  CaCl + H2
O
Are you satisfied with it?
17. When ammonium chloride was dissolved in water taken in a beaker, it become cold. What acutally
happened?
18. Enlist the main factors which promote corrosion.
19. Under what conditions, does a chemical reaction become endothermic?
20. What is rust?
21. To preserve food items, we flush them with nitrogen or keep them in refrigerator. Why?
22. Identify the most reactive and least reactive metal among the following : Al, K, Ca, Au.
23. Which of the following is a combination reaction and which is a displacement reaction?
(a) Cl2
+ 2Kl  2KCl + I2
(b) 2K + Cl2
 2KCl
24. Write the following chemical equations in balanced form:
(i) Na + O2
 Na2
O (ii) Fe + H2
O  Fe3
O4
+ H2
(iii) NH3
+ CuO  Cu + N2
+ H2
O (iv) KClO3
 KCl + O2

25. When hydrogen burns in oxygen, water is formed. When water is electrolysed, then hydrogen and oxygen
are produced. What types of chemical reactions have occurred in both the cases?
26. What happens carbon dioxide gas is bubbled through lime water (i) in small amount (ii) in excess.
27. Ornaments of gold do not get corroded. Assign reason.
28. Iron nails do not get rusted when kept in distilled water even for a long time. Assign reason.
29. Phophorus and chlorine react to form two compounds. Write balanced equations for the reactions.
30. When SO2
is dissolved in water, acid rain containing sulphric acid is formed. The acid rain attacks marble
statues and damages them. Write the balanced chemical equations for these.
31. In the reaction :
MnO2
+ 4HCl  MnCl2
+ 2H2
O + Cl2
(a) Name the substance oxidised (b) Name the oxidising agent
(c) Name the substance reduced (d) Name the reducing agent
32. Suggest two ways to check the rancidity of food articles.
33. Ashining brown coloured element ‘X’on heating in air becomes black in colour. Name the element ‘X’and
the black coloured compound formed.
34. Asolution of CuSO4
was kept in an iron pot.After a few days, the pot developed some holes in it. How will
you account for this?
35. A silver spoon is kept immersed in an aqueous copper sulphate solution. What change will take place?
36. Why does not copper liberate hydrogen on reacting with dilute sulphuric acid?
37. Which of the following are chemical changes?
(a) Digestion of food (b) Liquefication of air (c) Ripening of fruit
(d) Dissolution of sulphur in carbon disulphide (e) Freezing of water
38. In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper
metal. Write down the reaction involved.
39. A, B and C are three elements which undergo chemical reactions according to following equations.
A2
O3
+ 2B  B2
O3
+ 2A
3CSO4
+ 2B  B2
(SO4
)3
+ 3C
3CO + 2A  A2
O3
+ 3C
Answer the following questions.
(a) Which element is the most reactive?
(b) Which element is the least reactive?
40. On mixing the solution of lead (II) nitrate and potassium iodide prepared in water,
(i) Write the chemical reactions involved in the balanced form
(ii) What is the colour of the precipitate? Name the precipitate.
41. Which types of reactions are represented by the following equations?
(i) CaO + CO2
 CaCO3
(ii) Mg + CuSO4
 MgSO4
+ Cu
(iii) CH4
+ 2O2
 CO2
+ 2H2
O (iv) NH4
NO2
 N2
+ 2H2
O
42. Identify the substance oxidised and substance reduced in the following reactions:
(i) ZnO + C  Zn + CO (ii) 2Na(s) + O2
(g)  2Na2
O (s)
(iii) CuO (s) + H2
(g)  Cu (s) + H2
O (l)

43. What information is conveyed by the following equation?
CaCO3
(s) 
 
heat CaO (s) + CO2
(g)
(Given: atomic mass of Ca = 40.C = 12, O = 16)
44. How is rusting of iron caused? Suggest some ways to prevent rusting of iron.
45. Write the balanced chemical equations with state symbols for the following reactions:
(a) Iron filings react with steam to produce iron (III) oxide and hydrogen gas.
(b) Magnesium reacts with nitrogen upon heating to form magnesium nitride.
(c) Ethane burns in oxygen to form carbon dioxide and water.
(d) Sodiumhydrogen solution (in water) reacts with hydrochloric acid (in water) to formsodium chloride (in
water) and water.
46. Write the balanced equation for the following reactions:
(a) Calcium hydroxide + Carbon dioxide  Calcium carbonate + Water
(b) Lead + Copper chloride  Lead chloride + Copper
(c) Barium chloride + Sodium sulphate  Barium sulphate + Sodium chloride
(d) Zinc + Silver nitrate  Zinc nitrate + Silver
47. Transfer the following into chemical equations and balance them:
(a) Hydrogen combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium chloride and precipitate of barium sulphate.
(d) Sodium metal reacts with water to give soidum hydroxide and hydrogen gas.
(e) Sodium chloride is electrolysed in molten state to form sodium and chlorine.
EXERCISE-II
SECTION-A
1. Give an example of a double displacement reaction other than the one given in Activity 1.10 of NCERT.
2. Which of the statements about the reaction below are incorrect?
2PbO (s) + C(s)  2Pb (s) + CO2
(g)
(i) Lead is getting reduced
(ii) Carbon dioxide is getting oxidised
(iii) Carbon is getting oxidised
(iv) Lead oxide is getting reduced
(A) (i) and (ii) (B) (i) and (iii) (C) (i), (ii) and (iii) (D)All
3. Fe2
O3
+ 2Al  Al2
O3
+ 2Fe
The above reaction is an example of a
(A) comobination reaction (B) double displacement reaction
(C) decomposition reaction (D) displacement reaction

4. What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.
(A) Hydrogen gas and iron chloride are produced
(B) Chlorine gas and iron hydroxide are produced
(C) No reaction takes place (D) Iron salt and water are produced
5. Why should a magnesium ribbon be cleaned before burning in air?
6. Write a balanced chemical equation with sate symbols for the following reactions
(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and
the solution of sodium chloride.
(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce so-
dium chloride solution and water.
7. A solution of a substance ‘X’ is used for white washing.
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
8. Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
9. Identify the substance that are oxidised and the substances that are reduced in the following reactions.
(i) 4Na (s) + O2
(g)  2Na2
O(s)
(ii) CuO (s) + H2
(g)  Cu(s) + H2
O (I)
10. What is a balanced chemical equation? Why should chemical equations be balanced?
11. Balance the following chemical equation
(A) HNO3
+ Ca(OH)2
 Ca(NO3
)2
+ H2
O
(B) NaOH + H2
SO4
 Na2
SO4
+ H2
O
(C) NaCl + AgNO3
 AgCl + NaNO3
(D) BaCl2
+ H2
SO4
BaSO4
+ HCl
12. In the refining of silver, the recovery of silver from silver nitrate solution involved displacment by copper
metal. Write down the reaction involved.
13. What do you mean by a precipitation reaction? Explain by giving examples.
14. Why do we apply paint on iron articles?
15. Oil and fat contatining food items are flushed with nitrogen. Why?
16. Write the balanced equation for the following chemical reactions.
(i) Hydrogen + Chlorine  Hydrogen chloride
(ii) Barium chloride +Aluminium sulphate  Barium sulphate +Aluminiumchloride
(iii) Sodium + Water  Sodium hydroxide + Hydrogen
17. Why is the amount of gas collected in one of the test tubes inActivity 1.7 of NCERT double of the amount
collected in the other? Name this gas.
18. What does one mean by exothermic and endothermic reactions? Give examples.
19. Why is repiration considered an exothermic reaction? Explain.
20. Why are decomposition reactions called the opposite of combination reactions? Write equations for these
reactions.
21. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light and
electricity.

22. What is the difference between displacement and double displacement reactions? Write equations for
these reactions.
23. Explain the following in terms of gain or loss of oxygen with two examples each .
(i) Oxidation (ii) Reduction
24. Ashiny brown coloured element ‘X’on heating in air becomes black in colour. Name the element ‘X’and
the black coloured compound formed.Write equation also?
25. Explain the following terms with one example each.
(a) Corrosion (b) Rancidity
26. Translate the following statements into chemical equations and then balance them.
(A) Hydrogen gas combines with nitrogen to form ammonia
(B) Hydrogen sulphide gas burns in air to give water and sulphur dioxide
(C) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a percipitate of barium
sulphate
(D) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas
27. Write the balanced chemical equation for the following and identify the type of reaction in each case.
(A) Potassium bromide (aq) + Barium iodide (aq)  Potassium iodide (aq) + Barium bromide (s)
(B) Zinc carbonate (s)  Zinc oxide (s) + Carbon dioxide (g)
(C) Hydrogen (g) + Chlorine (g)  Hydrogen chloride (g)
(D) Magnesium (s) + Hydrochloric acid (aq)  Magnesium chloride (aq) + Hydrogen (g)
SECTION-B
 Balance the following equation
1. Pb(NO3
)2
(s) 
 PbO (s) + NO2
(g) + O2
(g)
2. Na (s) + O2
(g) — Na2
O (s)
3. MnO2
+ Al — Mn + Al2
O3
4. Fe2
O3
+ CO — Fe + CO2
5. Al + CuCl2
— AlCl3
+ Cu
6. C2
H6
+ O2
— CO2
+ H2
O
7. NH3
+ O2
— NO + H2
O
8. K2
Cr2
O7
+ KOH — H2
O + K2
CrO4
9. Na2
CO3
+ HCl — NaCl + H2
O + CO2
10. As2
O3
+ H2
S — As2
S3
+ H2
O
11. KI + H2
O2
— KOH + I2
12. Zn(NO3
)2
— ZnO + NO2
+ O2
13. NaOH + H2
SO4
— Na2
SO4
+ H2
O
14. NH3
+ O2
— N2
+ H2
O
15. SO2
+ H2
S — H2
O + S
16. H2
S + O2
— SO2
+ H2
O
17. Al(OH)3 
 Al2
O3
+ H2
O

18. Al2
(SO4
)3
+ NaOH — Al(OH)3
+ Na2
SO4
19. NH3
+ O2
— NO + H2
O
EXERCISE-III
SECTION-A
  Fill in the blanks
1. The compound YBa2
Cu3
O7
, which shows superconductivity, has copper in oxidation state __________
assume that the rare earth element yttrium is in its usual +3 oxidation state.
2. The oxidation number of carbon in CH2
O is ______________.
3. The brown ring complex compound in formulated as [Fe(H2
O)5
(NO)+
]SO4
. The oxidation number of iron
is __________.
4. The oxidation number of phosphorus in Ba(H2
PO2
)2
is: _______________.
5. The oxidation number of sulphur in S8
, S2
F2
, H2
S respectively are ________________.
2. Balance the following equation by ion-electron method.
1. –
4
MnO (aq) + I–
(aq) — MnO2
(s) + I2
(s) (In basic medium)
2. –
4
MnO (aq) + SO2
(g) — Mn2+
(aq) + –
4
HSO (aq) (In acidic medium)
3. H2
O2
(aq) + Fe2+
(aq) — Fe3+
(aq) + H2
O (l) (In acidic medium)
4. –
2
7
2O
Cr (aq) + SO2
(g) — Cr3+
(aq) + –
2
4
SO (aq) (In acidic medium)
5. H2S + HNO3 — H2SO4 + NO2 + H2O
6. N 
3
O + Bi — Bi3+
+ NO2 (in acidic medium)
7. Al + N 
3
O — Al(OH 
4
) + NH3
(in basic medium)
3. Find the oxidation number or oxidation state of underlined elements.
(i) CH2O (ii) Mg3N2 (iii) NH2OH (iv) ICl3
(v) KO2 (vi) H2S2O8
SECTION-B
 Multiple choice question with one correct answers
1. A reaction in which, under equilibrium conditions, both the reactants and products are present is called,
(A) reversible (B) Irreversible (C) endothermic (D) exothermic
2. The reaction, H2
+ Cl2
— 2HCl is,
(A) an oxidation reaction (B) a reduction reaction
(C) a combination reaction (D) an isomoerisation reaction
3. When sodium metal is dropped into water, it gets
(A) oxidised (B) reduced (C) remain unchanged (D) hydrolysed
4. Fatty foods become rancid because of which one of the following?
(A) Oxidation (B) Reduction (C) Hydrogenation (D) Corrosion

5. Which one of the following reactions is an example of thermal decomposition?
(A) CaCO3
(s) — CaO (s) + CO2
(g) (B) 2HOCl (aq) — O2
(g) + HCl (g)
(C) 2AgCl (I) — 2Ag (s) (g) + Cl2
(g) (D) 2H2
O (I) — 2H2
(g) + O2
(g)
6. In the reaction 2Al + Fe2
O3
— Al2
O3
+ 2Fe which one is oxidized?
(A) Al (B) Fe (C) Fe2
O3
(D) none
7. Oily and fatty food items are flushed with nitrogen gas because of which one of the following reasons?
(A) Nitrogen reacts with oils and fats and thus prevents oxidation
(B) Nitrogen is inert and excludes a direct contact of air with oily and fatty food items
(C) Nitrogen helps in the decomposition of food items and makes them tasty.
(D) The given statements is wrong.
8. Which one of the following is used in the white washing of walls?
(A) Calcium oxide mixed with water (B) Calcium carbonate mixed with water
(C) Calcium sulphate mixed with water (D) Sodium chloride mixed with water
9. Dissolving sugar in water is an example of:
(A) Physical change (B) Chemical change (C) Redox reaction (D) None of these
10. Heat is evolved during:
(A) Endothermic reaction (B) Displacement reaction
(C) Combustion reaction (D) Combination reaction
11. In an electroytic cell where electrolysis is carried, anode has:
(A) Positive charge
(B) Negative charge
(C) Connected to negative terminal of the battery
(D) None of these is correct
12. A change is said to be a chemical change when:
(A) energy change occurs (B) new substances are formed
(C) the change cannot be easily reversed (D) all statements are correct
13. Copper displaces which of the following metals from its salt solution:
(A) ZnSO4
(B) FeSO4
(C) AgNO3
(D) NiSO4
14. Which of the following is an example of displacement reaction?
(A) 2KClO3
— 2KCl + 3O2
(B) 2H2
+ O2
— 2H2
O
(C) Zn + 2HCl — ZnCl2
+ H2
(D) N2
+ 3H2
— 2NH3
15. A redox reaction is one in which :
(A) both the substances are reduced
(B) both are the substances are oxidised
(C) an acid is neutraliesd by the base
(D) one substance is oxidised while the other is reduced
SECTION-C
 Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.

(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true
1. Assertion: In HClO3
oxidation number of Cl is -1.
Reason: Oxygen is more electropositive then fluorine.
2. Assertion: Oxidation state of carbon in its compounds is some time +4.
Reason: An element has a fixed oxidation state.
SECTION-D
 Match the following (one to one)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. Only One entries of column-I may have the matching with the same entries of column-
II and one entry of column-II Only one matching with entries of column-I.
(Match the oxidation state of the underlined elements.)
1. Column I Column II
(A) XeF4
(P) –3
(B) N +
4
H (Q) +4
(C) XeO2
F2
(R) +5
(D) Na3
PO4
(S) +6
EXERCISE-IV
SECTION-A
1. In the reaction, H2
S + Cl2
— 2HCl + S, the electrons are transfered from,
(A) S2–
to S (B) Cl2
to HCl (C) S2–
to Cl2
(D) H2
S to S
2. Which one of the following is not double displacment reaction?
(A) Pb (NO3
)2
(aq) + 2Kl (aq) — PbI2
(s) + 2KNO3
(aq)
(B) CuSO4
(aq) + H2
S (aq) — CuS (s) + H2
SO4
(aq)
(C) Na2
CO3
(s) + 2HCl (aq) — 2NaCl (aq) + H2
CO3
(aq)
(D) CaO (s) + CO2
(g) — CaCO3
(s)
3. Which one of the following reactions is an example of photodecomposition?
(A) CaCO3
(s) — CaO (s) + CO2
(g) (B) NH4
Cl (s) — NH3
(g) + HCl (g)
(C) 2H2
O (I) — 2H2
(g) + O2
(g) (D) 2AgCl (s) — 2Ag (s) + Cl2
(g)
4. Which one of the following reactions is an example of electrical decomposition?
(A) CaCO3
(s) — CaO (s) + CO2
(g) (B) 2HOCl (aq) — O2
(g) + HCl (g)
(C) 2H2
O (I) — 2H2
(g) + O2
(g) (D) 2AgCl (s) — 2Ag (s) + Cl2
(g)
5. A shiny - brown substance X on heating in air turns black and a new compound Y is formed. Name the
substance X and black compound Y.

(A) X = Fe and Y = FeO (B) X = Cu and Y = Cu(OH)2
(C) X = Cu and Y = CuO (D) X = Al and Y = AlO3
6. A substance which oxidises itself and reduces other is known as:
(A) Oxidising agent (B) Reducing agent (C) Both of these (D) None of these
7. The reaction between lead nitrate and potassium iodide present in aqueous solutions is an example
(A) Decomposition reaction (B) Displacement reaction
(C) Double displacement (D) Neutralisation reaction
8. Oxidation is a process which involves
(A)Addition of oxygen (B) Removal of hydrogen
(C) Loss of electrons (D) All are correct
9. Aluminium oxide reacts with sulphuric acid to form
(A)Aluminium sulphate and hydrogen (B)Aluminium sulphate and oxygen
(C) Aluminium suphate and water (D)Aluminiumsulphate and sulphur dioxide
10. Chemically rust is:
(A) hydrated ferrous oxide (B) hydrated ferric oxide
(C) only ferric oxide (D) none of these
SECTION-B
 Multiple choice question with one or more than one correct answers
1. In the reaction,
4 2 2 3
P H PO PH


  (Alkaline medium)
which statement is correct
(A) P4
is oxidized (B) It is a decomposition reaction
(C) P4
is reduced (D) It is not a redox reaction
2. Which of the following can act both as oxidising agent and reducing agents?
(A) HNO2
(B) H2
O2
(C) H2
S (D) SO2
3. In the reaction:
–
2
3
2O
2S + I2
— –
2
6
4O
S + 2I–
(A) –
2
3
2O
S gets reduced to –
2
6
4O
S (B) –
2
3
2O
S gets oxidised to –
2
6
4O
S
(C) I2
gets reduced to I–
(D) I2
gets oxidised to I–
4. When a chemical species loses one or more electrons, it is said to have been
(A) oxidised (B) reduced (C) decomposed (D) act as reducing agent
5. Which statements are correct for the reaction CuO + H2
— Cu + H2
O
(A) CuO is reduced (B) H2
is oxidized
(C) CuO is reduced and H2
is oxidized (D) Both CuO and H2
are oxidized
6. Which of the following is not balanced equation?
(A) Fe + Cl2
— FeCl3
(B) Mg + CuSO4
— MgSO4
+ Cu
(C) NaOH + HCl — NaCl + H2
O (D) N2
O3
— N2
+ 3O2
7. Which of the following reactions is not correct?
(A) Zn + CuSO4
— ZnSO4
+ Cu (B) 2Ag +Cu (NO3
)2
— 2AgNO3
+ Cu
(C) Fe + CuSO4
— FeSO4
+ Cu (D) Cu + H2
SO4
— CuSO4
+ H2

8. In the reaction PbO + C — Pb + CO
(A) PbO is reduced (B) C acts as an oxidising agent
(C) C acts as a reducing agent (D) This reaction does not represent redox reaction
9. In the reaction:
Zn + FeSO4
— ZnSO4
+ Fe
(A) Zn gets oxidised (B) Fe gets reduced
(C) Zn is an oxidising agent (D) Zn and Fe are both oxidised
10. Which of the following are example of decomposition reaction?
(A) CaCO3
(s) — CaO (s) + CO2
(g)
(B) Ca(OH)2
(s) — CaO + H2
O (l)
(C) CuSO4
. 5H2
O (s) — CuSO4
(s) + 5H2
O (g)
(D) 2KClO3
(s) — 2KCl (s) + 3O2
(g)
SECTION-C
Comprehensions
Passage-1
Oxidation number is the charge assigned to an atom of a molecule or ion according to some arbitrary rules.
In neutral molecules, the sum of the oxidation numbers of all the atoms present is zero while in a simple or
complex ion it is equal to the net charge on ion. In some cases, the oxidation number may be even frac-
tional.Although sometimes, a particular element may have same valency and oxidation state but these are
based upon different concepts. The number of oxidation states available for a particular element are nor-
mally more than the valencies.
1. A brown complex has the formula:
[Fe(H2
O)5
NO]SO4
. The oxidation number of iron is:
(A) +1 (B) +2 (C) +3 (D) 0
2. In which compound, Mn exhibits highest oxidation state?
(A) MnO2
(B) Mn3
O4
(C) K2
MnO4
(D) MnSO4
3. In which of the following pairs, there is the maximum difference in the oxidation state of the underlined
elements?
(A) NO2
and N2
O4
(B) P2
O5
and P4
O10
(C) N2
O and NO (D) SO2
and SO3
Passage-2
A redox reaction consists of oxidation and reduction half reactions. There is a loss of electrons in oxidation
and the species which loses electrons is reducing agent. Its oxidation number increases during oxidation.
Similarly there is a gain of electrons during reduction and the species which gains reduction and the species
which gains electrons is an oxidising agent. Its oxidation number decreases during reduction. The number
of electrons released during oxidation is equal to number of electrons gained during reduction.
1. The reaction:
2H2
O (l) — 4H+
(aq) + O2
+ 4e–
(A) an oxidation reaction (B) a reduction reaction
(C) a redox reaction (D) a hydrolysis reaction
2. Which of the following involves transfer of five electrons?
(A) (MnO4
)–
— Mn2+
(B) (CrO4
)2–
— (Cr)3+
(C) (MnO4
)2–
— MnO2
(D) (Cr2
O7
)2–
— 2Cr

3. In I2
 I
+ 1
3
IO
alaklyne medium
(A) 0 to 1 & 0 to +5 (B) 5 to –1 & o to 5
(C) –5 to 0 & 0 to 1 (D) –7 to –1 & 0 to +7
SECTION-D
 Match the following (one to many)
entries of column-II. One or more than one entries of column-I may have the matching with the same entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
(A) Double displacment (P) CuO + H2 
 

heat Cu + H2
O
(B) Decomposition (Q) Na2
SO4
(aq) + BaCl2
(aq) 
 BaSO4
(aq) + 2NaCl (aq)
(C) Precipitation (R) CaCO3 
 

heat CaO + CO2
(D) Redox (S) NaOH + HCl NaCl + H2
O
*****

Answers
EXERCISE-II
Section-A
2. (A) 3. (D) 4. (A)
Section-B
1. Pb(NO3
)2
(s) 
 PbO (s) + 2NO2
(g) +
2
1
O2
(g)
2. 2Na (s) +
2
1
O2
(g) — Na2
O (s)
3. 3MnO2
+ 4Al — 3Mn + 2Al2
O3
4. Fe2
O3
+ 3CO — 2Fe + 3CO2
5. 2Al + 3CuCl2
— 2AlCl3
+ 3Cu
6. C2
H6
+
2
7
O2
— 2CO2
+ 3H2
O
7. 2NH3
+
2
5
O2
— 2NO + 3H2
O
8. K2
Cr2
O7
+ 2KOH — H2
O + 2K2
CrO4
9. Na2
CO3
+ 2HCl — 2NaCl + H2
O + CO2
10. As2
O3
+ 3H2
S — As2
S3
+ 3H2
O
11. 2KI + H2
O2
— 2KOH + I2
12. 2Zn(NO3
)2
— 2ZnO + 4NO2
+ O2
13. 2NaOH + H2
SO4
— Na2
SO4
+ 2H2
O
14. 2NH3
+
2
3
O2
— N2
+ 3H2
O
15. SO2
+ 2H2
S — 2H2
O + 3S
16. H2
S +
2
3
O2
— SO2
+ 2H2
O
17. 2Al(OH)3 
 Al2
O3
+ 3H2
O
18. Al2
(SO4
)3
+ 6NaOH — 2Al(OH)3
+ 3Na2
SO4
19. 2NH3
+
2
5
O2
— 2NO + 3H2
O
EXERCISE-III
Section-A
1. 1. +7/3 2. O 3. 2 4. +1 5. 0, +1, -2

2. 1. O
2H
MnO
I
2I
4H
MnO 2
2
2
4 



 


2. 2
4
2
2
4 2Mn
5HSO
O
2H
H
5SO
2MnO 








3. O
2H
2Fe
O
H
2H
2Fe 2
3
2
2
2



 


4. O
H
3SO
2Cr
2H
3SO
O
Cr 2
2
4
3
2
2
7
2 



 



5. H2S + 8HNO3 — H2SO4 + 8NO2 + 4H2O
6. Bi + 3N 
3
O + 6H+
— Bi3+
+ 3NO2 + 3H2O
7. 8Al + 3N 
3
O + 18H2
O + 5OH–
— 8Al(OH 
4
) + 3NH3
3. (i) 0 (ii) +2 (iii) –1 (iv) +3 (v) –1/2 (vi) +6
Section-B
1. (A) 2. (C) 3. (A) 4. (A) 5. (A)
6. (A) 7. (B) 8. (A) 9. (A) 10. (C)
11. (A) 12. (B) 13. (C) 14. (C) 15. (D)
Section-C
1. (D) 2. (C)
Section-D
1. (A)-(Q), (B)-(P), (C)-(S), (D)-(R)
EXERCISE-IV
Section-A
1. (C) 2. (D) 3. (D) 4. (C) 5. (C)
6. (B) 7. (C) 8. (D) 9. (C) 10. (B)
Section-B
1. (A,C) 2. (A,B,D) 3. (B,C) 4. (A,D) 5. (A,B,C)
6. (A,D) 7. (B,D) 8. (A,C) 9. (A,B) 10. (A,B,D)
Section-C
Passage-1
1. (A) 2. (C) 3. (D)
Passage-2
1. (A) 2. (A) 3. (A)
Section-D
1. (A)-(Q,S), (B)-(R), (C)-(Q), (D)-(P)
*****

“This exercise is optional”
SECTION-A
1. How many moles of electrons weigh one kilogram
(A) 6.022 × 1023
(B)
31
1
10
9.108

(C)
54
6.022
10
9.108
 (D)
8
1
10
9.108 6.022


2. Oxidation number of C in C3
O2
in
(A) 4 (B) 2 (C) 1.33 (D) 1.34
3. Which of the following has a highest mass
(A) 1 g atom of C (B) ½ mole of CH4
(C) 10 ml of water (D) 3.011 × 1023
atoms of oxygen
4. 2CuI  Cu + CuI2
, the reaction is
(A) Redox (B) Neutralization (C) Oxidation (D) Reduction
5. When copper turnings are added to silver nitrate solution, a blue coloured solution is formed after some
time, it is because copper
(A) Displaces silver from the solution (B) Gains of electrons
(C) Increases in the valence of negative part (D) Decrease in the valence of positive part
6. Hydrogen sulphide (H2
S) is a strong reducing agent. Which of the following reactions shows its reducing
action
(A) Cd(NO3
)2
+ H2
S  CdS + 2HNO3
(B) CuSO4
+ H2
S  CuS + H2
SO4
(C) 2FeCl3
+ H2
S  2FeCl2
+ 2HCl + S (D) Pb(NO3
)2
+ H2
S  PbS + 2CH3
COOH
7. Cu + xHNO3
 Cu(NO3
)2
+ yNO2
+ 2H2
O
The value of x and y are
(A) 3 and 5 (B) 8 and 6 (C) 4 and 2 (D) 7 and 1
8. Ferrous sulphate on heating produces
(A) Ferric oxide (B) Carbon dioxide (C) Oxygen (D) Water
9. The reactions in which rearrangement of the atoms present in molecules of the reactants occurs are known
as
(A) Combination (B) Isomerization
(C) Polymerization (D) Decomposition
10. A reaction in which rate of the reaction decrease with rise in temperature
(A) exothermic reaction (B) endothermic reaction (C) both (D) None
Ans.
1. (D) 2. (C) 3. (A) 4. (A) 5. (A)
6. (C) 7. (C) 8. (A) 9. (B) 10. (A)

SECTION-B
 Multiple choice question with one or more than one correct answers
1. When Cesium metal react with H2
O it form
(A) CsOH (B) H2
(C) CsO (D) Cs2
O
2. Which of the following reaction would not result in a displacement reaction?
(A) I2
+ NaBr  (B) Cl2
+ NaI  (C) Br2
+ NaCl (D) F2
+ NaBr 
3. A solution of KOH is mixed with a solution of MgI2
. Predict what happens?
(A) The four ions involved are K+
, OH–
, Mg2+
, I–
(B) An exchange of ions in the reactants produces the compounds
(C) It form Mg(OH)2
which is soluble in water
(D) It is a precipitation reaction also
4. In the reaction C2
H5
OH + O2
 CO2
+ H2
O
(A) Oxidation of carbon (B) Carbon act as oxidizing agent
(C) Reduction of oxygen (D) Oxygen is act as reducing agent
5. aK4
Fe(CN)6
+ bCe(NO3
)4
+ cKOH  dCe(OH)3
+ eFe(OH)3
+ fH2
O(g) + gK2
CO3
+ hKNO3
(A) a = 0, b = 61, d = 0 (B) b = 61, g = 6
(C) a = 0 b = 61, d =250 (D) d = 61, e = 0, f = 36
6. In the reaction FeS2
+ O2
 Fe2
O3
+ SO2
(A) Increase in oxidation number is of 22 (B) Decrease in oxidation number of 22
(C) Oxidation of iron (D) Reduction of oxygen
7. In sodium amalgam and brass the oxidation of sodium and copper is
(A) Na = 0, Hg = 0 (B) Cu= 0, Sn = 0 (C) Na = 1, Hg = 0 (D) Cu=2, Hg=0
8. aCu3
P + bH+
+ cCr2
O7
2–
 dCu2+
+ eH3
PO4
+ fCr3+
+ gH2
O
(A) a = 0, b = 24, c = 11 (B) a = 5, b = 124, c = 11
(C) a = 6, d = 50, e = 6 (D) d = 18, e = 6, f = 22
9. H2
O2
+ I2
 HIO3
+ H2
O
(A) Total change in oxidation number of iodine is 10
(B) Total change in oxidation number of iodine is 5
(C) Total change in oxidation number of oxygen is 2
(D) Total change in oxidation number of oxygen is 1
10. Find correct statement
(A) Oxidation number of oxygen can be zero, +2, –1, –2
(B) Oxidation number of fluorine can be zero and –1
(C) Oxidation number of fluorine is only –1
(D) Oxidation number of sodium in sodium amalgam is zero
Ans.
1. (A,B) 2. (A,C) 3. (A,B,D) 4. (A,C) 5. (A,B,D)
6. (A,B,C,D) 7. (A,B) 8. (B,D) 9. (A,C) 10. (A,B,D)

SECTION-C
 Comprehensions
Passage-1
In ion electron method the two equations describing oxidation and reduction in the redox reaction are
separated and completely balanced. The number of electrons gained and lost in each half reaction are
equalized and finally the half-reactions are added to give the overall balanced equation. The use of half-
reaction permits us to balance equations using only the principles of atom and charge conservation.
1. 2 2 3
2 7 2 4 2 2
Cr O C O H Cr CO H O
   
  
  
Half reaction for this equation is
(A) 2
2 7
Cr O 
+ 14H+
+ 9e–
 Cr3+
+ 7H2
O (B) 2
2 4
C O 
 CO2
(C) 2
2 7
Cr O 
+ 14H+
+ 6e–
 2Cr3+
+ 7H2
O (D) 2
2 4
C O 
2CO2
+ 4e–
2. Electron involved in the reaction are
(A) 6 electrons are gained in reduction (B) 9 electrons are lossed in oxidation
(C) 9 electron are gained in reduction (D) 6 electrons are lossed in reduction
3. How many moles of water molecules at product side
(A) 7 (B) 14 (C) 12 (D) 1
Passage-2
The chemical equations gives the following quantitative information. It tells us about
(i) The number of molecules or atom of reactants and products taking part in the reaction.
(ii) The number of moles of each substance involved in the reaction.
(iii) The mass of each substance involved in the reaction
(iv) Mass-mass, mass-volume, volume-volume relationships between the reactants and products
1. 1.84 g of a mixture of CaCO3
and MgCO3
was heated to a constant weight. The constant weight of the
residue was found to be 0.96 m. Calculate the percentage composition of the mixture?
(A) % of CaCO3
= 54.34%, % of MgCO3
= 45.66%
(B) % of CaCO3
= 54.34%, % of MgCO3
= 54.66%
(C) % of CaCO3
= 45.34%, % of MgCO3
= 54.66%
(D) None of the above
2. Calculate the weight of lime(CaO) that can be prepared by heating 200 kg of limestone (CaCO3
) which is
95% pure?
(A) weight of CaO = 106.4 kg (B) weight of CaO = 160.4 kg
(C) weight of CaO = 601.4 kg (D) weight of CaO = 106.98 kg
3. What is the number of moles of Fe(OH)3
that can be prodcued by allowing 1 mole of Fe2
S3
, 2 moles of H2
O
and 3 moles of O2
to react
2Fe2
S3
+ 6H2
O + 3O2
 4Fe(OH)3
+ 6S
(A) 1.34 (B) 1.43 (C) 3.14 (D) 4.31
Ans.
Passage-1
1. (C) 2. (A) 3. (A)
Passage-2
1. (A) 2. (A) 3. (A)

SECTION-D
 Match the following (one to many)
entries of column-II. One or more than one entries of column-I may have the matching with the some entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
(A) KMnO4
+ HCl (P) Hg
(B) NH3
+ Hg2
Cl2
(Q) KCl
(C) KMnO4
+ H2
O2
(R) H2
O
(D) HCl+WO3
+SnCl2
(S) HgNH2
Cl
(A) The number of moles of electrons are (P) 6
required to reduce one mole of
2
2 7
Cr O 
to Cr3+
ion
(B) The compound of chlorine in which (Q) KClO3
its oxidation state is +5
(C) Compound having oxidation number of (R) HClO3
chlorine is not –1
(D) Possible oxidation number of S in its (S) +4
compounds
Ans.
1. (A)-(Q,R), (B)-(P,S), (C)-(R), (D)-(R)
2. (A)-(P), (B)-(Q,R), (C)-(Q,R), (D)-(P,S)
SECTION-E
 Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.
(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true
1. Assertion: SO2
and CaOCl2
both are bleaching agents
Reason: Both are reducing agents
2. Assertion: White P reacts with caustic soda, the products are PH3
and NaH2
PO2
. This reaction is an
example of disproportionation reaction.
Reason: In disproportionation reactions one and the same substnce may act simultaneously as an oxidising
agent and on a reducing agent.
3. Assertion: The reaction of ammonia solution with Calomel is a disproportionation reaction in which a
mixture Hg(II) amido chloride and mercury are formed.
Reason: In a disproportionation reaction species under reaction is neither oxidized nor reduced.
Ans.1. (C) 2. (A)
3. (C)

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