In classical terms, oxidation entails adding oxygen or an electronegative element, or removing hydrogen or an electropositive element. In the electronic concept, oxidation is described as the process where an atom or ion loses electrons.
What is Reduction?
Reduction encompasses adding hydrogen or an electropositive element, or removing oxygen or an electronegative element. According to the electronic concept, reduction is the process wherein an atom or ion gains electrons.
Classical Idea of Oxidation and Reduction reactions:
Oxidation reactions involve:
1. Addition of oxygen:
C + O2 → CO2 (oxidation of carbon)
2. Addition of electronegative element:
Fe + S → FeS (oxidation of Iron)
3. Removal of hydrogen:
H2S + Br2 → 2 HBr + S (oxidation of sulphide)
4. Removal of electropositive elements:
2 KI + H2O2 → I2 + 2 KOH (oxidation of iodide)
Oxidising agent is a substance which brings about oxidation. In the above examples O2, S, Cl2, Br2, and H2O2 are oxidising agents.
Reduction reactions involve:
1. Addition of hydrogen:
N2 + 3 H2 → 2NH3 ( reduction of nitrogen)
2. Addition of electropositive element:
SnCl2 + 2HgCl2 → SnCl4 + Hg2Cl2 ( reduction of mercuric chloride)
3. Removal of oxygen
ZnO + C → Zn + CO (reduction of zinc oxide)
4. Removal of electronegative element
2FeCl3 + H2 → 2FeCl2 + 2HCl (reduction of ferric chloride)
Reducing agent is a substance which brings about reduction. In the above examples H2, HgCl2 and C are Reducing agents.
Calculation Of Oxidation Number:
The total oxidation number of every compound is equal to Zero. Then for ions the Oxidation number is equal to the charge and same with radicals.
For example the Oxidation State of Sodium as natural occuring sodium is 0, Na = 0.
Then for Sodium ion is +1, Na+.
Examples:
1) Find the Oxidation Number of the following;
a. Na in NaCl
b. S in Na2SO4
c. K in KCl
d. S in H2SO4
Solutions:
a. NaCl
Let's call the Oxidation number of Na = x
Cl = -1
x + (-1) = 0
x-1=0
x= +1
The Oxidation number of Na in NaCl is +1
b. Na2SO4
Let's call the Oxidation number of S = x
Na =+1
O = -2
(+1)2+x+(-2)4=0
2+x-8=0
x=8-2
x=+6
The Oxidation Number of S in Na2SO4 is +6
For questions c and d solve and send response here.