chemistry form 5

1. CHAPTER 1
REDOX
EQUILIBRIUM

2. what WILL YOU LEARN?

3. 1.1 OXIDATION
& REDUCTION
PART 1

4. Why do apple slices turn brown after being cut?
Apples are rich in iron.
When you cut an apple into slices, its
cells are damaged.
The damaged cells exposed to the air
and allows the oxygen to react with the
iron and an enzyme called polyphenol
causing iron oxide to form .
Essentially, the apple rusts!

5. Tips for Preventing Apples From
Browning
1.Slice the fruit in water.
2.Brush or dip sliced apples in lemon juice.
3.Soak cut fruit in ginger ale.
4.Soak the slices in salt water
5.Sprinkle with ascorbic acid powder.
6.Wrap a rubber band around a sliced
apple put back together.
How do you keep cut apples from turning brown?

6. The definition of redox
reaction
A redox reaction is a chemical reaction in
where oxidation and reduction occur
simultaneously.

7. Oxidation and reduction can
be defined in terms of:
1. Loss or gain of oxygen, O
2. Loss or gain of hydrogen, H
3. Transfer of electrons
4. Change in oxidation number

8. Oxidation is a process that
involves:
1. Gain of oxygen, O
2. Loss of hydrogen, H
3. Loss of electrons
4. Increases of oxidation number

9. Reduction is a process that
involves:
1. Loss of oxygen, O
2. Gain of hydrogen, H
3. Gain of electrons
4. Decreases of oxidation number

10. 1. Oxidation and reduction in
term of loss or gain oxygen
Oxidation
Is a chemical reaction
in which a substance
gains oxygen
Reduction
Is a chemical reaction
in which a
substance loses
oxygen

11. 2Mg (s) + CO2(g)  2Mg O (s) + C(s)
Oxidation (gain of oxygen)
Reduction (Loss of oxygen)

12. Reduction (loss of oxygen)
CuO (s) + H2(g)  Cu (s) + H2O (l)
Oxidation
(gain of oxygen)

13. OXIDISING AGENT AND
REDUCING AGENT
• The oxidizing agent is the material that’s reduced
or undergoes reduction.
• The reducing agent is the material that’s oxidized
or undergo oxidation.

14. Reduction (loss of oxygen, O)
2FeO(s) + C(s)  2Fe (s) + CO2 (g)
(oxidizing agent) (reducing agent)
oxidation (gain of oxygen,O)

15. • 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.

16. ACTIVITY 1 A

19. 2. Oxidation and reduction in
term of loss or gain hydrogen
Oxidation
Is a chemical reaction
in which a substance
loses hydrogen
Reduction
Is a chemical
reaction in which a
substance gain
hydrogen

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

24. 3. Oxidation and reduction in
term of transfer of electron
Oxidation
Is a chemical reaction
in which a substance
loses electron.
Reduction
Is a chemical reaction
in which a
substance gain
electron.

26. • Since the discovery of electrons by J.J
Thomson in 1897, chemists have started to
use the concept of electron transfer in
explaining redox reactions.
• Primarily, redox reactions can be explained
through the concept of electron transfer.

27. • Transfer of electrons in a redox reaction can
be elaborated in the following half equations:

28. Oxidation and reduction in terms of
electron transfer
The combustion of Zink, Zn to form
Zink oxide, ZnO
Oxidation (loss electrons) Zn2+
2Zn(s) + O2(g) 2ZnO(s)
O2-
Reduction (gain of electrons)

29. 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 = ?

30. Transfer of electrons at a
distance
When the reducing and oxidizing agents are separated by
an electrolyte in a U-tube as shown in Figure below, the
transfer of electrons occurs through the connecting wires
and can be detected by a galvanometer.

31. Transfer of electrons at a
distance
Negative terminal : The electrode at which electrons are released by
the reducing agent.
Positive terminal : The electrode at which electrons are accepted by
the oxidizing agent.

32. Transfer of electrons at a
distance
4) The reducing agent loses its electrons and hence
undergoes oxidation. The electrode at which electrons
are released by the reducing agent is called the
negative terminal.
5) The electrons then flow through the connecting wires to
the oxidizing agent. The electrode at which electrons
are accepted by the oxidizing agent is called the
positive terminal.
6) As the oxidizing agent accepts the electrons, it
undergoes reduction.
7) The electrolyte allows the movement of ions to take
place, thus completing the electric circuit. This ensures
a continuous flow of electrons in the external circuit.

33. EXPERIMENT

34. EXPERIMENT
PROCEDURE
1. A U-tube is clamped to a retort stand.
2. Dilute sulphuric acid is poured into the U-tube until its levels are 6 cm away
from the mouths of the U-tube.
3. Using a dropper, 0.5 mol dm iron(Il) sulphate solution is carefully added to
one of the arms of the U-tube until the layer of iron(Il) sulphate solution
reaches the height of 3 cm.
4. In a similar manner as in step 3, 0.2 mol dm-3 acidified potassium
manganate(Vll) solution is added to the other arm of the U-tube.
5. The electrodes are connected to a galvanometer as shown in Figure 3.7.

35. EXPERIMENT
PROCEDURE
6. Based on the deflection of the galvanometer, the electrodes that act as the
positive terminal and negative terminal are determined.
7. The set-up is left aside for 30 minutes. Any change is observed.
8. Using a clean dropper, 1 cm3 of iron(ll) sulphate solution is drawn out and
placed in a test tube. Then, a few drops of 0.2 mol dm-3 potassium
thiocyanate solution are added to the test tube. Any change is observed.
9. Steps 1 to 7 are repeated using 0.5 mol dm3 potassium iodide solution and
0.2 mol dm-3 acidified potassium dichromate(VI) solution to replace the
iron(lI) sulphate solution and acidified potassium manganate(Vll) solution.
Step 8 is repeated to test the potassium iodide solution with 1% starch
solution.

36. D. Transfer of electrons at a
distance

37. EXPERIMENT
OBSERVATION / RESULT
1. Solutions used: lron(ll) sulphate solution and acidified potassium
manganate(Vll) solution
Observation Inference
(a) The electrode in the iron(lI) sulphate solution acts
as the negative terminal while the electrode in the
acidified potassium manganate(VlI) solution acts
as the positive terminal.
Electrons flow from iron(ll) sulphate solution to
acidified potassium manganate(VlI) solution.
(b) lron(ll) sulphate solution changes from pale green
to yellow. It gives blood-red colouration with
potassium thiocyanate solution.
At the end of the reaction. Iron(lll) ions are
present .
lron(ll) ions have changed to iron(lll) ions.
(c) The purple acidified potassium manganate(Vll)
solution decolourises.
Manganate(VlI) ions that give the solution its
purple colour are used up in the reaction.

38. EXPERIMENT
OBSERVATION / RESULT
2. Solutions used: Potassium iodide solution and acidified potassium
dichromate (VI) solution
Observation Inference
(a) The electrode in the potassium iodide solution acts
as the negative terminal, whereas the electrode in the
acidified potassium dichromate(VI) solution acts as the
positive terminal.
Electrons flow from potassium iodide solution to
acidified potassium dichromate(VI) solution.
(b) The colourless potassium iodide solution turns
brown. It gives a dark blue colouration with starch
solution.
At the end of the reaction, iodine is present.
Iodide ions have changed to iodine.
(c) Potassium dichromate(Vl) solution changes colour
from orange to green.
Dichromate(Vl) ions have changed to chromium
(lll) ions.

39. EXPERIMENT
OBSERVATION / RESULT
2. Solutions used: Potassium iodide solution and acidified potassium dichromate (VI)
solution
3. Oxidation : Half Equation
Iodide ion iodine molecules
2I- (aq)  I2 (aq) + 2e-
(colourless) (brown)
4. Reduction : Half Equation
Dichromate(Vl) ion Chromium(lll) ion
Cr2O7
2- (aq) + 14H+(aq) + 6e-  2Cr3+ (aq) + 7H2O (l)
(green) (orange)
Overall ionic equation:
Cr2O7
2- (aq) + 14H+(aq) + 6I- (aq)  2Cr3+ (aq) + 3I2 (aq) + 7H2O (l)

40. Laboratory experiment

41. Laboratory experiment

43. Overall ionic equation

44. Examples of oxidising agents

45. Examples of reducing agents

47. QUESTIONS 1
C

48. QUESTION 2
D

49. 4. Oxidation and reduction in
term of change in oxidation
number
Oxidation
*
Is a chemical
reaction
in which a substance
increases its
oxidation number.
Reduction
*
Is a chemical
reaction in which
a substance
decreases its
oxidation number.

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

51. Exercise

52. Answer

53. Determining of oxidation number

54. What is Oxidation Number?
 The oxidation number or oxidation state
is the charge of the elements in a
compound if the transfer of electrons
occurs in an atom to form chemical
bonds with other atoms.

58. H in compound H in metal hydrides
+1 -1
Example : H2O and NH3 Example : sodium hydride,
NaH

59. Oxygen

61. Exercise 2

62. The Oxidation Number and Naming of
Compounds According
to the IUPAC Nomenclature

63. Naming compound using the IUPAC nomenclature.
Copper(II)
sulphate,
CuSO4
Potassium
manganate(VII),
KMnO4
Zinc
chloride,
ZnCl2
How these compounds get their names?
Do you realize it?

64. According to the IUPAC nomenclature, Roman numerals
are used to indicate the oxidation number of the metals in
their compounds.
Fe(NO3)2 Iron (II) nitrate
Oxidation nu. = +2
Fe(NO3)3 Iron (III) nitrate
Oxidation nu. = +3

65. Mg(NO3)2 Magnesium nitrate
Oxidation nu. = +2

66. Table 1.4 shows the naming of compounds containing metals
that have more than one oxidation
number according to the IUPAC nomenclature.

67. Calculating the oxidation number of
elements.
i. Write down the fixed oxidation number of
elements
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.

68. 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.

69. Example 2
Calculate the oxidation number of chromium, Cr in
the chromate (VI) ion, CrO4
2-
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

70. PERLIS TRIAL 2020

71. Exercise 3

72. Exercises
• 1. Calculate the oxidation number of
underlined element.
• a) KClO3
• b) Na2S2O3
• c) P2O7
4-

73. Exercises
• 2. Identify the name according to the
IUPAC naming system for each of the
following compounds.
• a) CuO
• b) Cu2O
• c) MnO2
• d) Mn2O7
• e) K2Cr2O7
• f) KMnO4

74. Oxidation and reduction
in terms of oxidation number
Reduction (decrease in oxidation number)
Oxidation involves an increase in oxidation number.
Reduction involves decrease in oxidation number.
Reducing Agent : ?
Oxidizing Agent : ?
Oxidation (increase in oxidation number)
0 +2
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
+2 0

75. Summarizing
OXIDATION REDUCTION
+ oxygen - oxygen
- hydrogen + hydrogen
- electron + electron
+ oxidation number - Oxidation number
Reducing Agent Oxidating Agent
REDOX

76. THANK YOU !