3.1 Analysing redox reactions

Oxidation and reduction can be
defined in terms of:

Loss or gain of
oxygen

Loss or gain of
hydrogen

Transfer of electrons

Change in oxidation
number

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 a
substance loses
in which a substance
oxygen
gains oxygen

Reduced (loss of oxygen, O)

2FeO(s) + C(s)  2Fe (s) + CO2 (g)

(oxidizing agent) (reducing

agent)

oxidized (gain of oxygen,O)

• FeO is reduced as it loses oxygen, O.
This means oxidizing agent are reduced
in redox reaction.

• C is oxidized as it gains oxygen, O.
Therefore, reducing agents are oxidized
in redox reaction.

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

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

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

Oxidation (loss electrons) Zn2+

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

1 hydrides, where it is -1.

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

2 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 The sum of the oxidation
numbers of all the numbers of all the

elements in the formula elements in the formula
of a compound must be of a polyatomic ion
must be equal to the
zero. 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

Oxidation (increase in oxidation number)

0 +2

Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)

+2 0

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?

Closure

• Match the terms with their correct
definition.

• Determine the oxidation, reduction, or
redox from the equations.

Example
A1(s) + CuSO4(aq)  A12(SO4)3(aq) + Cu(s)

In this redox reaction, the Aluminium atom A1 loses

electrons to form Aluminium ions, Al3+.

Aluminium atom are oxidized to form Alminium ion A13+, the
half-reaction for oxidation is represented by the half equation,

A1(s) A13+(aq) + 3e

At the same time, the copper(II) ion, Cu2+ gain electrons to
form copper atom, Cu.

The half-reaction for reduction is represented by the half
equation,

Cu2+(aq) + 2e  Cu(s)

The ionic equation of the redox reaction is obtained by adding
the two half equations.

2x[A1  A13+ + 3e] …… Half-Equation 1
3x [Cu2+ + 2e  Cu] …… Half-Equation 2

-------------------------------------------------------------
2A1 + 3Cu2+  2A13+ + 3Cu … Ionic

equation

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In this reaction, Aluminium is the

reducing agent the oxidizing agent

is copper (II) sulphate.