C12 Notes S Redox

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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(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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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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(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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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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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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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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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


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


equation of oxidation*


equation of reduction*

Ionic equation

Al Cu

2+

SO42-

CuSO

4

2-Al

3+

water

sodium


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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


equation of oxidation*


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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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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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.


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.


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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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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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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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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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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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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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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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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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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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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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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(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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(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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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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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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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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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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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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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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(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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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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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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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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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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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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(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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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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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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(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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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.


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

C12 Notes S Redox

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