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Mg2+

O2-

KMnO4



Redox reactions are a chemical reactions involving oxidation

and reduction occurring simultaneously.

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 .



Mg + PbO MgO + Pb

Gain of oxygen ---- oxidation

Loss of oxygen ---- reduction

PbO – oxidising

agent (experiences

reduction



H2 S + Cl

2 2HCl + S

Loss of hydrogen --- oxidation

Gain of hydrogen --- reduction

Cl2 – oxidising agent

( undergoes reduction )

--- chlorine oxidises

hydrogen sulphide to

sulphur

H2S – reducing agent

( undergoes oxidation)

--- hydrogen sulphide

reduces chlorine to

hydrogen chloride



B Transfer of Electrons

oxidation involves the loss (releases) of electrons

reduction involves a gain (receives) in electrons

2Na(s) + Cl2 (g) 2NaCl(s)

Na Na+ + e ( loss of electron)

Cl2 + 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 .



Determine the oxidation and reduction process , oxidising and

reducing agent that occurs in the reactions below .

(a) Mg + PbO MgO + Pb

(b) Anode : Cu Cu2+ + 2e- ; Cu2+ + 2e- Cu

(c) 2CuO + C 2Cu + CO2

(d) Fe2 O3 + 3C 2Fe + 3CO

(e) Mg + CuO MgO + Cu

Solution :



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

Mg2+ , Ca+2 is +2

O2- , S2- is -2

O.N for atom or molecule in a neutral elements are zero ( 0 )

example : O2 , N2 , Na , Mg, Br 2 is 0 .

2FeCl2 + Cl2 2FeCl3

Oxidation number decreases (0 → -1)

Oxidation number increases( +2 → +3)

reduction

oxidation

+2 +3 -10-1O.N

Reducing agent ---

iron (II) chloride

Oxidising agent

--- chlorine gas

C : Change in Oxidation Number ( O.N )



(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) ClO3- , oxidation number of chlorine is X

X + 3(-2) = -1

X - 6 = -1

X = +5

(ii) The oxidation number of S in MgSO4

+2 + X + 4 (-2) = 0

+2 + X - 8 = 0

X = + 6

the total oxidation number for dichromate (VI) ion, Cr 2O72- is -2 ,

manganate (VII) , MnO4-1 is -1

The charge of

chlorate

Oxidation

number of S



Test Yourself :

Calculate the oxidation number of the following elements :

(a) Manganese , Mn in potassium manganate (VII) , KMnO4

(b) Manganese, Mn in manganate(VII) ion, MnO4-

(c) Chromium, Cr, in potassium dichromate(VI), K2Cr 2O7

(d) Cromium, Cr,in chromate(VI) ion, CrO4

2-

(e) Iron in iron(II) chloride , FeCl2

(f) Iron in iron(III) chloride , FeCl3

(g) Carbon , C in sodium carbonate, Na2CO3

In each of the cases above, the oxidation number of each element is

represented by the value of X .



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 :

(i) Fe 2+ Fe 3+ + e-

(O.N ) +2 +3

(ii) 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 oxidationnumber



Example :

Half equation :

Fe2+ Fe3+ + e- -------- (1) X 2

Br 2 + 2e- 2Br - -------- (2)

Ionic equation :

2Fe 2+ + Br 2 2Fe3+ + 2Br -

Combined

Redox reactions need to shown in the form of :

half - equations

ionic equations

Fe2+

Fe3+

Cl-

Cl2

Br - I-

Br 2

Green solution

Brown solution

Yellow solution

Colourless solution

MnO4- Purple solution

Cr 2O7 2-

Cu2+

Orange solution

Blue solution

Cr 3+

I2



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 Mg2+ + 2e- ------------(1)

O.Number 0 +2 (oxidation)

Magnesium reduces oxygen to oxide ion .

Oxygen atom receives electrons to form oxide ion .

Half- equation : O2 + 2e- O2- ------------(2)

O.Number 0 -2 (reduction )

The overall equation : (1) + (2)

2Mg + O2 2MgO

Observation :-

Combusts with a white shiny flame

A white solid is formed

Oxidising agent : Oxygen

Reducing agent : Magnesium Reactants



B. The change of iron(II) ions to iron(III) ions ( Fe2+ Fe3+)

Bromine water oxidises iron(II) ion , Fe2+ to iron(III) ion , Fe3+

Iron(II) ion releases electron to form iron(III) ion.

Half-equation : Fe2+ Fe3+ + 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, Fe2+

• Ionic Equation : 2 Fe2+ + Br 2 2Fe3+ + 2Br -

• other oxidising agent : Cl2 ,KMnO4 ,K2Cr 2O7 ,HNO3 concentrated,H2O2

C om b i n e d



C : The change of iron(III) ion, Fe3+ to iron(II) ion , Fe2+

Iron (III) ion oxidises Zn atom to zinc ion , Zn2+

Iron(III) ion receives electron to form iron(II) ion

The brown iron(III) chloride solution changes to green .

Half-equation : Fe3+ + e- Fe2+ --------(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 Zn2+ + 2e- -------(2)

O. Number 0 +2 ( oxidation )

The ionic equation : 2Fe3+ + Zn 2Fe2+ + Zn2+

Other reducing agent : metals that are more electropositive than iron

// SO2 , H2 S gas // Na2 SO3 , SnCl2 solution

combined



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 & CuSO4 solution .

Zn more electropositive than copper .

Zn releases two electron to form zinc ion , Zn2+

Zn reduces copper(II) ion ,Cu2+ to copper , Cu

Copper(II) ion oxidises Zn to zinc ion ,Zn2+

K, Na , Ca , Mg , Al , Zn , Fe , Sn , Pb , H , Cu , Hg , Ag , Au

Most electropositive Least electropositive



Observation :

• The blue CuSO4 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 Zn2+ + 2e- ----------(1)

O . Number 0 +2 ( oxidation )

Cu2+ + 2e Cu ----------(2)

O. Number +2 0 ( reduction )

Ionic equation : Zn + Cu2+ Zn2+ + Cu

Zn ------ reducing agent

Copper(II) ion ------- oxidising agent

Zn loses

electrons &

is oxidised

to Zn2+

Cu2+

receives

electrons &is reduced to

Cu



E : Displacement of Halogens from Halide Solutions

Halogen ----- Group 17

examples : Cl2 ( chlorine) ------- yellow

Br 2 ( bromine) ------- brown

I2 (iodine) ------- yellow or brown

can be differentiated by shaking the solution with a little CCl4

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



Chlorine water react with sodium bromide solution

Chlorine water , Cl2 oxidises bromide ion, Br - to bromine ,Br 2

Bromide ion , reduces chlorine , Cl2 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 , Cl2 receives electrons to form chloride ion , Cl-

Half-equation : Cl2 + 2e- 2Cl- ------- (2)

0 -1 reduction

yellow decolourised(colourless)

Ionic equation : (1) + (2)

Cl2 + 2Br - Br 2 + 2Cl-

Chlorine ----- oxidising agent

Bromide ion ----- reducing agent

Chlorine displaces bromine from the sodium bromide solution.



Confirmatory Test for the Bromine, Chlorine and Iodine

By adding and shaking the halogen solution in tetrachloromethane

(CCl4 ) liquid

SolutionColour in water

Colour in

CCl4 Concentrated Dilute

Iodine Brown Yellow Purple

Bromine Brown Yellow Brown

Chlorine Light greenish

yellow

Colourless Colourless



F : Transfer of Electrons at a Distance

a) If two chemicals are separated at a distance by an electrolyte solution

in a U-tube

b) acts as a salt bridge .

c) used to separate two solutions but allows ions to pass (flow) through

to complete the circuit .

d) examples : H2SO4 , KNO3 , Na2SO4 solution

The electrons that are released from reducing agent (negative electrode) will

flow out through outer circuit to the oxidising agent ( positive electrode)

reductionoxd



The Reaction Between Bromine Water and Iron(II) Sulphate solution

Iron(II) ion, Fe2+ releases electron & is oxidised to iron(III) ion , Fe3+

Fe2+ Fe3+ + e-1 ---------- (1)

O. Number + 2 +3 ( oxidation)

The green solution ,(Fe2+) changes to brown, Fe3+

The electrons that are released collect at the carbon electrode that is

immersed in FeSO4

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 :- 2Fe2+ + Br 2 2Fe3+ + 2Br -1

O. Number +2 0 +3 -1



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)

T t Y lf



Test Yourself

The figure shows a U-tube redox cell .

(a) Write a summary of the redox reaction for the reaction between Iron(II)

sulphate, FeSO4 solution and the acidified potassium manganate (VII) ,

KMnO4 solution.

(b) Can dilute sulphuric acid be replaced with dilute hydrochloric acid ?

Give the reason for your answer .

S l ti



(a) 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 (-) : Fe2+ Fe3+ + e- ----------- ( oxidation)

Electrode (+) : MnO4- + 8H+ + 5e- Mn2+ + 4H2O (reduction)

Ionic Equation : 5Fe2 + MnO4- + 8H+ Fe3+ + Mn2+ + 4H2O

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 :



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 throughan 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 :

At th ti l t d ( d )



At the negative electrode( anode) :

Zn is more electropositive than copper

Zn has more tendency to releases two electrons to form zinc ion,Zn2+

Zn rod acts as the negative electrode .

Zn Zn2+ + 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, Cu2+ receives two electrons to form copper atom, Cu &

undergoes reduction process .

Cu2+

+ 2e Cu ------ reduction process Copper(II) ion oxidises(mengoksidakan) zinc, Zn to zinc ion, Zn2+

Cu2+ ----------- oxidising agent

Zinc reduces(menurunkan) copper(II) ion,Cu2+ to copper atom, Cu

Zn ------------- reducing agent



Overall Ionic Equation :

Zn + Cu2+

Zn2+

+ Cu

Observation :

cathode – the blue copper(II) sulphatesolution 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) / Zn2+(aq) // Cu2+ (aq) / Cu(s)

0 +2

+2 0

Oxidation

Reduction



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 , Fe2+

Fe Fe2+ + 2e-1 ------- oxidation

O. Number 0 +2

Iron is corroded .

If magnesium loses electrons to form magnesium ion Mg2+ ,

magnesium is corroded.

Mg Mg2+ + 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

i

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http://localhost/var/www/apps/conversion/tmp/scratch_2/rusting.swf




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 Fe2+ + 2e ------- oxidation

The electrons flows to the edge(pinggir) of the water droplet through

the iron

S t a g e

1

corrosion





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 Fe2O3 .3H2O (rust)O2 in the air S t a g e

4

Iron(II) ion , Fe2+ & hydroxide ion , OH- combine to form iron(II)

hydroxide ( green solid )

Fe2+ (aq) + 2OH-1 (aq) Fe(OH)2 (s) S t a g e

3

At the edge of the water droplet ( cathode / positive terminal )

Electrons are received by oxygen & water to form OH ions through

reduction

O2 + 2H2O + 4e 4OH-

------ Reduction

S t a g e

2

Iron rusting

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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 willnot corrode .

the corrosion of iron can be accelerated by the presence of electrolytes

such as acid & salt solution.

K

NaCa

Mg

Al

Zn

Fe

SnPb

H

Cu

Hg

Ag

Au

Example : Electrochemical Corrosion of Metals

More

easilycorroded

Difficult to

be corroded

Tendency for

corrosion increases



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

Reactivity Series (R S) of Metals



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

Th i i f C b i h R S



K

Na

Ca

M

g

Al

C

Zn

Fe

Sn

Pb

Cu

Ag

A

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 + CO2

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 notbe seen when carbon react with metal oxide Y is heated .

The position of Carbon in the R. S.

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