Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014

8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014

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Oxidation and Reduction in Terms ofOxygen/Hydrogen Transfer

Understanding Oxidation and ReductionOxidation and reduction can be understood from theaspect of:

Losing or gaining oxygen Losing or gaining hydrogen Transferring of electron Changing of oxidation number


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Oxidation and Reduction in Terms of Hydrogen Transfer Oxidation is the process of losing hydrogen. Reduction is the process of gaining hydrogen. For example, ethanol can be oxidised to ethanal:

Ammonia loses hydrogen. Ammonia is oxidised to becomenitrogen. This is an oxidation process.Bromine gains hydrogen. Bromine is reduced to becomehydrogen bromide. This is a reduction process.


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Example:Combustion of Magnesium in Air

Magnesium is oxidised to become magnesium oxide.Displacement of copper(II) oxide by Carbon

Copper(II) oxide is reduced to become copper metal Carbon is oxidised to become carbon dioxide.
http://3.bp.blogspot.com/-Igx3oeUxUeA/UrRPr5vHhYI/AAAAAAAAD1E/dUAzghEtL4I/s1600/Redox2-18.png


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Displacement of lead(II) oxide by Zinc

Zinc is oxidised to become zinc oxide. Lead(II) oxide is reduced to become lead metal

Reaction between Magnesium and Steam

Water is reduced to become hydrogen gas. Magnesium is oxidised to become magnesium oxide.


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Hydrogen Sulphide Reacts with Chlorine

Hydrogen sulphide is oxidised to become sulphur.

Chlorine is reduced to become hydrogen peroxide.Copper(II) oxide Reacts with Ammonia

Copper(II) oxide is reduced to become copper(II) metal. Ammonia is oxidised to become nitrogen gas.


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Oxidation and Reduction in Terms of Electron Transfer Oxidation occurs when a reactant loses electron. Reduction occurs when a reactant gain of electron.

Example:CuO + Mg Cu + MgO

In this reaction,

copper (II) ion incopper(II) oxide gains2electrons to form coppermetal.Half equation:

Cu2++ 2eCuThis is a reduction process.

Magnesium metal loses2electrons to formmagnesium ions inmagnesium oxide.Half equation:

MgMg2++ 2eThis is a oxidation process.


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Summary Magnesium is oxidised because it loses2 electrons

to form magnesium ion, Mg2+. Copper(II) oxide is reduced because copper(II) ion

gains 2 electrons to form copper metal, Cu.


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Oxidation States (Oxidation Numbers)

Oxidation state shows the total number of electrons

which have been removed from an element (a positiveoxidation state) or added to an element (a negativeoxidation state) to get to its present state.

Oxidation State of Some Elements

1. The oxidation state of an element iszero.

Element OxidationState

Mg 0

H2 0

Br2 0


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2. For a simple ion with single atom, the oxidation state isequal to the charge.

Ion Oxidation State

Cu2+ +2

Br- -1

O2- -2

Al3+ +3


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3. Some elements almost always have the same oxidationstates in their compounds:

Example 1:The oxidation state of oxygen is always -2except peroxide, which is -1.

Compound Oxidation state of oxygenH2O -2

H2SO4 -2

ZnO -2

KClO3 -2

H2O2 -1


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Example 2:The oxidation state of hydrogen is always +1except hydride, which is -1.

Compound Oxidation state ofhydrogen

NH3 +1

HCl +1

NaOH +1

MgH2 -1

NaH -1


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4. The sum of the oxidation states of all the atoms ormolecule in a neutral compound is zero.

Ion Sum of Oxidation State

H2O 0

CO2 0

NH3 0


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5. The sum of the oxidation states of all the atoms in an ionis equal to the charge on the ion.

Ion Sum of Oxidation StateNO3- -1

CO32- -2

PO43- -3

NH4+ +1


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Working Out the Unknown Oxidation State of anElement in A Compound

The sum of the oxidation state of each element in a

compound are equal to the charge of the compound. This rule can be used to find the unknown oxidation

number of an element is a compound.

Example 1

Find the oxidation state of all the elements in aChlorate(V), ClO3-ion.

Answer:Oxidation number of O = -2

Oxidation number of Cl = xx + 3(-2) = -1

x = -1 + 6 = +5


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Example 2Find the oxidation state of all the elements in aPotassium manganate(VII), KMnO4ion.

Answer:Oxidation number of K = +1

Oxidation number of O = -2

Oxidation number of Mn = x

(+1) + x + 4(-2) = 0x = -1 + 8x = +7


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Example 3Find the oxidation state of all the elements in anAmmonium ion, NH4+ion.

Answer:Oxidation number of H = +1

Oxidation number of N = x

x + 4(+1) = +1x = +1 - 4x = -3


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Using Oxidation States in Naming Compounds

1. You will have come across names like iron(II)sulphate and iron(III) chloride. The (II) and (III)are the oxidation states of the iron in the twocompounds: +2 and +3 respectively. That tells youthat they contain Fe2+and Fe3+ions.

Example

Formula Name of the compound

FeCl2 Iron(II) chloride

FeCl3 Iron (III) chloride

MnO2 Manganese(IV) oxide

Mn(NO3)2 Manganese (II) nitrate

PbCl2 Lead(II) chloride


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2. Transition metals always show difference oxidationstate as shown in the table below.

Metal Oxidation state

Fe +2, +3

Cu +1, +2

Mn +2, +4, +6, +7

3. Non-metal elements (except fluorine) usually have

more than one oxidation state.


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Oxidation and Reduction in Term of Changes ofOxidation State

Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state

1. Another way to determine oxidation and reduction is

to see the change of the oxidation state after areaction.2. An atom is said to be oxidised when its oxidation

state increases.3. An atom is said to be reduced when its oxidation

state decreases.


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Example:

1. The magnesium's oxidation state has increased by 2,from 0 to +2. Therefore, it has been oxidised.

2. The hydrogen's oxidation state has decreased by 1,from +1 to 0. Therefore it has been reduced.

3. The chlorine is in the same oxidation state on bothsides of the equation - it hasn't been oxidised nor

reduced.


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Example:

There is no change of oxidation state for allelements. This isn't a redox reaction.


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Example:

1. In this example, we can see that the oxidation state ofchlorine has increased and also decreased.

2. Chlorine is oxidised and reduced, at the same time.3. This is a good example of

a disproportionation reaction. A disproportionationreaction is one in which a single substance is both

oxidised and reduced.


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Redox Reaction

A redox reaction is a chemical reaction that involvesreduction and oxidation that occurs simultaneously.

1. A redox reaction is a chemical reaction that involvesreduction and oxidation that occurs simultaneously.

2. In a redox reaction, both reduction and oxidation aregoing on side-by-side.

3. As we have learned, oxidation and reduction can bedefined in terms of

loss or gain of oxygen loss or gain of hydrogen transfer of electrons change in oxidation number


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The redox reaction that you need to know in SPMincludes

1. Redox reaction in aqueous solution changing of iron(II) ions to ions(III) and vice

versa displacement of hydrogen

displacement of halogens transfer of electrons at a distance

2. Electrochemistry3. Corrosion of metals4. Combustion of metals5. Rusting of iron6. Extraction of metal


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Oxidising Agents and Reduction Agents

1. In a redox reaction, compound that is reduced is

oxidizing agent. An oxidising agent is substance whichoxidises something else.2. Inversely, compound that is oxidised is reducing

agent. A reducing agent reduces something else.


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Example:

1. In this reaction, iron(III) oxide is reduced.

Therefore it is the oxidising agent. It has oxidisedcarbon monoxide to become carbon dioxide.

2. The carbon monoxide is oxidised. Therefore it actas the reducing agent. It has reduced iron(III)

oxide to become iron metal.

E l


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Example:

1. copper(II) oxide is reduced, hence it is theoxidising agent.

2. carbon is oxidised, hence it is the reducing agent.


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Commonly Used Oxidising and Reducing Agent andTheir Half Equation

Oxidising Agent

Acided Potassium Manganate (VII)MnO4-+ 8H+ + 5e Mn2++ 4H2O

Acided Potassium Dicromate (VI)Cr2O72-+ 14H++ 6e 2Cr3++ 7H2O

Hydrogen Peroxide

H2O2+ 2H++ 2e 2H2O

Concentrated Nitric AcidNO3-+ 4H++ 3e NO + 2H2O


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Reducing Agent

Sulphur DioxideSO2+ 2H2O SO42-+ 4H++ 2e

Hydrogen Sulphide

H2S 2H++ S + 2e

Sodium Sulphite AqueousSO32-+ H2O SO42-+ 2H++ 2e

Tin(II) Chloride AqueousSn2+ Sn4++ 2e


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Forming Ionic Equation from the Half Equation

1. Ionic equation of a redox reaction can beformed from the half equations of thereaction.

2. When writing the ionic equation, make

sure that the number of electrons inboth the oxidation reaction andreduction reaction are balance.

Example:


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Example:

Redox reaction between potassium iodide and potassiummanganate (VII)

Half equations2I-I2+ 2e -------(1)MnO4-+ 8H++ 5eMn2++ 4H2O -------(2)

To make the number of electrons in both chemical equationequal

(1) x 510I-5I2+ 10e

(2) x 2 2MnO4-+ 16H++ 10e2Mn2++ 8H2OIonic EquationsAdd the 2 equations together. Exclude the electrons.

10I-+ 2MnO4-+ 16H+5I2+ 2Mn2++ 8H2O


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Example:Redox reaction between iron(II) sulphate and potassiumdicromate(VI)

Half equationsFe2+Fe3++ e -------(1)Cr2O72-+ 14H++ 6e2Cr3++ 7H2O -------(2)


(1) x 66Fe2+6Fe3++ 6e

(2) Cr2O72-+ 14H++ 6e2Cr3++ 7H2OIonic EquationsAdd the 2 equations together. Exclude the electrons.

6Fe2++ Cr2O72-+ 14H+6Fe3++ 2Cr3++ 7H2O

Ex mpl :


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Example:Redox reaction between iron(III) nitrate and sulphur dioxidegas

Half equationsFe3+Fe2++ e -------(1)SO2+ 2H2OSO42-+ 4H++ 2e -------(2)


(1) x 22Fe3+2Fe2++ 2e

(2) SO2+ 2H2OSO42-+ 4H++ 2eIonic EquationsAdd the 2 equations together. Exclude the electrons.

2Fe3++ SO2+ 2H2O2Fe2++ SO42-+ 4H+

Example:


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Example:Redox reaction between iron(III) chloride and hydrogensulphide gas

Half equationsFe3+Fe2++ e -------(1)H2S2H++ S + 2e -------(2)

To make the number of electrons in both chemical equationequal(1) x 2

2Fe3+2Fe2++ 2e(2)

H2S2H++ S + 2eIonic EquationsAdd the 2 equations together. Exclude the electrons.

2Fe3++ H2S2Fe2++ 2H++ S


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Oxidation Reduction RedoxReaction

OxidisingAgent

ReducingAgent

A gain inoxygen

A loss ofoxygen

Transfer ofoxygen

Donor ofoxygen

Acceptor ofoxygen

A loss ofhydrogen

A gain inhydrogen

Transfer ofhydrogen

Acceptor ofhydrogen

Donor ofhydrogen

A loss ofelectron

A gain ofelectrons

Transfer ofelectron

Acceptor ofelectrons

Donor ofelectrons

An increase inoxidationnumber

A decrease inoxidationnumber

A change inoxidationnumber

Its oxidationnumberdecreases

Its oxidationnumberincreases

R d ti i l ti


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Redox reaction in aqueous solution

Change of Iron(II) Ion to Iron(III) Ion

1. Iron shows two oxidation numbers, that is +2and +3.

2. The aqueous solution of iron(II) ion Fe2+is light

green in colour. The aqueous solution ofiron(III) ion Fe3+is brown in colour.3. The change of iron(II) ion for iron(III) ion is

an oxidation process. This can be done bymixing an oxidation agent.


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Example

1. 2 cm of iron(II) sulphate solution is poured into a test

tube.2. Bromine water is added drop by drop into the solution

until no further changes are observed.3. The mixture is then shaken and warmed gently.

4. The observation is recorded.

Procedure:

Ob ti


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Observation:

1. The brown colour of bromine water turns colourless.2. The colour of the solution changes from light green to

yellowish brown.Half Equations:

Fe2+Fe3++ e

Br2+ 2e2Br-Ionic Equation

2Fe2+

+ Br2

2Fe3+

+ 2Br-

E l ti


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Explanation:

1. The light green colour of iron(II) sulphate solutionturns brown because iron(II) ions Fe2+are oxidise

2. to become iron(III) ion, Fe3+

.3. Iron(II) ion, Fe2+undergoes oxidation by releasingelectron to form iron(III) ion, Fe3+.

4. The brown colour of bromine water turn colourlessbecause bromine molecules are reduced to becomebromide ions.

5. Bromine molecules receive electrons andundergoesreduction to form bromide ion, Br.

Oxidising agent: Bromine waterReducing agent: Iron(II) ions Fe2+

Confirmation Test


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f m2 cm of solution of the product is filled into a test tube.

Test 1: Dilute sodium hydroxide solution (NaOH) is then

added into the test tube until excess.Result : Brown precipitate formed. The precipitate does notdissolve in excess sodium hydroxide solution.

Test 2 : Dilute ammonium hydroxide solution(NH4OH)/ammonia aqueous (NH3) is then added into the testtube until excessResult :Brown precipitate formed. The precipitate does notdissolve in excess ammonium hydroxide solution /ammonia

aqueous.

Test 3: 2cm of potassium thiocyanate is added into thetest tube.

Result : Red blood solution formed.

Other oxidation agents that get to replace bromine water


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1) Chlorine waterHalf EquationsCl2+ 2e2Cl-

2) Acidic potassium manganate (VII)Half EquationsMnO4-+ 8H+ + 5e Mn2++ 4H2O

3) Potassium dichromate (VI)Half EquationsCr2O72-+ 14H++ 6e 2Cr3++ 7H2O

4) Hydrogen peroxide

Half EquationsH2O2+ 2H++ 2e 2H2O5) Concentrated nitric acid

Half EquationsNO3-+ 4H++ 3e NO + 2H2O

Change of Iron(III) Ion to Iron(II) Ion


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g f ( ) ( )

1. The change of iron(III) ion for iron(II) ion is a reduction process.This can be done by mixing a reducing agent.

Procedure:

1. 2 cm of iron(III) sulphate solution is poured into atest tube.2. Half spatula of zinc powder is added into the solution.3. The mixture is then shaken and warmed gently.4. The observation is recorded.

b


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Observation:

1. Zinc powder dissolves.

2. The brown coloured of iron(III) sulphate solutionturn light green.

Half Equations:

Fe3+Fe2++ eZn Zn2+ + 2e

Ionic Equation

2Fe3++ Zn2Fe2++ Zn2+


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Confirmation Test


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Confirmation Test2 cm of solution of the product is filled into a test tube.

Test 1: Dilute sodium hydroxide solution (NaOH) is thenadded into the test tube until excess.Result : Dirty green precipitate formed. The precipitate doesnot dissolve in excess sodium hydroxide solution.

Test 2 : Dilute ammonium hydroxide solution(NH4OH)/ammonia aqueous (NH3) is then added into the testtube until excessResult : Green precipitate formed. The precipitate does not

dissolve in excess ammonium hydroxide solution /ammoniaaqueous.

Test 3: 2cm of potassium thiocyanate is added into the testtube.

Result : No change observed



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g g p

1) MagnesiumHalf Equations

Mg Mg2+ + 2e2) Sulphur dioxide gas

Half EquationsSO2+ 2H2O SO42-+ 4H++ 2e

3) Hydrogen sulphide gasHalf EquationsH2S 2H++ S + 2e

4) Solution of sodium sulphide, Na2SO3Half Equations

SO32-+ H2O SO42-+ 2H++ 2e5) Solution of tin(II) Chloride

Half EquationsSn2+ Sn4++ 2e

Displacement of Metal


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Displacement of Metal

1. The electrochemical series is a series of

arrangement of metals according to the order ofthe tendency of the metal to lose electrons toform positive ions.

2. Elements which placed higher in the

electrochemical series are more electropositive act as strong reducing agent can be oxidised easily

3. The metal ions are weak oxidising agents becausethey do not have a tendency to gain electrons.


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4. In a reaction of displacement of metal, a metal which isplaced on the top of the electrochemical series (more

electropositive) can displace other metals that lie below it (less electropositive) from its salt solution.5. So,

there is a transfer of electrons from a moreelectropositive metal to the ions of a metal which is lesselectropositive.

The more electropositive metal acts as a reducing agent.The metal experiences oxidation and is oxidised to ametal ion.

The metal ion which is less electropositive acts as anoxidising agent. This ion experiences reduction and isreduced to a metal.


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ExamplesReaction between magnesium and copper(II) sulphate solution

Mg + CuSO4MgSO4+ Cu

Observation:

The blue colour of copper(II) sulphate solution turn colourless.

Half Equation:MgMg2++ 2eCu2++ 2eCu

Ionic Equation:

Mg + Cu2+Mg2++ Cu


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Displacement of Halogen From Halide Solution


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Displacement of Halogen From Halide Solution

1.Halogens are elements in Group 17 of the Periodic Table ofElements. They are fluorine, chlorine, bromine, iodine,astatine.2.All halogens tend to accept one electron to form negativeions. For instance,

3.The ions of halogen are called halide.4.The electronegativity of halogens decreases down thegroup, as shown in the chart below:Cl2+ 2e2C1-

5.Halogen which placed higher in group 17 are more electronegative act as strong oxidising agent can be reduced easily


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6. The halogen which is at a higher position in group 17 (more electronegative and reactive) can displace a halogenthat is below it (less electronegative and less reactive)

from its solution of halide ions.7. When the displacement reaction of the halogen occurs:

transfer of electrons from the halide ions which are

positioned further down in group 17 to halogenswhich are positioned further up occurs. Halogens which are positioned further up in group 17

act as oxidation agents. These halogens undergo reduction and are reduced

to halide ions. The halide ions which are positionedfurther down in group 17 act as reducing agents.These ions undergo oxidation and are oxidised tohalogens.

Example


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Example

Reaction between Chlorine water and Potassium Bromidesolution.

Procedure

1. A few drops of chlorine water are added to 2 cm ofpotassium bromide solution.2. 2 cm of tetrachloromethane is then added into the

mixture and shaken.3. The colour of the tetrachloromethane (the lower layer)

is recorded.

Observation:


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The colourless tetrachloromethane turn brown. This indicatesthat bromine is presence in the solution.

Half Equation:2Br-Br2+ 2eCl2+ 2Cl-Cl2

Ionic Equation:

Cl2+ 2Br-

2Cl-

+ Br2Note:

1. Chlorine is more electronegative than bromine.2. In the reaction, chlorine molecules displaced bromide ions

in the solution.3. Bromide ion is oxidised by releasing electrons to become

bromine.4. Chlorine molecule is reduced by receiving electrons from

bromide ion.

Oxidising agent: ChlorineReducing agent: Bromide

Identifying Halogens in a Solution


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fy g g

1. If halogen or halide ions are involved, usually aconfirmatory test is required to testify the presence of

the halogen or halides. The tetrachloromethane can beused in the halogen confirmatory test.

2. This can be done by shaking the aqueous halogen solution ina little tetrachloromethane. Two layer; are formed.

Tetrachloromethane which is denser forms a layer belowwhile the aqueous solution forms a layer on top. So, thehalogen present can be confirmed by the colour of thehalogen in tetrachloromethane.

3. The identity of chlorine, bromine and iodine cannot beconfirmed through the colour of its aqueous solutionbecause its colour changes depending on its concentration.

4. Halide solutions in water are colourless. For example,solution of NaCl, NaBr, NaI, KCl, KBr and KI are colourless.

H l C l f l i C l f h l i


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Halogen Colour of aqueous solution Colour of halogen intetrachloromethane

Chlorine Pale yellow and almost

colourless

Pale yellow and almost colourless

Bromine Brown or yellowish brown oryellowdepending on concentration

Brown or reddish yellow or yellowdepending on concentration

Iodine Brown or yellowish brown oryellowdepending on concentration

Purple

Transfer of Electrons from One Point to Another


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1. A redox reaction occurs when a solution of an oxidisingagent is mixed with a solution of a reducing agent.

2. If the solution of the oxidising agent and the solution ofthe reducing agent are separated by an electrolyte in aU-tube, the redox reaction will still take place but thetransfer of electrons will occur through the connectingwire.

3. The reducing agent undergoes oxidation with the


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. g g gloss of electrons. So, the carbon electrode that isimmersed in the solution of the reducing agentbecomes the negative cell terminal.

4. Electrons flow through the connecting wire to thecarbon electrode that is immersed in the oxidisingagent solution. The carbon electrode acts as thepositive cell terminal.

5. The oxidising agent undergoes reduction with theacceptance of electrons.

Guide to Solve Problems Related to the Transfer ofElectrons Through a Distance

1. Oxidation always happens at anode.2. Reduction always happens at cathode.3. Anode is the negative terminal.

4. Cathode is the positive terminal.

Reaction Between Potassium Dichromate(VI) and Potassium


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Iodide

Step 1: Identifying the Oxidising Agent andReducing Agent

Oxidising Agent: Potassium dichromate(VI)Reducing Agent:Potassium Iodide

Step 2: Determining the Oxidation and Reduction Processd P di i h Ob i


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and Predicting the ObservationOxidationThe reducing agent (Potassium Iodide) undergoes oxidation

2I---> I2+ 2e

Observation: The colourless solution turn yellow/orange.

Note: Potassium iodide is colourless whereas iodine is yellow or orange in colourwhen dissolve in water.

ReductionThe oxidising agent undergoes reduction

Cr2O72-

+ 14H+

+ 6e

2Cr3+

+ 7H2O

Observation: The orange colour of the solution turn green.

Note: Dichromate(VI) ion is orange in colour whereas chromium(III) ion is green in

colour

Step 3: Identifying the Anode and CathodeEl t d P A d


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Electrode P: AnodeElectrode Q: Cathode

Note: Oxidation occurs at anode whereas reduction occurs at cathode.

Step 4: Determine the positive and Negative Terminal

Positive Terminal: Electrode QNegative Terminal: Electrode P

Note: Anode is the negative terminal whereas cathode is the positive terminal.

Step 5: Determine the Direction of Flow of Electrons.

From electrode P to electrode Q.

Note: Electrons flow from the negative terminal to the positive terminal through

the wire.

Reaction Between Iron(II) Sulphate and Bromine Water


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Step 1: Identifying the Oxidising Agent and ReducingAgent

Oxidising Agent: Bromine waterReducing Agent: Iron(II) sulphate

Step 2: Determining the Oxidation and Reduction Processnd P dictin th Obs ti n


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and Predicting the Observation

Oxidation

The reducing agent undergoes oxidationFe2++ 2e --> Fe3+

Observation: The green colour of iron(II) sulphate solution

turn brown.Note: Iron(II) ion is green in colour whereas iron(III) ion is brown in colour.

ReductionThe oxidising agent undergoes reduction

Br2+ 2e --> 2Br-

Observation: The yellow/orange colour of bromine waterbecome colourless.

Note: Bromine is yellow/orange in water whereas bromide is colourless.

Step 3: Identifying the Anode and CathodeElectrode P: Anode


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Electrode P: AnodeElectrode Q: Cathode

Note: Oxidation occurs at anode whereas reduction occurs at cathode.

Step 4: Determine the positive and Negative Terminal

Positive Terminal: Electrode QNegative Terminal: Electrode P

Note: Anode is the negative terminal whereas cathode is the positive terminal.

Step 5: Determine the Direction of Flow of Electrons.

From electrode P to electrode Q.

Note: Electrons flow from the negative terminal to the positive terminal through

the wire.

Corrosion as a Redox Reaction


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1. Corrosion of metal = metal loses its electrons to formpositive ions.Example

Corrosion of ironFe Fe2++ 2e

Corrosion of MagnesiumMgMg2++ 2e

2. The higher the position of a metal in the electrochemicalseries, the more electropositive (reactive) the metal is.The metal has a greater tendency to give away electrons to

form the metal ion, that is the metal is more easilycorroded.3. For instance, the metal magnesium corrodes more easily

than copper because magnesium is more electropositivethan copper.

4 C i f t l i d ti th t l


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4. Corrosion of metal is a redox reaction as the metalloses electrons to oxygen and water, which act as theoxidising agents to receive the electrons.

5. Corrosion of iron is also called rusting.Fe (s)Fe2+(aq) + 2e

6. Rusting of iron can only occur if both oxygen and waterare present.

Rusting of Iron


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1. In the rusting of iron, iron acts as the reducing agent andoxygen gas as the oxidising agent.

2. The process or rusting of iron can be explained by usingfigure below.

3. When the surface of the iron is exposed to water droplets,the center of the water droplets undergoes the process of


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the center of the water droplets undergoes the process ofoxidation and is known as the anode.

4. The edge of the water droplets undergoes a process of

reduction and is known as the cathode. (The edge of thewater droplets acts as the cathode because of theconcentration of soluble oxygen is higher on the edge ofthe water droplets than in the center.)

5. At the anode, the metal iron undergoes oxidation to formthe iron(II) ion with the loss of electrons.Fe Fe2++ 2e

6. Electrons that are free at the anode flow through themetal iron to the cathode area where soluble oxygen in the

water accepts electrons to form hydroxide ions.O2+ 2H2O + 4e4OH-

7. The iron(II) ions are then combines with the hydroxide ionto form iron(II) hydroxide.

Fe2+

+ OH-

Fe(OH)2

8. Iron(II) hydroxide is then oxidised by oxygen to formi n(III) h d xid


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iron(III) hydroxide.

4Fe(OH)2+ 2H2O + O24Fe(OH)3

9. The iron(III) hydroxide is then decomposed to formhydrated iron(III) oxide, Fe2O3xH2O by oxygen in theair.

4Fe(OH)3Fe2O3xH2O

10. The hydrated iron(III) oxide is brown in color and isknown as rust.

11. The overall equation for the rusting of iron is

4Fe + 3O2+2xH2O2Fe2O3xH2O


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Controlling Rusting

St t C t l R ti


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Steps to Control RustingThe use of a metal which is less electropositive

1. The plating of iron with a thin layer of a metal which is lesselectropositive such as tin, silver or copper will prevent the ironunderneath it to react with water and air, and so prevents the ironfrom rusting.

2. However, the rusting of iron will occur faster if the protectivelayer is scratched. This is because iron is more electropositivethan tin, silver or copper. The plating of iron by tin is used a lot inthe making of tinned food.

The use of a metal which is more electropositive


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1. Metals which are more electropositive are used as asacrificial metal to prevent corrosion on metals whichare less electropositive. The metal which is moreelectropositive corrodes and acts as the anode.

2. The less electropositive acts as the cathode and isprotected from corroding. This method is known as thecathode protection or electrochemical protection.

Cover by paint, oil and grease


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A layer of paint, oil/grease, or plastic that is used to coverthe surface of iron from contact with air and water in the

atmosphere. So, rusting can be avoided. For example,1. Protection by a layer of paint usually is used for iron

and steel objects like cars, ships, bridges and otherthings which do not undergo friction during use.

2. Protection by a layer of oil/grease is used for part ofmachinery that move.3. Protection by plastic is used for daily items at home

like clothes hangers and fencing.4. Alloying of iron can prevent rusting. For example, when

iron is alloyed with chromium and nickel to formstainless steel, the layer of chromium oxide that ishard, strong and difficult to crack on the surface ofthe iron alloy prevents the iron from rusting.

Series of Reactivity of Metals


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1. When a metal reacts with oxygen to form a metaloxide, a redox reaction occurs.Metal + OxygenMetal oxide

2. In this reaction, Metal is oxidised to metal ions. The oxidation

number of the metal increases.

Oxygen is reduced to the oxide ion. The oxidationnumber of oxygen decreases from 0 to -2.

Metal acts as a reducing agent while oxygen actsas an oxidising agent.

3. Different metals burn in oxygen with different ratedepending on their differing activeness.4. The more reactive metal towards oxygen, the

brighter and faster the combustion of the metal.

5. The arrangement of the metal according to the


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5. The arrangement of the metal accord ng to thetendency of reaction with oxygen to form the metaloxide is known as the reactivity series of metals.

6. Figure above shows the experiment is conduct tobuild the reactivity series of metal. Table belowshows the result of the experiment.

Metal Observation


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Magnesium Combust quickly with a bright white shiny flame. Awhite powder is formed.

Zinc A bright flame spread slowly. The powder produced isyellow when hot, and white when it is cold.

Iron Embers spread slowly. A reddish brown powder is

formed.

Lead Red hot embers slowly. The powder produced is brownwhen hot, and yellow when it is cold.

Copper Embers burn at a very slow rate. A black powder isformed.


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7. Oxygen that is used in combustion of other metals is


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7. Oxygen that is used in combustion of other metals isprovides by heating solid potassium manganate(VII),also can be obtained from

a) heating the mixture potassium chlorate (V) andmanganese (IV) oxide (catalyst)

2KClO3(s)2KCl (s) + 3O

2(g)

b) heating potassium nitrate

2KNO3(s)2KNO2(s) + O2(g)

Position of Carbon in The Series of Reactivity of Metals


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1. The position of carbon in the series of reactivity ofmetals can be determined based on:

the ability of metals to take away oxygen fromcarbon oxide, that is carbon dioxide. the ability of carbon to take away oxygen from

metal oxides

Experiment 1 1. If the metal can take away oxygenfrom carbon dioxide then the metal


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from carbon dioxide, then the metalis more reactive than carbon.Metal + carbon dioxidemetal oxide

+ carbon2. On the other hand, if the metal

cannot take away oxygen from carbondioxide, then the metal is less

reactive than carbon.

Example

2Mg (s) + CO2(g)2MgO (s) + C (s)

Conclusion: Magnesium is morereactive than carbon.

Experiment 2 1. If carbon can take away oxygen frommetal oxide, then the carbon is more


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metal oxide, then the carbon is morereactive than the metal.Carbon + metal oxide metal +

carbon dioxide2. On the other hand, if carbon cannottake away oxygen from metal oxide,then the carbon is less reactive thanthe metal.

Example

C + 2CuO2Cu + CO2

Conclusion: Copper is less reactivethan carbon.

The chart below show the position of carbon in a reactivityseries.


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.

Position of Hydrogen in The Reactivity Series of Metals


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1. The position of hydrogen in the reactivity series of

metal can also be determined based on its ability todisplace oxygen from metal oxides.

2. If hydrogen is more reactive than a metal, it candisplace oxygen from metal oxide, and reduces the

metal oxide to its metal.3. Hydrogen + metal oxide metal + water4. Conversely, if hydrogen cannot remove oxygen from

metal oxide, hydrogen is less reactive than the metal inthe reactivity series of metal.

5. Hydrogen can reduce iron (II) oxide, Fe2O3to formiron, Fe and water.

Fe2O3+ 3H22Fe + 3H2O

6. However, hydrogen cannot reduce zinc oxide, ZnO.7 Therefore hydrogen is below zinc but above iron in


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7. Therefore hydrogen is below zinc but above iron inthe reactivity series of metals.

8. The diagram below shows the set-up of apparatus

used to determine the position of hydrogen in thereactivity series of metal.

9. The chart below shows the position of carbon andhydrogen in the reactivity series of metal base on theirability to attract oxygen to form oxide.


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Application of The Reactivity Series of Metals in TheExtraction of Metals


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Extraction of Metals

1. The method that is used in the extraction of metalfrom its ore depends on the position of the metal in thereactivity series of metals.

metals that are located lower than carbon in thereactivity series of metals can be extracted using

the reduction of metal oxide by carbon in a blastfurnace.

metals that are located higher than carbon in thereactivity series of metals are extracted by using

electrolytic melting of metal compounds.2. The chart below give the method of extraction ofcertain metals by referring to the reactivity series.


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Extraction of Iron


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1. Iron is extracted from its ore, that is hematite (Fe2O3)and magnetite (Fe3O4) through reduction by carbon in

the form of carbon in a blast furnace.

2. The mixture of iron ore, carbon, and limestone is enteredinto a blast furnace through the top of the furnace.

3. Hot air is then put in through the lower part of thefurnace

4. Limestone (calcium carbonate) is disintegrated by hot air

into calcium oxide and carbon dioxide gas.CaCO3(s)CaO (s) + CO2(g)

5. Carbon also burns in hot air to produce carbon dioxide gas

C (s) + O2(g)

CO2(g)

6. Carbon dioxide that is produced reacts with excesscarbon to produce carbon monoxide gas which is a type


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carbon to produce carbon monoxide gas which is a typeof reducing agent.

CO2+ C2CO

7 Carbon and carbon monoxide then reduces the iron ore to


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7. Carbon and carbon monoxide then reduces the iron ore tomelted iron which flows to the lower part of the furnace.

2Fe2O3(s) + 3C(s)4Fe(s) + 3CO2(g)Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)Fe3O4(s) + C(s)3Fe(s) + 2CO2(g)Fe3O4(s) + 4CO(g)3Fe(s) + 4CO2(g)

8. In the blast furnace, calcium oxide that is produced fromthe disintegration of calcium carbonate, reacts withforeign matter such as sand (silicon dioxide) in the iron oreto form slag.

CaO (s) + SiO2(s)

CaSiO3(s) (slag)


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9. The melted dross flows down to the bottom part of the

furnace and floats on the layer of melted iron.10. The melted iron and dross then are taken out from the

furnace separately.11. The melted iron is cooled in a mould to form cast iron,

while the dross is used to make the foundation forroads and houses.


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5. Stanum(IV) oxide in the ore is reduced to tin by the


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( ) yreducing agent carbon and carbon monoxide.

SnO2 (s) + 2C(s)Sn (s) + 2CO (g)

SnO2 (s) + C(s)Sn (s) + CO2(g)SnO2 (s) + 2CO(s)Sn (s) + 2CO2(g)

6. The melted tin that is formed collects at the base of

the furnace and then is channeled out into a mould toform tin ingot.

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Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014

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