This document provides information on calculating oxidation numbers and discusses redox reactions. It contains: 1) Five rules for determining oxidation numbers of elements in compounds and ions. These rules state that oxidation numbers of elements in their elemental forms are zero, ions take the charge of the ion, oxygen is -2 except in peroxides, hydrogen is +1 except in hydrides, and the sum of oxidation numbers equals the net charge. 2) Examples of using the rules to calculate oxidation numbers in various compounds and ions, including transition metal compounds that use Roman numerals. 3) A statement that oxidation is an increase in oxidation number and reduction is a decrease in oxidation number during a redox reaction.

1. HELLO!!
WELCOME TO
CHEMISTRY
CLASS
HOW TO CALCULATE
OXIDATION NUMBER??

2. -1
Redox Reactions
Rules in determining the oxidation number of an element
Element Oxidation number
Ions Oxidation number
Compound Oxidation number of oxygen
Mg 0
Na+ +1
HNO3
Zn
O2
Cl2
N2
0
0
0
0
Ca2+
Al3+
Br-
S2-
+2
+3
-1
-2
H2SO4
CO2
K2Cr2O7
H2O2
-2
-2
-2
-2
Rule 1
The oxidation number of atoms in
elements is zero.
Rule 2
The oxidation number of a
monoatomic ion is the charge
carried by the ion.
Rule 3
The oxidation number of oxygen
in its compound is -2 except for
peroxide where its oxidation
number is -1.
Chapter 3 Oxidation And Reduction
Chemistry Form 5

3. Redox Reactions
Compound Oxidation number of hydrogen
Species Net charge
HNO3 +1
HNO3
H2SO4
HClO3
CH4
NaH
+1
+1
+1
-1
CO2
ClO-
3
SO2-
4
0
0
-1
-2
Rules in determining the oxidation number of an element
Rule 4
The oxidation number of
hydrogen in its compound is +1
except for hydrides where its
oxidation number is -1.
Rule 5
The sum of oxidation numbers is
equal to the net charge of the
species.
Chapter 3 Oxidation And Reduction
Chemistry Form 5

4. Redox Reactions
Worked Example (1)
H2S2O7 Net charge = 0
Determine the oxidation number of the underlined element in H2S2O7
(Assuming the oxidation number of the element is p)
(Oxidation number of S) = p
(Oxidation number of O) = -2
2(+1) + 2p + 7(-2) = 0
+2 + 2p - 14 = 0
2p - 12 = 0
2p = +12
p = +6
(Oxidation number of H) = +1
Chapter 3 Oxidation And Reduction
Chemistry Form 5

5. i) HCO-
3
The net charge is -1
(+1) + p + 3(-2) = -1
+1 + p - 6 = -1
p – 5 = -1
p = +4
Redox Reactions
(a) CO2
The net charge is 0
p + 2(-2) = 0
p – 4 = 0
p = +4
(b) HNO3
The net charge is 0
(+1) + p + 3(-2) = 0
+1 + p – 6 = 0
p – 5 = 0
p = +5
(c) H2SO4
The net charge is 0
2(+1) + p + 4(-2) = 0
+2 + p – 8 = 0
p – 6 = 0
p = +6
(d) NH3
The net charge is 0
p + 3(+1) = 0
p = -3
(e) H2C2O4
The net charge is 0
2(+1) + 2p + 4(-2) = 0
+2 + 2p – 8 = 0
2p – 6 = 0
2p = +6
p = +3
(f) S2O8
2-
The net charge is -2
2p + 8(-2) = -2
2p – 16 = -2
2p = +14
p = +7
(g) PO4
3-
The net charge is -3
p + 4(-2) = -3
p – 8 = -3
p = -3 + 8
= +5
(h) NO-
2
The net charge is -1
p + 2(-2) = -1
p – 4 = -1
p = -1 + 4
= +3
Worked Examples - Continued
Chapter 3 Oxidation And Reduction
Chemistry Form 5

6. Redox Reactions
Transition elements have more than one oxidation number. We use Roman
Numeral to denote the oxidation number of the element in a compound.
(a) CuO
The net charge is 0
p + (-2) = 0
p – 2 = 0
p = +2
Copper(II) oxide
(b) Cu2O
The net charge is 0
2p + (-2) = 0
2p = +2
p = +1
Copper(I) oxide
(c) MnO
p + (-2) = 0
p = +2
Manganese(II) oxide
(d) Mn2O3
2p + 3(-2) = 0
2p - 6 = 0
2p = +6
p = +3
Manganese(III) oxide
(e) MnO2
p + 2(-2) = 0
p – 4 = 0
p = +4
Manganese(IV) oxide
(f) FeCl2
p + 2(-1) = 0
p – 2 = 0
p = +2
Iron(II) chloride
Worked Examples (2)
Chapter 3 Oxidation And Reduction
Chemistry Form 5

7. Redox Reactions
(g) FeCl3
p + 3(-1) = 0
p – 3 = 0
p = +3
Iron(III) chloride
(h) V2O5
2p + 5(-2) = 0
2p – 10 = 0
2p = +10
p = +5
Vanadium(V) oxide
(i)Cr2O7
2-
The net charge is -2.
2p + 7(-2) = -2
2p – 14 = -2
2p = +12
p = +6
Dichromate(VI) ion
(j) VO2
+
Net charge is +1
p + 2(-2) = +1
p – 4 = +1
p = +3
Vanadate(III) ion
(k) MnO4
2-
The net charge is -2
p + 4(-2) = -2
p – 8 = -2
p = +6
Manganate(VI) ion
(l) MnO4
-
The net charge is -1
p + 4(-2) = -1
p – 8 = -1
p = +7
Manganate(VII) ion
Worked Examples (2) - Continued
Chapter 3 Oxidation And Reduction
Chemistry Form 5

8. REDOX REACTION IN
TERMS OF CHANGE IN
OXIDATION NUMBER
● When the oxidation number
of an element increases, the
element undergoes
oxidation
● When the oxidation number
of an element decreases,
the element undergoes
reduction