Oxidation-Reduction Reaction

Oxidation-Reduction Reaction

Oxidation Number

 Oxidation is any reaction in which an atom or ion loses electrons. On the other hand, reduction is defined as any reaction in which an atom or ion gains electrons. As oxidation involves removal of electrons, it has also been termed de-electronation. Similarly reduction may be referred to as electronation.
Change of ferrous (Fe2+) to ferric (Fe3+), stannous (Sn2+) to stannic (Sn4+), manganate (MnO2-4) to permanganate (MnO4) are all cases of oxidation as each one of them involves loss of electrons.
            Fe2+ = Fe3+ + 1 electron
            Sn2+ = Sn4+ + 2 electron
            MnO2-4 = MnO4 + 1 electron
 Similarly formation of mercurous ions (Hg22+) from mercuric ions (Hg22+) and formation of chlorine ions (Cl) from chlorine atoms are cases of reduction as both involve a gain of electrons.
            Hg22+ + 2 electrons = Hg22+
                    Cl + 1 electron = Cl
 In order to keep track of electron shifts in oxidation-reduction reaction, it is convenient to use the concept of oxidation state of various atoms involved in these reactions. The oxidation number is defined as the formal charge which an atom appears to have when electrons are counted in accordance with the following rules:
  1.  Electrons shared between two unlike atoms are counted with more electronegative atom. For example, the electron pair shared between H and Cl in H+1:Cl-1 is counted with more electronegative Cl. As a result of it hydrogen having lost share in the electron pair appears to have +1 charge and chlorine appears to have -1 charge. Hence oxidation numbers of H and Cl are +1 and -1 respectively. 
  2. Electrons shared between two like atoms are divided equally between the two sharing atoms. For example, in hydrogen molecule, H:H, the electron pair is equally shared between the two atoms. Thus both the atoms appears to have no charge i.e. oxidation number of hydrogen is zero in hydrogen molecule.

In the molecule of water given in the margin, the two electron pairs shared between oxygen and the two hydrogen atoms are counted with the more electronegative atom. Hence in water, oxidation number of each H is +1 and that of the O atom is -2.

H+1: O-2: H+1
Counting of electron like this is very hard. The following operation rules derived from the above will be very convenient:
1) In the elementary or uncombined state, the atoms are assigned an oxidation number zero.
2) In compounds, the oxidation number of fluorine is always -1.
3)     In compounds, the group IA elements (Li, Na, K, Rb, Cs and Fr) have an oxidation number +1 and the group IIA elements (Be, Mg, Ca, Sr, Ba, and Ra) have an oxidation number +2.
4) Oxidation number of hydrogen in compounds is generally +1 except in metallic hydrides wherein its oxidation number is -1.
5) In compounds, the oxidation number of oxygen is generally -2 except in FeO wherein oxidation number of fluorine is -1 and that of oxygen is +2. In hydrogen peroxide molecule the electron pair shared between O and H is counted with O but the other electron pair shared between two O atoms is equally shared. The number of electrons counted with each O is, therefore, seven (i.e. one more than its own electrons). The oxygen atom, therefore, appears to have -1 charge or its oxidation number in H2O2 is -1.
6) In neutral molecules, the sum of oxidation numbers of all the atoms is zero.
7) For complex ions (charged species), the sum of the oxidation numbers of all the atoms is equal to the net charge on the ion.

With the help of above rules, we can calculate the oxidation number of an atom present in a molecule or complex ion.

Example: What is the oxidation number of S in (a) H2SO4  (b) H2S2O7 and (c) Na2S2O3?

a) Let the oxidation number of S be x.
Sum of oxidation numbers of various atoms in H2SO4
= 2× (+1) + x + 4× (-2)
= 2+x-8 = x-6
This sum must be zero (rule 6). Hence
X – 6 = 0
x = 6
Or Oxidation number of S in H2SO4 = +6
b) Sum of oxidation numbers of various atoms in H2S2O7
= 2 × (+1) + 2x + 7 × (-2)
= 2 + 2x – 14 = 2x – 2
Putting 2x – 12 = 0 as above, we have
 X = +6.
 Oxidation number of S in H2S2O7 = +6
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  1. Anonymous says:

    thanks for help me

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