Chemistry redox reaction

1. CHAPTER 3CHAPTER 3
OxidationOxidation
andand
ReductionReduction

2. 2
(A) Redox Reactions, RR(A) Redox Reactions, RR
After this lesson, you should be able to:
• State what oxidation is
• State what reduction is
• Explain what redox reaction is
• State what oxidizing agent is
• State what reducing agent is
• Calculate the oxidation number of an element in a
compound
• Relate the oxidation number of an element to the name of its
compound using the IUPAC nomenclature
• Explain with examples oxidation and reduction processes in
terms of the change in oxidation number
• Explain with examples oxidation and reduction processes in
terms of electron transfer
• Explain with examples oxidizing agents and reducing agents
in redox reactions
• Write oxidation and reduction half-equations and ionic
equations.

3. 3
What is Redox Reaction?What is Redox Reaction?
• Redox reaction are chemical
reactions involving oxidation
and reduction occurring
simultaneously.

4. 4
Explanation ofExplanation of
Redox Reaction, RRRedox Reaction, RR
Redox reactions can be explained in term of:
– Loss or gain of oxygen
– Loss or gain of hydrogen
– Transfer of electrons
– Changes in oxidation number

5. 5
Explanation of RR based on
Loss or Gain of Oxygen
• Oxidation is a chemical reaction in which oxygen
is added to a substance
• Reduction is defined as the loss of oxygen from a
substance
• The substance that causes oxidation is called
oxidizing agent (oxidant)
• The substance that causes reduction is called the
reducing agent (reductant)

6. 6
2CuO(s) + C(s)  2Cu(s) + CO2(g)
• CuO is reduced to Cu
• C is oxidized to CO2
• CuO acts as oxidizing agent (oxidant)
• C acts as reducing agent (reductant)
Gains oxygen
(oxidation)
Loses oxygen
(reduction)
Example 1:

7. 7
Explanation of RR based on
Loss or Gain of Hydrogen
• Oxidation is the loss of hydrogen from a
substance
• Reduction is the gain of hydrogen from a
substance

8. 8
H2S(g) + Cl2 (g)  S(s) + 2HCl(g)
• H2S is oxidized to S
• Cl2 is reduced to HCl
• Cl2 acts as oxidizing agent (oxidant)
• H2S acts as reducing agent (reductant)
Gains hydrogen
(reduction)
Loses hydrogen
(oxidation)
Example 2:

9. 9
Practice A1:
Study the following equations and identify the
oxidized substances, reduced substances, oxidant
and reductant.
a)2HBr(aq) + Cl2(l)  2HCl(aq) + Br2(l)
b)Mg(s) + PbO(s)  MgO(s) + Pb(s)
c)CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
d)Fe3O4(s) + 4CO(g)  3Fe(s) + 4CO2(g)
e)PbS(s) + 4H2O2(aq)  PbSO4(s) + 4H2O(l)

10. 10
Explanation of RR based on
Transfer of Electrons
• Oxidation is the loss of electrons
• Reduction is the gain of electrons
• Oxidizing Agent is electron acceptors
• Reducing Agent is electron donors

11. 11
Zn(s) + Cu2+
(s)  Zn2+
(aq) + Cu(s)
Loss e- Zn(s)  Zn2+
(aq) + 2e-
Gain e- Cu2+
(aq) + 2e-  Cu(s)
-------------------------------------------------------------
Zn(s) + Cu2+
(s) + 2e-  Zn2+
(aq) + Cu(s) + 2e-
----------------------------------------------------------------------
Ionic Eq. Zn(s) + Cu2+
(s)  Zn2+
(aq) + Cu(s)
=========================================
Gains electron
(reduction)
Loses electron
(oxidation)
Example 3:

12. 12
Practice A2:
Study the following redox reactions
a) Cu(s) + 2Ag+
(aq)  Cu2+
(aq) + 2Ag(s)
b) Cl2(g) + 2Br−
(aq)  2Cl−
(aq) + Br2(l)
c) Ca(s) + 2HCl(aq)  CaCl2(aq) + H2(g)
d) 2Na(s) + Cl2(g)  2NaCl(s)
e) 2Fe2+
(aq) + Br2(aq)  2Fe3+
(aq) + 2Br−
(aq)
For each of the above reaction above,
i) Write the half equations
ii) Identify the
– Oxidized substance
– Reduced substance
– Oxidizing agent
– Reducing agent based on the transfer of electrons

13. 13
What is an Oxidation Number?
• The oxidation number or oxidation
state of an element is the charge that
the atom of the element would have if
complete transfer of electron occurs.

14. 14
Rules in assigning Oxidation Number
Rule 1:Rule 1:
• 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.

15. 15
Rules in assigning Oxidation
Number
Rule 2:Rule 2:
•The oxidation number of monoatomic ion is equal
to its charge
For example:
Ion Na+
Mg2+
Al3+
Br−
S2−
N3−
Oxidation
Number
+1 +2 +3 −1 −2 −3

16. 16
Rules in assigning Oxidation Number
Rule 3:Rule 3:
•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

17. 17
Rules in assigning Oxidation Number
Rule 4:Rule 4:
•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

18. 18
Rules in assigning Oxidation Number
Rule 5:Rule 5:
•The oxidation number of fluorine in all its
compound is −1.
•The oxidation number of other halogens
(Cl, Br, I) in their compounds is −1 except
when they combine with more electronegative
elements such as oxygen or nitrogen.

19. 19
Rules in assigning Oxidation Number
Rule 6:Rule 6:
•The sum of the oxidation numbers of all the
elements in the formula of a compoundformula of a compound must
be zero.
•The sum of the oxidation numbers of all the
elements in the formula of a polyatomic ionformula of a polyatomic ion
must be equal to the charge of the ion.

20. 20
Calculation of Oxidation Number, ON
Example 1:
Determine the oxidation number of nitrogen in NH3.
Assume that the oxidation number of nitrogen is X
The ON of H in NH3 is +1 (rule 3)
The sum of ON of all atoms = 0 (rule 6)
Thus , x + 3(+1) = 0
x + 3 = 0
x = −3
NH3
x +1

21. 21
Calculation of Oxidation Number, ON
Example 2:
Determine the oxidation number of copper in Cu2O
Assume that the oxidation number of copper is X
The ON of O in Cu2O is −2 (rule 4)
The sum of ON of all atoms = 0 (rule 6)
Thus , 2x + (−2) = 0
2x −2 = 0
x = +1
Cu2O
x −2

22. 22
Calculation of Oxidation Number, ON
Example 3:
Determine the oxidation number of sulphur in SO4
2−
Assume that the oxidation number of sulphur is X
The ON of O in SO4
2−
is −2 (rule 4)
The sum of ON of all atoms = −2 (rule 6)
Thus , x + 4(−2) = −2
x −8 = −2
x = +6
SO4
2−
x −2

23. 23
Calculation of Oxidation Number, ON
Example 4:
Determine the oxidation number of manganese in
MnO4
−
Assume that the oxidation number of manganese is X
The ON of O in MnO4
−
is −2 (rule 4)
The sum of ON of all atoms = −1 (rule 6)
Thus , x + 4(−2) = −1
x −8 = −1
x = +7
MnO4
−
x −2

24. 24
Practice A3:
Determine the oxidation number of the underlined
elements in the following compound
a) CO2 b)MgF2 c) H3PO4 d) V2O5
e) CO f) NH4
+
g) SO3h) ClO4
-
i) N2O j) H2O2 k) S2O3
2−
l) CrO4
2−
m) Cr2O7
2−
n) Al2O3 o) BrO3
−
p)VO2
q) PbO2
2−
r) NO3
−
s) NO2
−
t) CO3
2−
u) HCl v) HClO w) HClO2 x)ClO2
y) HClO3 z)HClO4

25. 25
Oxidation Number and IUPAC Nomenclature
Formula of Compound
Oxidation Number of
Underlined Metal
IUPAC name
FeCl2 +2 Iron(II) chloride
FeCl3 +3 Iron(III) chloride
CuCl +1 Copper(I) chloride
CuSO4 +2 Copper(II) sulphate
Mn(NO3)2 +2 Manganese(II) nitrate
MnO2 +4 Manganese(IV) oxide
K4Fe(CN)6 +2
Potassium
hexacyanoferrate (II)
K3Fe(CN)6 +3
Potassium
hexacyanoferrate (III)

26. 26
Common Names and IUPAC Names for some compound
Molecular
Formula
ON of
metal
Common Name IUPAC name
Na2SO3 +4 Sodium sulphite Sodium sulphate(IV)
Na2SO4 +6 Sodium sulphate Sodium sulphate(VI)
NaNO2 +3 Sodium nitrite Sodium nitrate(III)
NaNO3 +5 Sodium nitrate Sodium nitrate(V)
HNO2 +3 Nitrous acid Nitric(III) acid
HNO3 +5 Nitric acid Nitric (V) acid
H2SO4 +6 Sulphuric acid Sulphuric (VI) acid

27. 27
Explanation of RR based on
The Changes in Oxidation Number
• An increase in oxidation number
indicates Oxidation
• A decrease in oxidation number
indicates Reduction

28. 28
2Mg(s) + O2 (s)  2MgO(s)
Oxidation 0 0 +2 −2
Number
Decrease in
oxidation number
Increase in
oxidation number
Example 4:
0
+2
-2
Mg2+
MgO2
O2-
• Increase in
Oxidation
Number
• (Oxidation)• decrease in
Oxidation
Number
• (Reduction)

29. 29
Explanation
• The oxidation number of Magnesium
increases from 0 to +2.
• Magnesium undergoes oxidation to
magnesium ion
• The oxidation number of Oxygen decrease
from 0 to −2
• Oxygen undergoes reduction to oxide ion
• Magnesium acts as reducing agent
• Oxygen acts as oxidizing agent

30. 30
Practice A4
a) 2H2 + O2 2H2O
b) 2Na + Br2  2NaBr
c) Pb + 2Ag+
Pb2+
+ 2Ag
d) Zn + 2HCl  ZnCl2 + H2
Explain the above redox reactions based on the
changes in oxidation number. Your explanation
should includes:
i) oxidized and reduced substance in each
reaction. Give reason for your answer.
ii) oxidizing agent and reducing agent in each
reaction. State what happens to them

31. 31
Summary on the Definition of
Oxidation and Reduction
OxidationOxidation ReductionReduction
Gain of Oxygen Loss of Oxygen
Loss of Hydrogen Gain of Hydrogen
Loss of Electrons Gain of Electron
Increase in Oxidation
Number
Decrease in Oxidation
Number

32. 32
Examples of Redox Reaction
1. Combustion
2. Extraction of Metals
3. Corrosion of Metals
4. Electrochemistry ( Reaction happen in
Electrolytic Cell and Voltaic Cell)
5. Change of Fe2+
to Fe3+
and vice versa
6. Displacement of Metal from its salt solution
7. Displacement of halogen from its halide solution
8. Transfer of electrons at a distance

33. 33
Examples of Non Redox Reaction
• Neutralization
• Precipitation Reaction

34. 34
Change of Fe2+
to Fe3+
& vice versa
• Iron metal (Fe) exhibits two oxidation numbers,
i.e. +2 and +3
• Fe2+
ion can be easily converted into Fe3+
ion.
• Fe3+
ion can also be easily converted into Fe2+
ion.
Fe2+
Fe3+
Loss of
electron
Gain of
electron
oxidation
reduction

35. 35
Oxidation of Fe2+
to Fe3+
Procedure:
1. Pour 2cm3
of freshly prepared iron(II) sulphate,
FeSO4 solution into a test tube.
2. Using a dropper, add bromine water drop by
drop until no further changes are observed.
3. Warm the test tube gently
4. Add NaOH solution slowly into the test tube
until it excess.
5. Record the observation.

36. 36
Oxidation of Fe2+
to Fe3+
Solution Used Observations
FeSO4
+
Br2
Reddish brown bromine water was
decolourized.
•Green FeSO4 solution turn brown.
•When NaOH solution was added, a brown
precipitate formed.
•The precipitate is insoluble in excess
NaOH

37. 37
Half Equation:
Oxidation : Fe2+
(aq)  Fe3+
(aq) + e-
Reduction : Br2(aq) + 2e-  2Br−
(aq)
------------------------------------------------------------------------------------
Ionic Equation :
2Fe2+
(aq) + Br2(aq)  2Fe3+
(aq) + 2Br−
(aq)
===========================================================
Oxidation of Fe2+
to Fe3+

38. 38
Reduction of Fe3+
to Fe2+
Procedure:
1. Pour 2cm3
of iron(III) sulphate, Fe2(SO4)3
solution into a test tube.
2. Add half a spatula of zinc powder to the
solution. Shake the mixture until no further
changes are observed.
3. Filter the mixture.
4. Add NaOH solution slowly into the filtrate until
in excess.
5. Record the observation.

39. 39
Reduction of Fe3+
to Fe2+
Solution Used Observations
Fe2(SO4)3
+
Zn
Part of Zn powder dissolved.
Brown Fe2(SO4)3 solution turn green. When
NaOH solution was added to the filtrate, a
green precipitate was formed.
The precipitate is insoluble in water.

40. 40
Reduction of Fe3+
to Fe2+
Half Equation:
Oxidation : Zn(s)  Zn2+
(aq) + 2e-
Reduction : Fe3+
(aq) + e-  Fe2+
(aq)
--------------------------------------------------------------------------
Ionic Equation :
Zn(s) + 2Fe3+
(aq)  Zn2+
(aq) + 2Fe2+
(aq)
============================================

41. 41
Displacement of Metals
• A metal displacement reaction involves a metal
and the salt solution of another metal.
• Displacement Reaction took place if any of these
observation is obtained:
– a deposition of solid occurs at the bottom of the test
tube.
– a change in colour of the salt solution
– a decrease in the amount or size of the metal used
– Test tube becomes hotter

42. 42
Displacement of Metals
• A more electropositive
metal can displace a less
electropositive metal from
its aqueous salt solution.
• A less electropositive
metal cannot displace a
more electropositive metal
from its aqueous salt
solution

43. 43
Displacement of Metals
A more electropositive
metal is located at higher
position in the
electrochemical series, ES.
A less electropositive metal
is located at lower position
in the electrochemical
series, ES.
KK
NaNa
CaCa
MgMg
AlAl
ZnZn
FeFe
SnSn
PbPb
CuCu
HgHg
AgAg
More
electropositive
Less
electropositive

44. 44
Displacement of Metals
Examples:
Fe(s) + CuSO4(aq)  Cu(s) + FeSO4(aq)
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
Mg(s) + FeSO4(aq)  Fe(s) + MgSO4(aq)
Cu(s) + FeSO4(aq)  No reaction
Zn(s) + MgSO4(aq)  No reaction

45. 45
Displacement of Metals
Zinc displaces copper metal from copper(II) sulphate solution
Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq)
Half Equation: Zn(s)  Zn2+
(aq) + 2e- (Oxidation)
Cu2+
(aq) + 2e-  Cu(s) (Reduction)
Overall ionic equation:
Zn(s) + Cu2+
(aq)  Cu(s) + Zn2+
(aq)

46. 46
Displacement of Metals
Observation:
(a) Brown copper metal deposited
(b) The colour of the solution changes from
blue to colourless.
(c) The temperature of the mixture increases.
( all displacement reaction are exothermic)

47. 47
Displacement of Halogen
• A more reactive halogen can displace a less
reactive halogen from its aqueous halide salt
solution.
• A less reactive halogen cannot displace a more
reactive halogen from its aqueous halide salt
solution.
Cl, Br, I
Less reactiveMore reactive

48. 48
Colour of halogen
Halogen
Colour in
aqueous solution
Colour in
1,1,1-trichlorethane
Chlorine Pale yellow Colourless
Bromine Brown Brown
Iodine Brown Purple

49. 49
Displacement of halogen
Procedure:
1. 1cm3
of aqueous potassium iodide solution, 1cm3
of
bromine water and 1cm3
of 1,1,1-trichloroethane are added
into a test tube, labelled A. The mixture is shaken.
2. Step 1 is repeated by adding 1cm3
of aqueous potassium
bromide solution, 1cm3
of chlorine water and 1cm3
of
1,1,1-trichloroethane are added into a test tube, labelled B.
The mixture is shaken.

50. 50
Displacement of halogen
Result:
Test tube
Colour of
CH3CCl3
Inference
A Purple
Iodine
displaced
B Brown
Bromine
displaced

51. 51
Discussion
Test Tube A
Cl2(aq) + 2KBr(aq)  Br2(aq) + 2KCl(aq)
Cl2(aq) + 2Br−
(aq)  Br2(aq) + 2Cl−
(aq)
ON : 0 −1 0 −1
• Bromine, Br2 dissolves in CH3CCl3 to give a brown colour
• Chlorine, Cl2 is reduced. Reducing agent are the Br−
ions.
• Bromide ions are oxidized. Oxidizing agent is Cl2.

52. 52
Discussion
Test Tube B
Br2(aq) + 2KI(aq)  I2(aq) + 2KBr(aq)
Br2(aq) + 2I−
(aq)  I2(aq) + 2Br−
(aq)
ON : 0 −1 0 −1
• Iodine, I2 dissolves in CH3CCl3 to give a purple colour
• Bromine, Br2 is reduced. Reducing agent are the I−
ions.
• Iodide ions are oxidized. Oxidizing agent is Br2.

53. 53
Transfer of Electron at a Distance I
At electrode X
Iodide ions lose electron and are oxidized to brown iodine.
2I−
(aq)  I2(aq) + 2e-
ON: -1 0 (oxidation)
Dilute sulphuric acid

54. 54
The colour of the solution changes from colourless to brown.
The e- released by the iodide ion flow from electrode X to
electrode Y along the connecting wires.
At electrode Y
The bromine molecules surrounding the electrode Y accept
the e- and are reduced to bromide ions.
Br2(aq) + 2e-  2Br−
(aq)
ON: 0 -1 (Reduction)
The colour of the solution changes from brown to colourless.
Overall ionic equation:
Br2(aq) + 2I−
(aq)  2Br−
(aq) + I2(aq)
oxidant reductant

55. 55
Transfer of Electron at a Distance II
Dilute sulphuric acid
• At electrode X
• Each Iron(II) ion loses an electron and is oxidized to
brown iron(III) ion.
• Fe2+
(aq)  Fe3+
(aq) + e-
• ON: +2 +3 (oxidation)

56. 56
The colour of the solution changes from green to brown.
The e- released by the iron(II) ion flow from electrode X to
electrode Y along the connecting wires.
At electrode Y
The bromine molecules surrounding the electrode Y accept
the e- and are reduced to bromide ions.
Br2(aq) + 2e-  2Br−
(aq)
ON: 0 -1 (Reduction)
The colour of the solution changes from brown to colourless.
Overall ionic equation:
Br2(aq) + Fe2+
(aq)  2Br−
(aq) + Fe3+
(aq)
oxidant reductant

57. 57
Transfer of Electron at a Distance III
• At electrode X
• Each Iron(II) ion loses an electron and is oxidized to
brown iron(III) ion.
• Fe2+
(aq)  Fe3+
(aq) + e-
• ON: +2 +3 (oxidation)
XX YY

58. 58
The colour of the solution changes from green to yellow/
brown.
The e- released by the iron(II) ion flow from electrode X to
electrode Y along the connecting wires.
At electrode Y
The manganate(VII) ion, MnO4
−
gathered at the electrode Y
accept the e- and are reduced to manganese(II) ion, Mn2+
.
MnO4
−
(aq) + 8H+
(aq) + 5e-  Mn2+
(aq) + 4H2O(l)
ON: +7 +2 (Reduction)
The colour of the solution changes from purple to colourless.

59. 59
Overall ionic equation:
anode 5[Fe2+
(aq)  Fe3+
(aq) + e-]
cathode MnO4
−
(aq) + 8H+
(aq) + 5e-  Mn2+
(aq) + 4H2O(l)
………………………………………………………………………
5Fe2+
(aq) + MnO4
−
(aq) + 8H+
(aq)  5Fe3+
(aq)+ Mn2+
(aq) + 4H2O(l)
Substance oxidized : iron(II) ion, Fe2+
Substance reduced : manganese(VII) ion, MnO4
−
Oxidizing agent : manganese(VII) ion, MnO4
−
Reducing agent : iron(II) ion, Fe2+

60. 60
Practice A5
The figure above shows an experiment on the transfer of electron at a
distance
a) Identify the
i) oxidizing agent ii) reducing agent
iii) positive electrode iv) negative electrode
b) Explain the changes at the
i) negative electrode ii) positive electrode
A B
K2Cr2O7 (aq)
+
H2SO4(aq)
FeSO4(aq)
H2SO4(aq)