REDOX REACTIONS Class 11 Notes

Redox Reactions Class 11th 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).

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CLASSICAL CONCEPT OF OXIDATION AND REDUCTION

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)  +  O_2(g)  →  CO_2(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)

ELECTRONIC CONCEPT OF OXIDATION AND REDUCTION

OXIDATION

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

For example,

Na   →   Na⁺   +   e^-

 

REDUCTION

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

For example,

Cl₂   +   2e^-   →   2Cl^-

 

OXIDIZING AGENT OR OXIDANT

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

 

REDUCING AGENT OR REDUCTANT

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 OR OXIDATION STATE

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^-     S^2-

↑      ↑        ↑       ↑      ↑

+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

Problems

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 PtCl₆^2-

PtCl₆^2-  is ionic species.

: .  Sum of oxidation number of all atom of PtCl₆^2-  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 PtCl₆^2-  is + 4.

 

 

KEY POINTS

Oxidation

·         Loss of electron.

·         Increase in oxidation number of oxidized species.

 

Reduction

·         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.

 

BALANCING OF REDOX REACTIONS

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.

H₂SO_4(aq)  +  C_(s)    →   CO_2(g)    +    SO_2(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 H₂SO_4(aq)    +    C_(g)     →    CO_2(g)     +    2SO_2(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.

 

Step VIII

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)    →   MnO_2(s)    +    ClO_2(aq)  

Solution :

Step I

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

                 

                              Gain of 2e^-  (Reduction)

                       |                                              ↓

Mn²⁺_(aq)  +  ClO^-_3(aq)    →   MnO_2(s)    +    ClO_2(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  

 

 

PROBLEMS FOR PRACTICE

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⁺  +  SO₄^2-   (Acidic medium)

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

c) S₂O₃^2-  +  I₂  →  S₄O₆^2-  +  I^-   (Basic Medium)

d) SeO₃^2-  +  Cl₂  →  SeO₄^2-  +  Cl^-  (Basic Medium)

 

 

APPLICATIONS OF REDOX REACTIONS

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.