1. REDOX

Oxidation and reduction can be
defined in terms of:

WHAT IS REDOX REACTION?

A redox reaction is a reaction in which
oxidation and reduction occur at the same
time.
 The oxidizing agent is the material that’s
reduced
 The reducing agent is the material that’s
oxidized

Oxidation (gain of oxygen)

2Mg (s) + CO2(g)  2Mg O (s) + C(s)
Reduction
(loss of oxygen)

Reduction (loss of oxygen)

CuO (s) + H2(g)  Cu (s) + H2O (l)

Oxidation
(gain of oxygen)

1. Oxidation and reduction in

term of loss or gain oxygen

Oxidation Reduction

Is a chemical Is a chemical
reaction reaction in which

in which a a substance loses
substance oxygen
gains oxygen

FeO is reduced (loss of oxygen, O)

2FeO(s) + C(s)  2Fe (s) + CO2 (g)
(FeO : oxidizing agent)

(C : reducing agent)

C is oxidized (gain of
oxygen,O)

2. Oxidation and reduction in

term of loss or gain hydrogen

Oxidation Reduction

Is a chemical Is a chemical
reaction reaction in which

in which a a substance
substance gain hydrogen

loses hydrogen

NH3 Oxidation (loss of hydrogen, H)
2NH3(g) + 3Br2(g)  N2 (g) + 6HBr (g)

Br2 : Reduction (gain of hydrogen, H)

3. Oxidation and reduction in
term of transfer of electron

Oxidation Reduction

Is a chemical Is a chemical
reaction reaction in which

in which a a substance
substance accepts electron.

donates electron.

Oxidation and reduction in terms of
electron transfer

The combustion of Zinc, Zn to form
Zinc oxide, ZnO

Zn : Oxidation (loss electrons)

2Zn(s) + O2(g) 2ZnO(s)

O2 : Reduction (gain of electrons)

Zn loses electrons to form zinc ion,
Zn2+ in zinc oxide, ZnO
Zn(s) Zn2+ (s) + 2e

Oxygen, O gains the electrons to form
oxide ions, O in zinc oxide, ZnO

O2 + 4e 2O2- (s)

Oxidation is a loss of electron and

reduction is a gain of electrons.

Oxidizing Agent = ?

Reducing Agent = ?

4. Oxidation and reduction in term
of change in oxidation number

Oxidation Reduction

Is a chemical Is a chemical
reaction reaction in which

in which a a substance
substance decreases its
increases its
oxidation oxidation
number. number.

Oxidation Number

Oxidation number or oxidation
state of an element is the

imaginary charge of an atom if
it exists as an ion.

• The oxidation number of an atom in
its elemental state is zero.

• For example, the oxidation number of
each atom in Mg, Cu, Na, H2, O2, Cl2 and
P4 is zero.

• The oxidation number of a monoatomic
ion is equal to its charge.

Ion Na+ Mg2+ Cl- N3-
-1 -3
Oxidation +1 +2

number

• The oxidation number of hydrogen in a
compound is always +1 except in metal
hydrides, where it is -1.

• For example, the oxidation number of H in
H2O and NH3 is +1. However, the
oxidation number of H in sodium hydride,
NaH is -1.

• The oxidation number of oxygen in a
compound is always -2 except in
peroxides.

• For example, the oxidation number of O in
H2O and MgO is -2.

• However, the oxidation number of O in
hydrogen peroxide, H2O2 is -1

• The oxidation number of
fluorine in all its compounds is
-1.

• The oxidation number of other
halogens (chlorine, bromine
and iodine) in their compounds
is -1 except when they
combine with more
electronegative elements such
as oxygen or nitrogen

• The sum of the oxidation numbers of all
the elements in the formula of a
compound must be zero.

• The sum of the oxidation numbers of all
the elements in the formula of a
polyatomic ion must be equal to the
charge of the ion.

Calculating the oxidation number of
elements.

i. Write down the fixed oxidation number of
elements (refer to the rules // page 118 text
book table 3.1)

ii. Multiply each oxidation number by the
subscript of the element in the chemical
formula.

iii. Write the mathematical equation for the sum of
oxidation numbers.

iv. Solve the mathematical equation to determine
the unknown oxidation number.

Example 1
Calculate the oxidation number of Mn
in potassium manganate(VII), KMnO4

Solution K Mn O

Rules (I) +1 X -2

Rules (II) 1(+1) 1 (x) 4(-2)

Rules (III) 1(+1)+1(x) + 4(-2) = 0
+1 + x – 8 = 0
X = +8 – 1

=+7

Rules (IV) The oxidation number of Mn is +7.

Example 2

Calculate the oxidation number of chromium, Cr

in the chromate (VI) ion, CrO42-

Solution Cr O

Rules (I) x -2
Rules (II) x 4(-2)
Rules (III) X + 4(-2) = -2
X – 8 = -2
Rules (IV) X = -2 + 8
= +6

The oxidation number of Cr is +6
Try to solve the task 3.2 // page 120 in your text book.

Oxidation and reduction
in terms of oxidation number

Zn : Oxidation (increase in oxidation number)

0 +2

Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)

+2 0

Cu2+ : Reduction (decrease in oxidation number)

Oxidation involves an increase in oxidation number.
Reduction involves decrease in oxidation number.

Reducing Agent : ?
Oxidizing Agent : ?

Summarizing

OXIDATION REDUCTION

+ oxygen - oxygen

- hydrogen + hydrogen

- electron + electron

+ oxidation number - Oxidation number

Reduction Agent Oxidation Agent

REDOX

Naming compound using the IUPAC nomenclature.

Zinc Potassium Copper(II)
chloride, manganate(VII), sulphate,
CuSO4
ZnCl2 KMnO4

How these compounds get their names?
Do you realize it?