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Showing posts with label Inorganic Chemistry Notes. Show all posts
Showing posts with label Inorganic Chemistry Notes. Show all posts

Friday, 21 January 2022

Inorganic Chemistry

Inorganic Chemistry

inorganic chemistry

What is Inorganic Chemistry ?

In simple words we can say that, "Inorganic chemistry is the the branch of chemistry that deals with inorganic compounds".

Definition of Inorganic Chemistry- Inorganic chemistry is the branch of chemistry, which deals with the study of inorganic compounds and organometallic compounds. Inorganic chemistry includes study of all chemical compounds excluding organic compounds (where, organic compounds are carbon based compounds, generally having carbon-hydrogen bonds i.e. C-H bonds.

Example of Inorganic Compound

Salt Magnesium Chloride (MgCl2) is an ionic Inorganic Compound which contains magnesium-cations Mg2+ and chloride-anions Cl

Branches of Inorganic Chemistry

Descriptive Inorganic Chemistry
Descriptive inorganic chemistry based on the arrangement and ordering of compounds based on their properties.

Coordination Chemistry

Coordination chemistry is the branch of Inorganic chemistry which deals with the study of coordination compounds or coordination complexes. A coordination compound or coordination complex contains of a central atom or ion, which is generally metallic and is known as coordination center, and get surrounded by group of bounding molecules or ions, these surrounding molecules and ions are known as ligands.

Materials Science

Material science deal with the study of materials. Study of metals, semiconductors, polymers, ceramics, nanomaterials, biomaterials and many other form of materials comes under material science. A material scientist study about any material to find out its structure, properties and use this in certain applications.

Bioinorganic Chemistry

Bioinorganic Chemistry is the branch of inorganic chemistry which deal with study of different metals in biological system. Bioinorganic chemistry deal in both natural field (like study and application of metalloproteins etc.) and artificial field (like study, application and development of medicines). Bioinorganic chemistry is the mixture of biochemistry and inorganic chemistry.

Organometallic Chemistry

Organometallic chemistry is the branch of inorganic chemistry which deal with the study of chemical compounds having at least one bond between a Metal and carbon atom of organic compound. Organometallic compounds are broadly used in research, industrial chemical reactions, catalysis to increase the rates of reactions.

Cluster Chemistry

Cluster chemistry is the branch of inorganic chemistry which deals with the study of clusters. Cluster forms when atoms or molecules bound together to form large size substances whose size is intermediate in between the size of a molecule and a bulk solid. Example of cluster- fullerene cluster from carbon atoms and borane cluster from boron atoms.

Thursday, 23 December 2021

Hydrogen Class 11 Notes

 Hydrogen Class 11th Chemistry Notes

Hydrogen is the first element in the periodic table and is the lightest element. It exists as a diatomic molecule (H2) and is called dihydrogen. It has one proton and one electron and it has electronic configuration 1s1. This electronic configuration is responsible for dual nature of hydrogen i.e. hydrogen shows resemblance with alkali metals as well as halogens.

Resemblance of hydrogen with alkali metals

Like alkali metals –

·         Hydrogen exhibits electropositive character i.e. it has tendency to lose electron.

·         The normal valency of hydrogen is 1.

·         Hydrogen forms halides, oxides and sulphides.

·         Hydrogen acts as strong reducing agent.


Resemblance of hydrogen with halogens

Like halogens-

  • Hydrogen exhibits electronegative character by accepting one electron to complete its valance shell.
  • Hydrogen shows same order of ionisation energy (IE1) as that of halogens.  
  • Hydrogen (H2) is a gas in molecular state similar to F2 and Cl2.
  • Hydrogen shows -1 oxidation state in its compounds with more electropositive element.



As the hydrogen resembles both the alkali metals and the halogens and also differs from them. Also hydrogen exhibits some unique behavior therefore the position of hydrogen is uncertain and it is placed separately in periodic to table.


Hydrogen has three isotopes, protium (11H), deuterium (12H) and tritium (13H).

            i.            Protium or ordinary hydrogen (11H).

 It is the most abundant isotope of hydrogen (99.985%). Its nucleus has one proton and no neutron (mass no. = 1).

          ii.             Deuterium or heavy hydrogen (12H or D).

It is less abundant and is present in heavy water (D2O). Its nucleus has one proton and one neutron (mass no. = 2).

        iii.            Tritium (13H or T)

It is the rarest isotope of hydrogen and is radioactive in nature. Its nucleus has one proton and two neutrons (mass no. = 3).



Laboratory methods

1.      Action of dilute HCl or H2SO4 on granulated zinc.

Zn    +     2HCl          ZnCl2    +    H2  


2.      Action of dilute H2SO4 on magnesium ribbon.

Mg    +     H2SO4         MgSO4    +    H2  


3.      Action of water on sodium hydride.

NaH    +     H2O          NaOH    +    H2  


4.      By action of KOH of scrap aluminum or silicon. (Uyeno’s method)

2Al    +     2KOH    +     2H2O          2KAlO2    +    3H2  


Commercial method

1) From water gas (Bosch process)

Water gas (CO + H2) is mixed with the steam and the gaseous mixture is passed over heated catalytic mixture of ferric oxide (Fe2O3) and chromium oxide (Cr2O3) at 773K.

H2     +    CO    +    H2O      Fe2O3 + Cr2O3 (773K)     CO2   +   2H2

  Water gas         Steam

 2) From Steam (Lanes process)

When superheated steam is passed over iron filling heated to 1023 – 1073K hydrogen is formed.

3Fe  +   4H2O(steam)      (1023 – 1073)K  (Fe)   Fe2O4   +   4H2(g)  


3) Hydrocarbon steam process

Hydrogen is prepared by the action of steam of hydrocarbon at 1270K.

CH4   +   H2O       1270K     CO   +   3H2


4) By electrolysis of water

Electrolysis of acidified water using platinum electrodes gives dihydrogen. Here dihydrogen is librated at the cathode while dioxygen is librated at anode.



1) Dihydrogen reacts with halogens (X2) to give hydrogen halide (HX).

    H2   +   X2       2HX    ( X = F, Cl, Br, I )


2) Dihydrogen reacts with dioxygen to forms water. This reactions is highly exothermic.

    2H2   +   O2      catalyst or heating     2H2O . ΔH = -285.9 KJ.mol-1


3) With dinitrogen dihydrogen forms ammonia. This process is coiled Hober process of    

    manufacture of ammonia.         

    3H2   +   N2     673K, Fe catalyst, 200atm     2NH3


4) Dihydrogen combines with many metals at high temperature to yield the corresponding                             hydrides                                                                                                                                                                                                                                                                                                 

   2Na  +  H2     2NaH    (sodium hydride)


5) It acts as reducing agent and thus reduces certain oxides of metals.

    ZnO   +   H2      Zn   +   H2O

    Fe2O4   +   4H2      3Fe   +   4H2O

6) Hydrogenation of vegetable oils using nickel as catalyst gives edible facts (margarine and Vanaspati ghee)

    Vegetable oil   +   H2      Finely divided Ni, 450K, 8-10atm    Solid fat



·         Dihydrogen is used in the preparation of ammonia by Haber’s process.

·         It is used in the hydrogenation of vegetable oils.

·         It is used for the manufacture of metal hydrides.

·         It is used as rocket fuel in space research.

·         Dihydrogen is used in fuel cell for generating electrical energy.

·         It is used in the atomic hydrogen torch and oxy hydrogen torches for cutting and welding.



The binary compounds of hydrogen with other elements are called hydrides. These hydrides have the formula EHx or EmHn (E = Element). These are classified into three types.

a) Ionic or saline hydrides

b) Covalent or molecular hydrides

c) Metallic or nonstoichiometric or interstitial hydrides


a) Ionic or saline hydrides

These are formed by the combination of hydrogen with metals which have low electro negativity values and are electropositive with respect to hydrogen. These includes elements of S block element. Ionic hydrides are prepared by the direct combination of metals.

e.g. LiH, KH, CaH2, MgH2 etc.


b) Covalent or molecular hydrides

These hydrides are formed by the combination of elements of comparatively higher electro –negativity as p-block element. The bonds formed are mostly covalent in character. Covalent hydrides are generally volatile.

e.g. H2O, CH4, NH3, HF etc.


c) Metallic or nonstoichiometric or interstitial hydrides

Metallic hydrides are formed by many d - block and f - block  elements. Elements of group 7, 8, 9 of d – block do not form hydrides and this is referred as hydride gap. Metallic hydrides are non – stoichiometric and show electric conductance. In these hydrides, hydrogen occupies interstices in the metal lattice producing distortion without any change in its type. Therefore, they are termed as interstitial hydrides.

e.g. SCH2, TiH2, VH, ZrH2



Water is essential to all forms of life. It is most common, abundant and easily obtainable of all chemical compounds. It is regarded as universal solvent. Water (H2O) is hydride of oxygen. In nature, water exist in three physical states, water (liquid), Ice (Solid) and water vapour (gas).


Structure of water

Water molecule is a bent molecule with bond angle of 104.5ᵒ and O-H bond length is 95.7pm.

Structure of water - Hydrogen class 11th chemistry notes for free download

Structure of Ice

Ice is the solid form of water. It has a highly ordered three dimensional hydrogen bonded structure. In ice each oxygen atom is surrounded tetrahedrally by four other hydrogen atoms at a distance of 276pm. The strength of hydrogen bonding in ice is more than liquid water. There is empty space in crystal of ice due to hydrogen bonding. This makes density of ice lower than liquid water, hence ice floats on water.

Structure of ice

Chemical properties of water

1) Dissociation of water

Water dissociates into its ion.

H2O   +    H2O             H3O+       +        OH-

                                 Hydronium ion


2) Amphoteric nature

Water has the ability to act as on acid as well as base. Such kind of behavior is known as amphoteric nature.

 H2O     +      NH3             OH-         +        NH4+

 H2O     +      HCl             H3O+       +        Cl-



3) Oxidizing and reducing nature

Water can act both as an oxidizing and reducing agent in its chemical reactions. With active metals water behaves as an oxidizing agent while with highly electronegative element, it acts as a reducing agent.

2Na    +    2H2O        2NaOH    +    H2

        Oxidizing agent

2F2    +    2H2O        4HF    +    O2

        Reducing agent


4) Formation of Hydrates

Water has strong ability to form compounds with some metal salts known as hydrates. This hydrates can be classified into three types:

i) Coordinated water


ii) Interstitial water


iii) Hydrogen – bonded water

[Cu(H2O)4]2+SO42-.H2O     in




Heavy water is deuterium oxide (D2O). It was discovered by Urey in 1932. The reaction of D2O are slightly slower than H2O. Ordinary water contains one part of heavy water in 6000 part of it. It is used as a moderator in nuclear reactors. It is used as tracer compound in determining the mechanism of many organic reactions.



Natural water contains dissolved salts. Depending upon its behavior towards soap solution, water may be classified as hard water and soft water.

a) Soft water : 

Water which produces lather with soap solution readily is called soft water. For example : distilled water, rain water, and demineralized  water.


b) Hard water : 

Water which does not produces lather with soap solution readily is called hard water. For example : sea water, river water, well water and tap water.

The hardness of water is due to the presence of the bicarbonates, chlorides and sulphates of calcium and magnesium.


Isotopic varieties of water

Ordinary water contains 18 different kinds of water. Such as variety is possible due to the different isotopic forms of hydrogen and oxygen. Hydrogen has three isotopes H(protium), D(Deutrium), T(tritium) and oxygen also has three isotopes 16O, 17O, 18O. These isotopes of hydrogen and oxygen combine to give 18 different kinds of water from which H2O is most abundant.




Hydrogen peroxide was discovered by Thenard in 1818. It is an important chemical used in pollution control treatment of domestic and industrial effluents.


Preparation of H2O2

1) In laboratory H2O2 is prepared by acidifying barium peroxide and removing excess water by evaporation under reduced pressure.

BaO2.8H2O   +   H2SO4       BaSO4   +    H2O2   +    8H2O


2) By the action of dilute acid on sodium peroxide (Merck’s method)

Na2O2   +   H2SO4       NaSO4   +    H2O2


3) By bubbling CO2 through a paste of BaO2

BaO2   +    H2O   +   CO2       BaCO3   +    H2O2


4) By the action of phosphoric acid on BaO2

3BaO2    +   2H3PO4       Ba3(PO4)2   +    3H2O2  


5) Industrially H2O2 is prepared by the auto – oxidation of 2 – Ethyl anthraquinol.

      2-ethyl anthraquinol     O2 (air) (H2/Pd   2-ethyl anthraquinol    +   H2O2


Structure of  H2O2

It is non–planar open book (Skew) structure. The bond length and bond angle are slightly different in gas and solid phase due to hydrogen bonding. Structure of  H2O2 HYDROGEN PROXIDE.


Chemical Properties of H2O2

1) It decomposes rapidly on heating in presence of finely divided metals, such as Ca, Fe, Cu, Au, Ag, Pt, MnO2, Carbon, dust, light, etc.

H2O2   +   H2O2        2H2O    +    O2   (ΔH = -196KJ)


2) It acts as an oxidizing as well as reducing agent in both acidic and alkaline or basic medium. Some example are given below

i) Oxidizing action in acidic medium

PbS   +   4H2O2        PbSO4   +   4H2O

ii) Reducing action in acidic medium

HOCl    +   H2O2        H3O+   +    Cl-    +   O2

iii) Oxidizing action in alkaline medium

Mn2+   +   H2O2      2Fe3+   +    2OH-

iv) Reducing action in alkaline medium

Cl2   +   H2O2   +    2OH-      2Cl-   +    2H2O    +    O2


3) Hydrogen peroxide acts as a bleaching agents due to the release of nascent oxygen.

H2O2          H2O     +     O


4) Hydrogen peroxide undergoes addition reaction with alkenes to form glycols.

Hydrogen peroxide undergoes addition reaction with alkenes to form glycols.

Uses of H2O2

·         It is widely used in environmental chemistry.

·         It is used as a antiseptic and is sold in the market as perhydrol.

·         It is used in daily life as a hair bleach and as mild disinfectant.

·         It is used as a bleaching agent and to manufacture chemicals like sodium perborate and percarbonate.


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Thursday, 17 December 2020


Redox Reactions Class 11th Chemistry Notes

Redox Reaction Class 11 Chemistry Notes

Chemical reactions are generally classified into three types:

1) Precipitation reactions

2) Acid – base neutralization

3) Redox reactions

Redox reaction is a very important group of reactions. These reactions are involved in large number of processes in nature, biological and industrial.

 Definition of Redox Reaction 

Redox or reduction -oxidation reactions are the chemical reactions in which the oxidation state of atoms is get changed. → Redox reactions are conducted by the transfer of electrons from one chemical species (reducing agent) undergoing oxidation (losing electrons) to another chemical species (oxidizing agent) undergoes reduction (gains electrons).

Don't recollect it yet, don't worry !!! just Learn more about Redox Reactions Class 11th Notes below→

Video on Redox Reaction Class 11th Chemistry Notes


Oxidation is a reaction which involves the addition of oxygen or any other electronegative element or the removal of hydrogen or any electropositive element from the substance.

For example,

C(s)  +  O2(g)  →  CO2(g)  (oxidation of carbon)


Reduction reaction involves the addition of hydrogen or any electropositive element or removal of oxygen or any electronegative element. It is just the reverse of oxidation.

For example,

Cl2  +  H2  →  2HCl   (reduction of Cl)



An oxidation is defined as the loss of one or more electron from a substance.

For example,

Na   →   Na+   +   e-



A reduction is defined as the gain of one or more electrons by a substance.

For example,

Cl2   +   2e-   →   2Cl-



Oxidizing agent is a substance which accepts or gain electrons and causes the oxidation of other substance.



Reducing agent is a substance which donates electrons and causes reduction of other substance.


REDOX REACTION (oxidation – reduction reaction)

The reactions in which oxidation and reduction reactions occur simultaneously are called redox reaction.

In redox reactions, electrons are transferred from one substance to the other substance, therefore these reactions are electron transfer reactions.

For example,

                Loss of 2e- (oxidation)

             |                                ↓                        

          Zn    +    Cu2+    →    Zn2+   +    Cu

                             |                                 ↑

                               Gain of 2e-  (Reduction)


Oxidation number of an element in a compound is defined as the number of electrical charges it carries when all other atoms are removed from it as ions.


Rules to assign oxidation numbers

1)      The oxidation number of an atom in free uncombined elemental state is zero.

For example,

H2, Ca, Cl2, O3, S8, P4, and so on has oxidation number zero.


2)      The oxidation number of an atom in a monoatomic ion is equal to its charge.

For example,


K+     Ba2+     Cr3+     Br-     S2-

↑      ↑        ↑       ↑      ↑

+1     +2        +3       -1       -2


3)      When hydrogen atom is bounded to non – metals, its oxidation number is +1. And when it is bounded to metals, its oxidation number is -1.

For example,

[O – H]-         H – O – H          Li – H         H – Ca – H

-2    +1         +1   -2   +1         +1   -1        -1    +2   -1


4)      The oxidation number of oxygen is usually -2 in all of its compounds except in peroxide or peroxide ion where it has oxidation number of -1.

For example,

Ca – O        H – O – O – H          [O – O]2-

+2   -2        +1   -1   -1   +1          -1    -1


5)      The oxidation number of F is -1 in all of its compounds. The other halogens Cl, Br and I usually exhibit oxidation number of -1 in their halide compounds. However when Cl, Br and I are bonded to oxygen then they exhibit oxidation number of +1.

For example,

H – F        K – Br       Cl – O – Cl         H – O – Cl

+1 -1       +1   -1       +1   -2   +1         +1   -2   +1


6)      The algebraic sum of the oxidation number of all atoms in a neutral molecule is zero and for the polyatomic ion it is equal to net charges of the ion.

For example,

In H – Cl

Oxidation number of H  +   oxidation number of Cl  = 0

=  +1  -1  =  0


Using above rules we can calculate oxidation number of an atom in any molecule or ion


Determine the oxidation number of

a)      N in HNO3

HNO3 is neutral molecule

: .  Sum of oxidation number of all atom of HNO3 is equal to zero.

[ O.N. of H ]  +  [ O.N. of N ]  +  [ 2 X O.N. of O ] = 0

=   +1  +  [O.N. of N ]  +  [ 2 X  (-2) ]  =  0

=   +1  +  [O.N. of N ]  -  4   =   0

=   O. N  of  N   - 3  =  0

     O. N  of  N    =  + 3

: .  Oxidation number of N in HNO3 is +3.


b)     Pt in PtCl62-

PtCl62-  is ionic species.

: .  Sum of oxidation number of all atom of PtCl62-  is equal to -2.

[ O.N. of Pt ]  +  [ O.N. of Cl ] X 6  =  -2

=  [O.N. of Pt ]  +  [ 6 X  (-1) ]  =  -2

=  O.N. of Pt  =  - 2  +  6   =  + 4

: .  Oxidation number of Pt in PtCl62-  is + 4.





·         Loss of electron.

·         Increase in oxidation number of oxidized species.



·         Gain of electron.

·         Decrease in oxidation number of reduced species.


Redox reaction

·         Oxidation and reduction occur simultaneously together.

·         Simultaneous loss and gain of electron.

·         Simultaneous increase and decrease in the oxidation number.


Oxidizing agent (oxidant)

·         Causes oxidation of other.

·         Accept electron.

·         It self undergoes reduction.


Reducing agent (Reductant)

·         Causes reduction of other.

·         Donates electron.

·         It self undergoes oxidation.



Two important methods are used for systematic balancing of redox reactions.


A] Oxidation number method

Step I

Write the unbalanced net equation for the redox reaction. Assign the oxidation number to all the atoms in both, the reactant and the product.


Step II

Identify the atoms that have changed oxidation number. Draw bracket to connect atoms of the elements that are oxidized and reduced.


Step III

Calculate the increase or decrease in the oxidation number per atom of molecule. If increase and decrease in oxidation number is not equal, then equalize them by multiplying with a suitable number.


Step IV

Balance hydrogen and oxygen. For balancing oxygen atoms add water molecules to the side containing less oxygen atoms. Balancing of hydrogen atoms depends upon the medium acidic or basic.

i) For acidic solutions, add H+ ions to the side deficient in hydrogen atoms.

ii) For basic solutions, add H2O molecule to the side deficient in hydrogen atoms and simultaneously add equal number of OH- ions on the other side of the equation.


Step V

Check the balance equation to make sure that the reaction is balanced with respect to both the number of atoms of each element and charge.




Illustrative example


Balance the following redox equation by oxidation number method.

H2SO4(aq)  +  C(s)    →   CO2(g)    +    SO2(g)   +    H2O(l)

Solution :

Step I

The oxidation number of all the atoms are as follows :

  H2SO4    +    C     →    CO2     +    SO2   +    H2O

   ↑  ↑

+1 +6 -2         0           +4 -2          +4 -2       +1 -2


Step II

                  Less of 4e-(oxidation)

                         |              ↓

  H2SO4    +    C     →    CO2     +    SO2   +    H2O

     +6             0              +4             +4       

        |                                                 ↑

               Gain of 2e-  (Reduction)



Step III

Increase in oxidation number :

C (O)   →   C (+4), net increase = +4

Decrease in oxidation number :

S (+6)   →   S (+4), net decrease = -2


To make total increase and decrease equal the net decrease must be multiplied by 2. Hence the coefficient 2 is required on both the side for S.

 2 H2SO4    +    C     →    CO2     +    2SO2   +    H2O


Step IV

Balance the equation for O atoms by adding H2O molecule to the side with less O atoms.

2 H2SO4    +    C     →    CO2     +    2SO2   +    H2O

H atoms have already been balanced.


Step V

In reaction the number of atoms and charge is balanced, thus the balanced reaction is

2 H2SO4(aq)    +    C(g)     →    CO2(g)     +    2SO2(g)   +    H2O(l)




B] lon-electron method  (Half reaction method)


Step I

Write the unbalanced equation for the redox reaction and assign the oxidation numbers to all the atoms in the reactants and the products.


Step II

Divide the equation into two half equation. One half is oxidation in which oxidation number of oxidized species increases and other half is reduction in which oxidation number of reduced species decreases.


Step III

Balance the atoms except O and H in each half equation balance O atom by adding H2O to the side with less O atoms.


Step IV

Balance  H atoms by adding H+ ions to the side with less H atoms.


Step V

Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation and to the left of the reduction half equation.


Step VI

Multiplying half equations by suitable factors to equalize the number of electrons in the two half equations. Then add two half equations and cancel the number of electrons on both the sides of the equation.


Step VII

If the reaction occurs in basic medium then add OH- ions, equal to the number of H+ ions, on both the side of the equation, H+ and OH- ion on same side combine to give H2O molecule.



Check the equation is balanced for both, the atoms and charges.



Illustrative example

Balance the following redox equation by ion electron method. The reaction occurs in acidic medium.

Mn2+(aq)  +  ClO-3(aq)    →   MnO2(s)    +    ClO2(aq)  

Solution :

Step I

Write the unbalanced equation and assign oxidation numbers to all atoms.


                              Gain of 2e-  (Reduction)

                       |                                              ↓

Mn2+(aq)  +  ClO-3(aq)    →   MnO2(s)    +    ClO2(aq)  

+2             +5 -2                +4  -2            +4  -2

 |                                           ↑     

         Less of 4e-(oxidation)


Step II

Divide the equation into two half equation such that one half is oxidation and other half is reduction.

Oxidation half equation :

Mn2+   →   MnO2

Reduction half equation :

ClO-3     →     ClO2


Step III

Balance the each half equations for O atoms by adding H2O to the side with less O atoms.

Oxidation  :  Mn2+   +    2H2O   →   MnO2

Reduction  :  ClO-3     →     ClO2    +    H2O


Step IV

Balance H atoms by adding H+ ions to the side with less H atoms

Oxidation :   Mn2+   +    2H2O   →   MnO2   + 4H+

Reduction :   ClO-3    + 2H+     →     ClO2    +   H2O


Step V

Balance the charge on both side of equation by adding appropriate number of electron.

Oxidation :   Mn2+   +    2H2O   →   MnO2   +   4H+    +    2e-

Reduction :   ClO-3    +     2H+    +    e-     →     ClO2    +   H2O



Step VI

Multiply reduction half equation by 2 to equalize the number of electrons in two half equations. Then add two half equations.

  Mn2+     +    2H2O   →   MnO2   +     4H+    +     2e-

  2ClO-3    +     4H+    +     2e-     →     2ClO2    +   2H2O

      Mn2+    +     2ClO-3   →    MnO2   +    2ClO2  


Step VII

The reaction has equal number of atoms and charge, thus the reaction is balanced.

      Mn2+    +     2ClO-3   →    MnO2   +    2ClO2  




Balance the following redox equations by oxidation number method.

a) Ag  +  NO-3  →  Ag+  +  NO2   (Acidic medium)

b) Fe2+  +  BrO-3   →   Fe3+  +  Br-   (Acidic Medium)

c) MnCl2  +  HO-2  →  Mn(OH)3  +  Cl-   (Basic Medium)

d) Mn2+  +  H2O2  →  MnO2  +  H2O  (Basic Medium)


Balance the following redox equations by ion - electron method.

a) SO2  +  Fe3+  →  Fe2+  +  SO42-   (Acidic medium)

b) 2HgO  →   2Hg  +  O2   (Acidic Medium)

c) S2O32-  +  I2  →  S4O62-  +  I-   (Basic Medium)

d) SeO32-  +  Cl2  →  SeO42-  +  Cl-  (Basic Medium)




1) Combustion

Burning of a substance by oxidation with oxygen in air is called combustion. Combustion involves redox reactions.


2) Bleaching

Decolourization or lightening of coloured materials uses redox reaction and is called bleaching.

3) Batteries

The electricity produced in batteries or galvanic cells is due to redox reactions occurring in them.


4) Metallurgy

The extraction and purification of metals uses redox reactions in different steps.


5) Corrosion

The corrosion is the destruction of metals by oxidation.

For example,

Rusting of iron is its oxidation by oxygen of air in presence of moisture.

4Fe    +    3O2       H2O   →    2Fe2O3 . H2O

                                         Corroded iron


6) Respiration

The process of breathing and using oxygen is biological redox reactions and is called respiration. you read these notes on CHEMISTRYNOTESINFO & this is 11th class chemistry notes


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