Suggested Teaching Scheme - STMGSSintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book2Banswer.pdf · Suggested Teaching Scheme ... Reactions in simple chemical cells and the electrochemical

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Suggested Teaching Scheme

The following suggested teaching schemes are for teachers’ reference only. Teachers may revise them based on the time-tabling arrangement of their own schools.

Scheme 1: Chemistry to be studied in Secondary 3, 4, 5 and 6In many schools, the Chemistry curriculum is studied in Secondary 3, 4, 5 and 6. Although the distribution of periods varies from school to school, the total number of periods for the curriculum is generally around 4�6. A possible distribution of periods is as follows:

A possible distribution of periods for S3, S4, S5 and S6

S3 S4 S5 S6

Number of teaching weeks per year 28 28 28 �6

Number of periods per week 2 5 5 5

Total number of periods per year 56 �40 �40 80

Total number of periods for the curriculum 4�6

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S3(56 periods)

Topic � Planet Earth �2

Topic 2 Microscopic World I 44

S4(�40 periods)

Revision on laboratory safety �

Topic 3 Metals 39

Topic 4 Acids and Bases 45

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 4�

Topic 6 Microscopic World II �4

S5(�40 periods)


Topic 7 Fossil Fuels and Carbon Compounds 32

Topic 8 Chemistry of Carbon Compounds 45

Topic 9 Chemical Reactions and Energy �3

Topic �0 Rate of Reaction �6

Topic �� Chemical Equilibrium �8

Topic �2 Patterns in the Chemical World �5

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3


2

Topic 5 Redox Reactions, Chemical Cells and Electrolysis

Scheme 2: Chemistry to be studied in Secondary 4, 5 and 6In some schools, the Chemistry curriculum is studied in Secondary 4, 5 and 6. The total number of periods for the curriculum is generally around 360. A possible distribution of periods is as follows:

A possible distribution of periods for S4, S5 and S6

S4 S5 S6

Number of teaching weeks per year 28 28 �6

Number of periods per week 5 5 5

Total number of periods per year �40 �40 80

Total number of periods for the curriculum 360

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S4(�40 periods)

Topic � Planet Earth 8

Topic 2 Microscopic World I 3�

Topic 3 Metals 32

Topic 4 Acids and Bases 36

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 33

S5(�40 periods)


Topic 6 Microscopic World II �3

Topic 7 Fossil Fuels and Carbon Compounds 29

Topic 8 Chemistry of Carbon Compounds 4�

Topic 9 Chemical Reactions and Energy �2

Topic �0 Rate of Reaction �5

Topic �� Chemical Equilibrium �6

Topic �2 Patterns in the Chemical World �3

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

3


Suggested number of periods for Topic 5

Chemistry forTotal number

of periodsSuggested number of periods for each unit

S3–S6(Scheme �)

4�

Unit �8 Chemical cells in daily lifeUnit �9 Simple chemical cellsUnit 20 Oxidation and reductionUnit 2� Oxidation and reduction in chemical cellsUnit 22 Electrolysis

44

�85

�0

S4–S6(Scheme 2)

33

Unit �8 Chemical cells in daily lifeUnit �9 Simple chemical cellsUnit 20 Oxidation and reductionUnit 2� Oxidation and reduction in chemical cellsUnit 22 Electrolysis

43

�448

4


Teaching Plan

An extension of the study of the reactivity of metals in Topic 3 leads to the study of electron flow in the external circuit when two different metals are immersed in an electrolyte. This is a chemical cell, a device in which chemical energy is converted into electrical energy. Students will study the characteristics of common primary cells and secondary cells in Unit �8. Reactions in simple chemical cells and the electrochemical series of metals are discussed in Unit �9.

To enhance students' understanding of the chemistry involved in a chemical cell, the concept of redox reactions is introduced in Unit 20. Students will carry out investigations involving common oxidizing and reducing agents. They will also learn how to write chemical equations for redox reactions.

With the concepts related to redox reactions, students will study reactions occurring in more complicated chemical cells (such as lead-acid accumulator and fuel cell) in Unit 2�.

Unit 22 introduces electrolysis, the use of electricity to bring out chemical reactions. Students will study reactions that occur during electrolysis. In addition, students should be able to predict products in electrolysis according to the different factors affecting the preferential discharge of ions.

The concepts of redox reactions have a number of applications in industrial chemistry and daily life. Through searching for information and critically reading articles about electrochemical technology, students should appreciate the contribution of chemical knowledge to technological innovations, which in turn improve our quality of life. Students should also be able to assess the environmental impact and safety issues associated with these technologies.

Organization of the topic

Redox Reactions, Chemical Cells and

Electrolysis

Unit 18Chemical cells in daily life

Unit 20Oxidation and reduction

Unit 21Oxidation and reduction

in chemical cells

Unit 22Electrolysis

Unit 19Simple chemical cells

5

Teaching Plan

Unit 18 Chemical cells in daily life

Section Key point(s)Suggested task(s) for

studentsRemark

Total number of period = 1

�8.� Electricity from chemical reactions

• What a chemical cell is• Magnesium-copper

chemical cell

�8.2 Different types of chemical cells

• Primary cells and secondary cells

�8.3 Terms related to chemical cells

• Battery• Negative and positive

electrodes• Electrolyte• Cell capacity• Discharge• Service life, cycle life

and shelf life


�8.4 Zinc-carbon cell • Zinc casing as the negative electrode

• Carbon rod as the positive electrode

• Ammonium chloride as the electrolyte

• Refer to the following

website for an animation illustrating how chemical cells power a flashlight:

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/animations/flashlightV8.html

(accessed July 20�4)

�8.5 Alkaline manganese cell

• Zinc as the negative electrode

• Manganese(IV) oxide as the positive electrode

• Potassium hydroxide as the electrolyte

Continued on next page

6



studentsRemark

�8.6 Silver oxide cell • Zinc as the negative electrode

• Silver oxide as the positive electrode


• • Application manuals

of different types of cells from a manufacturer

http://data.energizer. com/Static. aspx?Name= AppManuals

• An online cell guidebook

http://support. radioshack.com/ support_tutorials/ batteries/batgd-b.htm



�8.7 Lithium ion cell • Lithium atoms lying between graphite sheets as the negative electrode

• Lithium metal oxide as the positive electrode

• Lithium salt as the electrolyte

�8.8 Nickel metal hydride (NiMH) cell

• Hydrogen absorbing alloys as the negative electrode

• Nickel(II) hydroxide as the positive electrode



7

Teaching Plan


studentsRemark

�8.9 Lead-acid accumulator

• Lead plates as the negative electrode

• Lead plates coated with lead(IV) oxide as the positive electrode

• Sulphuric acid as the electrolyte


�8.�0 Choosing a chemical cell for a particular use

• Aspects to consider when choosing a chemical cell for a particular use

• Decision Making

�8.�� Environmental impact of using chemical cells

• Heavy metal components of chemical cells are toxic

• Advantages of using secondary cells

8


Unit 19 Simple chemical cells


studentsRemark


�9.� Reactions in simple chemical cells

• How electrons flow in the external circuit of a simple chemical cell

• Ionic half-equation

• Activity �9.� — Building simple chemical cells


websites for animations illustrating the chemical reactions occurring in a zinc-copper chemical cell:

– http://group.chem. iastate.edu/

Greenbowe/sections/ projectfolder/ animations/CuZncell. html

– http://www.mhhe. com/physsci/

chemistry/ essentialchemistry/

flash/galvan5.swf• The following website

allows students to build computer simulated chemical cells using given materials:

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/electroChem/volticCell.html



9

Teaching Plan


studentsRemark

�9.2 Lemon cells made from different metal couples

• Measuring the voltages of lemon cells formed when coupling other metals with copper

• Comparing the tendency of metals to form ions based on voltages of the lemon cells

Total number of periods = 2 (Scheme 1), total number of period = 1 (Scheme 2)

�9.3 The electrochemical series of metals

• Introducing the electrochemical series of metals

• Comparing the electrochemical series with the reactivity series

• Practice �9.�

�9.4 Improving simple chemical cells

• Separating a simple chemical cell into two half-cells

• Activity �9.2 — Determining the order of three metals in the electrochemical series

• Chemical cell with a

salt bridge

�9.5 The role of a salt bridge

• Functions of a salt bridge

• Practice �9.2

Total number of periods = 1

�9.6 The Daniell cell • Structure of a Daniell cell

• Reactions occurring in a Daniell cell


website for an animation illustrating the chemical reactions occurring in a Daniell cell:

http://www.physics-chemistry-interactive-flash-animation.com/chemistry_interactive/daniell_cell.htm


�0


Unit 20 Oxidation and reduction



studentsRemark


20.� Defining oxidation and reduction in terms of gain and loss of oxygen

• Describing oxidation and reduction in terms of gain and loss of oxygen

• Redox reaction

20.2 Defining oxidation and reduction in terms of loss and gain of hydrogen

• Describing oxidation and reduction in terms of loss and gain of hydrogen


20.3 Defining oxidation and reduction in terms of loss and gain of electrons

• Describing oxidation and reduction in terms of electron transfer

20.4 Oxidizing agent and reducing agent

• Introducing oxidizing agent and reducing agent

• Defining oxidizing and reducing agents in terms of electron transfer


20.5 Relative strength of reducing and oxidizing agents

• The trend of reducing power of metals in the electrochemical series

• The trend of oxidizing power of metal ions in the electrochemical series

• Practice 20.�


20.6 Oxidation numbers • The concept of oxidation number

• Rules for assigning oxidation numbers to elements in different species

• Practice 20.2

��

Teaching Plan


studentsRemark


20.7 Defining oxidation and reduction in terms of changes in oxidation numbers

• Using changes in oxidation number to identify redox reactions

20.8 Using oxidation numbers to identify the oxidizing agent and reducing agent in a redox reaction

• Using changes in oxidation number to identify the oxidizing agent and reducing agent in a redox reaction

• Practice 20.3

20.9 Advantages and disadvantages of using the concept of oxidation number

• Advantages and disadvantages of using the oxidation number concept

20.�0 The Stock system of naming compounds

• Naming cations• Naming polyatomic

anions


20.�� Common oxidizing and reducing agents

• Chemical changes of common oxidizing and reducing agents

• Ionic half-equations representing the chemical changes

• Activity 20.� — Investigating redox

reactions• Practice 20.4


20.�2 Balancing redox equations using ionic

half-equations

• How to balance redox equations using ionic half-equations

20.�3 Balancing redox equations using oxidation number method

• How to balance redox equations using oxidation number method

• Practice 20.5


20.�4 The electrochemical series and the relative oxidizing / reducing power of common oxidizing / reducing agents

• Introducing a detailed version of the electrochemical series

• Predicting the feasibility of a redox reaction using the electrochemical series


�2



studentsRemark


20.�5 Chlorine as an oxidizing agent

• Action of aqueous chlorine on potassium bromide solution

• Action of aqueous chlorine on potassium iodide solution

• Reaction with sodium hydroxide solution

• Activity 20.2 — Ranking halogens

according to their oxidizing power

• Practice 20.6

• Reaction of aqueous

chlorine with halide solutions


20.�6 Nitric acid of different concentrations as oxidizing agents

• Oxidizing property of concentrated and dilute nitric acids

• Activity 20.3 — Investigating the

action of nitric acid of different concentrations on metals

• Practice 20.7

• Comparing the action

of nitric acid of different concentrations on magnesium and copper


20.�7 Concentrated sulphuric acid as an oxidizing agent

• Action of concentrated sulphuric acid on metals

• Action of concentrated sulphuric acid on non-metals

• Action of concentrated sulphuric acid on halides

• Activity 20.4 — Investigating the

properties of concentrated sulphuric acid

• Practice 20.8

• Action of concentrated

sulphuric acid on zinc and copper

• Action of concentrated

sulphuric acid on halides


20.�8 Aqueous sulphur dioxide as a reducing agent

• Action of aqueous sulphur dioxide on some common oxidizing agents

• Test for sulphur dioxide gas

• Practice 20.9 • Investigating the chemical properties of sulphur dioxide (I) (microscale experiment)

• Investigating the

chemical properties of sulphur dioxide (II) (microscale experiment)

�3

Teaching Plan

Unit 21 Oxidation and reduction in chemical cells


studentsRemark


2�.� Oxidation and reduction in a simple chemical cell

• Reactions occurring at the electrodes of a chemical cell

• Practice 2�.�

2�.2 Redox reactions in a zinc-carbon cell

• Redox reactions in a zinc-carbon cell

• Two main disadvantages of the cell

• Refer to the following website for an animation illustrating the chemical reactions occurring in a zinc-carbon cell:

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/animations/ZnCbatteryV8web.html



2�.3 Redox reactions in simple chemical cells with inert electrodes

• Redox reactions in simple chemical cells set up using

– acidified K2Cr2O7(aq) and FeSO4(aq)

– KI(aq) and Br2(aq)

• Practice 2�.2 • Redox reactions in

simple chemical cells (using carbon electrodes)


2�.4 Fuel cells • How a hydrogen-oxygen fuel cell works

• Advantages and limitations of hydrogen-oxygen fuel cells

• Decision making — Lithium or hydrogen

powered vehicles• Chemistry magazine —

Oxygen absorbers for packaged foods

• Fuel cell• Refer to the following

websites for animations illustrating how a fuel cell works:

– http://www.lanl.gov/ orgs/mpa/mpa��/

animation.htm – http://www.hk-phy.

org/energy/ commercial/print/cell_

is_print_e.html (accessed July 20�4)

�4


Unit 22 Electrolysis


studentsRemark


22.� Electrolysis: chemical reactions from electricity

• Terms commonly used in electrolysis

22.2 Comparing a chemical cell and an electrolytic cell

• Function• Direction of electron

flow• Reactions at electrodes

22.3 Electrolysis of molten sodium chloride using carbon electrodes

• Illustrating the chemical changes brought out by electricity using electrolysis of molten sodium chloride as an example

22.4 Some knowledge related to aqueous electrolytes

• Dissociation of water• Dissociation of acids in

water• Dissolving electrolytes in

water


22.5 Electrolysis of aqueous solutions of ionic compounds

• Electrolysis of acidified water using platinum electrodes

• Electrolysis of very dilute sodium chloride solution using carbon electrodes

• Activity 22.� — Investigating the

electrolysis of acidified water

• Practice 22.�

• Electrolysis of acidified

water•

Refer to the following website for an animation illustrating the chemical reactions occurring during the electrolysis of acidified water:

http://my.rsc.org/video/ 288 (accessed July 20�4)• Electrolysis of very

dilute sodium chloride solution using carbon electrodes


�5

Teaching Plan


studentsRemark


22.6 Factors affecting the order of discharge of ions during the electrolysis of aqueous solutions

• The position of ions in the electrochemical series

• The effect of concentration of ions in the solution

• The nature of the electrodes

22.7 The position of ions in the electrochemical series and the order of discharge of ions

• Order of discharge of cations

• Order of discharge of anions

• Activity 22.2 —Investigating factors affecting the order of discharge of ions during electrolysis — position of ions in the electrochemical series


22.8 The effect of concentration of ions in the solution and the order of discharge of ions

• Electrolysis of dilute or concentrated sodium chloride solution using carbon electrodes

• Activity 22.3 — Investigating factors affecting the order of discharge of ions during electrolysis — effect of concentration of ions in the solution

• Practive 22.2

• Electrolysis of

concentrated sodium chloride solution using

carbon electrodes


22.9 The nature of electrodes and the order of discharge of ions

• Electrolysis of dilute copper(II) sulphate solution using carbon electrodes

• Electrolysis of dilute copper(II) sulphate solution using copper electrodes

• Electrolysis of concentrated sodium chloride solution using a mercury cathode

• Activity 22.4 — Investigating factors affecting the order of discharge of ions during electrolysis — effect of the nature of electrodes

• Practice 22.3• Practice 22.4

• Manufacture of chlorine

by electrolysis of brine — Mercury electrolytic cell


�6



studentsRemark


22.�0 Industrial uses of electrolysis

• Refining of copper• Electroplating

• Activity 22.5 — Electroplating with nickel

• Practice 22.5

• Electroplating with

nickel•

Electroplating

22.�� Environmental impact of the electroplating industry

• Pollutions due to acids, alkalis, compounds of heavy metals and cyanides

• Methods to control pollution from the electroplating industry

�7

Teaching Notes

Teaching Notes


page 42N2

Oxidation number of sulphur in different substances

Examination questions often ask about the oxidation number of sulphur in different substances.

SubstanceOxidation number of S

in the substance

H2S –2

Na2S2O3 +2

SO2 +4

NaHSO3 +4

Na2SO4 +6

H2S2O7 +6

page 42N3

Oxidation number of nitrogen in different substances

Examination questions often ask about the oxidation number of nitrogen in different substances.

SubstanceOxidation number of N

in the substance

NH3 –3

N2O +�

NO +2

HNO2 +3

NO2 +4

HNO3 +5

�8


page 44N4

Using oxidation numbers to identify unfamiliar redox reactions

Examination questions often list equations of unfamiliar reactions and ask students to determine whether the reactions involve oxidation and reduction. Students may use changes in oxidation numbers to identify the redox reactions. Examples : 0 +4 +2

Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l) –� –� 0 –2

H2O2(aq) + H2SO4(aq) + 2KI(aq) K2SO4(aq) + I2(aq) + 2H2O(l) –3 +5 +� NH4NO3(s) N2O(g) + 2H2O(l) +3 –2 +2 0

Fe2(SO4)3(aq) + H2S(g) 2FeSO4(aq) + S(s) + H2SO4(aq) +5 –2 –� 0

2KClO3(s) 2KCl(s) + 3O2(g) +2 +6 +3 +4

2FeSO4(s) Fe2O3(s) + SO3(g) + SO2(g) +4 –� +2 0

MnO2(s) + 4HCl(aq) MnCl2(aq) + Cl2(g) + 2H2O(l)

page 46N5

Using oxidation numbers to identify the oxidizing agent and reducing agent in an unfamiliar redox reaction

Examination questions often list equations of unfamiliar reactions and ask students to identify the oxidizing agents (i.e. species being reduced) or the reducing agents (i.e. species being oxidized).

Examples:

+4 0

SO2 + 2Mg 2MgO + S SO2 is the oxidizing agent, it is being reduced

+4 0

SO2 + 2H2S 3S + 2H2O SO2 is the oxidizing agent, it is being reduced

–3 0

2NH3 + 3CuO 3Cu + N2 + 3H2O NH3 is the reducing agent, it is being oxidized

0 +2

Zn + 2AgNO3 Zn(NO3)2 + 2Ag Zn is the reducing agent, it is being oxidized

+2 0

CuSO4 + Zn ZnSO4 + Cu CuSO4 is the oxidizing agent, it is being reduced

+2 +4

Fe2O3 + 3CO 2Fe + 3CO2 CO is the reducing agent, it is being oxidized

–� 0

Fe2(SO4)3 + 2KI 2FeSO4 + K2SO4 + I2 KI is the reducing agent, it is being oxidized

�9

Teaching Notes

page 51N7

Distinguishing between Fe2+(aq) ions and Fe3+(aq) ions

Examination questions often ask about methods to distinguish between Fe2+(aq) ions and Fe3+(aq) ions:

• observing their colours;

• treating with acidified potassium permanganate solution;

• treating with dilute sodium hydroxide solution / dilute aqueous ammonia;

• treating with concentrated nitric acid.

Colour

Fe2+(aq) ions are pale green in colour

while Fe3+(aq) ions are yellow-brown in colour.

Acidified potassium permanganate solution

Fe2+(aq) ions can decolorize acidified potassium permanganate solution

but Fe3+(aq) ions cannot.

MnO4–(aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Dilute sodium hydroxide solution / dilute aqueous ammonia

Fe2+(aq) ions give a green precipitate with dilute sodium hydroxide solution / dilute aqueous ammonia.

Fe3+(aq) ions give a reddish brown precipitate with dilute sodium hydroxide solution / dilute aqueous ammonia.

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

Concentrated nitric acid

Fe2+(aq) ions give a brown gas with concentrated nitric acid

but Fe3+(aq) ions do not.

Refer to N16 for further details.

20


page 59N8

Oxidizing / reducing action of hydrogen peroxide

Examination questions may ask about the oxidizing / reducing action of hydrogen peroxide, an unfamiliar reagent. Students need to write the ionic half-equations for the chemical changes based on the given reactants and products.

Hydrogen peroxide can act as an oxidizing agent:

H2O2(aq) + 2H+(aq) + 2e– 2H2O(l)

and as a reducing agent:

H2O2(aq) O2(g) + 2H+(aq) + 2e–

Acidic solutions favour the oxidizing action.

e.g. oxidizing Fe2+(aq) ions to Fe3+(aq) ions

H2O2(aq) + 2Fe2+(aq) + 2H+(aq) 2Fe3+(aq) + 2H2O(l)

oxidizing I–(aq) ions to I2(aq)

H2O2(aq) + 2I–(aq) + 2H+(aq) I2(aq) + 2H2O(l)

Alkaline solutions favour the reducing action.

e.g. reducing Cl2(aq) to Cl–(aq) ions

H2O2(aq) + Cl2(aq) O2(g) + 2Cl–(aq) + 2H+(aq)

page 66N15

Similarity in chemical properties of halogens

Examination questions aften ask about the similarity in chemical properties of halogens.

Examples:

• Both Cl2(aq) and Br2(aq) can oxidize SO32–(aq) ions to SO4

2–(aq) ions.

X2(aq) + SO32–(aq) + H2O(l) 2X–(aq) + SO4

2–(aq) + 2H+(aq)

• Both Cl2(aq) and Br2(aq) can react with Fe2+(aq) ions to give Fe3+(aq) ions.

X2(aq) + 2Fe2+(aq) 2X–(aq) + 2Fe3+(aq)

• Both Cl2(aq) and Br2(aq) can react with I–(aq) ions to give I2(aq).

X2(aq) + 2I–(aq) 2X–(aq) + I2(aq)

• Both Cl2(aq) and Br2(aq) can undergo disproportionation in alkalis.

X2(g) + 2OH–(aq) X–(aq) + OX–(aq) + H2O(l)

Refer to N22 for the gradual change in chemical properties of halogens / halide ions.

2�

Teaching Notes

page 67N16

Action of concentrated nitric acid on metals, iron(ll) salts and sulphites

1 Metals

Concentrated nitric acid oxidizes most metals, for example, magnesium and copper. However, it has no reaction with iron and aluminium. Concentrated nitric acid renders them completely passive. This is due to the formation of a layer of oxide on the metal surface.

2 Iron(II) salts

Concentrated nitric acid oxidizes iron(ll) salts to iron(lll) salts. Nitrogen monoxide is produced and this reacts with oxygen in air to form brown nitrogen dioxide gas.

3Fe2+(aq) + NO3–(aq) + 4H+(aq) 3Fe3+(aq) + NO(g) + 2H2O(l)

2NO(g) + O2(g) 2NO2(g)

3 Sulphites

Concentrated nitric acid oxidizes sulphites to sulphates.

SO32–(aq) + 2H+(aq) + 2NO3

–(aq) SO42–(aq) + 2NO2(g) + H2O(l)

22


page 68N17

Comparing the properties of dilute nitric acid and dilute sulphuric acid / dilute hydrochloric acid

Examination questions often ask students to compare the properties of different acids.

Comparing the properties of dilute HNO3(aq) and dilute H2SO4(aq)

Observation

PropertyDilute HNO3(aq) Dilute H2SO4(aq)

Reaction with alkali salt and water are produced

Reaction with carbonate (or hydrogencarbonate)

carbon dioxide gas is given off

Action on litmus solution give a red colour

Reaction with zinc a brown gas is given off a colourless gas is given off

Reaction with barium chloride solution

no white precipitate forms a white precipitate forms

Titration with NaOH(aq) more NaOH(aq) is needed to reach the end point for H2SO4(aq) than HNO3(aq)

Acid

Comparing the properties of dilute HNO3(aq) and dilute HCl(aq)

Observation

PropertyDilute HNO3(aq) Dilute HCl(aq)





Reaction with copper a brown gas is given off no gas is given off

Reaction with silver nitrate solution

no white precipitate forms a white precipitate forms

Acid

23

Teaching Notes

page 69N18

Comparing the properties of concentrated sulphuric acid and dilute sulphuric acid

Examination questions often ask students to compare the properties of different acids.

Comparing the properties of concentrated H2SO4(l) and dilute H2SO4(aq)

Observation

PropertyConcentrated H2SO4(l) Dilute H2SO4(aq)





Oxidizing property can oxidize copper no reaction with copper

Non-volatility non-volatile, can displace other volatile acids (e.g. HCl(g) and HNO3(g)) from their salts

not non-volatile

Acid

page 69N20

Preparation of sulphur dioxide in the laboratory

In the laboratory, we can prepare sulphur dioxide by heating copper turnings with concentrated sulphuric acid. The gas is dried by passing through concentrated sulphuric acid. The gas is collected by downward delivery because it is denser than air.

Cu(s) + 2H2SO4(l) CuSO4(aq) + SO2(g) + 2H2O(l)

24


page 69N21

Distinguishing between concentrated nitric acid and concentrated sulphuric acid

Copper can be used to distinguish between concentrated nitric acid and concentrated sulphuric acid.

Copper gives a brown gas (NO2) with concentrated nitric acid.

Copper gives a colourless gas (SO2) with concentrated sulphuric acid.

page 69N22

Gradual change in chemical properties of halogens / halide ions

Examination questions often ask about the gradual change in chemical properties of halogens / halide ions.

Examples:

• order of increasing oxidizing power: I2 < Br2 < Cl2

– Aqueous chlorine can displace bromine from potassium bromide solution and displace iodine from potassium iodide solution.

– Aqueous bromine can displace iodine from potassium iodide solution but cannot displace chlorine from potassium chloride solution.

• order of increasing reducing power: I– > Br– > Cl–

– NaCl(s) is not oxidized by concentrated H2SO4(l);

– NaBr(s) reacts with concentrated H2SO4(l) to give Br2(l) and SO2(g);

– NaI(s) reacts with concentrated H2SO4(l) to give I2(s) and H2S(g).

25

Teaching Notes


page 85N1

Redox reactions occurring in unfamiliar chemical cells

Examination questions often show unfamiliar chemical cells and ask students to write ionic half-equations for redox reactions occurring in the cells based on their knowledge on oxidation and reduction.

Examples:

• A chemical cell made up of an aluminium can, a carbon rod and household bleach

At the aluminium can

The aluminium can undergoes oxidation to give tetrahydroxylaluminate ions, [Al(OH)4]–(aq).

Al(s) + 4OH–(aq) [Al(OH)4]–(aq) + 3e–

At the carbon rod

The hypochlorite ions, OCl–(aq) (in the household bleach) undergo reduction to give chloride ions and hydroxide ions.

OCl–(aq) + H2O(l) + 2e– Cl–(aq) + 2OH–(aq)

• A chemical cell made up of copper electrodes and copper(ll) sulphate solution of different concentrations (concentration cell)

26


In the set-up, electrons flow in such a direction that the concentration of Cu2+(aq) ions in each half-cell becomes the same eventually, i.e. electrons flow from X to Y in the external circuit.

At electrode X

Cu(s) Cu2+(aq) + 2e–

At electrode Y

Cu2+(aq) + 2e– Cu(s)

• A sodium-sulphur cell operating at about 370 ºC (above the melting points of sodium and sulphur)

At electrode A

Na(l) Na+(l) + e–

At electrode B

S(l) + 2e– S2–(l)

page 89N4

Predicting chemical changes occurring in a chemical cell based on the oxidizing power of the reagents involved

Examination questions may ask students to predict the chemical changes that would occur in a chemical cell based on the oxidizing power of the reagents involved.

Example: consider the following chemical cell:

27

Teaching Notes

� To tackle the question, first identify the oxidizing agent and the reducing agent.

The question gives the information that Br2(aq) is a stronger oxidizing agent than Fe3+(aq) ion.

Hence it can be deduced that Br2(aq) acts as the oxidizing agent and Fe2+(aq) ion acts as the reducing agent. The following chemical changes would occur:

Fe2+(aq) Fe3+(aq) + e–

Br2(aq) + 2e– 2Br–(aq)

2 Based on the chemical changes above, it is possible to identify the negative electrode (anode) and the positive electrode (cathode) of the chemical cell.

As oxidation occurs at electrode X, thus it is the negative electrode, i.e. the anode.

As reduction occurs at electrode Y, thus it is the positive electrode, i.e. the cathode.

3 Based on the above information, it is possible to decide that the direction of electron flow in the external circuit is from electrode X to electrode Y.

page 89N5

A simple chemical cell involving KI(aq) and Fe2(SO4)3(aq)

Fe2(SO4)3(aq) is a strong oxidizing agent than I2(aq).

28


Hence the following changes occur in the chemical cell:

2I–(aq) I2(aq) + 2e–

Fe3+(aq) + e– Fe2+(aq)

If some starch solution is added to the concentrated potassium iodide solution, a blue colour would appear after some time. This is because iodine forms a complex with starch.


page 116N8

Electrolysis of dilute copper(II) sulphate solution using carbon anode and copper cathode

Besides using carbon electrodes / copper electrodes, examination questions may ask about the electrolysis of dilute copper(ll) sulphate solution using carbon anode and copper cathode.

At the anode

4OH–(aq) O2(g) + 2H2O(l) + 4e–

At the cathode


The blue colour of the solution fades gradually because the concentration of copper(II) ions in the electrolyte decreases.

Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and sulphate ions remain in the solution. Thus the solution eventually becomes sulphuric acid.

29

Suggested Answers

Suggested Answers

page 1

� Rusting refers to the corrosion of iron ONLY.

2 Fe2O3•xH2O(s)

3 Reactivity series

Unit 18 Chemical cells in daily life

Decision Making page 12

There is no right or wrong answer to this question. Students may choose according to one of the following criteria:

• Choose alkaline manganese cell in every case for the sake of convenience. This saves the trouble of replacing dry cells frequently and recharging of cells.

• Choose based on the ‘cost-effectiveness’ of the different types of cell.

• Choose rechargeable nickel metal hydride cell if dry cells are used for a long time every day.

Type of cell

Operating cost

Torch(HKD per hour)

Camera flash unit (HKD per time)

Portable CD player (HKD per hour)

Radio(HKD per hour)

Zinc-carbon 4 0.�3 0.67 0.040

Alkalinemanganese

3.53 0.03 0.50 0.043

It is more cost effective to use alkaline manganese cell for torch, camera flash unit and portable CD player, while it is more cost effective to use zinc-carbon cell for radio.

pages 16–19Unit Exercise

� a) electrical

b) electrolyte

c) primary cell

d) zinc-carbon cell

e) alkaline manganese cell

f) silver oxide cell

30


g) lithium ion cell

h) nickel metal hydride cell

i) lead-acid accumulator

2 a) voltmeter; digital multimeter

b) battery

c) negative; positive

d) electrolyte

e) service

f) shelf

3

Chemical cellRechargeable

or not?

Material of negative electrode

Material of positive electrode

Material of electrolyte

Maximum voltage (V)

Zinc-carbon cell

No zinc carbonammonium

chloride1.5

Alkaline manganese cell

No zincmanganese(IV)

oxidepotassium hydroxide

1.5

Silver oxide cell No zinc silver oxidepotassium hydroxide

1.5

Lithium ion cell Yeslithium atoms lying between graphite sheets

lithium metal oxide

lithium salt dissolved in an organic solvent

3.7

Nickel metal hydride cell

Yeshydrogen-absorbing

alloy

nickel(II) hydroxide

potassium hydroxide

1.2

Lead-acid accumulator

Yes leadtitanium plates

coated with lead(IV) oxide

sulphuric acid

2

4 A Option B — No electrons flows in the external circuit as both electrodes are made of copper.

Options C and D — No electrons flows in the external circuit as distilled water and ethanol do NOT conduct electricity at all.

5 B

6 C

3�

Suggested Answers

7 A

8 C Option C — Nickel metal hydride cells have a relatively high rate of self-discharge.

9 D

�0 D

�� B (�) The electrolyte of a zinc-carbon cell is ammonium chloride.

(3) A zinc-carbon cell shows poor performance in high-drained devices.

�2 B (2) The maximum voltage of both alkaline manganese cell and zinc-carbon cell is �.5 V.

�3 B

�4 D

�5 A (2) The electrolyte of a nickel metal hydride cell is potassium hydroxide.

(3) A nickel metal hydride cell shows good performance in high-drained devices.

�6 a) In any aircraft or spacecraft.

b) Any one of the following:

• Life-support systems required during a power failure, in a remote area or in an ambulance.

• Medical aids that cannot run from a normal power supply, such as heart pacemakers and bionic ears.

• Other sensitive and medical instrumentation used in a remote area.

c) Any one of the following:

• Inside any computer or other electronic system.

• Inside the human body.

�7 a) Anode – lithium atoms lying between graphite sheets

Cathode – lithium metal oxide

b) A lithium salt dissolved in an organic solvent

c) light weight

high energy density

32


(d) Any one of the following:

• Electric razors

• Electric toothbrushes

• Medical equipment

• Portable DVD players

• PDAs

• Laptop computers

�8 Answers for the HKALE question are not provided.

�9 a) �.5 V


• Use alkaline manganese cells

reasons — these cells have a steady voltage during discharge;

— there is no need to recharge the used cells.

• Use nickel metal hydride cells

reasons — more cost effective as these cells can be recharged over 500 times;

— the physical volume of discarded cells in landfills can be reduced.

Unit 19 Simple chemical cells

Practice

P19.1 page 24

a) Zinc

b) Form zinc to copper

c) i) Cu2+(aq) + 2e– Cu(s)

ii) Zn(s) Zn2+(aq) + 2e–

d) The voltage of the cell becomes zero. This is because sugar solution contains no mobile ion and thus does not allow ionic conduction between the two electrodes.

33

Suggested Answers

P19.2 page 26

a) Metal X

b) i) P is salt bridge.

A salt bridge serves two important functions:

– It completes the circuit by allowing ions to move from one half-cell to the other.

– It provides ions that can move into the half-cells to prevent the build-up of charge in the solutions which would cause the reaction to stop.

ii) Metal X

iii) From metal X to metal Y


� a) electrolyte

b) positive

c) electrons

d) negative

e) positive

2 a) electrons; zinc ions; Zn(s) Zn2+(aq) + 2e–

b) electrons; silver; Ag+(s) + e– Ag(s)

c) zinc; silver

d) negative; positive

3 C

4 D Zinc forms ions more readily than silver does. Thus, electrons flow from zinc to silver in the external circuit.

5 D Option A — The silver strip acts as the positive electrode, i.e. the cathode.

Option B — The magnesium strip acts as the negative electrode, i.e. the anode.

Option C — Silver ions in beaker X gain electrons and form silver atoms.

Ag+(aq) + e– Ag(s)

Thus, the concentration of silver ions in beaker X decreases.

34


Option D — There is a build up of negative charge in beaker X due to the NO3–(aq) ions that

remained. K+(aq) ions migrate from the salt bridge into beaker X to offset any buildup of negative charge.

e– e–

Mg2+Ag+

electron flow NO3– ions from the salt bridge

K+ ions from the salt bridge

6 B

7 B Option A — The zinc strip acts as the negative electrode, i.e. the anode.

Option B — Electrons flow from the zinc strip to the copper container in the external circuit.

Option C — The porous pot does NOT provide ions for charge balance.

Option D — Copper(II) ions gain electrons and form copper atoms. The concentration of copper(II) ions in the copper(II) sulphate solution decreases. Thus, the blue colour of the solution fades.

8 A (3) Zinc atoms lose electrons and form zinc ions.

Zn(s) Zn2+(aq) + 2e–

This does not affect the colour of the dilute sulphuric acid.

9 C (2) Electrons move from the zinc plate to the copper plate in the external circuit.

(3) The difference between the tendencies for magnesium and copper to form ions is greater than between zinc and copper. Thus, the voltage of the cell would increase if the zinc plate is replaced by a magnesium plate.

35

Suggested Answers

�0 A (�) Magnesium is more reactive than nickel. Thus, magnesium atoms will lose electrons and form magnesium ions.

Mg(s) Mg2+(aq) + 2e–

Thus, concentration of magnesium ions in beaker X increases.

(2) Nickel(II) ions in the nickel(II) sulphate solution gain electrons and form nickel atoms.

Ni2+(aq) + 2e– Ni(s)

Thus, the mass of the nickel electrode increases gradually.

(3) The salt bridge provides ions that can move into the half-cells to prevent the build-up of charge in the solutions.

�� a) Fe

b) Ag

c) Iron and silver.

They are furthest apart in the electrochemical series.

�2 a) Cadmium

b) Cd(s) + Ni2+(aq) Cd2+(aq) + Ni(s)

c) From the cadmium strip to the nickel strip

d) When the cell operates, Cd2+(aq) ions will form at the cadmium strip.

NO3–(aq) ions in the salt bridge will migrate into cadmium half-cell to offset any buildup of Cd2+(aq)

ions.

�3 a) • It completes the circuit by allowing ions to move from one half-cell to the other.

• It provides ions that can move into the half-cells to prevent the build-up of charge in the solutions which would cause the reaction to stop.

b) iron < X < Y

Metal X is more reactive than iron. Thus, X forms ions more readily than iron does.

Metal Y is more reactive than metal X. Thus, Y forms ions more readily than X does.

c) i) Fe2+(aq) + 2e– Fe(s)

X(s) X2+(aq) + 2e–

ii) The green colour of solution fades out, because concentration of Fe2+ decreases.

iii) Voltage increased, because copper forms ions less readily than iron does.

36


�4 a)

b) Electrons flow from the zinc rod to the copper can in the external circuit.

c) No current would flow through the conducting wires. This was because the glass beaker would not allow ions to move between the two solutions.

d) The reading on the ammeter would decrease.

�5 a) Electrons flow from sodium and electrode B via external circuit towards electrode A and nickel(II), because sodium is more reactive than nickel.

b) Na(l) Na+(l) + e–

c) To melt the electrolyte and mobilize the ions.

d) This cell provides a high voltage.

�6 Answers for the HKCEE question are not provided.


Practice

P20.1 page 41

� a) An oxidation is involved as Zn(s) loses electrons.

b) A reduction is involved as Fe3+(aq) gains electrons.

c) An oxidation is involved as I–(aq) loses electrons.

37

Suggested Answers

2 Cu(s) Ag+(aq)

Ionic half-equation for chemical change that occurs for the species

Cu(s) Cu2+(aq) + 2e– Ag+(aq) + e– Ag(s)

Whether the species undergoes oxidation or reduction?

oxidation reduction

P20.2 page 43

a) Oxidation number of O = –2

Suppose the oxidation number of N in NO is x.

x + (–2) = 0 x = +2

\ the oxidation number of N in NO is +2.

b) Oxidation number of H = +�

Suppose the oxidation number of N in NH3 is x.

x + (+�) x3 = 0 x = –3

\ the oxidation number of N in NH3 is –3.

c) Oxidation number of Na = +�

Oxidation number of O = –2

Suppose the oxidation number of C in Na2CO3 is x.

[(+�) x 2 + x + (–2) x 3] = 0 x = +4

\ the oxidation number of C in Na2CO3 is +4.

d) Cu(OH)2 consists of Cu2+ ion and OH– ions.

Oxidation number of Cu = charge on the ion = +2

e) Oxidation number of Fe = charge on the ion = +3

f) Oxidation number of O = –2

Suppose the oxidation number of Cr in Cr2O72– is x.

[(x) x 2 + (–2) x 7] = –2 x = +6

\ the oxidation number of Cr in Cr2O72– is +6.

38


P20.3 page 47

� a) from +5 decrease to +4

b) from 0 increase to +2

c) from +4 decrease to +2

2 a) Yes.

The oxidation number of Al increases from 0 to +3, so Al(s) is reducing agent.

The oxidation number of H decreases from +� to 0, so H+(aq) is oxidizing agent.

b) No.

The oxidation number of Cr, O, H remain +6, –2 and +� respectively.

c) Yes.

The oxidation number of Mg increases from 0 to +2, so Mg(s) is reducing agent.

The oxidation number of Cu decreases from +2 to 0, so Cu2+(aq) is oxidizing agent.

P20.4 page 52

Acidified potassium dichromate solution

Iron(II) sulphate solution

Product formed when it reacts Chromium(III) Iron(III)

Ionic half-equationCr2O7

2–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l) Fe2+(aq) Fe3+(aq) + e–

As an oxidizing agent or a reducing agent?

oxidizing agent reducing agent

P20.5 page 59

� 3Zn(s) + �4H+(aq) + Cr2O72–(aq) 3Zn2+(aq) + 7H2O(l) + 2Cr3+(aq)

2 a) Reduction: MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)

Oxidation: SO32–(aq) + H2O(l) SO4

2–(aq) + 2H+(aq) + 2e–

b) 5SO32–(aq) + 6H+(aq) + 2MnO4

–(aq) 5SO42–(aq) + 3H2O(l) + 2Mn2+(aq)

c) Reducing agent: SO32–(aq)

Oxidizing agent: MnO4–(aq)

d) Purple permanganate solution is becomes colourless.

39

Suggested Answers

P20.6 page 66

Add the sample solution to bromine solution.

Potassium iodide solution changes the solution brown, but potassium chloride solution does not change.

P20.7 page 68

Acid

Action of acid on magnesium

Action of acid on copper Property shown by the acid (acidic / oxidizing

property)Name of gas given off, if any

Name of gas given off, if any

Very dilute nitric acid yes, hydrogen no, — acidic property

Dilute nitric acid yes, nitrogen monoxide yes, nitrogen monoxide oxidizing property

Concentrated nitric acid yes, nitrogen dioxide yes, nitrogen dioxide oxidizing property

P20.8 page 71

Action of acid on

Dilute sulphuric acid Concentrated sulphuric acid


Property shown by the acid (acidic /

oxidizing property)


Property shown by the acid (acidic /

oxidizing property)

Sodium carbonate carbon dioxide acidic property carbon dioxide acidic property

Zinc hydrogen acidic property sulphur dioxide oxidizing property

Copper no reaction acidic property sulphur dioxide oxidizing property

P20.9 page 73

Add bromine solution / iodine solution / acidified potassium permanganate / acidified potassium dichromate

Sodium sulphite solution changes the bromine solution / iodine solution / acidified potassium permanganate colourless, but sodium sulphate solution does not change.

Sodium sulphite solution changes acidified potassium dichromate green, but sodium sulphate solution does not change.

40



� a) reduction

b) oxidation

c) oxidation

d) oxidation

e) reducing

2Species that causes Species that is

Reducing agent oxidation / reduction oxidized / reduced

Oxidizing agent oxidation / reduction oxidized / reduced

3

power ofmetal ion increasing

power ofmetal

decreasing

Metal ion MetalK+(aq) + e–

Ca2+(aq) + 2e–

Na+(aq) + e–

Mg2+(aq) + 2e–

Al3+(aq) + 3e–

Zn2+(aq) + 2e–

Fe2+(aq) + 2e–

Pb2+(aq) + 2e–

2H+(aq) + 2e–

Cu2+(aq) + 2e–

Ag+(aq) + e–

Au+(aq) + e–

K(s)Ca(s)Na(s)Mg(s)Al(s)Zn(s)Fe(s)Pb(s)H2(g)Cu(s)Ag(s)Au(s)

weak

agents

strong

agents

strong

agents

weak

agents

reducing

oxidizing

oxidizing

oxidizing reducing

reducing

4 a)Oxidizing agent Product when reduced Ionic half-equation

Aqueous chlorine Cl–(aq) Cl2(g) + 2e– 2Cl–(aq)

Aqueous bromine Br–(aq) Br2(aq) + 2e– 2Br–(aq)

Potassium permanganate in acidic solution

Mn2+(aq)MnO4

–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)

Potassium dichromate in acidic solution

Cr2O72–(aq) 2Cr3+(aq)

Cr2O72–(aq) + 14H+(aq) + 6e–

2Cr3+(aq) + 7H2O(l)

Iron(III) sulphate solution Fe3+(aq) Fe2+(aq) Fe3+(aq) + e– Fe2+(aq)

4�

Suggested Answers

b) Oxidizing agent Product when oxidized Ionic half-equation

Sodium sulphite solution SO42–(aq)

SO32–(aq) + H2O(l)

SO42–(aq) + 2H+(aq) + 2e–

Iron(II) sulphate solution Fe2+(aq) Fe3+(aq) Fe2+(aq) Fe3+(aq) + e–

Potassium bromide solution 2Br–(aq) Br2(aq) 2Br–(aq) Br2(aq) + 2e–

Potassium iodide solution 2l–(aq) l2(aq) 2l–(aq) l2(aq) + 2e–

5 B

6 A

7 C

8 C

9 C

�0 A

�� B

�2 A

�3 A

�4 B

�5 D

�6 A

�7 D

�8 a) Oxidation number of O = –2 Suppose the oxidation number of S in SO3 is x. x + (–2) x 3 = 0 x = +6

\ the oxidation number of S in SO3 is +6.

42


b) Oxidation number of O = –2 Suppose the oxidation number of Mn in MnO4

– is x. x + (–2) x 4 = –� x = +7

\ the oxidation number of Mn in MnO4– is +7.

c) Oxidation number of H = +� Oxidation number of O = –2 Suppose the oxidation number of C in HCO3

– is x. (+�) + x + (–2) x 3 = –� x = +4

\ the oxidation number of C in HCO3– is +4.

d) Oxidation number of H = +� Oxidation number of O = –2 Suppose the oxidation number of S in H2S2O7 is x. (+�) x 2 + 2x + (–2) x 7 = 0 2x = +�2 x = +6

\ the oxidation number of S in H2S2O7 is +6.

e) PbSO4 consists of Pb2+ ion and SO42– ion.

Oxidation number of Pb = charge on the ion = +2

f) Oxidation number of Ca = +2 Suppose the oxidation number of H in CaH2 is x. (+2) + (x) x 2 = 0 x = –�

\ the oxidation number of H in CaH2 is –�.

g) Oxidation number of H = +� Oxidation number of O = –2 Suppose the oxidation number of Al in [Al(OH)4]

– is x. x + [(–2) + (+�) x 4] = –� x = +3

\ the oxidation number of Al in [Al(OH)4]– is +3.

h) Oxidation number of H = +� Suppose the oxidation number of O in H2O2 is x. (+�) x 2 + (x) x 2 = 0 x = –�

\ the oxidation number of O in H2O2 is -�.

i) Oxidation number of Cl = –� Suppose the oxidation number of Co in [CoCl4]

2– is x. x + [(–�) x 4] = –2 x = +2

\ the oxidation number of Co in [CoCl4]2– is +2.

43

Suggested Answers

j) Oxidation number of O = –2 Suppose the oxidation number of V in VO2

+ is x. x + 2x(–2) = +� x = +5

\ the oxidation number of V in VO2+ is +5.

�9 a) This is not a redox reaction

because the oxidation numbers of all elements remain unchanged in the reaction. +� –� +� +6 –2 +� +� +6 –2 +� –�

NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)

b) The oxidation number of K increases from 0 to +� while that of H decreases from +� to 0.

Therefore it is a redox reaction. 0 +� +� 0

2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

The oxidation number of H decreases, i.e. H2O is being reduced.

c) The oxidation number of N increases from –3 to 0 while that of Cu decreases from +2 to 0.

Therefore it is a redox reaction. –3 +2 0 0

2NH3(g) + 3CuO(s) 3Cu(s) + N2(g) + 3H2O(g)

The oxidation number of Cu decreases, i.e. CuO is being reduced.

d) The oxidation number of Zn increases from 0 to +2 while that of Ag decreases from +� to 0.

Therefore it is a redox reaction. 0 +� +2 0

Zn(s) + 2AgNO3(aq) Zn(NO3)2(s) + 2Ag(s)

The oxidation number of Ag decreases, i.e. AgNO3 is being reduced.

e) The oxidation number of Br increases from –� to 0 while that of Cl decreases from 0 to –�.

Therefore it is a redox reaction. –� 0 –� 0

2KBr(aq) + Cl2(aq) 2KCl(aq) + Br2(aq)

The oxidation number of Cl decreases, i.e. Cl2 is being reduced.

20 a) 2Cr2O72–(aq) + 6I–(aq) + �4H+(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)

b) 2MnO4–(aq) + �0Cl–(aq) + �6H+(aq) 2Mn2+(aq) + 5Cl2(g) + 8H2O(l)

c) 2Fe2+(aq) + H2O2(aq) + 2H+(aq) 2Fe3+(aq) + 2H2O(l)

44


2� a) The brown iodine solution becomes colourless.

This is because brown iodine is reduced to colourless iodide ions.

l2(aq) + SO32–(aq) + H2O(l) 2l–(aq) + SO4

2–(aq) + 2H+(aq)

b) A yellow-brown colour develops.

This is due to the formation of bromine in the reaction.

Cl2(aq) + 2Br–(aq) 2Cl–(aq) + Br2(aq)

c) A brown solution results.

This is due to the formation of iodine in the reaction.

2MnO4–(aq) + �0I–(aq) + �6H+(aq) 2Mn2+(aq) + 5I2(aq) + 8H2O(l)

d) The orange acidified potassium dichromate solution turns green.

This is because the orange dichromate ions are reduced to green chromium(III) ions.

Cr2O72–(aq) + �4H+(aq) + 6Fe2+(aq) 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

e) The magnesium ribbons dissolves

and a brown gas is given off.

This is because nitrate ions are reduced to brown nitrogen dioxide gas.

Mg(s) + 2NO3–(aq) + 4H+(aq) Mg2+(aq) + 2NO2(g) + 2H2O(l)

f) The zinc granules dissolves

and a colourless gas is given off. This gas gives a brown gas when mixed with air.

This is because nitrate ions are reduced to colourless nitrogen monoxide gas. This gas gives brown nitrogen dioxide gas when mixed with air.

3Zn(s) + 2NO3–(aq) + 8H+(aq) 3Zn2+(aq) + 2NO(g) + 4H2O(l)

g) When concentrated sulphuric acid is added to sodium bromide, reddish brown fumes are given off.

This is because hydrogen bromide is first produced in the reaction.

NaBr(s) + H2SO4(l) NaHSO4(s) + HBr(g)

The hydrogen bromide is then oxidized to bromine.

2HBr(g) + H2SO4(l) Br2(s) + SO2(g) + 2H2O(l)

22 a) Experiment number Experiment details Colour seen within the organic solvent

� addition of Cl2(aq) to I–(aq) ions purple

2 addition of Cl2(aq) to Br–(aq) ions orange

3 addition of Br2(aq) to Cl–(aq) ions orange

45

Suggested Answers

b) Cl2(aq) + 2Br–(aq) 2Cl–(aq) + Br2(aq)

c) Addition of Br2(aq) to I–(aq) ions

23 a) A reducing agent causes a decrease in the oxidation number of an element of another species.

b) i) 2OCl–(aq) + 4H+(aq) + 2e– Cl2(g) + 2H2O(l)

ii) 2Cl–(aq) Cl2(aq) + 2e–

c) Cl–(aq) + 2H+(aq) + OCl–(aq) Cl2(g) + H2O(l)

d) Steamy fumes

Purple fumes

Black solid

e) i) 2KClO3(s) 2KCl(s) + 3O2(g)

ii) The oxidation number of chlorine decreases from +5 to –�.

The oxidation number of oxygen increases from –2 to 0.

24 Answers for the HKCEE question are not provided.


26 a) Any one of the following:

• Add concentrated sulphuric acid to each solid separately.

Sodium chloride gives steamy fumes.

Sodium iodide gives purple fumes that condense to a black solid.

• Dissolve each solid in water. Then add aqueous chlorine to each solution separately.

The sodium iodide solution turns brown.

There is no observable change for the solution of sodium chloride.


• Add copper to each acid separately.

When copper reacts with dilute nitric acid, a colourless gas is given off. The gas gives a brown gas when mixed with air.

There is no observable change for dilute sulphuric acid.

• Add barium chloride solution to each acid separately.

Dilute sulphuric acid gives a white precipitate.

There is no observable change for dilute nitric acid.

46


c) Any one of the following:

• Acidified potassium permanganate solution

Fe2+(aq) ions can decolorize acidified potassium permanganate solution

but Fe3+(aq) ions cannot.

• Dilute sodium hydroxide solution / dilute aqueous ammonia

Fe2+(aq) ions give a green precipitate with dilute sodium hydroxide solution / dilute aqueous ammonia.

Fe3+(aq) ions give a reddish brown precipitate with dilute sodium hydroxide solution / dilute aqueous ammonia.

• Concentrated nitric acid

Fe2+(aq) ions give a brown gas with concentrated nitric acid

but Fe3+(aq) ions do not.



Practice

P21.1 page 86

a) At the iron electrode: Fe2+(aq) + 2e– Fe(s)

At the chromium electrode: Cr(s) Cr3+(aq) + 3e–

b) Electrons flow from the chromium electrode to the iron electrode in the external circuit.

c) The chromium electrode is the anode

because oxidation occurs here.

P21.2 page 89

� a) Bromine is a stronger oxidizing agent.

b) Beaker X: 2I–(aq) I2(aq) + 2e–

Beaker Y: Br2(aq) + 2e– 2Br–(aq)

c) The reddish brown colour of aqueous bromine gradually fades out.

d) Electrons flow from the potassium iodide solution to the bromine solution in the external circuit.

2 a) Oxidation, because C of CH3CH2OH loses H and gains O.

b) Anode. It is because the conversion of ethanol to ethanoic acid is an oxidation.

47

Suggested Answers

c) The higher concentration of ethanol, the higher voltage / current.

Decision Making page 92

Lithium or hydrogen powered vehicles

xxxxxxxx

Chemistry Magazine page 93

Oxygen absorbers for packaged foods

� Oily food substances (e.g. mooncakes) are often packed with oxygen absorbers.

2 The packet should not be air-tight as its role is to absorb oxygen from the air. In addition, the presence of tiny holes on the packet can confirm that it is not air-tight.

3 Substances in enclosed containers and metals should not be heated in a microwave oven.


� a) loses

b) gains

c) cathode

d) oxidation

e) reduction

f) electrons

2 a)

hydrogen oxygen

unused oxygenunused hydrogenand steam

hot concentratedpotassium hydroxide

solution

48


b) H2(g) + 2OH–(aq) 2H2O(l) + 2e–

O2(g) + 2H2O(l) + 4e– 4OH–(aq)

c) X

3 C

4 C

5 D The following ionic half-equations represent the changes that occur in the chemical cell:

Fe2+(aq) Fe3+(aq) + e–

Cr2O72–(aq) + �4H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)

6 B Options A and C — Oxidation occurs at electrode X while reduction occurs at electrode Y. Thus, electrode X acts as the anode.

Option D — Electrons flow from electrode X to electrode Y in the external circuit.

7 D

8 A As the green colour of the solution of Ni2+(aq) gradually fades out, it can be deduced that Ni2+(aq) ions undergo reduction and form Ni(s).

(�) Cadmium undergoes oxidation and forms Cd2+(aq) ions. Electrons flow from cadmium electrode X to nickel electrode Y.

(2) Reduction occurs at nickel electrode Y.

Ni2+(aq) + 2e– Ni(s)

(3) Ni2+(aq) ions gain electrons and form Ni(s) atoms. This results in a build up of negative charge in nickel half-cell. K+(aq) ions migrate from the salt bridge into the nickel half-cell to offset any buildup of negative charge.

9 C

�0 B

�� a) The colour change occurs because yellow-brown iron(III) ions gain electrons and change to green iron(II) ions.

Fe3+(aq) + e– Fe2+(aq)

49

Suggested Answers

b) i) A brown colour appears around the carbon electrode.

This is because colourless iodide ions lose electrons and change to iodine.

2I–(aq) I2(aq) + 2e–

ii) The iodide ions are oxidized,

because they lose electrons / the oxidation number of iodine increases from –� to 0.

c) Electrons flow from the potassium iodide solution to the iron(III) sulphate solution in the external circuit.

�2 a) The purple colour of permanganate fades gradually

MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)

b) i) SO32–(aq) + H2O(l) SO4

2–(aq) + 2H+(aq) + 2e–

ii) The sulphite ions are oxidized to sulphate ions

because the oxidation number of sulphur increases from +4 to +6.

c) Electrons flow from the carbon electrode in beaker Y to the carbon electrode in beaker X.

�3 a) i) Carbon rod

ii) 2NH4+(aq) + 2e– 2NH3(aq) + H2(g)

b) i) Zinc case

ii) Zn(s) Zn2+(aq) + 2e–

c) Hydrogen is produced and collected on the surface of the positive electrode. Since hydrogen is a poor conductor of electricity, the accumulation of hydrogen at the positive electrode may hinder further reactions and decrease the current of the cell. Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen.

d) If a current is drawn from the cell rapidly, the gaseous product cannot be removed fast enough. The voltage drops as a result.

e) The zinc case undergoes oxidation to give zinc ions in the cell reaction.

f) The materials inside the cells do not decompose even after a long time. These materials may combine with other compounds and form harmful substances which pollute the environment.

g) There is a slow direct reaction between the zinc electrode and ammonium ions.

After some time, the zinc case becomes too thin and the paste leaks out. This may cause damage to electric appliances.

50


�4 a) From Y to X.

b) X is cathode.

Reduction occurs at X.

(c) Ag2O(s) + H2O(l) + Zn(s) 2Ag(s) + Zn(OH)2(s)

�5 Answers for the HKASLE question are not provided.


Practice

P22.1 page 107

a) i) 2H+(aq) + 2e– H2(g)

ii) At electrode X, hydrogen ions were preferentially discharged to form hydrogen gas.

Water dissociated continuously to replace the hydrogen ions discharged. Thus, there was a build-up of hydroxide ions near electrode X. The solution there became alkaline.

The universal indicator turned blue.

b) i) 4OH–(aq) O2(g) + 2H2O(l) + 4e–

ii) At electrode Y, hydroxide ions were preferentially discharged to form oxygen gas.

Water dissociated continuously to replace the hydroxide ions discharged. Thus there was a build-up of hydrogen ions near electrode Y. The solution there became acidic.

The universal indicator turned red.

P22.2 page 112

a) Ions attracted to electrode X: Na+(aq), H+(aq)

Ions attracted to electrode Y: I–(aq), OH–(aq)

b) i) A colourless gas is given off.

A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions are preferentially discharged (reduced) to form hydrogen gas.

2H+(aq) + 2e– H2(g)

ii) A brown colour develops around electrode Y.

The concentration of iodide ions in the solution is much greater than that of hydroxide ions.

lodide ions are preferentially discharged to form iodine. The brown colour is due to the iodine formed.

2I–(aq) I2(aq) + 2e–

5�

Suggested Answers

P22.3 page 116

a) Ions attracted to copper cathode: Cu2+(aq), H+(aq)

Ions attracted to carbon anode: SO42–(aq), OH–(aq)

b) i) At carbon anode:

A hydroxide ion is a stronger reducing agent than sulphate ions.

Therefore hydroxide ions were preferentially discharged to form oxygen gas.

At copper cathode:

A copper(II) ion is a stronger oxidizing agent than a hydrogen ion.

Therefore copper(II) ions are preferentially discharged to form a deposit of copper on copper cathode.

ii) At carbon anode:

4OH–(aq) O2(g) + 2H2O(l) + 4e–

At copper cathode:


c) Copper(II) ions and hydroxide ions are consumed in the electrolysis process.

Hydrogen ions and sulphate ions remain in the solution.

Thus the solution eventually becomes sulphuric acid.

d) The copper(II) sulphate solution gradually fades out

because the copper ions become copper deposited on cathode, but no copper(II) ions formed at anode.

P22.4 page 119

a) Carbon rod is anode because Cl–(aq) loses electrons there.

b) At carbon rod: 2Cl–(aq) Cl2(aq) + 2e–

At mercury electrode: Na+(aq) + Hg(l) + e– Na/Hg(l)

c) After electrolysis, the mercury contains sodium. The sodium reacts with water to produce sodium hydroxide, which is alkali and turns phenolphthalein red.

2Na(l) + 2H2O(l) 2NaOH(aq) + H2(g)

52


P22.5 page 122

a) To make the knob a conductor of electricity for the nickel-plating process.

b) Ni2+(aq) + 2e– Ni(s)

c)


� a) anions

b) oxidation

c) cathode

d) cations

2

SolutionMaterial of Product at Change in the

solutionanode cathode anode cathode

Dilute sulphuric acid platinum platinumoxygen

gashydrogen

gasbecomes moreconcentrated

Very dilute sodium chloride carbon carbonoxygen

gashydrogen

gasbecomes moreconcentrated

Concentrated sodium chloride carbon carbonchlorine

gashydrogen

gasbecomes sodium

hydroxide solution

Concentrated sodium chloride carbon mercurychlorine

gassodium becomes more dilute

Dilute copper(II) sulphate carbon carbonoxygen

gascopper

becomes sulphuric acid

Dilute copper(II) sulphate copper carboncopper(II)

ionscopper remains the same

Dilute copper(II) sulphate copper coppercopper(II)

ionscopper remains the same

53

Suggested Answers

3

4

o x i d a t i o n

a

i

o

n

n

d l

e

c

t

r

o

p

l

a

e l e c t r o d e

c a t h o d e

a

b

o

n

c a t i o n

e

e

e

t

r

o

l

y

l

a

m

t

e

r

m

r e d u c t i o n

i

n

g

5 D

6 C

54


copper deposit

Option A and D — Copper(II) ions are preferentially discharged to form a deposit of copper on the cathode.


Thus, the concentration of Cu2+(aq) ion in the solution decreases.

Option C — Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and sulphate ions remain in the solution. Thus, the solution eventually becomes sulphuric acid. The pH of the solution decreases.

4OH–(aq) O2(g) + H2O(l) + 4e–

7 D At the anode

Hydroxide ions are preferentially discharged (oxidized) to form oxygen gas.

4OH–(aq) O2(g) + 2H2O(l) + 4e–

Water dissociates continuously to replace the hydroxide ions discharged at the anode.

H2O(l) H+(aq) + OH–(aq)

Thus, there is a build-up of hydrogen ions. The pH around the anode decreases.

At the cathode

Hydrogen ions are preferentially discharged (reduced) to form hydrogen gas.

2H+(aq) + 2e– H2(g)

Water dissociates continuously to replace the hydrogen ions discharged at the cathode. Thus, there is a build-up of hydroxide ions. The pH around the cathode increases.

8 A

9 D

�0 A

55

Suggested Answers

�� B

Option A — The impure copper acts as the anode.

Option B — Iron, zinc and copper in the impure copper undergo oxidation at the anode.

Zn(s) Zn2+(aq) + 2e–

Fe(s) Fe2+(aq) + 2e–

Cu(s) Cu2+(aq) + 2e–

Option C — Copper(II) ions undergo reduction at the cathode.

Option D — During the refining process, the copper is gradually transferred from the anode to the cathode. The concentration of copper(II) ions in the electrolyte and the color intensity of the electrolyte decreases gradually. This is because at the anode, iron and zinc readily dissolve as ions while at the cathode, copper(II) ions are always preferentially discharged.

�2 C

Option A — The nickel are discharged to form a deposit of nickel on the surface of the iron bolt.

Option C — The nickel(II) ions are reduced at the cathode.

Option D — The net effect is the transfer of nickel from the anode to the iron bolt. Thus, the concentration of nickel(II) ions in the solution remains the same.

56


�3 A (2) and (3) — Copper(II) ions are preferentially discharged to form a deposit of copper on the cathode.

Thus, the concentration of Cu2+(aq) ions on the solution decreases.

�4 B

�5 C

�6 A

�7 a) To increase the electrical conductivity of pure water.

b) Carbon / platinum

c) At the cathode: 2H+(aq) + 2e– H2(g)

At the anode: 4OH–(aq) O2(g) + 2H2O(l) + 4e–

d) 2:�

e) When hydrogen ions and hydroxide ions are discharged, more water molecules dissociate. The net effect is that water is decomposed.

The number of hydrogen ions and sulphate ions from the sulphuric acid remain the same. The concentration of sulphuric acid increases at the end as water is consumed in the electrolysis.

�8 Answers for the HKCEE question are not provided.

�9 Answers for the HKDSE question are not provided.

20 a) i) electrons

ii) ions

b) Cu(s) Cu2+(aq) + 2e–

c) Exactly 2 mol dm–3

2� a) At the carbon electrode X

The sulphate ions and hydroxide ions are attracted to electrode X.

A hydroxide ion is a stronger reducing agent than a sulphate ion.

Therefore hydroxide ions are preferentially discharged.

At the carbon electrode Y

The copper(II) ions and hydrogen ions are attracted to electrode Y.

A copper(II) ion is a stronger oxidizing agent than a hydrogen ion.

Therefore copper(II) ions are preferentially discharged to form a deposit of copper on electrode Y.

57

Suggested Answers

b) At the carbon electrode X

4OH–(aq) O2(g) + 2H2O(l) + 4e–

At the carbon electrode Y


c) The solution becomes sulphuric acid.

This is because copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and sulphate ions remain in the solution.

22 a) i) At the anode

2Cl–(aq) Cl2(g) + 2e–

ii) At the cathode

Na+(aq) + e– + Hg(l) Na/Hg(l)

b) i) Sodium hydroxide and hydrogen

ii) The sodium formed dissolves in the mercury cathode to form an alloy called sodium amalgam.

The sodium amalgam then moves towards the water. The sodium reacts with water to form sodium hydroxide and hydrogen.

2Na/Hg(l) + 2H2O(l) 2NaOH(aq) + H2(g) + 2Hg(l)

c) Sodium ions and chloride ions are consumed in the electrolysis. Thus the sodium chloride solution becomes more and more dilute.


24 a) i) Ag+(aq) + e– Ag(s)

ii) Oxygen.

Oxygen relights a glowing splint.

iii) 4OH–(aq) O2(g) + 2H2O(l) + 4e–

b) Use the block of impure silver as the anode and the pure silver as the cathode.

Use silver nitrate solution as the electrolyte.

Ag(s) Ag+(aq) + e–

25 Answers for the HKDSE question are not provided.

58


pages 137–145Topic Exercise

� C

2 D

3 B

4 C

5 B

6 B

7 C

8 A (2) Alkaline manganese cells are NOT rechargeable.

(3) Alkaline manganese cells of different sizes have the same voltage.

9 C

�0 A

�� B

�2 A

�3 A

�4 C

�5 C

�6 a) C in CO2: +4

C in C6H�2O6: 0

b) CO2 is oxidizing agent, because oxidation number of C decreases from +4 to 0.

CO2 and H2O are reducing agents, because oxidation number of O increases from –2 to 0.

59

Suggested Answers

�7 a) Solution goes brown.

b) iodine and potassium chloride

Cl2(g) + 2I–(aq) 2Cl–(aq) + I2(aq)

c) Chlorine is a more powerful oxidizing agent than bromine.

d) Organic layer goes purple.

�8 a) 2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)


• Sulphuric acid oxidizes hydrogen iodide to iodine;

• Hydrogen iodide reduces sulphuric acid;

• Phosphoric acid does not oxidize hydrogen iodide to iodine.

�9 a) Oxidation number of Br in BrO3– = +5

Oxidation number of Br in Br– = –�

The oxidation number of Br decrease / Br is reduced in the reaction

b) 6Fe2+(aq) + 6H+(aq) + BrO3–(aq) 6Fe3+(aq) + 3H2O(l) + Br–(aq)

c) To provide H+(aq) ions / acidic conditions


2� a) MH + OH– M + H2O + e–

b) Adsorbed on the surface of a solid or as a liquid under pressure.


23 a) The colour of the solution changes from orange to green.

Cr2O72–(aq) + �4H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)

b) i) The colour of the solution changes from colourless to brown gradually.

2I–(aq) I2(aq) + 2e–

ii) Iodide is oxidized, because the oxidation number of iodine increases from –� to 0.

c) From carbon electrode in potassium iodide solution to carbon electrode in acidified potassium dichromate solution.

60



25 a) Electrode A: 2H+(aq) + 2e– H2(g)

Electrode B: 4OH–(aq) O2(g) + 2H2O(l) + 4e–

b) At electrode A, hydrogen ions were preferentially discharged to form hydrogen gas.

Water dissociated continuously to replace the hydrogen ions discharged. Thus, there was a build-up of hydroxide ions near electrode A. The solution there became alkaline.

The universal indicator turned blue.

At electrode B, hydroxide ions were preferentially discharged to form oxygen gas.

Water dissociated continuously to replace the hydroxide ions discharged. Thus there was a build-up of hydrogen ions near electrode B. The solution there became acidic.

The universal indicator turned red.

c) As water is decomposes to hydrogen and oxygen, concentration of electrolyte increases.

26 a) Ions can move between the electrodes / ions are charged.

b) i) Electrons are gained.

ii) sodium hydroxide

iii) 2Cl– Cl2 + 2e–


28 xxxxxxxx

29 a) Any one of the following:

• Process 2 releases SO2(g) while

Process 3 converts SO2(g) to H2SO4(aq)

SO2(g) causes acid rain.

• Process 2 releases CO(g) while

Process 3 does not.

CO(g) is toxic.

b) Process 3 has a high electricity consumption.

This is justified because the product zinc is pure.

6�

Suggested Answers

c) Zn2+(aq) + 2e– Zn(s)

The cathode

d) The carbon in Process 2 is oxidized.

Any one of the following:

The reaction of zinc oxide with sulphuric acid in Process 3: neutralization.

The reaction of zinc carbonate in Process 1: Thermal decomposition


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Suggested Teaching Scheme - STMGSSintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book2Banswer.pdf · Suggested Teaching Scheme ... Reactions in simple chemical cells and the electrochemical