Oxidation is :
(i) the gain of oxygen
(ii) the loss of hydrogen
(iii) the loss of electrons (de-electronation)
(iv) the increase of O.N.
Reduction is :
(i) the loss of oxygen
(ii) the gain of hydrogen
(iii) the gain of electron
(iv) the decrease in O.N.
Fe → Fe2+ + 2e−
O.N. = 0 +2
∎ Fe loses electrons
∎ There is increase in O.N.
∎ Hence Fe is said to be oxidised
∎ It is a source of electrons hence it can act as a reducing agent (R.A.)
∎ Any species that can be oxidised is a reducing agent (R.A.)
Cu2+ + 2e → Cu
O.N. = + 2 0
∎ Cu2+ gains electron
∎ There is decrease in O.N.
∎ Hence Cu2+ is said be reduced.
Since it can gain electrons, hence it can act as an oxidising agent (O.A.)
∎ Any species that can be reduced is an oxidising agent (O.A.), or oxidant.
Above examples represent half reactions.
A complete reaction showing oxidation and reduction together is called a Redox reaction.
Now that you are clear on what is oxidation − reduction, we are now in the right position to know how to balance a redox (oxidation – reduction) reaction.
This is important because many of the problems of stoichiometry would be based on such redox reactions.
Oxidation State / Oxidation Number
There are several chemical reactions in which oxidation – reduction takes place.
Oxidation refers to a reaction in which electrons are removed from an atom and reduction refers to a reaction in which electrons are added to an atom.
To describe these changes, the concept of oxidation state becomes necessary.
For ionic species, the charge on each ion is said to be the oxidation state for that atom.
For example in NaCl, Na exists as Na+ and Cl exists as Cl−.
Therefore the oxidation state of Na in NaCl is +1 and that of Cl− is −1.
But in covalent molecules, the charge on an atom would be so small that sometimes it becomes impossible to calculate the exact charge on each atom of a molecule.
Therefore, the Oxidation State (O.S.) or Oxidation Number (O.N.) is defined as the charge, an atom would have in a molecule if all the bonds associated with this atom in the molecule are considered to be completely ionic.
For example in H2O there are two O−H bonds. If we assume both the O−H bonds to be completely ionic, then each H would possess a charge of +1, while O possess a charge of −2.
This is because oxygen is more electronegative than hydrogen. On the other hand, in H2O2 there are two O−H bonds and one O−O bond (H−O−O−H).
Considering each O−H bond to be ionic both the oxygen atoms acquire a charge of −1 and both the H , +1.
This is because O−O bond can not be assumed to be ionic as both the atoms have the same electronegativity.
To calculate the oxidation state of an element in a molecule you need not always know the structure of the molecule.
There are certain set of rules used to assign oxidation states in polyatomic molecules:
1. The O.S. of all elements (in any allotropic form) is zero.
2. O.S. of oxygen is −2 in all its compounds except peroxides like H2O2 and Na2O2 where it is −1 and superoxides like KO2 where it is −1/2.
3. O.S. of hydrogen is +1 in all of its compounds except those with the metals where it is −1.
4. O.S. of all alkali metals is +1 and alkaline earth metals is +2 in all their compounds.
5. O.S. of all the halogens is −1 in all their compounds except where they are combined with an element of higher electronegativity. Since fluorine is the most electronegative of all elements, its O.S. is always −1.
Illustration : Calculate the O.S of all the atoms in the following species:
(i) ClO− (ii) NO2– (iii) NO3– (iv) CCl4 (v) K2CrO4 (vi) KMnO4
(i) In ClO– , the net charge on the species is −1 and therefore the sum of the oxidation states of Cl and O must be equal to −1.
Oxygen will have an O.S. of −2 and if the O.S. of Cl is assumed to be ‘ x ‘ then x − 2 should be equal to −1.
∴ x is + 1
(ii)NO2− : (2 × −2) + x = −1 (where ‘ x ‘ is O.S. of N)
∴ x = +3
(iii) NO3– : x + (3 × −2) = −1 (where ‘ x ‘ is O.S. of N)
∴ x = +5
(iv) In CCl4, Cl has an O.S. of -1
∴ x + 4 × −1 = 0
∴ x = +4 (where ‘ x ‘ is O.S. of C)
(v) K2CrO4 : K has O.S. of +1 and O has O.S. of −2 and let Cr has O.S. ‘ x ‘ then, 2 × +1 + x + 4 × −2 = 0
∴ x = + 6
(vi) KMnO4 : +1 + x + (4 × −2) = 0
∴ x = +7 (where x is O.S. of Mn).