Oxidation & reduction

1. Mg
2+
O
2-
KMnO
4

2. Redox reactions are a chemical reactions involving
simultaneously (serentak) oxidation and reduction processes
Oxidation & reduction :
 a addition (gain) or loss (elimination) of oxygen or hydrogen
 accepting(receives) or donating of electrons
 change in oxidation number
A Redox Reactions :
 Oxidation is the process of gaining oxygen & elimination of
hydrogen
 reduction is the process of losing oxygen & addition of hydrogen
 oxidising agent is the substance which experiences reduction &
receives electrons .
 reducing agent is the substance which experiences oxidation &
donates electrons .

3. Mg + PbO MgO + Pb
Gain of oxygen ---- oxidation
Loss of oxygen ---- reduction
PbO – oxidising
agent (experiences
reduction
Mg –
reducing
agent
( experiences
oxidation)

4. H
2
S + Cl
2
2HCl + S
Loss of hydrogen --- oxidation
Gain of hydrogen --- reduction
Cl
2
– oxidising agent
( undergoes reduction )
--- chlorine oxidises
hydrogen sulphide to
sulphur
H
2
S – reducing agent
( undergoes oxidation)
--- hydrogen sulphide
reduces chlorine to
hydrogen chloride
Redox reaction

5. B Transfer of Electrons
 oxidation involves the loss (releases) of electrons
 reduction involves a gain (receives) in electrons
2Na(s) + Cl
2
(g) 2NaCl(s)
Na Na
+
+ e ( loss of electron)
Cl
2
+ 2e
-
2Cl
-
(gain of electron)
Oxidation Process
Reduction Process
Chlorine
–
oxidising
agent
Sodium –
reducing
agent
 metals are oxidised & its loss their electrons to form cations
 non- metal are reduced & its receive electrons to form anions .

7. Determine the oxidation and reduction process , oxidising and
reducing agent that occurs in the reactions below .
•Mg + PbO MgO + Pb
• Anode : Cu Cu
2+
+ 2e
-
; Cu
2+
+ 2e
-
Cu
• 2CuO + C 2Cu + CO
2
• Fe
2 O
3 + 3C 2Fe + 3CO
• Mg + CuO MgO + Cu
Solution :

8.  oxidation is the increase in oxidation number
 reduction is the decrease in the oxidation number
 O.N of ions is same value to the charge of the ion.
 Ex : Na
+
, K
+
, H
+
is +1
Mg
2+
, Ca
+2
is +2
O
2-
, S
2-
is -2
 O.N for atom or molecule in a neutral elements are zero ( 0 )
 example : O
2 , N
2 , Na , Mg, Br
2 is 0 .
2FeCl
2
+ Cl
2
2FeCl
3

Oxidation number decreases (0 → -1)
Oxidation number increases( +2 → +3)
reduction
oxidation
+2 +3-10-1
O.N
Reducing agent ---
iron (II) chloride
Oxidising agent
--- chlorine gas
C : Change in Oxidation Number ( O.N )

9.  (i) The total oxidation number of all the atoms is equal to the charge
on the ion .
 (ii) the total oxidation numbers for all atoms in neutral compound is
zero .

Example : (i) ClO
3
-
, oxidation number of chlorine is X
X + 3(-2) = -1
X - 6 = -1
X = +5
(ii) The oxidation number of S in MgSO
4
+2 + X + 4 (-2) = 0
+2 + X - 8 = 0
X = + 6
 the total oxidation number for dichromate (VI) ion, Cr
2O
7
2-
is -2

,
 manganate (VII) , MnO
4
-1
is -1
The charge of
chlorate
Oxidation
number of S

10. Test Yourself :
Calculate the oxidation number of the following elements :
•Manganese , Mn in potassium manganate (VII) , KMnO
4
•Manganese, Mn in manganate(VII) ion, MnO
4
-
•Chromium, Cr, in potassium dichromate(VI), K
2
Cr
2
O
7

•Cromium, Cr,in chromate(VI) ion, CrO
4
2-
•Iron in iron(II) chloride , FeCl
2
•Iron in iron(III) chloride , FeCl
3
•Carbon , C in sodium carbonate, Na
2
CO
3
In each of the cases above, the oxidation number of each element is
represented by the value of X .

11. The oxidising agent is the substance that
 receives electrons
 experiences a reduction(pengurangan) in the oxidation number .
The reducing agent is the substance that :
 loses electrons
 experiences an increase (penambahan) in the oxidation number .
Example :
• Fe
2+
Fe
3+
+ e
-
(O.N ) +2 +3
• Br
2
+ 2e
-
2Br
-
(O.N) 0 -1
Make sure that you add
the electron on the side
of the half equation that
has the bigger oxidation
number

12. Example :
Half equation :
Fe
2+
Fe
3+
+ e
-
-------- (1) X 2
Br
2
+ 2e
-
2Br
-
-------- (2)
Ionic equation :
2Fe
2+
+ Br
2
2Fe
3+
+ 2Br
-
Combined
Redox reactions need to shown in the form of :
 half - equations
 ionic equations
Fe
2+

Fe
3+
Cl
-
Cl
2
Br
-
I
-
Br
2
Green solution
Brown solution
Yellow solution
Colourless solution
MnO
4
-
Purple solution
Cr
2
O
7

2-
Cu
2+
Orange solution
Blue solution
Cr
3+

I
2

13. A . Redox Reaction ( The combustion of Magnesium in oxygen)
 Oxygen oxidises magnesium to magnesium ion .
 Magnesium releases electrons to form Magnesium ion .
 Half –equation : Mg Mg
2+
+ 2e
-
------------(1)
 O.Number 0 +2 (oxidation)
 Magnesium reduces oxygen to oxide ion .
 Oxygen atom receives electrons to form oxide ion .
 Half- equation : O
2 + 2e
-
O
2-
------------(2)
 O.Number 0 -2 (reduction )
 The overall equation : (1) + (2)
2Mg + O
2
2MgO
 Observation :-
 Combusts with a white shiny flame
 A white solid is formed
 Oxidising agent : Oxygen
 Reducing agent : Magnesium
combined
Reactants

14. B. The change of iron(II) ions to iron(III) ions ( Fe
2+
Fe
3+
)
 Bromine water oxidises iron(II) ion , Fe
2+
to iron(III) ion , Fe
3+
 Iron(II) ion releases electron to form iron(III) ion.
 Half-equation : Fe
2+
Fe
3+
+ e
-
------------(1)X 2
 O. Number +2 +3 ( oxidation)
 Iron(II) ion reduces bromine ,Br
2
to bromide ion , Br
-
 Bromine receives electron to form bromide ion , Br
-1

 Half-equation : Br
2
+ 2e
-
2Br
-
------------(2)
 O.Number 0 -1 ( reduction )
 Observation :
• The green iron(II) sulphate solution changes to brown .
• The brown coloured bromine water is decolourised .
• Oxidising agent : Bromine water , Br
2
• Reducing agent : Iron(II) ion, Fe
2+
•

Ionic Equation : 2 Fe
2+
+ Br
2
2Fe
3+
+ 2Br
-
• other oxidising agent : Cl
2 ,KMnO
4 ,K
2Cr
2O
7 ,HNO
3 concentrated,H
2O
2
Combined

15. C : The change of iron(III) ion, Fe
3+
to iron(II) ion , Fe
2+

 Iron (III) ion oxidises Zn atom to zinc ion , Zn
2+
 Iron(III) ion receives electron to form iron(II) ion
 The brown iron(III) chloride solution changes to green .
 Half-equation : Fe
3+
+ e
-
Fe
2+
--------(1) X 2
 O.Number +3 +2 ( Reduction)
 Iron(III) ion -------- oxidising agent ( oxidation number decrease) .
 Zn reduces iron(III) ion to iron(II) ion .
 Zinc atom releases electrons to form zinc ion , Zn
 Zinc powder dissolves .
 Zinc metal -------- reducing agent ( oxidation number increase)
 Half-equation : Zn Zn
2+
+ 2e
-
-------(2)
 O. Number 0 +2 ( oxidation )
 The ionic equation : 2Fe
3+
+ Zn 2Fe
2+
+ Zn
2+
 Other reducing agent : metals that are more electropositive than iron
// SO
2
, H
2
S gas // Na
2
SO
3
, SnCl
2
solution
combined

16. D : The Displacement (penyesaran) of Metal from its Salt Solution
 The element is more electropositive in the E.S, the higher the tendency
(kecenderungan) to release electrons to form positive ions .
 More electropositive , oxidised more easily & act as a reducing agent
 The higher the position in the E.S. can displace other elements that are
lower in the E.S .
The displacement reaction between Zn & CuSO
4
solution .
 Zn more electropositive than copper .
 Zn releases two electron to form zinc ion , Zn
2+
 Zn reduces copper(II) ion ,Cu
2+
to copper , Cu
 Copper(II) ion oxidises Zn to zinc ion ,Zn
2+
K, Na , Ca , Mg , Al , Zn , Fe , Sn , Pb , H , Cu , Hg , Ag , Au
Most electropositive Least electropositive

17.  Observation :
• The blue CuSO
4
solution fades or becomes colourless .
• A brown solid is formed .
• The Zn piece is corroded or dissolves .
• Copper is displaced by zinc from the copper(II) sulphate solution .
 Half-equations : Zn Zn
2+
+ 2e
-
----------(1)
 O . Number 0 +2 ( oxidation )
Cu
2+
+ 2e Cu ----------(2)
 O. Number +2 0 ( reduction )
 Ionic equation : Zn + Cu
2+
Zn
2+
+ Cu
 Zn ------ reducing agent
 Copper(II) ion ------- oxidising agent
Zn loses
electrons &
is oxidised
to Zn
2+
Cu
2+
receives
electrons &
is reduced to
Cu

18. E : Displacement of Halogens from Halide Solutions
 Halogen ----- Group 17
 examples : Cl
2
( chlorine) ------- yellow
Br
2
( bromine) ------- brown
I
2 (iodine) ------- yellow or brown
 can be differentiated by shaking the solution with a little CCl
4
 Halogens are reduced to halide ions
 Halogen ----- oxidising agent
 The more reactive halogen can displace less reactive halogens from
its halide solutions.
 Group 17 :
Flourine
Chlorine
Bromine
Iodine
Solution
Reactivity decreases,
higher act as a
oxidising agent

19. Chlorine water react with sodium bromide solution
 Chlorine water , Cl
2
oxidises bromide ion, Br
-
to bromine ,Br
2
 Bromide ion , reduces chlorine , Cl
2 to chloride ion , Cl
-
 Bromide ion, Br
-
releases electrons to form bromine ,Br
2
 Half-equation : 2Br
-
Br
2
+ 2e
-
------ (1)
-1 0 oxidation

colourless brown
 Chlorine , Cl
2
receives electrons to form chloride ion , Cl
-
 Half-equation : Cl
2
+ 2e
-
2Cl
-
------- (2)
0 -1 reduction
yellow decolourised(colourless)
 Ionic equation : (1) + (2)
Cl
2 + 2Br
-
Br
2 + 2Cl
-
 Chlorine ----- oxidising agent
 Bromide ion ----- reducing agent
 Chlorine displaces bromine from the sodium bromide solution.

20. Confirmatory Test for the Bromine, Chlorine and Iodine
 By adding and shaking the halogen solution in tetrachloromethane
(CCl
4
) liquid
Solution
Colour in water
Colour in
CCl
4
Concentrated Dilute
Iodine Brown Yellow Purple
Bromine Brown Yellow Brown
Chlorine Light greenish
yellow
Colourless Colourless

21. F : Transfer of Electrons at a Distance
•If two chemicals are separated at a distance by an electrolyte solution
in a U-tube
• acts as a salt bridge .
• used to separate two solutions but allows ions to pass (flow) through
to complete the circuit .
• examples : H
2SO
4 , KNO
3 , Na
2SO
4 solution
The electrons that are released from reducing agent (negative electrode) will
flow out through outer circuit to the oxidising agent ( positive electrode)
reduction
oxd

22. The Reaction Between Bromine Water and Iron(II) Sulphate solution
 Iron(II) ion, Fe
2+
releases electron & is oxidised to iron(III) ion , Fe
3+
Fe
2+
Fe
3+
+ e
-1
---------- (1)
O. Number + 2 +3 ( oxidation)
 The green solution ,(Fe
2+
) changes to brown, Fe
3+

 The electrons that are released collect at the carbon electrode that is
immersed in FeSO
4

 It act as the negative terminal .
 Bromine ,Br
2
receives electron & is reduced to bromide ion, Br
-1
 Half equation : Br
2
+ 2e
-
2Br
-1
--------------(2)
brown colour decolourised
O.Number 0 -1 (reduction)
 the carbon electrode in bromine water act as the positive terminal
 the ionic equation :- 2Fe
2+
+ Br
2
2Fe
3+
+ 2Br
-1

O. Number +2 0 +3 -1

23.  oxidising agent ----- Bromine water , Br
 reducing agent ----- Iron (II) ion, Fe
 The galvanometer needle is deflected because the movement of
electrons from the negative electrode to the positive electrode produces an
electric current .
Reduction
Oxidation
(Positive terminal)
(Negative terminal)

24. Test Yourself
The figure shows a U-tube redox cell .
•Write a summary of the redox reaction for the reaction between Iron(II)
sulphate, FeSO
4
solution and the acidified potassium manganate (VII) ,
KMnO
4
solution.
• Can dilute sulphuric acid be replaced with dilute hydrochloric acid ?
Give the reason for your answer .

26. •Observation :
Electrode (-) :
The green coloured iron(II) nitrate solution changes to brown
Electrode (+) :
The purple coloured acidified potassium manganate(VII) solution is
decolourised .
Half equation :
Electrode (-) : Fe
2+
Fe
3+
+ e
-
----------- ( oxidation)
Electrode (+) : MnO
4
-
+ 8H
+
+ 5e
-
Mn
2+
+ 4H
2
O (reduction)
Ionic Equation : 5Fe
2
+ MnO
4
-
+ 8H
+
Fe
3+
+ Mn
2+
+ 4H
2O
Oxidising agent : manganate(VII) ion
Reducing agent : Iron (II) ion .
Confirmatory test for the product( Fe ) that is formed.
Add sodium hydroxide solution, a brown precipitate is formed .
(b) Can . Hydrochloric acid also allows the transfer of ions to occur .
Solution :

27. Redox Reaction in a simple voltaic cells
 The porous pot ( pasu berliang) --- to separate the two solutions but allows the ions
to flow through it to complete circuit .
 the transfer of electrons occur from reducing agent to the oxidising agent through an
outer circuit .
 The negative electrode ( anode) ----- metal which is more electropositive in the E.S.
 The positive electrode ( cathode) ----- metal which is less electropositive in the E. S.
 electron flows from the negative electrode to the positive electrode .
 two types of Daniell cell that uses a porous pot :

28. At the negative electrode( anode) :
 Zn is more electropositive than copper
 Zn has more tendency to releases two electrons to form zinc ion,Zn
2+
 Zn rod acts as the negative electrode .
 Zn Zn
2+
+ 2e ------ oxidation process occurs
 The electrons will flows from the zinc rod to the copper rod through the
outer circuit
 an electric current is produced .
At the cathode :
 copper ion, Cu
2+
receives two electrons to form copper atom, Cu &
undergoes reduction process .
 Cu
2+
+ 2e Cu ------ reduction process
 Copper(II) ion oxidises(mengoksidakan) zinc, Zn to zinc ion, Zn
2+

 Cu
2+
----------- oxidising agent
 Zinc reduces(menurunkan) copper(II) ion,Cu
2+
to copper atom, Cu
 Zn ------------- reducing agent

29. Overall Ionic Equation :
Zn + Cu
2+
Zn
2+
+ Cu
Observation :
 cathode – the blue copper(II) sulphate
solution becomes fade/ colourless
--- a brown solid forms at the copper rod //
the copper rod thickens // the mass of the
copper will increases.
 anode ---- the zinc rod dissolves /
corrodes/ becomes thinner(menipis)
 Cell symbol :
Zn(s) / Zn
2+
(aq) // Cu
2+
(aq) / Cu(s)
0 +2
+2 0
Oxidation
Reduction

31. G . Corrosion of Metals
 occur when a metal loses electrons & is oxidised to form the metal ion .
 the metal is corroded
 example : Iron loses electrons to form iron(II) ion , Fe
2+

 Fe Fe
2+
+ 2e
-1
------- oxidation
 O. Number 0 +2
 Iron is corroded .
 If magnesium loses electrons to form magnesium ion Mg
2+
,
 magnesium is corroded.
 Mg Mg
2+
+ 2e
-
------- oxidation
 The metals is more electropositive in E.S. , corrode much easier .
 because the metals more tendency to release electrons to form metal ions
 Example : Al corrodes more easily compared to copper .
 because Al is more electropositive than copper .
 the rusting requires water and oxygen
Metal corrosion

32. RUSTING OF IRON
 At the end of the water droplet ( Anode / negative terminal )
 the iron , Fe loses electrons and is oxidised to iron(II) ion, Fe
 Iron(II) ion dissolves in water
 Iron is corroded .
 Fe Fe
2+
+ 2e ------- oxidation
 The electrons flows to the edge(pinggir) of the water droplet through
the iron
Stage 1
corrosion

33.  Iron(II) hydroxide , Fe(OH)
2
is then oxidised by oxygen to form
hydrated Iron(III) oxide, (brown solid ) or rust .
 equation : Fe(OH)
2 Fe
2O
3 .3H
2O (rust)
O
2
in the air
Stage 4
 Iron(II) ion , Fe
2+
& hydroxide ion , OH
-
combine to form iron(II)
hydroxide ( green solid )
 Fe
2+

(aq)
+ 2OH
-1

(aq)
Fe(OH)
2 (s)Stage 3
 At the edge of the water droplet ( cathode / positive terminal )
 Electrons are received by oxygen & water to form OH ions through
reduction
 O
2
+ 2H
2
O + 4e 4OH
-
------ Reduction
Stage 2
Iron rusting

34.  is a process that occurs when two metals come into contact(bersentuhan)
with an electrolyte .
 the more electropositive metal will donate(release) electrons & is corroded
 If the iron comes into contact with metal that is more electropositive ,it will
not corrode .
 the corrosion of iron can be accelerated by the presence of electrolytes
such as acid & salt solution.
K
Na
Ca
Mg
Al
Zn
Fe
Sn
Pb
H
Cu
Hg
Ag
Au
Example : Electrochemical Corrosion of Metals
More
easily
corroded
Difficult to
be corroded
Tendency for
corrosion increases

35. Example : The effect of rusting when iron comes into contact with other metals
( Mg, Cu , Zn , Sn)
Hypothesis : Iron is protected from rusting when it comes in contact with more
electropositive metals, but rusts when it contact with less electropositive metals .
 Potassium hexacyanoferrate(III) solution ---- to detect the presence of iron(II) ions .
 Iron nail is corroded, the dark blue colour will be seen in the solid agar .
 Phenolphthalein will turn pink in colour if OH
-
ions are present.
 The gas bubbles formed are hydrogen gas .
 The iron nail in test tube B is not corroded , while zinc is corroded because Zn is more
electropositive than iron .
 In test tube A ,the iron nail is corroded because Iron more electropositive than copper .
B
A
rusting

36. PREVENTION THE
RUSTING OF IRON
Coating a layer of
metal such as Al or
Sn on food tins
Applying paint,oil or grease
on surface such as engine
Wrapping the iron with
a layer of plastic . Ex: hangers
Applying a coat of Al
such as car bumpers
or water pipes
Iron sheet used as house
roofs Are galvanised
with a layer of zinc
Iron is alloyed with other
metals such as chromium
or nickel to produce
stainless steel
Huge iron construction structure such
as bridges protected from corrosion by
using sacrificial metals(logam korban)
such as Mg & Zn

37. Reactivity Series (R.S) of Metals
A : Metals with Oxygen
 Metal is heated in oxygen to produce
metal oxide .A more reactive metal will displace
a less reactive metal from its oxide.
 Observation :
 Mg ----- burn very rapidly & vigorously with a very bright flame
------ metal oxide colour : white powder ( Hot & cold )
 Zinc ---- burns rapidly , glows brightly
----- metal oxide colour : yellow when hot & white when it is cold .
 Iron ----- burns rapidly, glows less brightly than Zn
----- metal oxide colour brown when hot & cold
 Cu ----- very slow reaction
----- metal oxide colour black ( hot & cold )
 Pb ----- burns slowly
----- metal oxide colour : brown when hot & yellow when colour

38. K
Na
Ca
M
g
Al
C
Zn
Fe
Sn
Pb
Cu

Ag
Au
Reactivity
decreases
Example : The reaction between Lead(II) oxide with Carbon
Observation : burn brightly
: produces a grey solids
Inference : Carbon is more reactive than Lead
Equation : PbO + C Pb + CO
2
If carbon is more reactive than metal X , a flame or glows(baraan)
can be seen.
If carbon is less reactive than metal Y ,the flame or glows will not
be seen when carbon react with metal oxide Y is heated .
The position of Carbon in the R. S.

40. PREPARED BY
PN ZAINAB BINTI AYUB
22 JULY 2008