chemistry form 5 chapter 3 oxidation and reduction 2014
TRANSCRIPT
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
1/97
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
2/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
3/97
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
4/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
5/97
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 -
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
6/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
7/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
8/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
9/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
10/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
11/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
12/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
13/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
14/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
15/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
16/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
17/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
18/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
19/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
20/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
21/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
22/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
23/97
Example:
There is no change of oxidation state for allelements. This isn't a redox reaction.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
24/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
25/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
26/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
27/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
28/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
29/97
Example:
1. copper(II) oxide is reduced, hence it is theoxidising agent.
2. carbon is oxidised, hence it is the reducing agent.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
30/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
31/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
32/97
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:
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
33/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
34/97
Example:Redox reaction between iron(II) sulphate and potassiumdicromate(VI)
Half equationsFe2+Fe3++ e -------(1)Cr2O72-+ 14H++ 6e2Cr3++ 7H2O -------(2)
To make the number of electrons in both chemical equationequal
(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 :
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
35/97
Example:Redox reaction between iron(III) nitrate and sulphur dioxidegas
Half equationsFe3+Fe2++ e -------(1)SO2+ 2H2OSO42-+ 4H++ 2e -------(2)
To make the number of electrons in both chemical equationequal
(1) x 22Fe3+2Fe2++ 2e
(2) SO2+ 2H2OSO42-+ 4H++ 2eIonic EquationsAdd the 2 equations together. Exclude the electrons.
2Fe3++ SO2+ 2H2O2Fe2++ SO42-+ 4H+
Example:
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
36/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
37/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
38/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
39/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
40/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
41/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
42/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
43/97
Other oxidation agents that get to replace bromine water
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
44/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
45/97
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+
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
46/97
Confirmation Test
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
47/97
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
Other oxidation agents that get to replace bromine water
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
48/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
49/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
50/97
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
51/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
52/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
53/97
Displacement of Halogen From Halide Solution
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
54/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
55/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
56/97
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:
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
57/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
58/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
59/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
60/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
61/97
. 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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
62/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
63/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
64/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
65/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
66/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
67/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
68/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
69/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
70/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
71/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
72/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
73/97
Controlling Rusting
St t C t l R ti
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
74/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
75/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
76/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
77/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
78/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
79/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
80/97
7. Oxygen that is used in combustion of other metals is
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
81/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
82/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
83/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
84/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
85/97
.
Position of Hydrogen in The Reactivity Series of Metals
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
86/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
87/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
88/97
Application of The Reactivity Series of Metals in TheExtraction of Metals
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
89/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
90/97
Extraction of Iron
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
91/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
92/97
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
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
93/97
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)
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
94/97
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.
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
95/97
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
96/97
5. Stanum(IV) oxide in the ore is reduced to tin by the
-
8/12/2019 Chemistry Form 5 Chapter 3 Oxidation and Reduction 2014
97/97
( ) 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.