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7/31/2019 C12 Notes S Redox
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction1
CHAPTER 12 : OXIDATION & REDUCTION
REDOX REACTIONS
1.1 The meaning ofREDOXreaction
1. REDOXreaction A reaction where both oxidation and reductionoccur at the same time.
2. Oxidation and reduction can be defined in four ways :
OXIDATION REDUCTION Example :
(i) Loss ofoxygen and
gain inoxygenGain of oxygen Loss of oxygen
2Mg + O2 2MgO
CuO + H2 Cu + H2O
(ii) Loss ofhydrogen and
gain inhydrogenLoss of hydrogen Gain of hydrogen N2 + 3H2 2NH3
(iii)Loss ofelectrons and
gain in electrons
Loss of electron Gain of electronMg Mg
2++ 2e
Cl + e Cl
-
(iv) Change inoxidation
number
Increase in
oxidation numberDecrease in oxidation
numberZn + Cu
2+ Zn
2++ Cu
3. (i) Oxidizing agent :
a substance thatoxidized another substance.
the oxidizing agent is reduced in the process.
(ii) Reducing agent :
a substance thatreduced another substance.
the reducing agent is oxidized in the process.
1.2 Example ofREDOXreactions
(i) In terms of gain/loss ofoxygen
Example 1 :
Explanation :
Magnesium, Mg gains oxygen and it is oxidized. Magnesium has reduced copper(II) oxide, CuO.
Magnesium, Mg is reducing agent.
Copper(II) oxide , CuO loses oxygen and it is reduced. Copper(II) oxide has oxidized magnesium, Mg.
Copper(II) oxide is oxidizing agent.
Magnesium is oxidised (gain of oxygen)
Copper(II) oxide is reduced (loss of oxygen)
Mg + CuO MgO + Cu
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction2
(ii) In terms of gain/loss ofhydrogen
Example 2 :
Explanation :
Hydrogen sulphide, H2S loses hydrogen and it is oxidized. Hydrogen sulphide, H2S has reduced
chlorine, Cl2. Hydrogen sulphide, H2S is reducing agent.
Chlorine, Cl2 gains hydrogen and it is reduced. Chlorine, Cl2 has oxidized hydrogen sulphide, H2S.
Chlorine, Cl2 is an oxidizing agent.
Example 3 :
Explanation :
Copper(II) oxide, CuO loses oxygen and it is reduced. Copper(II) oxide, CuOhas oxidized ammonia,
NH3. Copper(II) oxide, CuO is oxidizing agent.
Ammonia, NH3 loses hydrogen and it is oxidized. Ammonia, NH3 has reduced copper(II) oxide.
Ammonia, NH3 act as reducing agent.
(iii)In terms of gain/loss ofelectron
Example 4 :
Hydrogen sulphide is oxidised(loss of hydrogen)
Chlorine is reduced (gain of hydrogen)
H2S + Cl2 S + 2HCl
Copper(II) oxide is reduced (loss of oxygen)
Ammonia is oxidised ( lossof hydrogen)
3CuO + NH3 3Cu + 3H2O + N2
2Na + Cl2 2NaCl
(i)
(ii)
NaCl is an ionic compound. It exist as
Na+ ions and Cl- ions.
NaCl Na+ + Cl-
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction3
Explanation :
Sodium, Na with an electron arrangement of 2.8.1 loses an electron to form sodium ion, Na+ in
sodium chloride, NaCl. Sodium is oxidized.
Na Na+
+ e
Chlorine, Cl2gains electrons to form chloride ions. Cl- in sodium chloride, NaCl. Chlorine is
reduced.
Cl2 + 2e
2Cl-
Sodium, Na has reduced chlorine, Cl2.
Sodium, Na is a reducing agent.
Chlorine, Cl2 has oxidized sodium, Na.
Chlorine, Cl2 is an oxidizing agent.
Example 5 :
Explanation :
Magnesium atom, Mg loses two electrons to form magnesium ion, Mg2+.
Mg Mg2+ + 2e
Magnesium is oxidized.
Copper(II) ion, Cu2+gains two electrons to form copper atom, Cu.
Cu2+ + 2e Cu
Copper(II) ion, Cu2+ is reduced.
Magnesium, Mg has reduced copper(II) ion, Cu2+
.
Magnesium, Mg is a reducing agent.
copper(II) ion, Cu2+
has oxidized Magnesium, Mg.
copper(II) ion, Cu2+
is an oxidizing agent.
Mg + CuSO4 MgSO4 + Cu
(i)
(ii)
(i)
(ii)
(i)
(ii)
MgSO4 is an ionic compound. It exist asMg2+ ions and SO4
2- ions.
MgSO4 Mg2+
+ SO42-
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction4
(iv)In terms of change inoxidation number
Generalrules to determine Oxidation Number
1. The oxidation number ofatoms and molecules of elements is zero.
Example :
Atom of Element Oxidation number
Copper, Cu 0
Sodium, Na 0
Iron, Fe 0
Helium,He 0
Molecule of Element Oxidation number
Hydrogen gas, H2 0
Oxygen gas , O2 0
Chlorine gas, Cl2 0
2. The oxidation number for hydrogen in most of itscompoundis +1.
H H2 HCl
Hydrogen
atom
Hydrogen
molecule
Hydrogen
ion
Oxidationnumber
0 0 +1
3. The oxidation number for oxygen in most of its compoundis -2.
O O2 MgO Na2O
Oxygen
atom
Oxygen
moleculeOxide ion Oxide ion
Oxidation
number0 0 -2 -2
4. The oxidation number for a simple ion is similar to the charge of the ion.
Example :
Ion Oxidation number Ion Oxidation number
Na+ +1 Cl
- -1
Cu2+
+2 Br- -1
Fe2+ +2 O2- -2
Fe3+ +3 S2- -2
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction5
5. The oxidation number of the ions from Group 1, 2 and 13 are fixed because the atom of these Groupsachieve stable octet electron arrangement by losing 1 e
-, 2e
-and 3e
-respectively. Therefore, the oxidation
number for these ions are :
Ion from element in Group Example Oxidation number
1 Na+ +1
2 Mg2+ +2
13 Al3+ +3
6. The sum of the oxidation numbers for elements in a compound is zero.
Example :
KMnO4
(+1) + x + 4(-2) = 0
x = +7
7. The sum of the oxidation numbers for elements in apolyatomic ion is equals to the charges of thepolyatomic ion.
Example :
SO42-
y + 4(-2) = -2
y = +6
8. The oxidation number of the Transition Elements and mostofnon-metal elements are vary from onecompound to another.
Example :
Formula NameOxidation
numberFormula Name
Oxidation
number
CuO Copper(II) oxide +2 FeCl2 Iron(II) chloride +2
Cu2O Copper(I) oxide +1 FeCl3 Iron(III) chloride +3
+1
x-2
y
-2
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction6
Activity 1:-
State the oxidation number of the underlined element in the following table :
Oxidation
number
Oxidation
number
(a) (i) MnO2 (c) (i) K2Cr2O7
(ii) KMnO4 (ii) Na2Cr2O3
(b) (i) CO32-
(d) (i) NH3
(ii) CO2 (ii) NO2
(iii) CO (iii) NH4+
(iv) NO3-
Nomenclature of ionic compounds using IUPAC Nomenclature
(i) Elements from Group 1, 2 and 13 have onlyone oxidation number, the names of the compounds are writtenwithout the oxidation number.
Example :
K2O is potassium oxide ; not potassium(I) oxide.
MgO is magnesium oxide ; not magnesium(II) oxide.
Al2Cl3 is aluminium chloride ; not magnesium(III) chloride.
(ii)Transition elements/metals and most of non metals compounds havemore than one oxidation number.The oxidation number of the element is written inRoman numeral, placed in a bracket behind the name of
the element.
Example :
Common name
Chemical
formula of
compound
Oxidation number
of transition metal/
non metal
IUPAC Name
Ferrous hydroxide Fe(OH)2 +2 Iron(II) hydroxide
Ferric hydroxide Fe(OH)3 +3 Iron(III) hydroxide
Lead monoxide PbO +2 Lead(II) oxide
Lead dioxide PbO2 +4 Lead(IV) oxide
Copper oxide CuO +2 Copper(II) oxide
Copper oxide Cu2O +1 Copper(I) oxide
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction7
Redox Reaction in terms of change in Oxidation Number
When the oxidation number of an element increases, the element is oxidized.
The element is a reducing agent.
When the oxidation number of an elementdecreases, the element is reduced.
The element is an oxidizing agent.
A reaction is not a redox reaction if no elements undergo a change in oxidation state.
Example :
Determine whether the reactions below are redox or not.
(a)
Oxidation number : +1 +5 -2 +1 -1 +1 -1 +1 +5 -2
This reaction is not a redox reaction because no changes of oxidation number of all elements in the
compounds of substances of reactants and products.
(b)
Oxidation number : 0 +1 +6 -2 +2 +6 -2 0
This reaction is a redox reaction because the oxidation number of magnesium and hydrogen in the
substances have changed.
Explanation :
Magnesium, Mg is oxidized because its oxidation number increases from 0 to +1.
Hydrogen ion, H+ is reduced because its oxidation number decreases from +1 to 0.
Oxidizing agent is hydrogen ion, H+.
Reducing agent is magnesium, Mg.
1.3 WritingEquations For Redox ReactionsEquations for redox reactions are :
(i) Chemical Equationfor the reaction.
(ii) Half equation for oxidation (loss of electron/increase in oxidation number).
Half equation for reduction (gain in electron/decrease in oxidation number).
(iii) Overall Ionic equation for redox reaction formed bycombining half equation for oxidation andhalf equation for reduction (thenumber of electrons in both the half equations mustcancel each other).
AgNO3 + NaCl AgCl + NaNO3
Mg + H2SO4 MgSO4 + H2
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction8
Example 1 :
Reaction Aluminium and copper(II) sulphate
Chemical Equationfor the reaction 2Al + 3CuSO4 Al2(SO4)3 + 3Cu
Half equation for oxidation Al Al3+ + 3e
Half equation for reduction Cu2+ + 2e Cu
Changing of the coefficient of the half
equation of oxidation*2Al 2Al
3+ + 6e
Changing of the coefficient of the half
equation of reduction*3Cu
2++ 6e 3Cu
Ionic equation 2Al + 3Cu2+
2Al3+
+ 3Cu
* Make sure that the number of electrons released in half equation for oxidation areequalto thenumber of electrons received in half equation for reduction.
Activity 2:-
1 Sodium metal react with water
Reaction Sodium and water
Chemical Equationfor the reaction
Half equation for oxidation
Half equation for reduction
Changing of the coefficient of the half
equation of oxidation*
Changing of the coefficient of the half
equation of reduction*
Ionic equation
Al Cu
2+
SO42-
CuSO
4
2-Al
3+
water
sodium
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction9
2 Copper(II) oxide react with dry hydrogen gas (Determine the empirical formula of copper(II) oxide)
Reaction Copper(II) oxide and hydrogen gas
Chemical Equationfor the reaction
Half equation for oxidation
Half equation for reduction
Changing of the coefficient of the half
equation of oxidation*
Changing of the coefficient of the half
equation of reduction*
Ionic equation
1.4 Analysing Redox Reactions in :
1.4.1 Displacement of metals from its salt solution.
1.4.2 Electrolytic and Chemical Cell (Voltaic cell)
1.4.3 Rusting of Iron //Corrosion of Metal
1.4.4 Displacement of halogen from its halide solution.
1.4.5 Change of : Fe2+ Fe3+and Fe3+ Fe2+
1.4.6 Transfer of electron at a distance.
1.4.7 Reactivity Series of Metals and Its Applications.
Copper(II) oxide
Dry hydrogen gas
[Refer toC7 :ELECTROCHEMISTRY]
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction10
1.4.1 Redox Reaction in the Displacement of Metals from its Salt solution
Electrochemical Series :
Electrochemical Series is an arrangement of elements according to their tendency to release/donate
electrons to form a positive ion (cation).
Electrochemical
Series
K
Na
Ca
Mg
Al
Zn
Fe
Sn
Pb
H
Cu
Ag
The higher the position of the metal in the Electrochemical Series, the greater is the tendency of the metal
atoms to donate/release electrons (become moreElectropositive).
Displacement of metals :
The metal which is higher in the electrochemical series will donate/lose/release its electrons to the metal
ion which is lower in the electrochemical series from its salt solution. The more electropositive metal
will be oxidized and dissolves. The less electropositive metal ion will be reduced and deposited.
MORE ELECTROPOSITIVE
Easier to release/donate electrons
to form a positive ion (cation)
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction11
Example : A piece of zinc metal plate is dipped in copper(II) sulphate solution.
zinc metal + copper(II) sulphate solution
Zn + CuSO4 Zn SO4 + Cu
Zn Zn2+ + 2e
Cu2+ + 2e Cu
Zn + Cu2+
Zn2+
+ Cu
Explanation :
In terms of gain/loss of electron
Zn is above Cu in the Electrochemical Series.
Zinc atom releases/donates/loses two electrons to form zinc ion, Zn2+.
Zinc is oxidized.
Copper(II) ion, Cu2+
gains/receives two electrons to form copper atom, Cu.
Copper(II) ion, Cu2+ is reduced.
Zinc has reduced copper(II) ion, Cu2+
. Zinc is the reducing agent.
Copper(II) ion, Cu2+
has oxidized zinc. Copper(II) ion is the oxidizing agent.
Half equation of oxidation : Zn Zn2+ + 2e
Half equation of reduction : Cu2+ + 2e Cu
Ionic equation : Zn + Cu2+ Zn
2++ Cu
In terms of change in oxidation numberZinc is oxidized as its oxidation number increases from 0 to +2.
As copper(II) ion, Cu2+ causes Zn to be oxidized, copper(II) ion, Cu2+ is the oxidizing agent.
Copper(II) ion, Cu2+
is reduced as its oxidation numberdecreases from +2 to 0.
As zinc , Zn causes copper(II) ion to be reduced, zinc, Zn is the reducing agent.
Zinc is more electropositive than copper
Zinc corrodes.
Copper is less electropositive
copper displaced.
Zn plate
CuSO4(aq)
Observation :
Zinc plate becomes
thinner // Part of zinc
plate dissolves
Brown solid is formed
on the surface of the
remaining part of the
zinc plate
The blue colour of
copper(II) sulphate
solution becomes
colourless
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction12
A
1.4.2 Redox Reaction in the Electrolyticand Chemical Cell (Voltaic cell)
Electrolysis :
The selected cation will gain/receive electrons and undergoes reduction at the cathode (negatively
charged electrode).
The selected anion will release/lose its electrons and undergoes oxidation at the anode (positivelycharged electrode)
Chemical cell / Voltaic cell :
The metal which is higher in the Electrochemical Series will become the negative terminal and release
electrons. The metal undergoes oxidation and dissolves. The metal which is lower in the
Electrochemical Series becomes the positive terminal. The ion that is selected for discharge in the
solution undergoes reduction and the mass of the positive terminal will increase.
The further the distance between two metals in the Electrochemical Series, the higher the voltage of the
chemical cell.
(a) Differences between an electrolytic cell and achemicalcell :
CHARACTERISTIC ELECTROLYTICCELL CHEMICAL CELL
Set up of apparatus
It has cells/ batteries It hasno cell
It has Ammeter It has Voltmeter
Same type of electrodes Different type of electrodes
V
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction13
Structure
Consist oftwo electrodes (usually
carbon / two similar or different metals)
connected to batteries and dipped in aelectrolyte using connecting wires.
Anode : the electrode that is connected
to the positive terminal of the battery.
Cathode : the electrode that is connected
to the negative terminal of the battery.
Consist oftwodifferent metalsdipped in anelectrolyte and
connected to a voltmeter /
galvanometer / bulb using connectingwires.
Negative terminal : the metal that ishigher in the Electrochemical Series
(more electropositive).
Positive terminal : the metal that is
lower in the Electrochemical Series
(less electropositive ).
Energy conversionElectrical energy tochemical energy
(involvesredox reaction)
Chemical energy toelectrical energy
(involvesredox reaction)
(b)Redox reaction in an electrolytic cell and a chemical cell.
Electrolytic cell Chemical cell
Anode CathodeNegative terminal
(Anode)
Positive terminal
(Cathode)
Transfer
ofelectron
Anion / Metal atom
loses its electrons and
acts as a reducing
agent.
e.g :
4OH- 2H2O + O2 + 4e
Cu Cu2+
+ 2e
Cation in the electrolyte
gains electrons and acts
as an oxidizing agent.
e.g :
2H+
+ 2e H2
Ag+ + e Ag
The more
electropositive metal
loses its electrons and
acts as a reducing
agent.
e.g :
Mg Mg2+
+ 2e
Zn Zn2+
+ 2e
The metal ion in the
electrolyte gains
electrons and acts as
an oxidizing agent.
e.g :
Cu2+
+ 2e Cu
Ag+ + e Ag
Redoxreaction
Anion / Metal atomundergoes oxidation
Cation undergoesreduction
The more
electropositive metal
undergoes oxidation
The metal ion in the
electrolyte undergoes
reduction
Oxidation occurs at theanode in an electrolytic cell and at thenegative terminalin achemicalcell.
the term anode is assigned for the electrode at which oxidation occurs,negative terminal anode Reduction occurs at thecathode in an electrolytic cell and at the positive terminal in achemicalcell.
the term cathode is assigned for the electrode at which reduction occurs,positive terminal cathode
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction14
Activity 3 :-
1 Diagram below shows the set up of the apparatus used for the electrolysis of potassium sulphate solution.
(a) State all the ions present in the electrolyte.
...............................................................................................................................................................................
(b) State the product formed at electrode X and Y.
X : ............................................................. Y : ...........................................................................................
(c) Write the half equation for the reaction at
(i) electrode X :
...................................................................................................................................................................
(ii) electrode Y :
...................................................................................................................................................................
(d) Name the substance which is
(i) oxidized : ......................................................................................................................................................
(ii) reduced : ......................................................................................................................................................
(e) Name the
(i) oxidizing agent : ...........................................................................................................................................
(ii) reducing agent : ............................................................................................................................................
(f) State the oxidation number of sulphur in sulphate ion?
...............................................................................................................................................................................
Carbon electrod X
potassium sulphate, K2SO4 solution
A
Carbon electrod Y
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction15
2 Table below shows the concentration of sodium chloride in solution X and Y.
Solution X Solution Y
0.001 mol dm-3
2.0 mol dm-3
Both solutions are electrolysed separately using carbon as electrodes.
(a) Write the half equation of the reaction that takes place at the anode for electrolysis of :
(i) solution X :
(ii) solution Y :
(b) Name the products formed at the cathode and anode.
(c) The products collected at the anode in the electrolysis of solutions X and Y are different.
Explain why.
..
..
(d) Name the substance oxidized in the electrolysis of :
(i) solution X :
(ii) solution Y :
(e) Name the substance reduced in the electrolysis of :
(i) solution X : .
(ii) solution Y : .
Product formed at :
Anode Cathode
Solution X
Solution Y
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction16
3 The diagram below shows the set up of an apparatus for an experiment.
(a) Which electrode is the positive terminal?
...
(b) Write the observations at the :
(i) negative terminal : .........................................................................................................................
(ii) positive terminal : ..........................................................................................................................
(c) Write the half equation for the reaction that takes place at the :
(i) negative terminal : .........................................................................................................................
(ii) positive terminal : ..........................................................................................................................
(d) Write the ionic equation for the reaction that takes place in the chemical cell above.
...
(e) Name the substance reduced in the above reaction.
...
(f) Name the substance that acts as a reducing agent in the above reaction.
...
(f) How will the voltmeter reading change if the magnesium electrode in the magnesium sulphate
solution is replaced by zinc electrode in zinc sulphate solution? Explain why?
...
...
...
V
Porous pot
Copper
Copper(II) sulphate solution
Magnesium
Magnesium sulphate solution
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction17
4 You are provided with the following materials and apparatus.
(a) By using all the materials and apparatus provided, draw the set up of the apparatus to produce
electricity from chemical reactions.
(b) (i) Name the solution that is used as a salt bridge.
...
(ii) What is the function of the salt bridge?
...
(c) Indicate on the set-up apparatus the following :
(i) negative terminal(ii) positive terminal
(iii) the direction of electron flow.
(d) Write the half equation for the reaction that occurs at :
(i) negative terminal : .
(ii) positive terminal :
(e) State the substance that oxidized in the above reaction.
........
(f) Name the substance that acts as an oxidizing agent in the above reaction.
........
Materials :
Zinc plate, copper plate, dilute sulphuric acid, zinc sulphate solution,
copper(II) sulphate solution.
Apparatus :Beakers, connecting wire, voltmeter, glass tube.
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction18
1.4.3 Redox Reaction in Corrosion of Metal [ Rusting of Iron]
Corrosion of metal isa redoxreaction in which a metal oxidized to its ion by losing electrons.
When metal corrodes, it usually forms a metal oxide coating.
Aluminium oxide, for example is non porous and firmly coated the metal. It will protect the aluminium
underneath from further corrosion. This further explain the resistance of aluminium to corrosion even though
it is higher in the Electrochemical Series ( electropositive metal). Other metals with similar property are zinc,
lead, nickel and chromium.
Rusting of iron/corrosion of iron.
Rusting of iron takes place when iron corrodes in the presence ofwater and oxygen.
It is a redox reaction whereby oxygen acts as an oxidizing agent while iron acts as a reducing
agent.
Mechanism of rusting of iron.
The surface of iron at A becomes an anode (negative terminal), the electrode at whichoxidationoccurs.
Iron atom, Fe loses electrons and is oxidized to form iron(II) ion, Fe2+
Half equation of oxidation: Fe Fe2+ + 2e
The electrons flow through iron to the edge of the water droplet at B, where the concentration of oxygenhere is higher.
The iron surface at B becomes cathode( positive terminal) , the electrode at which reduction occurs.
Oxygen, O2 gains electrons and is reduced to form hydroxide ions, OH-.
Half equation of reduction: O2 + 2H2O + 4e 4OH-
The iron(II) ion, Fe2+
produced combines with hydroxide ions, OH- to form iron(II) hydroxide.
Fe2+ + 2OH- Fe(OH)2
Ionic equation for rusting : 2Fe + O2 + 2H2O 2Fe(OH)2
O2 O2
B A
Iron
Water dropletO2
B
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction19
Iron(II) ion, Fe2+
isgreen but rust isbrown because iron(II) hydroxide , Fe(OH)2
undergoes
further oxidation by oxygen to form hydrated iron(III) oxide, Fe2O3 .xH2O (rust).
[xis an integer whereby the value varies ]
Iron(II) hydroxide , Fe(OH)2 is oxidised by oxygen to form iron(III) hydroxide, Fe(OH)3
Then, iron(III) hydroxide, Fe(OH)3 is decomposed to form hydrated iron(III) oxide (rust).
Rust is brittle, porous and not tightly packed. Thus, water and oxygen can penetrate the metal
underneath. Iron will undergo continuous corrosion.
Rusting of iron occursfaster in the presence ofacid or salt because when these substances
dissolve in water, the solutions becomes better electrolyte. An electrolyte will increase the
electrical conductivity of water.
Iron structures at coastal and industrial areas rust faster because of :
the presence of salt in the coastal breeze
the presence of acidic gases in industrial area such as sulphur dioxide, SO2 and nitrogendioxide, NO2.
Activity 4 :-
Draw a labelled diagram to show how the rusting of iron involved the ionization of iron and the flow of electron.
O2 O2
Iron
Water dropletO2
Anode (negative terminal)
Fe Fe2+ + 2e
Cathode (positive terminal)O2 + 2H2O + 4e 4OH
-
Fe2O3 . xH2O (rust)
Cathode (positive terminal)O2 + 2H2O + 4e 4OH
-
Fe2O3 . xH2O (rust)
e- e-
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction20
Controlling Metal Corrosion
(i) Metal corrosion can be controlled by using other metal.
K Na Ca Mg Al Zn Fe Sn Pb Cu Ag
Ease of releasing electron increases (more electropositive)
When iron is in contact with more electropositive metal for example zinc, rusting of iron isprevented.
Zinc, Zn loses electrons more easily than iron, Fe. Zinc corrodes oris oxidized instead of iron.
Half equation of oxidation : Zn Zn2+
+ 2e
The electrons that are released by zinc flow through the iron to the metal surface where there is
water and oxygen.
Half equation reduction : O2 + 2H2O + 4e 4OH-
When iron is in contact with less electropositive metal for example copper, rusting of iron is faster.
Iron, Fe loses electron more easily than copper, Cu. Hence, iron corrodes / rusts oris oxidizedinstead of copper.
The further apart the metals in the electrochemical series are, the faster the more electropositive
metal corrodes.
Activity 5 :-
Diagram below shows the use of zinc plates on an iron ship to prevent rusting.
(a) Explain how the zinc plates protect the iron ship from rusting.
(b) Write the half equation for the reaction in (a).
Iron ship
Zinc plate
Sea water
TAQ 8388
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction21
Activity 6 :-
1 The diagram shows the set up of the apparatus to study the effect of other metals on the rusting of iron nails.
(a) What is the function of :
(i) phenolphthalein?
(ii) potassium hexacyanoferrate(III) solution?
(b) State the observation for each test tube P, Q, R and S after a days.
(i) Test tube P
(ii) Test tube Q
(iii) Test tube R
(iv) Test tube S
Copper
Jelly + phenolphthalein + potassium hexacyanoferrate(III) solution
ZincMagnesium
P Q R S
Iron nailIron nail
Iron nail Iron nail
SPM
2008
Q9 (b)KNa
Ca
MgAl
Zn
Fe
SnPb
H
CuAg
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction22
(c) Based on the observations,
(i) state the metals that can prevent the rusting of iron nail
(ii) state the metal that can accelerate the rusting of iron nail.
(iii) arrange the four metals i.e. iron, zinc, magnesium and copper in ascending order of their
electropositivity.
(d) (i) State the type of reaction that takes place when iron rusts.
(ii) Write the half equation for the reaction in (d) (i).
(iii) What is the purpose of test tube R in this experiment?
2 The diagram below shows three iron nails that are coiled with stannum, metal Y, and metal Z
respectively and placed in three different beakers.
Each beaker is filled with aqueous sodium chloride. After a few days the following results are obtained.
Beaker Observation
A The iron nail rusts a little.
B The iron nail does not rust.
C The iron nail rusts a lot.
(a) Based on the observations, arrange the metals i.e. tin, Y and Z in a descending order of
their electropositivity.
.., .,
BA C
Tin Y Z
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction23
(b) Suggest one possible metal for
Y: . Z:
(c) Explain why the iron in beaker B does not rust.
(d) For the chemical changes that takes place in beaker A, write the :
(i) oxidation half equation :
(ii) reduction half equation :
(ii) overall ionic equation :
3
(c) If magnesium and iron are exposed to the atmosphere. Which metal will corrode faster?
Explain your answer.
.............................
.............................
(d) Why are the products made of aluminium self-protected from corrosion?
...................... .......
.............................
(e) Electroplating is one way to control the rusting of iron. Suggest two other ways to prevent iron from
rusting?
.............................
Metals will corrode when exposed to the atmosphere over a period of time.
The rate of corrosion depends on the position of the metal in the Electrochemical Series.
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction24
1.4.4 Redox Reaction InDisplacement of Halogen From Its Halide Solution.
a) Electronegativity Series ofHalogen and Halide
Themore electronegative halogen :
can attract electrons from halides that are less electronegative.
displaces less electronegative halogen from its halide solution
gains electrons and acts as an oxidizing agent.
undergoes reduction to form halide ions.
Example :
Chlorine molecule :
gain / lose two electrons.
undergo oxidation / reduction.
acts as an oxidizing / reducing agent.
The halide ions of the less electronegative halogen :
lose their electrons, undergo oxidation, acts as a reducing agent.
Example :
Iodide ion :
gain / lose electrons
undergo oxidation / reduction.
acts as a oxidizing / reducing reducing agent.
HALOGEN HALIDE
Electronegativityof halogens increases.
[A measurement of the
strength of an atom in
its molecule to attract
electrons towards its
nucleus to form
negative ions]
Cl2Bromine
molecule
Chlorine
waterCl
-
Chloride ion
(Potassiumchloridesolution)
The tendency for a
halide to become a
halogen increases.
Br2
Chlorine
molecule
Bromine
water
Br-
Bromide ion
(Potassium
bromidesolution)
I2
Iodine
molecule
Iodine
water
I-
Iodide ion
Potassium
iodide
solution)
Cl2 + 2e 2Cl-
2I- I2 + 2e
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction25
Activity 7 :-
1 Predict whether the following reactions occur or not.
[If the reaction occurs, put a tick( ) ; if no reaction occurs, put a cross ( X ) ]
Reactants / X Products
1 KI + Cl22 KI + Br2
3 KBr + Cl2
4 NaI + Br2
5 NaBr + I2
6 KCl + Br2
7 NaCl + I2
2 A few drops of chlorine water are added to 2 cm3 of potassium iodide solution and the
mixture is then shaken thoroughly.
Half equation forreduction : Cl2 + 2e- 2Cl
-
Half equation foroxidation : 2I- I2 + 2e
-
Ionic equation : Cl2 + 2I- 2Cl
- + I2
Explanation :
Explanation :
Chlorine molecules, Cl2gain / lose electrons to form chloride ions, Cl-.
Chlorine molecule is oxidized / reduced.
Iodide ions, I- gain / loseto form iodine molecule, I2.
Iodide ions are oxidized / reduced.
Iodide ions have oxidized / reduced chlorine molecule, Cl2 . Iodide ion is an oxidizing / reducing.
Chlorine molecules, Cl2 has oxidized / reduced iodide ions. Chlorine is a oxidizing / reducing.
Chlorine water
Potassium iodide solution
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction26
Two layers are formed when an aqueous solution of halogen is mixed with 1,1,1-trichloroethane.
The denser 1,1,1-trichloroethane will be at the bottom and less dense aqueous solution will be at the top.
Halogen Colour inaqueous solution Colour in 1,1,1 trichloroethane, CH3CCl3
Chlorine, Cl2 Pale yellow orcolourless Pale yelloworcolourless
Bromine, Br2Brown / yellowish brown / yellow
(depends on concentration)Brown / yellowish brown / yellow
(depends on concentration)
Iodine, I2Brown / yellowish brown / yellow
(depends on concentration)Purple
Note :
The colour of halogen cannot be differentiated in aqueous solution, especially bromine and
iodine. The presence of halogens is confirmed using 1,1,1 trichloroethane, CH3CCl3
aqueous solution layer
1,1,1-trichloroethane layer
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction27
Activity 8 :-
(a) How do you confirm the formation of iodine in the experiment?
(b)Write the half equation for the chemical change that takes place in :
(i) bromine water :
(ii) potassium iodide :
(iii) a reducing agent :
(iv) an oxidizing agent :
(c) Write ionic equation for redox reaction.
(d) Suggest halogen X that can replace bromine water so that iodine is also formed.
Iodine is formed when bromine water is added to potassium iodide solution.
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction28
1.4.5 Redox Reaction In The Change Of Fe2+
Fe3+
and Fe3+
Fe2+
(a) Changing of iron(II) ion, Fe2+ to iron(III) ion, Fe
3+
Iron (II) ion, Fe2+
undergoes oxidation by losing its electron to form iron(III) ion, Fe3+
Oxidation half equation : Fe2+
Fe3+
+ e
The substance added is an oxidizing agent such as bromine water, Br2.
Bromine molecule gains electrons and undergoes reduction to form bromide ion, Br-.
Reduction half equation : Br2 + 2e 2Br-
Observation for the change ofiron(II) ion, Fe2+ to iron(III) ion, Fe3+ :
(i) Iron(II) solution changes colour from pale green to yellow
(ii) Brown colour of bromine water decolourises
Confirmatory test for iron(III) ion, Fe3+
:
Add sodium hydroxide solution to the solution until excess.
A brown precipitate is formed. It is insoluble in excess sodium hydroxide solution.
Other oxidizing agents that can replace bromine water to change Fe2+
to Fe3+
are :
Oxidizing agent Reduction half equation
Chlorine water, Cl2 Cl2 + 2e 2Cl-
Acidifiedpotassium manganate(VII)
solution, KMnO4MnO4
- + 8H+ + 5e Mn2+ + 4H2O
Acidifiedpotassium dichromate(VI)solution, K2Cr2O7
Cr2O72- + 14H+ + 6e 2Cr
3+ + 7H2O
Hydrogen peroxide, H2O2 H2O2 + 2H+
+ 2e 2H2O
Bromine water
Iron(II) sulphate
solution
Heat
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction29
Activity 9 :-
1 Chlorine water is added drop by drop to 2 cm3
of iron(II) sulphate solution in a test tube. The test
tube is warmed gently.
(a) Write the :
Half equation ofoxidation: ..
Half equation ofreduction: ..
Overall ionic equation :
2 Diagram 2 shows the set- up of apparatus to investigate the reaction between iron(II) chloride solution
and potassium manganate(VII) solution through the transfer of electrons at a distance.
(a) What is the function of dilute sulphuric acid?
(b) In Diagram 2, draw the direction of the flow of electrons.
(c) (i) What is the colour change in the solution around electrode P?
..................................................................................................................................
(ii) Describe a chemical test to determine the product formed in the solution at electrode P.
..............
...............
Dilute sulphuric acid
G
Carbon electrode Q
Potassium
mangganate(VII)Iron(II) chloride solution
Carbon electrode P
DIAGRAM 2
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction30
(d)What is the substance that is being oxidized in the experiment? Explain why.
.....
.....
(e) Write the half equation for the reaction that occurs at electrode Q.
.....
(f) Suggest another reagent that can replace potassium manganate(VII) solution.
....................
(g)What is the change in oxidation number of manganese in the reaction?
......
(b)Changing of iron(III) ion, Fe3+
to iron(II) ion, Fe2+
Iron (III) ion, Fe3+
undergoes reduction bygaining one electron to form iron(III) ion, Fe2+
Oxidation half equation : Fe3+
+ e Fe2+
The substance added is an reducing agent such as zinc, Zn.
Zinc atom loses electrons and undergoes oxidation to form zinc ion, Zn2+.
Reduction half equation : Zn Zn2+ + 2e
Ionic equation : 3Zn + 2Fe3+
3Zn2+
+ 2Fe
Observation for the change ofiron(III) ion, Fe3+ to iron(II) ion, Fe2+ :
(i) Iron(III) solution changes colour from brown to pale green
(ii) Some zinc powder dissolved
Iron(III) sulphate solution
Zinc powder
Heat
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction31
Confirmatory test for iron(II) ion, Fe2+
:
Add sodium hydroxide solution to the solution until excess.
A green precipitate is formed. It is insoluble in excess sodium hydroxide solution.
Other reducing agents that can replace zinc to change Fe2+ to Fe3+ are :
Reducing agent Half equation of Oxidation
Magnesium, Mg Mg Mg2+ + 2e
Sulphur dioxide, SO2 SO2 + 2H2O SO42- + 4H+ + 2e
Hydrogen sulphide, H2S H2S 2H+
+ S + 2e
Sodium sulphite solution, Na2SO3 SO32-
+ H2O SO42-
+ 2H+
+ 2e
Activity 10 :-
1 The following is an equation represents a redox reaction.
Based on the redox reaction :
(a) Write half equation for :
(i) oxidation : H2S 2H+ + S + 2e // S2- S + 2e
(ii) reduction : Fe3+
Fe2+
+ e
(b)Name the substance that acts as :
(i) Oxidizing agent : Iron(III) ion, Fe3+
(ii) Reducing agent : Hydrogen sulphide, H2S //sulphide ion, S2-
(c) Change in oxidation ;
(i) Oxidation number ofiron isdecrease from +3 to +2
(ii) Oxidation number ofsulphur is increase from -2 to 0
2Fe3+
(aq) + H2S(g) 2Fe2+
(aq) + 2H+
(aq) + S(s)
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction32
1.3.6 Redox Reaction In term of Transfer of Electron at a Distance
1. Transfer of electron at a distance occurs when two solutions ofreducing agent and oxidizing agent
areseparated by an electrolyte in a U-tube.
2. Redox reaction occurs as a result of electron flow through an external circuit.
3. Electrons flow from the reducing agent (loseselectrons)to the oxidizing agent (gains electrons)through the connecting wires and can be detected by a galvanometer.
4. Carbon electrode that is immersed in a reducing agent is known as the negative terminal (anode).
5. Carbon electrode that is immersed in an oxidizing agent is known as the positive terminal (cathode).
6. The electrolyteallows the movement of ions andcompletes the electric circuit.
7. The diagram shows the set up of the apparatus for the electron transfer at a distance.
8. Reducing agent loses its electron and undergoesoxidation.
9. Oxidation agent gains its electron and undergoesreduction.
electrolyte
Carbon electrode
Oxidizing agentGains electron
Undergoesreduction
Reducing agentLoses electron
Undergoesoxidation
Carbon electrode_
G
+
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction33
Examples of common substances used asreducing agents are :
Substance Half equation for oxidation Observation/ Test
Potassium iodide,
KI
Potassium bromide,KBr
Iron(II) chloride //
Iron(II) sulphate
In U-tube cell / a chemical cell electrode that is immersedinthereducing agent becomes the negative
terminal or anode because the agent loses electrons and undergoes oxidation.
Oxidation agent gains the electrons and undergoes reduction.
Examples of common substances used asoxidizing agents are :
Substance Half equation for reductionObservation //
Confirmatory test
Acidified potassium
manganate(VII)MnO4
-+ 8H
++ 5e Mn
2++
4H2O
Acidifiedpotassium/
sodium dichromate(VI)Cr2O7
2- + 14H+ + 6e 2Cr3+ + 7H2O
Chlorine water Cl2 + 2e 2Cl-
Bromine water Br2 + 2e 2Br-
Iron(III) chloride,
Iron(III) sulphateFe3+ + e Fe
2+
- In a chemical cell/U-tube cellelectrode that is immersed in an aqueous solution of an oxidizing agent becomes
the positive terminal or anode because the agent gains electrons and undergoes reduction.
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction34
Complete the table below for the electron transfer at a distance
ReactantsOxidizing
agent
Reducing
agent
Half equation for
oxidation : losses of electron
negative terminal
Half equation for
reduction : gains of electron
positive terminal
Ionic equation
Diagram for the set up of
apparatus : The direction of electron flow
Positive / negative terminal
FeSO4(aq)and
Br2(aq)
KI(aq)and
KMnO4(aq)
FeSO4(aq)and
K2Cr2O7(aq)
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction35
1.3.7 Redox Reaction In The Reactivity Series Of Metals And Its Applications
(a) Reactivity series of metals is an arrangement of metals in accordance to the reactivity of theirreactions with oxygen to form metal oxides.
K
Na
Ca
Mg
Al
C
Zn
H
Fe
Sn
Pb
Cu
(b) The metals are arranged in the Reactivity Series by observing how vigorously they react with
oxygen. The metal at the top of the series burns most vigorously and most quickly in oxygen.
(c) Carbon reacts with oxygen to form carbon dioxide. Carbon is also an element in the Reactivity Series
of metals.
(d)Determination the position ofCarbon in Reactivity Series of metals :
(i) If carboncan remove oxygen from a metal oxide // (reduce the metal oxide to metal),
Carbon is above the metal in the Reactivity Series of metal.
(ii)Conversely, if carboncannot remove oxygen from metal oxide, carbon is less reactive than the
metal in the reactivity series of metal. Thus, no reaction will occur.
Carbon is below the metal in the Reactivity Series of metal.
Reactivity of metal towards oxygen increases.
The more reactive metal is able to remove oxygen from less reactive metal oxide.
The more reactive metal gains oxygen to form metal oxide and undergoes oxidation /
oxidation number of metal increases.
The less reactive metal oxide loses oxygen to form metal and undergoes reduction /
oxidation number of metal in the metal oxide decreases.
The more reactive metal has reduced the less reactive metal oxide and acts as reducing a
agent.
The less reactive metal cannot remove oxygen from more reactive metal oxide.
The empirical formula of metal oxide for amore reactive metal than hydrogen in the
reactivity series (such as magnesium) can be determined by heating the metal strongly in a
crucible.
The empirical formula of metal oxide for a less reactive metal than hydrogen (such as
copper) can be determined by passing dry hydrogen gas through strongly heated metal
oxides in a combustion tube.
Carbon + metal oxide metal + carbon dioxide
Carbon + metal oxide (no reaction)
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction36
(e) Determination the position ofHydrogen in Reactivity Series of metals :
(i) If hydrogencan remove oxygen from a metal oxide // (reduce the metal oxide to metal),
Carbon is above the metal in the Reactivity Series of metal.
(ii) If hydrogen is unable to remove oxygen from metal oxide, hydrogen is less reactive than the
matel in the Reactivity Series of metals.
(f) Application of the Reactivity Series
Metals that are located belowcarbon in the Reactivity Series (less reactive than carbon) can beextracted from their ores (metal oxides) using carbon.
Carbon is released as carbon dioxide gas after the reaction
(i) Carbon is widely used to extract iron (Fe), tin/stanum (Sn), zinc (Zn) and lead (Pb) from their
ores. Carbon is more reactive than these metals and act as reducing agent in the metal
extraction process.
(ii) The extraction is carried out in the a blast furnace. Hot air is pumped to the blast furnace to
burn carbon and carbon monoxide is produced . In the blast furnace, a series of chemical
reactions take place.
The extraction of iron, Fe :
C(s) + O2 (g) CO2(g)
C(s) + CO2(g) 2CO (g)
Both carbon, C and carbon monoxide, CO can remove oxygen from the ores
2Fe2O3(s) + C(s) 4Fe(s) + 3CO2(g)
Fe2O3(s) + CO(g) 2Fe(s) + 3CO2(g)
Iron , Fe is extracted from its ore, hematite ( Fe2O3)
Hydrogen + metal oxide metal + water
Note :
If metal X ismore reactive than metal Y, then metal X can remove oxygen from the oxide of metal Y.
metal X can reduce oxide of metal Y.
X + Y oxide X oxide + Y
X oxide + Y (no reaction)
Metal oxide + Carbon Metal + Carbon dioxide
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction37
The extraction of Tin, Sn :
SnO2(s) + C(s) Sn(s) + CO2(g)
SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)
Tin, Sn is extracted from its ore, cassiterite (SnO2).
Metals that are located abovecarbon in the Reactivity Series (more reactive than carbon)can be extracted from their ores (metal oxides) by electrolysis of molten ores using carbon
electrode.
Extraction of aluminium metal, Al from aluminium oxide, Al2O3 (bauksite)
[Refer to Chapter 6/Form 4 :Electrochemistry ]
The empirical formula of metal oxide :
for a more reactive metal than hydrogen in the Reactivity series (such as magnesium) can
be determined by heating the metal strongly in a crucible.
for a less reactive metal than hydrogen (such as copper) can be determined by passing
hydrogen gas through strongly heated metal oxides in a combustion tube.[Refer to Chapter 3/Form 4 : Chemical Formulae & Equations ]
Activity 11 :-
1 Determine whether the following reactions occur or not.
If the reaction occurs, mark and if not, mark X.
Reactants / X Chemical Equation
(a) Hydrogen + zinc oxide
(b) Magnesium oxide + carbon
(c) Copper + zinc oxide
(d) Aluminium + carbon dioxide
(e) Carbon + silver oxide
(f) Hydrogen oxide + copper
(g) Iron(II) oxide + hydrogen gas
(h) Magnesium + steam
(i) carbon dioxide + lead
(j) Iron + lead(II) oxide
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction38
2 Diagram below shows the set up of apparatus to investigate the reactivity of metal P, Q and R
towards oxygen. Metal P, Q and R is heated before hydrogen peroxide is poured into the conical
flask.
The observations are given below
Metal Observation
P Burns fairly bright. The residue is yellow when hot but turns white when cold.
Q Glows faintly. The residue is black.
R Burns vigorously with bright flame. The residue is white.
(a) Write chemical equation for the decomposition of hydrogen peroxide.
..
(b)What is the function of manganese(IV) oxide in the experiment?
..
(c) Based on the observations, arrange the metals P, Q and R in a ascending order of their reactivity.
..
(d)Name the P and Q.
Metal P :.
Metal Q :
(e) Chemical formula the oxide of metal R is R2O3. This oxide can be reduced by hydrogen gas to formetal R.
(i) Write an equation for the reaction between R2O3 with hydrogen.
(ii) Name the oxidation agent in this reaction.
Hydrogen peroxide
+ Manganese(IV) oxide
Heat
P Q R
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4541 CHEMISTRY Chapter 12
Chapter 12 Oxidation & Reduction39
(iii) Draw a labeled diagram of the apparatus that can used to conduct this experiment.
(f)(i) Name another substance that can replace hydrogen peroxide.
(ii) Write an equation to represent the reaction that occurs in (f) (i).
Activity 12 :-
REVISION :
1. REDOXREACTION is a reaction whereoxidationand .occurat the
time.
2. Oxidationinvolved :
losses of ..
gains of....
.of electrons.
. oxidation number.
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4541 CHEMISTRY Chapter 12
3. Reduction involved :
losses of ..
gains of....
. of electrons.
. oxidation number.
4.
Magnesium atom isoxidizedbecause :
magnesium atom is released .. electrons.
the oxidation number of magnesium is . from .. to ..
5.
Hydrogen sulphide, H2S is oxidizedto .
H2S acts as an . agent.
Chlorine gas,Cl2 is reducedto .
Cl2 undergoes reductionand acts as an . agent.
6.
(a) Zn Zn2+ + 2e
Zinc atom undergoes because its .. electrons.
Its oxidation number increases from to
Zinc acts as an ..agent
(b) 2H+ + 2e H2
Hydrogen ion undergoes because it .. electrons.
Its oxidation number from to
Hydrogen ion acts as an ..agent.
7.
Copper(II) ion, Cu2+
undergoes .
Cu2+
ion ... electrons.
Its oxidation number decreases from .to
Cu2+
ion acts as an.agent.
Mg Mg2+
+ 2e
H2S + Cl2 S + 2HCl
Zn + 2HCl ZnCl2 + H2
Cu2+
+ 2e Cu