1. OXIDATION AND REDUCTION
REDOX REACTIONS
Definition:
Oxidation and reduction in terms of:
i. loss or gain of oxygen
ii. loss or gain of hydrogen
iii. transfer of electrons
iv. changes in oxidation number
Oxidation:
- combination of substance with oxygen
- loss of hydrogen
- loss of electrons
- an increase in oxidation number
Reduction:
- removal of oxygen from substance
- gain of hydrogen
- gain of electrons
- a decrease in oxidation number
Chemical reactions involving
OXIDATION and REDUCTION
occurring simultaneously

2. Oxidant / Oxidizing agent:
 The substance that causes oxidation
Reductant / Reducing agent:
 The substance that causes reduction
Important
Neutralization and Precipitation are
NOT redox reactions
WHY? You tell me
[wait until we discuss the ‘oxidation number’]

3. A. Loss or gains of oxygen
Oxidation:
- combination of substance with oxygen
Reduction:
- removal of oxygen from substance
Example:
2CuO + C  2Cu + CO2
CuO loses its oxygen to form copper. Carbon gains the
oxygen to form carbon dioxide.
CuO causes the oxidation of carbon. Carbon causes the
reduction of CuO.
Material undergoes oxidation: Carbon, C
Material undergoes reduction: Copper(II) oxide, CuO
Oxidizing agent / oxidant : Copper(II) oxide, CuO
Reducing agent / reductant : Carbon, C

4. B. Loss or gains of hydrogen
Oxidation:
- loss of hydrogen
Reduction:
- gain of hydrogen
Example:
H2S + Cl2  S + 2HCl
H2S loses its hydrogen to form sulfur. Cl2 gains the
hydrogen to form HCl.
H2S causes the reduction of Cl2. Cl2 causes the oxidation
of H2S.
Material undergoes oxidation : Hydrogen sulphide, H2S
Material undergoes reduction : Chlorine, Cl2
Oxidizing agent / oxidant : Chlorine, Cl2
Reducing agent / reductant : Hydrogen sulphide, H2S

5. C. Tranfer of electrons
Oxidation:
- loss of electrons
Reduction:
- gain of electrons
Example 1: [Daniell cell]
Zn + Cu2+
 Zn2+
+ Cu
Electrons transfer from zinc to copper(II) ions.
Half reaction:
One zinc atom loses 2 electrons to form one zinc ion.
Zinc is oxidized to zinc ions.
Oxidizing half-equation: Zn  Zn2+
+ 2e-
Half reaction:
One copper(II) ion gains 2 electron to form one copper
atom. Copper(II) ion is reduced to copper.
Reduction half-equation: Cu2+
+ 2e  Cu

6. Copper(II) ions act as oxidizing agent because it
accepts electrons.
Zinc acts as reducing agent because it releases
electrons.
Oxidizing half-equation: Zn  Zn2+
+ 2e-
Reduction half-equation: Cu2+
+ 2e-
 Cu
[balanced the number of electron please]
Zn + Cu2+
+ 2e-
 Cu + Zn2+
+ 2e-
Thus;
Ionic Equation : Zn + Cu2+
 Zn2+
+ Cu
Chemical equation:
Zn + CuSO4  ZnSO4 + Cu
[note: the sum of the two half-equations gives the
ionic equation]
Cancel the electrons

7. D. Changes in oxidation number
Oxidation number?
Definition:
The oxidation number of an element is the
charge that the atom of the element would
have if complete transfer of electron occurs
Rules: pg 107 figure 3.1 (read)
Tips from the rules:
• the oxidation number for atom and molecule is
zero
• the oxidation number for monoatomic ion is
equal to its charge.
• the sum of oxidation numbers of all elements in
the compound is zero
• the sum of oxidation number of all elements in
polyatomic ions is equal to the charge of the ions
Oxidation:
- an increase in oxidation number
Reduction:

8. - a decrease in oxidation number
Example: [ I use the same example but from different perspective]
Zn + Cu2+
 Zn2+
+ Cu
[easy way to detect which substance is oxidized or reduced]
The oxidation number of zinc, Zn increases from
0 to +2. Zn undergoes oxidation to zinc ions, Zn2+
.
The oxidation number of copper(II) ions, Cu2+
decreases
from +2 to 0. Cu2+
undergoes reduction to copper, Cu.
Copper(II) ions, Cu2+
act as oxidizing agent.
Zinc, Zn acts as reducing agent.
Oxidizing half-equation: Zn  Zn2+
+ 2e-
Reduction half-equation: Cu2+
+ 2e-
 Cu
Ionic Equation : Zn + Cu2+
 Zn2+
+ Cu
[note: every redox reaction MUST have half equations
+2
0
Oxidation number
Zn
Zn2+
Cu
Cu2+
Oxidation
(loses electrons)
Reduction
(gains electrons)

9. and ionic equation]
Example:
2Mg + O2  2MgO
Describe the process……..
The oxidation number of magnesium, Mg increases
from 0 to +2. Mg undergoes oxidation to magnesium
ions, Mg2+
.
Oxidizing half-equation: Mg  Mg2+
+ 2e-
1 magnesium atom loses 2 electrons to from 1
mzgnesium ions. Mg undergoes oxidation to
magnesium ions, Mg2+
.
The oxidation number of oxygen, O2 decreases from
0 to -2. O2 undergoes reduction to oxide ions, O2-
.
+2
0
Oxidation number
Mg
Mg2+
O2-
O2
Oxidation (loses electrons)
Reduction (gains electrons)
-2

10. Reduction half-equation: O2 + 4e-
 2O2-
1 molecule of oxygen gains 4 electron to form 2 oxide
ions. O2 undergoes reduction to oxide ions, O2-
.
Oxygen, O2 act as oxidizing agent.
Magnesium, Mg acts as reducing agent.
Oxidizing half-equation: Mg  Mg2+
+ 2e-
(×2)
[it becomes: 2Mg  2Mg2+
+ 4e]
Reduction half-equation: O2 + 4e-
 2O2-
[balanced the number of electrons for both half equation]
2Mg + O2 + 4e-
 2Mg2+
+ 2O2-
+ 4e-
Thus;
Ionic Equation:
2Mg + O2  2Mg2+
+ 2O2-
or
2Mg + O2  2MgO
HW: pg 110 Learning task 3.1 Analyzing
Cancel the electrons

11. [a and b] Explain each of the reaction
Solution:
(a) (i) 2H2 + O2  2H2O [rule: pg 107]
The oxidation number of hydrogen, H2 increases from
0 to +1. H2 undergoes oxidation to hydrogen ion, H+
.
Oxidizing half-equation: H2  2H+
+ 2e-
1 molecule of hydrogen loses 2 electrons to form
2 hydrogen ions.
The oxidation number of oxygen, O2 decreases from
0 to -2. O2 undergoes reduction to oxide ions, O2-
.
Reduction half-equation: O2 + 4e-
 2O2-
1 molecule of oxygen gains 4 electrons to form
+1
0
Oxidation number
H2
H+
O2-
O2
Oxidation (loses electrons)
Reduction (gains electrons)
-2

12. 2 oxide ions.
Oxygen, O2 act as oxidizing agent.
Hydrogen, H2 acts as reducing agent.
Oxidizing half-equation: H2  2H+
+ 2e-
(×2)
Reduction half-equation: O2 + 4e-
 2O2-
[balanced the number of electrons for both half equation]
2H2 + O2 + 4e-
 4H+
+ 2O2-
+ 4e-
Thus;
The ionic equation
2H2 + O2  4H+
+ 2O2-
or
2H2 + O2  2H2O
Easy lah!
Prepared by;
Kamal Ariffin Bin Saaim
SMKDBL
Cancel the electrons