Chemical Reaction is the process which leads to the transformation of one set of chemical substances to other substances.

Classically, chemical reactions encompass changes that strictly involve the motion of electrons in forming and breaking of the chemical bonds.

The concept of electron transfer can easily explain the redox reaction in case of ionic substances. However, for covalent compounds oxidation and reduction or redox reactions are explained using  a new term oxidation number .

The term oxidation  was first used to mean the addition of oxygen to an element or compound, or the removal of hydrogen from a compound. Reduction meant the addition of hydrogen to an element or compound, or the removal of oxygen from a compound. Such definitions have been extended and now-a-days many oxidation-reduction, or redox, reactions are best interpreted in terms of transfer of electrons.

• Oxidation is defined as the loss of electrons by a chemical species (atom, ion or molecule).

• Reduction is the gain of electrons by a chemical species (atom, ion or molecule).

• An oxidising agent that chemical species which takes electrons thus  it is an electron acceptor.

• A reducing agent is the chemical species that gives electrons and thus acts as an electron donor

What are Redox Reactions?

• Redox reactions are those chemical reactions  in which both oxidation as well as reduction occur simultaneously.

• Oxidation and reductions go hand in hand. 

• The substance which undergo reduction is called oxidising agent while the substance which undergo oxidation is called reducing agent. One can say that the substance that causes the oxidation of any substance in reaction is called the oxidizing agent while the substance that causes the reduction  is called the reducing agent. 

• Neither reduction nor oxidation occurs alone.  Both of them occur simultaneously. Since both these reactions must occur at the same time they are often termed as "redox reactions". The oxidation or reduction portion of a redox reaction, including the electrons gained or lost can be determined by means of a Half-Reaction.

Classical Concept of Oxidation

The term oxidation was first used to describe chemical reactions in which oxygen was added to an element of a compound. The phenomenon of combustion was the earliest example of oxidation. Later on the term oxidation was extended to describe many more reactions which occurred without the use of even oxygen. Lets try to find out what is answer of the question, What is oxidation ! according to classical concept.

• Addition of Oxygen:  Reduction is a chemical reaction in which oxygen is removed from any chemical species (atom, ion or molecule).
  2Mg + O2 → 2MgO
  S + O2 →  SO2
  2CO + O2 →  2CO2
  Na2SO3 + H2O2 →  Na2SO4 + H2O

• Removal of Hydrogen : Oxidation is a chemical reaction in which hydrogen is removed from any chemical species (atom, ion or molecule). 
  H2S + Cl2 → 2HCl + S
  4HI + O2 →  2H2O + 2I2
  4HI + MnO2 →  MnCl2 + 2H2O + Cl2

• Addition of an Electronegative Element: Oxidation is a chemical reaction in which an electronegative element is added into any chemical species (atom, ion or molecule).
  Fe+ S →  FeS (Oxidation of iron)
  SnCl2 + Cl2 →  SnCl4 (Oxidation of stannous chloride)
  2Fe + 3F2 →  2FeF3 (Oxidation of iron)

• Removal of an Electropositive Element: Oxidation is a chemical reaction in which an electropositive element is removed from any chemical species (atom, ion or molecule). 
  2KI + H2O2 →  2KOH + I2  (Oxidation of potassium iodide)
  2K2MnO4 + Cl2 →  2KCl + 2KMnO4 (Oxidation of potassium manganate)
  2KI + Cl2 →  2KCl + I2 (Oxidation of potassium iodide)

• A substance which brings oxidation is known as oxidizing agent. 

Classical Concept of Reduction

• Removal of Oxygen:  Reduction  is a chemical reaction in which oxygen is removed from into any chemical species (atom, ion or molecule).
  CuO + C → Cu + CO   
  H2O + C →  CO + H2
  2CO + O2 →  2CO2
  Fe3O4 + 4H2 →  3F2 + 4H2O

• Addition of Hydrogen : Reduction is a chemical reaction in which hydrogen is added to any chemical species (atom, ion or molecule). 
  Cl2 + H2 →  2HCl
  S + H2 →  H2S
  C2H4 + H2 →  C2H6

• Removal of an Electronegative Element: Reduction is a chemical reaction in which an electronegative element is removed from any chemical species (atom, ion or molecule). 2HgCl2 + SnCl2 →  Hg2Cl2 + SnCl4 (Reduction of mercuric chlorine)
  2FeCl3 + H2 →  2FeCl2 + 2HCl (Reduction of ferric chloride)
  2FeCl3 + H2S →  2FeCl2 + 2HCl + S (Reduction of ferric chloride)

• Addition of an Electropositive Element: Reduction is a chemical reaction in which an electropositive element is added any chemical species (atom, ion or molecule). 
  HgCl2 + Hg →  Hg2Cl2 (Reduction of mercuric chloride)
  CuCl2 + Cu →  Cu2Cl2 (Reduction of cupric chloride)

• The substance which brings reduction is known as reducing agent. 

• A substance, which undergoes oxidation, acts as a reducing agent while a substance, which undergoes reduction, acts as an oxidizing agent.
  Mg, S, Cu, Na2SO3, H2S, HI, H2, C, KI are reducing agents, while O2, Cl2, F2, H2O2, MnO2, FeCl3, CuCl2, Fe3O4, CuO, etc., are oxidizing agents in the above examples.

• All oxidation and reduction reactions are complimentary of one another and occur simultaneously, one can not take place without the other.

• No single oxidation and no single reduction process is known. The simultaneously oxidation and reduction reactions are generally termed as redox reactions.  e.g., 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4  or  2Fe3+ + Sn2+ → 2Fe2+ + Sn4+
  In above example iron undergoes reduction from +3 to +2 and tin undergoes oxidation from +2 to +4.

Modern Concept of Oxidation & Reduction

According to the modern concept, loss of electrons is oxidation whereas gain of electrons is reduction.

• Examples of oxidation reactions are:
  Na → Na+ + e
  Zn → Zn2+ + 2e
  Fe2+ → Fe3+ + e
  Sn2+ → Sn4+ + 2e
  H2O2 → O2 + 2H+ + 2e
  2S2O32– → S4O62– + 2e
  [Fe(CN)6]4– → [Fe(CN)6]3– + e
  MnO42– → MnO4– + e

• Examples of reduction reactions are:-
  Cl2 + 2e → 2Cl
  S + 2e → S2–
  Cu2+ + 2e → Cu
  MnO4– + 8H+ + 5e → Mn2+ + 4H2O
  Cr2O72– + 14H+ + 6e → 2Cr3+ + 7H2O
  H2O2 + 2H+ + 2e → 2H2O

• Oxidation and reduction can be represented in a general way as shown below:

•  M4- → M3- → M2- → M- → M0 → M+ → M2+ → M3+ → M4+ 

• In a redox process the valency of the involved species changes. The valency of a reducing agent increases while the valency of an oxidizing agent decreases in a redox reaction. The valency of a free element is taken as zero.

• Decrease in valency When there is no change in valency it means there is no oxidation or reduction, e.g.,
  BaCl2 + H2SO4 → BaSO4 + 2HCl    or BaSO4 → Ba2+ + SO42–    (No change in valency)

Conclusion

• Oxidation is a process in which one or more electrons are lost or valency of the element increases.

• Reduction is a process in which one or more electrons are gained or valency of the element decreases.

• Oxidizing agent is a material which can gain one or more electrons, i.e., valency decreases.

• Reducing agent is a material which can lose one or more electrons, i.e., valency increases.

• Redox reaction involves two half reactions, one involving loss of electron or electrons (oxidation) and the other involving gain of electrons or electrons (reduction).

Summary

Question 1: Which of the following chemical reactions represent oxidation process?

a. Cl2 + 2e → 2Cl
b. S + 2e → S2–
c. 2S2O32– → S4O62– + 2e
d. MnO4– + 8H+ + 5e → Mn2+ + 4H2O

Question 2: According to the modern concept, oxidation is the process of 

a. gain of electrons

b. lose of electrons

c. addition of hydrogen

d. removal of hydrogen

Question 3: Which of the following conditions does not represent oxidation?

a. Removal of hydrogen.

b. Addition of oxygen.

c. Removal of electropositive element.

d. Addition of electrons.

Question 4: Which of the following statements is incorrect ?

a. Oxidation is a process in which one or more electrons are lost or valency of the element increases.

b. Reduction is a process in which one or more electrons are gained or valency of the element decreases.

c. Oxidizing agent is a material which can gain one or more electrons,

d. Reducing agent is a material whose valency decreases during reaction.

Q.1       Q.2         Q.3        Q.4

c              b            d             d

When an element is oxidised it must be acting as a reducing agent and it, therefore, loses electrons; when reduced, it gains electrons. The oxidation state or oxidation number of an element is the number of electrons it might be considered to have lost or gained.

• All elements in the elementary, uncombined state are given oxidation numbers of zero. When sodium, for example, is oxidised it loses one electron, and the Na+ ion is said to have an oxidation number of +1. Similarly, the Cu2+ and Al3+ ions have oxidation numbers of +2 and +3, whilst F- and O2- have oxidation numbers of - 1 and - 2. For simple ions, the oxidation number is equal to the ionic charge, e.g

• The oxidation number for an element in a covalent compound is  by taking the oxidation number to be equal to the charge that the element would carry, if all the bonds in the compound were regarded as ionic instead of covalent. In doing this, a shared pair or electrons between two atoms is assigned to the atom with the greater electronegativity. Or, if the two atoms are alike, the shared pair is split between the two, one electron being assigned to each atom. The resulting charges on the various atoms when the bonding electrons are so assigned are the oxidation numbers of the atoms.

• The sum of the oxidation numbers of all the atoms in an uncharged compound is zero. In case of an ion,  the algebraic sum of the oxidation numbers of all the atoms is equal to the charge on the ion.

• Oxidation number of any element in its elementary state is zero.

• Fluorine is the most electronegative element. Its oxidation number is always  -1 .

• Oxygen after fluorine is the second most electronegative element. It  shows an oxidation state of -2 in almost all the compounds excepts seroxides and superoxides.

• In peroxides (O2–), oxygen has oxidation number –1; in superoxides (O22–), oxygen has oxidation number –1/2; and in OF2, the oxygen has an oxidation number +2.

the oxidation numbers in the peroxide ion being calculated by splitting the shared pair equally between the two oxygen atoms

• In all compounds, except ionic metallic hydrides, the oxidation number of hydrogen is +1.

• In any compounds has more than two elements, the oxidation number of any one of them may have to be obtained by first assigning reasonable oxidation numbers to the other elements.

• Some elements like manganese, chromium, nitrogen and chlorine show variable oxidation states. 

• When an element is oxidised its oxidation number gets increased while reduction on any element decreases its oxidation number. Change in oxidation number can be used to decide whether an oxidation or a reduction has taken place. In the change from chloromethane to dichloromethane, for example,
  C  H3  Cl C  H2 Cl2
  -2 +1 -1 0 +1 -1
  The oxidation number of carbon is increased from -2 to 0. The carbon is therefore being oxidised.

Common Oxidising and Reducing Agents

• The functioning of some common oxidising and reducing agents is summarised below

Oxidising agent

Effective Change

Decrease in Oxidation Number

KMnO4 in acid solution

MnO4 -  → Mn2+

5

KMnO4 in alkaline solution

MnO4 - →  MnO2

3

K2Cr2O7 in acid solution

Cr2O72- →  Cr3+

3

dilute HNO3

NO3- →  NO

3

concentrated HNO3

NO3- →  NO2

1

concentrated H2SO4

SO42- →   SO2

2

manganese (IV) oxide

MnO2 → Mn2+

2

chlorine

Cl →  Cl-

1

chloric (I) acid

ClO-  →  Cl-

2

KlO3 in dilute acid

IO3-  →  I

5

KlO3 in concentrated acid

IO3-   →  I-

4

Reducing agent

Effective Change

Increase in Oxidation Number

iron (II) salts (acid)

Fe2+ →  Fe3+

1

tin (II) salts (acid)`

Sn2+ →  Sn4+

2

ethanedioates (acid)

C2O42- →  CO2

1

sulphites (acid)

SO32-  →SO42-

2

hydrogen sulphide

S2- → S

2

iodides (dilute acid)

I- → I

1

iodides (concentrated acid)

I- → I+

2

metals, e.g. Zn

Zn → Zn2+

2

hydrogen

H  → H+

1

 Example

Question: 

What is the oxidation number of Mn is KMnO4 ?

Solution:

Let the oxidation number  of Mn in KMnO4 be x.

We know that

Oxidation number of K = +1

Oxidation number = –2

(Oxidation number K) + (Oxidation number of  Mn) +4(Oxidation number of O) = 0

(+1)+(x)+4(-2) = 0

or x = +7

Oxidation State as Periodic PropertY

Oxidation state of an atom depends upon the electronic configuration of atom, a periodic property.

• IA group or alkali metals shows +1 oxidation state.

• II A group or alkali earth metals show +2 oxidation state

• The maximum normal oxidation state, shown by III A group elements is +3. These elements also show +2 to +1 oxidation states.

• Elements of IV A group show their max. and min. oxidation states +4 and -4 respectively.

• Non-metals show number of oxidation states, the relation between max. and mini. oxidation states for non metals is equal to maximum oxidation state - minimum oxidation state =8
  For example S has maximum oxidation number +6 as being VI Agroup element.

• Maximum oxidation number of an element is equal to group number – 8

• I Group elements always show +1 oxidation number.

• II Group elements always show +2 oxidation number.

• III Group elements always show +3 oxidation number.

• IV Group elements show -4 to +4 oxidation number.

• V Group elements show -3 to +5 oxidation number.

• VI Group elements show -2 to +6 oxidation number.

• VII Group elements show -1 to +7 oxidation number.

• Inert gases show zero oxidation number.

Application of Oxidation State

To compare the strength of acid and base

• Strength of acids increases with increase in oxidation number.

• Strength of base decreases with increase in oxidation number.

• If any compound is in maximum oxidation state , then it will act as oxidant only.

• If any compound is in minimum oxidation state, then it will act as reductant only.

• If the oxidation state is intermediate, then compound can act as both reductant as well as oxidant.

To determine molecular formula of any compound

Suppose that there are three atoms A, B, C and their oxidation number are 6, -1, -2 respectively. Then the molecular formula of compound formed by them will be AB4C  because

+6 = (-14)+(-2)

or +6 = -6

Oxidation State V/S Valency

Oxidation State

Valency

It represents the number of electrons which an atom of an element appears to have gained or lost when in the combined state.

It is the number of hydrogen or chlorine atoms or double the number of oxygen atoms that combine with one atom of the element

Oxidation number of an element may be different in different compounds.

Valency of an element usually remains fixed.

Oxidation state or oxidation number of an element may be positive, negative or zero.

Valency of an element is either positive or negative.

Oxidation state may have fractional values

Valency is always a whole number

Summary 

• The oxidation number (Ox.no.) of an atom in free elements is zero, no matter how complicated the molecule is, hydrogen in H2, sulphur in S8, phosphorus in P4, oxygen in O2 or O3, all have zero value of oxidation numbers. 

• The fluorine, which is the most electronegative element, has oxidation number –1 in all of its compounds.

• Oxidation number of oxygen is –2 in all compounds except in peroxides, superoxides and oxygen fluorides. In peroxides (O2–), oxygen has oxidation number –1; in superoxides (O22–), oxygen has oxidation number –1/2; and in OF2, the oxygen has an oxidation number +2.

• The oxygen number of hydrogen is +1 in all of its compounds except in metallic hydrides like NaH, BaH2, etc. Hydrogen is in –1 oxidation state in these hydrides.

• The oxidation number of an ion is equal to the electrical charge present on it.

• The oxidation number of IA elements (Li, Na, K, Rb, Cs and Fe) is +1 and the oxidation number IIA elements (Be, Mg, Ca, Sr, Ba and Ra) is +2.

•  Sum of oxidation number of all the atoms of a complex ion is equal to the net charge on the ion.

• Sum of the oxidation number of all the atoms present in a neutral molecule is zero.

• If any compound is in maximum oxidation state , then it will act as oxidant only.

• If any compound is in minimum oxidation state, then it will act as reductant only.

• If the oxidation state is intermediate, then compound can act as both reductant as well as oxidant.

• I Group elements always show +1 oxidation number.

• II Group elements always show +2 oxidation number.

• III Group elements always show +3 oxidation number.

• IV Group elements show -4 to +4 oxidation number.

• V Group elements show -3 to +5 oxidation number.

• VI Group elements show -2 to +6 oxidation number.

• VII Group elements show -1 to +7 oxidation number.

• Inert gases show zero oxidation number.

Question 1: Oxidation state of Mg in MgO is

a. -1
b. -2
c. +2
d. +1

Question 2: Which of the following oxidation states are not possible for O?

a. -1

b. -2

c. -1/2

d. +3/2

Question 3: Oxidation state of O in OF2 is..

a. +2

b. -2

c. -1

d. +1

Question 4: In the case of neutral molecules, the algebraic sum of the oxidation number of all the atoms present in the molecule is.

a. zero

b. one

c. two

d. three

Oxidation Number / State Method For Balancing Redox Reactions

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

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

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

• Add respective electrons on the right of oxidation reaction and on the left of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.

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

• Transfer the coefficient of the oxidizing and reducing agents and their products to the main equation.

• By inspection, arrive at the co-efficients of the species not undergoing oxidation or reduction.

Solved Examples

Question 1

Balance the equation :

KMnO4 + H2SO4 + FeSO4 →  K2SO4 + MnSO4 + Fe2(SO4)3 + H2O

Solution: 

 Mn+7 →  Mn+2 ; Fe+2 →  Fe2+3

 Mn+7 + 5e- →  Mn+2 (1)

2;      2Fe+2 →  Fe2+3 + 2e- (2)  5

2Mn+7 + 10e- →  2Mn+2;   10Fe2+2 →  5Fe2+3 + 10e-.

Therefore coefficient of KMnO4 is 2, that of MnSO4 is 2, that of FeSO4 is 10 and that of Fe2(SO4) is 5.

Therefore 2KMnO4 + 10FeSO4 →  2MnSO4 + 5Fe2(SO4)3

Coefficient of K2SO4 should be 1.

2KMnO4 + 10 FeSO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3

To balance SO42- so that left hand side must have 8H2SO4, thus H2O also becomes 8H2O.

 2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O

______________________________________________

Question 2:

Balance the equation :

MnO4- + C2O42–  + H+ →  CO2 + Mn+2 + H2O

Solution:

Mn+7 →  Mn+2 ;  C2 →  C+4

5e- + Mn+7 →  Mn+2 (1) ´2;  

C2 →  2C+4 + 2e-   (2) ´5

10e- + 2Mn+7 →  2Mn+2 ;  5C2 →  10C+4 + 10e-

Therefore 2MnO4- + 5C2O4-2 →  10CO2- + 2Mn+2

There must be 8H2O on the right to balance O. Therefore there must be 10H+ to balance H.

2MnO4- + 5C2O4-2 + 16H+ →  10CO2 + 2Mn+2 + 8H2O

Half-Reaction or Ion-Electron Method For Balancing Redox Reactions

This method involves the following steps :

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

• Balance the atoms in each half reaction separately according to the following steps:

  • First of all balance the atoms other than H and O.

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

  • Equalize the charge on both sides by adding a suitable number of electrons to the side deficient in negative charge.

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

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

Solved Example

Question 1:

Balance the redox reaction  Al + NO3- →  Al(OH)4- +NH3 in alkaline medium.

Solution: 

• Dividing the equation in two half reactions
  NO3-  →NH3  (Oxidation Half Reaction)
  Al + 4OH- →Al(OH)4- (Reduction Half Reaction)

• All the atoms other than O & H are balanced (N and Al in this case which are already balanced).
  NO 3- + H2O →NH3 + OH-
  Al + 4OH- →Al(OH)4-

• Then O and H are balanced using OH- and H2O
  NO3- + 6H2O →NH3 + 9OH-
  Al + 4 OH- → Al(OH) 4 -  

• Electrons need to be added in order to balance the charge.
  8 e- + NO 3 - + 6 H 2 O →NH 3 + 9 OH -
  Al + 4 OH- → Al(OH)4- + 3 e- 

• Loss and gain of electrons need to be made equal:
  For this, we need to multiply first equation by 3 and second equation by 8

• Adding both equations we get the balanced redox equation
  3NO 3 - + 18H 2 O+ 8Al + 5 OH - → 8Al(OH) 4 - +3NH 3  

____________________________________________________

Question 2: 

Balance the equation 

Cr2O7 2-(g) + SO2(aq) → Cr3+ + SO42-(aq) (in acidic medium) 

Solution:

• Dividing the equation in two half reactions
  SO2 →SO4 2-  (Oxidation half reaction)
  Cr2O7 2-→Cr3+  (Reduction half reaction)

• Balance the atoms other than O and H
  SO2 →SO42- 
  Cr2O7 2-→2Cr3+  

• O and H are balanced using H+ and H 2 O
  SO2  + 2H2O →SO4 2-  +4H+ 
  Cr2O7 2-+14H+→2Cr3+  +7H2O 

• Electrons need to be added to balance the charge.
  SO2  + 2H2O →SO4 2-  +4H+ +2e- ...............................(2)
  Cr2O7 2-+14H+  +6e- →2Cr3+  +7H  2O .........................(1)

• Loss and gain of electrons need to be made equal. For this, we need to multiply equation (2) by 3

• Adding both equations to get the balanced redox reaction
  Cr2O7 2-+2H+    +3SO 2  →2Cr3+  +7H2O

Question 1: Which of the following equations represents a balanced redox reaction?

a. KMnO4 + H2SO4 + FeSO4 →  K2SO4 + MnSO4 + Fe2(SO4)3 + H2O
b. 2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O
c. Cr2O7 2-(g) + SO2(aq) → Cr3+ + SO42-(aq)
d. Al + NO3- →  Al(OH)4- +NH3

Question 2: Ion-electron method for balancing Redox reactions is also known as

a. Oxidation number method

b. State method

c. Half reaction method

d. Half life method

Question 3: In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding 

a. H2O & H+

b. H2O & OH-

c. H+ & OH-

d. H2O only

Question 4: What is the number of electrons being transferred in the redox reaction 2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O?

a. Zero

b. One

c. Five

d. Ten

Types of Redox Reactions

Redox reactions are divided into two main types.

(i) Inter molecular Redox Reactions:

In such redox reactions, one molecule of reactant is oxidized whereas molecule of other reactant is reduced.

(ii) Intra molecular Redox Reactions:

One atom of a molecule is oxidized and other atom of same molecule is reduced then it is intra molecular redox reaction.

• The reaction taking place in batteries are redox reactions. Redox reactions  take place in the batteries such that electrons transferred can pass through some external circuit so that they produce electric current

• Digestion and metabolism of food which takes place in our body in order to supply us the energy required to perform work is also takes place through a series of redox reactions. 

• Ordinary bleach oxidise the  substances that stain fabric, this makes them colourless and  easier to remove from fabric.

Basic Terms Used 

Molecular Equations

When the reactant and products involved in a chemical change are written in molecular form in a chemical  equation, it is termed as molecular equation.

Example: MnO2 + 4HCl → MnCl2 + 2H2O +Cl2
In above example the reactant and products have been written in molecular forms, thus it is a molecular equation.

Ionic Equation

When the reactant and products involved in a chemical change are ionic compounds, these will be present in the form of ions in the solution. The chemical change is written in ionic forms in the chemical equation, it is termed as ionic equation. 

Example: MnO2 + 4H++ 4Cl- → Mn2++ 2Cl- + 2H2O +Cl2

In the above example, the reactant and products have been written in ionic forms, thus the equation is termed as ionic equation.

The rules to be followed for writing ionic equations are:

• All soluble ionic compounds involved in a chemical changes are expressed in ionic symbols and covalent substances are written in molecular form. H2O, NH3, NO2, NO, SO2, CO, CO2, etc., are expressed in molecular form. 

• The ionic compound which is highly insoluble is expressed in molecular form.

• The ions which are common and equal in number on both sides, i.e., spectator ions, are cancelled.

• Besides the atoms, the ionic charges must also be balanced on both the sides.
   

Spectator Ions

Species that are present in the solution but not take part in the reaction and are also omitted while writing the net ionic equation are called spectator ions or bystander ions.
Zn + 2H+ + 2Cl- → Zn2+ + 2Cl- + H2
In this reaction, ions are omitted and are called as spectator ions and appear on the reactant as well as product side.”
Oxidising Agent

The substance (atom, ions or molecules) that gain electrons and is thereby reduced to a low valency state is called an oxidising agent.
Reducing Agent

The substance that loses electrons and its valency thereby oxidised to a higher valency state is called a reducing agent.

Question 1: Which of the following statements regarding redox reactions is incorrect?

a. Oxidation is the process of  loss of electrons by a chemical species ( atom, ion or molecule).

b. Reduction is the process of  gain of electrons by an atom, ion or molecule.

c. A reducing agent is one that undergo reduction

d. A reducing agent gives electrons; 

Question 2: Which of the following rules to be followed for writing ionic equations is incorrect?

a. All soluble ionic compounds involved in a chemical changes are expressed in ionic symbols and covalent substances are written in molecular form. H2O, NH3, NO2, NO, SO2, CO, CO2, etc., are expressed in molecular form. 

b. The ionic compound which is highly insoluble is expressed in molecular form.

c. The ions which are common and equal in number on both sides are  written on product side.

d. Besides the atoms, the ionic charges must also be balanced on both the sides.

Question 3: Which of the following equations is not a redox reaction?

a. Fe2+ → Fe3+ + e-

b. MnO2 + 4HCl → MnCl2 + 2H2O +Cl2

c. Zn + 2H+ + 2Cl- → Zn2+ + 2Cl- + H2

d. MnO2 + 4H++ 4Cl- → Mn2++ 2Cl- + 2H2O +Cl2

Question 4: In the reaction MnO2 + 4HCl → MnCl2 + 2H2O +Cl2

Which of the following species acts as reducing agent?

a. MnO2 

b. HCl

c. Cl-

d. None of above

Question 1:

Find the oxidation number of 

a. S in SO42- ion

b. S in HSO3- ion

c. Pt in (PtCl6)2-

Solution:

a. Let the oxidation state of S in SO42 = x

We know that ox. no. of O = -2

So, x +  4(-2) = -2

or x =+6

b. Let the oxidation state of S in HSO3- = x

We know that ox. no. of O = -2

&  ox. no. of H = +1

So, +1 +x+3(-2) = -1

or x = +4

c. Let the oxidation state of Pt in (PtCl6)2- = x

We know that ox. no. of Cl = -1

So, x+ 6(-1) = -2

or x =  -2 + 6 = +4

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Question 2: 

What is the oxidation number of Mn is KMnO4 and of S in Na2S2O3?

Solution:

Let the Ox.no. of Mn is KMnO4 be x.

We know that

 Ox.no. of K = +1

Ox.no. of O = –2

So   
Ox.no. K + Os.No. Mn+4 (Ox.no.O) = 0

or   +1+ x + (4–2)  = 0

or +1+ x  – 8  = 0

or  x = +8 – 1 = +7

Hence, Ox.no. of Mn in KMnO4 is +7

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Question 3: 

What is the oxidation number of Cr in K2Cr3O7 ?

Solution:

Let the oxidation number of Fe be x.

We know that

Ox.no. of K =+1

Ox.no. of (CN)– = –1

So 4(Ox.no. K) + Os.No. Fe+6(Ox.no. CN–) = 0

or  4(+1)      + x  +  (6–1)    = 0

or     +4   + x   –  6   = 0

or       x = +6 – 4 = +2

The oxidation number of iron in K4Fe(CN)6 is +2.

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Question 4:

Which compound amongst the following has the highest oxidation number for Mn?

KMnO4, K2MnO4, MnO2 and Mn2O3

Solution:

KMnO4,    
+1+x –8 = 0            
ot x = + 7 

Ox.no. of Mn =- +7

K2MnO4,   

+2 +x –8 = 0             

x = +6                         

MnO2      

x – 4 = 0

or  x = +4        

Mn2O3        

2x – 6 = 0

 x = + 3                            

Thus, the highest oxidation number for the Mn is in KMnO4.

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Question 5:

Can oxidation number of any element in a compound ever be zero? Justify the answer.

Solution

Sometimes, oxidation numbers have such values which at first sight appear strange. For example, the oxidation number of carbon in cane sugar (C12H22O11), glucose (C6H12O6), dichloromethane, etc., is zero.

Cane Sugar (C12H22O11)                     Glucose (C6H12O6)

12×x+22×1+11(–2)=0                   6×x+12×1+6(–2)=0

12x+22 – 22 = 0                          6x+12–12=0

So x = 0                                     So x = 0

Dichloromethane (CH2Cl2)

x = 2 × 1 + 2(–1) = 0

x + 2 – 2 =0

So x = 0

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Question 6: 

Balance the equation Cr(OH)3 + IO3 - →  I - + CrO42-

Solution:

Oxidation : Cr(OH)3 → CrO42- Reduction : IO3- → I-

Balancing oxidation reaction : Balancing atoms other than H and O

Cr(OH)3 → CrO42-

Balancing O;     2OH- + Cr(OH)3 → CrO42- + H2O

Balancing H;     3OH- + 2OH- + Cr(OH)3 → CrO42- + H2O + 3H2O

Balancing charge; 5OH- + Cr(OH)3 → CrO42- + 4H2O + 3e- (1)

Balancing reduction reaction :

Balancing O;     IO3- + 3H2O → 6OH- + I-

Balance the charges 6e- + IO3- + 3H2O ® 6OH- + I- (2)

Multiply (1) by 2 and adding to (2)

2Cr(OH)3 + IO3- + 4OH- → 2CrO42- + 5H2O + I-.     

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Question 7: 

Balance the equation

Cr2O72- + C2O42- + H+ ® Cr+3 + CO2 + H2O

Solution:

Reduction : Cr2O72- →Cr+3 Oxidation : C2O42- → CO2

Balancing atoms other than O and H : Cr2O72- → 2Cr+3

Reaction is taking place in acidic medium.

Balancing O

Cr2O72- → 2Cr+3 + 7H2O

Balancing H

14H+ + Cr2O72- → 2Cr+3 + 7H2O

Balancing charges, we get reduction half-reaction.

 6e- + 14H+ + Cr2O72- →  2Cr+3 + 7H2O(1)

Balancing oxidation reaction :

Balancing C C2O42- →2CO2

Balancing O C2O42- → 2CO2

Balancing charges, we get oxidation half-reaction

C2O42- → 2CO2 + 2e-         (2)

Multiplying (2) by 3 and adding to (1) (To cancel out the electrons)

Cr2O72- + 14H+ + 3C2O42- → 2Cr+3 + 7H2O + 6CO2

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Question 8:

One mole of N2H4 loses 10 mole electrons to form a new compound Y. Assuming that all the N2 appears in new compound, what is the oxidation state of Nitrogen in Y? (There is no change in the oxidation state of H)

Solution:

N2H4 →  (Y) + 10e-

( Y contains all N atoms)

i.e.  (2N)x + 10e-

2x - (-4) = 10

x = +3

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