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
