Nilam Workbook - Ch.5 F4

MODULE • Chemistry FORM 4

UNIT

ELECTROCHEMISTRY

5

ELEKTROKIMIA

Concept Map / Peta Konsep Extraction of aluminium from molten aluminium oxide Pengekstrakan aluminium daripada aluminium oksida lebur

Anode / Anod: Non metal / Bukan logam

Cathode / Katod: Metal / Logam

Uses of electrolysis in industry Kegunaan elektrolisis dalam industri

Position of ion in the electrochemical series Kedudukan ion dalam siri elektrokimia

Concentration of ion Kepekatan ion

Products / Hasil

Molten / Lebur

Aqueous / Akueus

Type of electrolyte Jenis elektrolit

Electrolysis / Elektrolisis

ELECTROCHEMISTRY

Metal displacement reaction Tindak balas penyesaran logam

Example Contoh

Chemical cell Sel kimia Produce Menghasilkan

Electrochemical Series Siri Elektrokimia Uses / Kegunaan

Prediction of displacement reaction Meramal tindak balas penyesaran

91

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 91

Electrical energy ➝ Chemical energy Tenaga elektrik ➝ Tenaga kimia

Chemical energy ➝ Electrical energy Tenaga kimia ➝ Tenaga elektrik

Learning objective / Objektif pembelajaran

Potential difference (voltage) Beza keupayaan (voltan)

Predict voltage of chemical cell Meramal voltan sel kimia

Factors determining selective discharge of ions at electrodes Faktor mempengaruhi pemilihan nyahcas ion di elektrod

Compare and contrast Banding dan beza

Energy change Perubahan tenaga

Can be constructed with Boleh dibina dengan

Determine terminals of chemical cell Menentukan terminal sel kimia

Energy change Perubahan tenaga

Type of electrode Jenis elektrod

5

– Dry cell / Sel kering – Lead-acid accumulator Akumulator asid plumbum – Mercury cell / Sel merkuri – Alkaline cell / Sel alkali – Nickel cadmium cell Sel nikel kadmium

ELEKTROKIMIA

Electroplating of metals Penyaduran logam

UNIT

Uses of electrolysis in industry Kegunaan elektrolisis dalam industri

Purification of metals Penulenan logam

1 Understand the electrolyte and non-electrolyte properties Memahami sifat-sifat elektrolit dan bukan elektrolit 2 Analyse the electrolysis process of molten compounds Menganalisis proses elektrolisis sebatian lebur 3 Analyse electrolysis of aqueous solution Menganalisis elektrolisis larutan akueus 4 Assess the electrolysis process in the industry Menilai proses elektrolisis dalam industri 5 Analyse chemical cells Menganalisis sel kimia 6 Synthesis electrochemical series Mensintesiskan siri elektrokimia 7 Apply awareness and apply responsible attitude in handling the chemicals used in the electrochemical series Menerapkan kesedaran serta mengamalkan sikap bertanggungjawab dalam mengendalikan bahan kimia yang digunakan dalam siri elektrokimia © Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Electrolysis / Elektrolisis What is conductor?

Apakah konduktor?

chemical changes

Element that can conduct electricity in solid or molten state without any

, normally metals

and carbon.

Unsur yang boleh mengkonduksi arus elektrik dalam keadaan pepejal atau leburan tanpa

perubahan kimia

,

biasanya logam dan karbon.

What are electrolytes?

Apakah elektrolit?

Compounds

that

can

conduct

chemical changes

electricity

in

*molten

state

or

*aqueous

solution

and

undergo

.

Sebatian yang boleh mengkonduksikan arus elektrik dalam keadaan *lebur atau *akueus serta mengalami perubahan kimia

.

Example / Contoh: – Aqueous solution of ionic compound such as copper(II) sulphate solution and sodium chloride solution Larutan akueus bagi sebatian ion contohnya larutan kuprum(II) sulfat dan larutan natrium klorida – Aqueous solution of *acid or alkali such as hydrochloric acid (HCl) and ammonia solution (NH3).

Larutan akueus *asid atau alkali contohnya asid hidroklorik (HCl) dan larutan ammonia (NH3). ionic

– Molten oxide.

Leburan sebatian aluminium oksida.

compounds such as molten lead(II) bromide, molten sodium chloride and molten aluminium ion

contohnya leburan plumbum(II) bromida, leburan natrium klorida dan leburan

* HCl and NH3 are covalent compounds, exist in form of molecule without water but ionised in water.

(Explanation is in the next topic in acid and base) HCl dan NH3 adalah sebatian kovalen, tanpa air terdiri daripada molekul tetapi ianya terion dalam air (akan dijelaskan dalam tajuk seterusnya dalam asid dan bes) * Molten state: a solid that is heated until it melts *Aqueous solution: a solid that is dissolved in water Lebur: pepejal yang dipanaskan sehingga cair. *Akueus: pepejal yang larut di dalam air.

What are non-electrolytes substances?

Apakah bahan bukan elektrolit?

Compounds that cannot conduct electricity in solid, molten and aqueous solution.

Sebatian yang tidak boleh mengkonduksikan arus elektrik dalam keadaan pepejal, lebur dan larutan akueus.

Example / Contoh: Molten

UNIT

5

What is electrolysis?

Apakah elektrolisis?

covalent

Leburan sebatian Electrolysis is a

process

electric current Elektrolisis ialah dialirkan melaluinya.

What is electrolytic cell?

Apakah sel elektrolisis?

compound such as naphthalene, molten sulphur and liquid bromine.

kovalen

contohnya naftalena, sulfur lebur dan cecair bromin. whereby an electrolyte is decomposed to its constituent elements when

passes through it.

proses

penguraian elektrolit kepada unsur juzuknya apabila

An electrolytic cell is a set up of apparatus that contains two and produce a chemical reaction when connected to a

electrodes battery

Sel elektrolisis adalah susunan radas yang terdiri daripada dua elektrolit

elektrik).

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 92

arus elektrik

which are dipped in an

electrolyte

(source of electricity). elektrod

menghasilkan tindak balas kimia apabila disambungkan kepada

yang dicelup ke dalam

bateri

(sumber arus

92

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 The set up of typical electrolytic cell Susunan bagi sel elektrolisis yang biasa

e e Anode / Anod (Electrode / Elektrod)

e e

e

e electrons / elektron

Cathode / Katod (Electrode / Elektrod)

anions / anion cations / kation

e

Important terminology / Terminologi penting: 1 Electrode / Elektrod: – Conductor which is dipped into electrolyte which carries electric current in and out of electrolyte.

Konduktor yang dicelup dalam elektrolit yang mengalirkan arus elektrik ke dalam dan keluar daripada elektrolit.

2 Cathode / Katod:

– An electrode that is connected to the Elektrod yang disambung kepada

negative terminal

terminal negatif

of the battery in the electrolytic cell. bateri dalam sel elektrolisis.

3 Cations / Kation: Positive – ions which are attracted and move to the negatively charged electrode, positif katod Ion akan tertarik dan bergerak ke arah elektrod yang bercas 4 Anode / Anod:

– An electrode that is connected to the Elektrod yang disambung kepada

positive terminal

terminal positif

cathode negatif

of the battery in the electrolytic cell. bateri dalam sel elektrolisis.

5 Anions / Anion: – Negative ions which are attracted and move to the positively charged electrode, negatif anod Ion akan tertarik dan bergerak ke arah elektrod yang bercas anode

(a) Anions (negative ions) are attracted and move to the

.

. The anions release electrons to the

surface of anode and become neutral atoms or molecule. The anions are discharged at the anode. anod Anion (ion negatif) akan tertarik dan bergerak ke arah . Anion melepaskan elektron pada dinyahcaskan permukaan anod dan menjadi atom atau molekul. Anion di anod.

anode (b) Electrons flow from the to the external circuit . / Elektron mengalir dari litar luar penyambung dalam .

(c) Cations (positive ions) are attracted and move to the

.

positif

cathode

anod

through the connecting wire in the katod ke melalui wayar

cathode

. The cations receive electrons at the surface of cathode and become neutral atoms or molecules. The cations are discharged at the cathode. katod Kation (ion positif) akan tertarik dan bergerak ke arah . Kation menerima elektron pada dinyahcaskan permukaan katod dan menjadi atom atau molekul. Kation di katod.

➞ Electrolyte decomposed to its constituent elements. / Elektrolit terurai kepada unsur-unsur juzuknya.

Remark / Catatan: Electricity is conducted in electrolytic cell by: / Elektrik dikonduksi dalam sel elektrolisis dengan: (i) Free moving anion and cation in the electrolyte. / Anion dan kation yang bebas bergerak dalam elektrolit. (ii) Flow of electrons in the connecting wire. / Aliran elektron dalam wayar penyambung.

Explain the energy change in electrolysis.

Terangkan perubahan tenaga dalam elektrolisis.

Example of electrolytic cell.

Contoh sel elektrolisis.

The stages in electrolysis process are: / Peringkat dalam proses elektrolisis: – Electrons flow through the external circuit. / Elektron mengalir melalui litar luar.

– Chemical changes occur at the anode and cathode. / Perubahan kimia berlaku di anod dan katod.

Energy change in electrolysis: / Perubahan tenaga dalam elektrolisis: Chemical Energy – Electric Energy to / Tenaga Elektrik kepada (i)

(ii) Electrodes Elektrod Electrolyte Elektrolit

A

Tenaga Kimia (iii)

Electrolyte Elektrolit

Electrode Elektrod A

Electrolyte Elektrolit

Heat Panaskan

Electrolysis of molten electrolyte Elektrolisis elektrolit lebur

Electrodes Elektrod

Electrode Elektrod

Electrolysis of aqueous electrolyte (No gas released) Elektrolisis elektrolit dalam bentuk akueus (Tiada gas dibebaskan)

93

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 93

5

anode

UNIT

Explain how electrolysis process occur.

Terangkan bagaimana proses elektrolisis berlaku.

. .

Electrolysis of aqueous electrolyte (Gas is released) Elektrolisis elektrolit dalam bentuk akueus (Gas dibebaskan) © Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Formation of Free Moving Ions in the Electrolyte Pembentukan Ion Bergerak Bebas dalam Elektrolit What is ionisation equation?

Apakah persamaan pengionan?

Example of ionisation of molten electrolyte (a compound that is heated until it melts).

Contoh pengionan elektrolit dalam keadaan leburan (sebatian yang dipanaskan hingga lebur).

Example of the ionisation on an aqueous electrolyte (a compound that is dissolved in water): Contoh pengionan elektrolit dalam keadaan akueus (sebatian yang dilarutkan dalam air)

It is an equation to determine the ions present in molten or aqueous electrolyte.

Persamaan yang menunjukkan ion yang hadir dalam elektrolit sama ada dalam keadaan leburan atau akueus.

(i) Molten sodium chloride / Natrium klorida lebur:

NaCl(s/p)

(ii) Molten lead(II) bromide / Plumbum(II) bromida lebur: PbBr2(s/p)

Na+(l/ce) + Cl–(l/ce) Pb2+(l/ce) + 2Br–(l/ce) 2Na+(l/ce) + O2–(l/ce)

(iii) Molten sodium oxide / Natrium oksida lebur:

Na2O(s/p)

(iv) Molten aluminium oxide / Aluminium oksida lebur:

Al2O3(s/p)

2Al3+(l/ce) + 3O2–(l/ce)

(i) Sodium chloride solution / Larutan natrium klorida:

NaCl(aq/ak)

Na+(aq/ak) + Cl–(aq/ak)

H2O

H+(aq/ak) + OH–(aq/ak) Cu2+ + SO42–

(ii) Copper(II) sulphate solution / Larutan kuprum(II) sulfat: CuSO4(aq/ak)

(iii) Sulphuric acid / Asid sulfurik:

H+ + OH–

H2O

2H+ + SO42–

H2SO4(aq/ak)

H+ + OH–

H2O

Reactions at the Electrodes / Tindak Balas di Elektrod Define discharged of cation or anion Takrifkan nyahcas bagi kation atau anion

discharged receives (a) A cation is when it electrons at the cathode. dinyahcaskan menerima Kation apabila elektron di katod. discharged releases (b) An anion is when it electrons at the anode. dinyahcaskan melepaskan Anion apabila elektron di anod. (c) When ions are

Apabila ion neutral.

discharged

dinyahcaskan

, they become neutral , ianya akan menjadi

atoms

atom

or

molecules molekul

atau

. yang

UNIT

Remark / Catatan: The process of discharging results in the / Proses nyahcas menghasilkan: – Conduction of electricity by the electrolyte.

Pengkonduksian elektrik melalui elektrolit.

– Decomposition of electrolyte into its component elements.

Penguraian elektrolit kepada komponen unsurnya.

5 What is half equation?

Apakah persamaan setengah?

The equation representing reaction that take place at the anode and cathode involve ions and electrons.

Persamaan yang mewakili tindak balas yang berlaku di anod dan katod melibatkan ion dan elektron.

– Half equation at the anode: anions / metal atoms release electrons to produce neutral atom / molecules.

Persamaan setengah di anod: anion / atom logam melepaskan elektron untuk menghasilkan atom / molekul neutral.

Xn– X + ne – Half equation at the cathode: Cations receive electrons to produce neutral atom / molecule.

Persamaan setengah di katod: Kation menerima elektron untuk menghasilkan atom / molekul neutral.

Ym+ + me Y Example 1: Chloride ions release electrons to form chlorine molecule at anode.

Contoh 1: Ion klorida melepaskan elektron untuk membentuk molekul klorin di anod.

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 94

Step 1 Langkah 1

Write the formula of the reactant and products Tulis formula bagi bahan dan hasil tindak balas

Step 2 Langkah 2

Balance the number of atoms on the left and right, calculate the total charge Imbangkan bilangan atom di kiri dan kanan, hitung jumlah cas

2CI– CI2 Left / Kiri Right / Kanan 2(–1) = –2 0 (not balanced / tidak seimbang)

Step 3 Langkah 3

Balance the total charge by adding electrons, calculate the total charge Imbangkan jumlah cas dengan menambahkan elektron, hitung jumlah cas

2CI– CI2 + 2e Left / Kiri Right / Kanan 2(–1) = –2 0 + 2(–1)= –2 (balanced / seimbang)

CI–

CI2

94

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Example 2: Hydrogen ions receive electrons to form hydrogen molecule at the cathode.

Contoh 2: Ion hidrogen menerima elektron untuk membentuk molekul hidrogen di katod.

Write the formula of the reactant and products Tulis formula bagi bahan dan hasil tindak balas

Step 2 Langkah 2

Balance the number of atoms on the left and right, calculate the total charge Imbangkan bilangan atom di kiri dan kanan, hitung jumlah cas

2H+ H2 Left / Kiri Right / Kanan 2(+1) = +2 0 (not balanced / tidak seimbang)

Step 3 Langkah 3

Balance the total charge by adding electrons, calculate the total charge Imbangkan jumlah cas dengan menambahkan elektron, hitung jumlah cas

2H+ + 2e H2 Left / Kiri Right / Kanan + 2 + 2(–1) = 0 0 (balanced / seimbang)

1

4OH– release

Four hydroxide ions .

molekul

2H2O + O2 + 4e

four electrons to form two water molecules and one oxygen

melepaskan

Empat ion hidroksida

empat elektron membentuk dua molekul air dan satu

oksigen.

2 2Cl– release

Two chloride ions

3

4 Atom kuprum 5 Silver atom

melepaskan

Atom argentum 1 Two hydrogen ions Dua ion hidrogen 2 Silver ion

receive menerima

receives

Ion argentum 3 Copper(II) ion Ion kuprum(II) Write the equation of discharge of ion: Tuliskan persamaan nyahcas ion:

Ag

menerima

one electron to form

.

ion kuprum(II)

silver ion

.

.

ion argentum

satu elektron membentuk

.

H2

2H+ + 2e

molecule

two electrons to form one hydrogen

molekul

dua elektron membentuk satu

.

hidrogen.

Ag atom atom

satu elektron membentuk satu Cu2+ + 2e

. argentum.

Cu

two electrons to form one copper dua elektron membentuk satu

atom atom

. kuprum.

(i) Lead(II) ion to lead atom / Ion plumbum(II) kepada atom plumbum:

Pb2+ + 2e

(ii) Silver ion to silver atom / Ion argentum kepada atom argentum:

Ag + e

(iii) Iodide ion to iodine molecule / Ion iodida kepada molekul iodin:

2I

95

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 95

copper(II) ion

one electron to form one silver

menerima

bromin.

Ag+ + e

Ag+ + e

receives

.

Cu2+ + 2e

dua elektron membentuk

releases

molekul

dua elektron membentuk satu

two electrons to form

melepaskan

klorin.

molecule

two electrons to form one bromine

Cu

releases

Copper atom

.

Br2 + 2e

2Br–

melepaskan

Dua ion bromida

molecule

molekul

dua elektron membentuk satu

release

Two bromide ions

Cl2 + 2e

two electrons to form one chlorine

melepaskan

Dua ion klorida

What are the common half equation at the cathode?

Apakah persamaan setengah yang biasa di katod?

H2

5

molecule

H+

UNIT

What are the common half equation at the anode? (anion/metal atom releases electrons) Apakah persamaan yang biasa di anod? (anion/atom logam membebaskan elektron)

Step 1 Langkah 1

+

–

Pb Ag

I2 + 2e

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Electrolysis of Molten Electrolyte / Elektrolisis Elektrolit Lebur 1

Aim Tujuan

: To investigate the electrolysis of molten lead(II) bromide : Mengkaji elektrolisis plumbum(II) bromida lebur

Apparatus : Battery, switch, carbon electrodes, connecting wires, ammeter, crucible, tripod stand, pipe-clay triangle, Bunsen Radas burner.

Bateri, suis, elektrod karbon, wayar penyambung, ammeter, mangkuk pijar, segi tiga tanah liat, penunu Bunsen.

2

3

4

5

Material : Lead(II) bromide powder Bahan Serbuk plumbum(II) bromida

Procedure / Prosedur : (a) A crucible is filled with lead(II) bromide powder until it is half full. (b) Two carbon electrodes are placed into lead(II) bromide and connected to the batteries and ammeter using connecting wire as shown in the diagram. (c) Lead(II) bromide powder is then heated until it melts.

Lead(II) bromide (d) The observations at the anode and the cathode are recorded.

Plumbum(II) bromida (e) Both electrodes are taken out from the electrolyte and the molten lead(II) bromide are poured out of crucible carefully to observe the product at cathode. (a) Mangkuk pijar diisi dengan serbuk plumbum(II) bromida hingga separuh penuh.

Heat (b) Dua elektrod karbon dimasukkan ke dalam serbuk plumbum(II) bromida dan Panaskan disambung kepada bateri dan ammeter menggunakan wayar penyambung seperti yang ditunjukkan dalam rajah di atas. (c) Plumbum(II) bromida dipanaskan hingga lebur. (d) Pemerhatian di anod dan katod direkodkan. (e) Kedua-dua elektrod dikeluarkan dari elektrolit dan dituangkan keluar dengan berhati-hati dan perhatikan hasil yang terbentuk pada katod.

Observation / Pemerhatian : Electrode / Elektrod

UNIT

Anode / Anod

5

Cathode / Katod

Observation / Pemerhatian Brown gas is released.

Gas perang dibebaskan.

A shiny grey globule is formed at the bottom of cathode.

Titisan kelabu berkilat terbentuk di bahagian bawah katod.

Explain electrolysis of molten lead(II) bromide Terangkan elektrolisis plumbum(II) bromida lebur Draw a labelled diagram for the set up of apparatus and by using arrows to show the movement of particles that occur in lead(II) bromide and the direction of electron flow in the external circuit.

Lukiskan rajah berlabel susunan radas dengan menggunakan anak panah untuk menunjukkan pergerakan zarah yang berlaku dalam plumbum(II) bromida dan arah aliran elektron dalam litar luar.

List all the ions present in molten lead(II) bromide.

Senaraikan semua ion yang ada dalam plumbum(II) bromida lebur. © Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 96

Carbon electrodes Elektrod karbon

Lead(II) bromide Plumbum(II) bromida

Heat Panaskan

The ions present are lead(II) ions/Pb2+ and bromide ions/Br– Ion yang hadir adalah ion plumbum(II)/Pb2+ dan ion bromida/Br–

96

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Explain the electrolysis of molten lead(II) bromide Jelaskan elektrolisis plumbum(II) bromida lebur

(a) At the anode / Di anod: – Bromide ion/ Br– move to the anode.

Ion bromida/Br– bergerak ke arah anod.

– Bromide ion/ Br– releases one electron to form bromine atom at the anode.

Ion bromida/Br– membebaskan satu elektron untuk membentuk atom bromin di anod.

– Two bromine atoms combine to form bromine molecule.

– Dua atom bromin bergabung untuk membentuk molekul bromin.

– Half equation / Persamaan setengah : 2Br– Br2 + 2e – Brown gas is released.

Gas perang dibebaskan. (b) At the cathode / Di katod: – Lead(II) ions/Pb2+ move to the cathode.

Ion plumbum(II)/Pb2+ bergerak ke arah katod.

– Lead(II) ions/Pb2+ receive two electrons to form lead atom at the cathode.

Ion plumbum(II)/Pb2+ menerima dua elektron untuk membentuk atom plumbum di katod.

– Half equation / Persamaan setengah : Pb2+ + 2e Pb – Shiny grey solid is formed.

Pepejal kelabu berkilat terbentuk.

Factors that Affect the Electrolysis of an Aqueous Solution Faktor-faktor yang Mempengaruhi Elektrolisis Larutan Akueus How many types of cations and anions usually present in an aqueous solution? / Berapa jenis kation dan anion yang biasanya terdapat dalam suatu larutan akueus?

Two types of cations and two types of anions.

Dua jenis kation dan dua jenis anion.

State the sources of these ions.

Nyatakan sumber ion-ion ini.

– A cation and an anion from the dissolved substance.

Satu kation dan satu anion daripada bahan terlarut.

– Hydrogen(H+) ions and hydroxide(OH–) ions from water.

Ion hidrogen(H+) dan ion hidroksida(OH–) daripada air.

Solution Larutan Dilute potassium iodide solution Larutan kalium iodida cair Concentrated sodium chloride Natrium klorida pekat Copper(II) sulphate solution Larutan kuprum(II) sulfat Concentrated hydrochloric acid Asid hidroklorik pekat

What is selective discharge?

Apakah pemilihan nyahcas?

Ions from dissolved substance Ion daripada bahan terlarut

Ions from water Ion daripada air

K+, I–

H+, OH–

Na+ , Cl–

H+, OH–

Cu2+, SO42–

H+, OH–

H+, Cl–

H+, OH–

UNIT

Example: Contoh:

5

Remark / Catatan: 1 Water is a weak electrolyte that partially ionises to H+ and OH– ions.

Air ialah elektrolit lemah yang mengion separa kepada ion H+ dan OH–.

2 However, this small amount of ions can compete with the ions from the dissolve substance for the discharge.

Walau bagaimanapun, jumlah kecil ion ini boleh bersaing dengan ion daripada bahan terlarut untuk nyahcas.

When more than one type of ion are attracted towards the electrodes during electrolysis, only one type of ion is selected to be discharged at each electrode.

Apabila lebih dari satu jenis ion tertarik kepada elektrod semasa elektrolisis, hanya satu jenis ion yang dipilih untuk dinyahcas pada setiap elektrod.

Factors affecting the selective discharge of ions Faktor mempengaruhi pemilihan nyahcas ion 1

The selection of ion for discharge depends on three factors / Pemilihan ion untuk nyahcas bergantung pada tiga faktor: (a) The position of ions in the Electrochemical Series (normally in dilute solution and inert electrode).

Kedudukan ion dalam Siri Elektrokimia (biasanya dalam larutan cair dan elektrod lengai). (b) The concentration of electrolyte (normally in concentrated solution and inert electrode).

Kepekatan elektrolit (biasanya dalam larutan pekat dan elektrod lengai). (c) The types of electrode (when metal as electrode is used). / Jenis elektrod (apabila elektrod logam digunakan).

97

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 97

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 98

Sodium nitrate solution

Anode

Observation:

Cathode

Procedure: 1 1 mol dm–3 sodium nitrate solution is poured into electrolytic cell until the solution covers the carbon electrodes.

2 Test tubes are filled with sodium nitrate solution and inverted on the carbon electrodes as shown in the diagram.

3 The carbon electrodes are connected to the batteries and ammeter using connecting wires.

4 The electricity is passed through the electrolyte for 10 to 15 minutes.

5 The observation at the anode and cathode are recorded.

6 The collected gases are tested with wooden splinter: • Use glowing wooden splinter at anode.

• Use lighten wooden splinter at cathode.

Carbon electrodes

5

Factor: Position of Ions in Electrochemical Series Electrolysis of sodium nitrate solution using carbon electrode Aim : To investigate the electrolysis of 0.1 mol dm–3 sodium nitrate solution using carbon electrodes.

Apparatus : Electrolytic cell, test tubes, batteries, ammeter, carbon electrodes and connecting wires Materials : 1 mol dm–3 sodium nitrate solution, wooden splinter

UNIT Hydrochloric acid

0.0001 mol dm–3 hydrochloric acid 2 mol dm–3 hydrochloric acid

Electrolyte

Observation:

Observation Anode Cathode

Procedure: 1 0.0001 mol dm–3 hydrochloric acid is poured into the electrolytic cell until it covers both carbon electrodes.

2 Both test tubes are filled with 0.0001 mol dm–3 hydrochloric acid and the test tubes are inverted on carbon electrodes as shown in the diagram.

3 The carbon electrodes are connected to the batteries and ammeter using connecting wires.

4 The electricity is passed through the electrolyte for 10 to 15 minutes.

5 The collected gases are tested with wooden splinter: • Use glowing wooden splinter at anode.

• Use lighten wooden splinter at cathode.

6 All the observations are recorded.

7 Steps 1 to 6 are repeated using 2 mol dm–3 hydrochloric acid to replace 0.0001 mol dm–3 hydrochloric acid. The gas released at the anode is tested with the moist blue litmus paper.

Carbon electrodes

Factor: Concentration of Electrolyte Electrolysis of 0.0001 mol dm–3 hydrochloric acid and 2 mol dm–3 of hydrochloric acid using carbon electrodes Aim : To investigate the effect of concentration of electrolyte on the product of electrolysis at the anode.

Problem statement : How does the concentration of electrolyte affect the product at the anode during electrolysis?

Manipulated variable : Concentration of hydrochloric acid Responding variable : Product at the anode Constant variable : Hydrochloric acid / type of acid, carbon electrodes Hypothesis : When very dilute hydrochloric acid is used as electrolyte, the product at the anode is oxygen gas. When concentrated hydrochloric acid is used as electrolyte, chlorine gas is released at the anode.

Material : 0.0001 mol dm–3 hydrochloric acid, 2 mol dm–3 hydrochloric acid, blue litmus paper, carbon electrodes, wooden splinter Apparatus : Electrolytic cell, test tube, batteries, ammeter, connecting wires Carbon electrodes

Copper(II) sulphate solution

Copper(II) sulphate solution

Copper electrodes

Carbon Copper

Electrode

Anode

Observation Cathode

Electrolyte

Procedure: 1 Copper(II) sulphate solution is poured into a beaker until half full.

2 Two copper plates are cleaned using sand paper.

3 Copper electrodes are dipped into copper(II) sulphate solution and connected to the batteries using connecting wires as shown in the diagram.

4 The electricity is passed through the electrolyte for 10 to 15 minutes.

5 The observations at the anode, cathode and electrolyte are recorded.

Observation:

Electrolysis of copper (II) sulphate using copper electrodes

Procedure: 1 1 mol dm–3 copper(II) sulphate solution is poured into electrolytic cell until the solution covers the carbon electrodes.

2 Test tubes are filled with copper(II) sulphate solution and inverted on the carbon electrodes as shown in the diagram.

3 The carbon electrodes are connected to the batteries and ammeter using connecting wires as shown in diagram.

4 The electricity is passed through the electrolyte for 10 to 15 minutes.

5 The observation at the anode and cathode are recorded.

6 The gases collected at anode and cathode are tested with wooden splinter: • Use glowing wooden splinter at anode.

• Use lighten wooden splinter at cathode.

Electrolysis of the copper(II) sulphate using carbon electrodes

Factor: Type of Electrode Electrolysis of copper(II) sulphate solution using copper electrode and carbon electrode.

Aim : To investigate the effect of the types of electrodes on the product of electrolysis at the anode.

Problem statement : Do the types of electrodes affect the product of electrolysis at the anode?

Manipulated variable : Copper electrode and carbon electrode Responding variable : Product at the anode Constant variable : Copper(II) sulphate solution Hypothesis : When copper is used as electrode in electrolysis of copper(II) sulphate solution, copper(II) is produced at the anode. When carbon is used as electrode in electrolysis of copper(II) sulphate solution, oxygen gas is released at the anode.

Apparatus : Electrolytic cell, test tubes, batteries, ammeter, carbon electrodes, connecting wires, beaker.

Material : 1.0 mol dm–3 copper(II) sulphate solution, wooden splinter, copper metal plate, sand paper

Experiments To Determine The Selective Discharge Of Ions At The Electrode

MODULE • Chemistry FORM 4

98

11/10/2017 9:14 AM

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 99

Larutan natrium nitrat

99

Anod

Pemerhatian:

Katod

Prosedur: 1 Larutan natrium nitrat 1 mol dm–3 dituangkan ke dalam sel elektrolitik sehingga menutupi elektrod karbon.

2 Tabung uji diisi dengan larutan natrium nitrat dan diterbalikkan ke atas elektrod karbon seperti yang ditunjukkan dalam rajah.

3 Elektrod karbon disambung kepada bateri dan ammeter menggunakan wayar penyambung.

4 Arus elektrik dialirkan melalui elektrolit selama 10 hingga 15 minit.

5 Pemerhatian di anod dan katod direkodkan.

6 Gas yang terkumpul di anod dan katod diuji dengan kayu uji: • Gas di anod diuji dengan kayu uji berbara.

• Gas di katod diuji dengan kayu uji menyala.

Elektrod karbon

Faktor: Kedudukan Ion dalam Siri Elektrokimia Elektrolisis larutan natrium sulfat menggunakan elektrod karbon Tujuan : Mengkaji elektrolisis larutan natrium nitrat 0.1 mol dm–3 menggunakan elektrod karbon.

Radas : Sel elektrolitik, tabung uji, bateri, ammeter, elektrod karbon dan wayar penyambung Bahan : Larutan natrium nitrat 1 mol dm–3, kayu uji

Asid hidroklorik

UNIT

Asid hidroklorik 0.0001 mol dm–3 Asid hidroklorik 2 mol dm–3

Elektrolit

5

Pemerhatian Anod Katod

Prosedur: 1 Asid hidroklorik 0.0001 mol dm–3 dituangkan ke dalam sel elektrolitik hingga menutupi elektrod karbon.

2 Kedua-dua tabung uji diisi dengan asid hidroklorik 0.0001 mol dm–3 dan tabung uji diterbalikkan di atas elektrod karbon seperti yang ditunjukkan dalam rajah.

3 Elektrod karbon disambung kepada bateri dan ammeter menggunakan wayar penyambung.

4 Arus elektrik dialirkan melalui elektrolit selama 10 hingga 15 minit.

5 Gas yang terkumpul di anod dan katod diuji dengan kayu uji: • Gas di anod diuji dengan kayu uji berbara.

• Gas di katod diuji dengan kayu uji menyala.

6 Semua pemerhatian direkodkan.

7 Langkah 1 hingga 6 diulangi menggunakan asid hidroklorik 2 mol dm–3 menggantikan asid hidroklorik 0.0001 mol dm–3. Gas yang terbebas di anod diuji dengan kertas litmus biru lembap.

Pemerhatian:

Elektrod karbon

Faktor: Kepekatan Elektrolit Elektrolisis asid hidrklorik 0.0001 mol dm–3 dan asid hidroklorik 2 mol dm–3 menggunakan elektrod karbon Tujuan : Mengkaji kesan kepekatan elektrolit kepada hasil elektrolis di anod.

Pernyataan masalah : Bagaimanakah kepekatan elektrolit mempengaruhi hasil elektrolisis di anod?

Pemboleh ubah dimanipulasikan : Kepekatan asid hidroklorik Pemboleh ubah bergerak balas : Hasil di anod Pemboleh ubah dimalarkan : Asid hidroklorik / jenis asid, elektrod karbon Hipotesis : Apabila asid hidroklorik yang sangat cair digunakan sebagai elektrolit, hasil di anod adalah gas oksigen. Apabila asid hidroklorik pekat digunakan sebagai elektrolit, hasil di anod adalah gas klorin.

Bahan : Asid hidroklorik 0.0001 mol dm–3, asid hidroklorik 2 mol dm–3, kertas litmus biru, elektrod karbon, kayu uji Radas : Sel elektrolitik, tabung uji, bateri, ammeter, wayar penyambung.

Elektrod karbon

Larutan kuprum(II) sulfat

Larutan kuprum(II) sulfat

Elektrod kuprum

Karbon Kuprum

Elektrod

Pemerhatian: Anod

Pemerhatian Katod

Elektrolit

Prosedur: 1 Larutan kuprum(II) sulfat dituangkan ke dalam bikar hingga separuh penuh.

2 Dua kepingan kuprum dibersihkan dengan kertas pasir.

3 Kepingan kuprum dicelup dalam larutan kuprum(II) sulfat dan disambung kepada bateri serta ammeter menggunakan wayar penyambung seperti yang ditunjukkan pada rajah.

4 Arus elektrik dialirkan melalui elektrolit selama 10 hingga 15 minit.

5 Pemerhatian pada anod, katod dan elektrolit direkodkan.

Elektrolisis larutan kuprum(II) sulfat menggunakan elektrod kuprum

Prosedur: 1 Larutan kuprum(II) sulfat 1 mol dm–3 dituangkan ke dalam sel elektrolitik sehingga larutan melitupi elektrod karbon.

2 Tabung uji diisi dengan larutan kuprum(II) sulfat dan diterbalikkan ke atas elektrod karbon seperti yang ditunjukkan dalam rajah.

3 Elektrod karbon disambung kepada bateri dan ammeter menggunakan wayar penyambung seperti yang ditunjukkan dalam rajah.

4 Arus elektrik dialirkan melalui elektrolit selama 10 hingga 15 minit.

5 Pemerhatian di anod dan katod direkodkan.

6 Gas yang terkumpul di anod dan katod diuji dengan kayu uji: • Gas di anod diuji dengan kayu uji berbara.

• Gas di katod diuji dengan kayu uji menyala.

Elektrolisis larutan kuprum(II) sulfat menggunakan elektrod karbon

Faktor: Jenis Elektrod Elektrolisis larutan kuprum(II) sulfat menggunakan elektrod kuprum dan elektrod karbon Tujuan : Mengkaji kesan jenis elektrod ke atas hasil elektrolisis di anod.

Pernyataan masalah : Adakah jenis elektrod mempengaruhi hasil elektrolisis pada anod?

Pemboleh ubah dimanipulasikan : Elektrod kuprum dan elektrod karbon Pemboleh ubah bergerak balas : Hasil di anod Pemboleh ubah dimalarkan : Larutan kuprum(II) sulfat Hipotesis : Apabila kuprum digunakan sebagai elektrod dalam elektrolisis larutan kuprum(II) sulfat, kuprum(II) dihasilkan pada anod. Apabila karbon digunakan sebagai elektrod dalam elektrolisis larutan kuprum(II) sulfat, gas oksigen dibebaskan pada anod.

Radas : Sel elektrolitik, tabung uji, bateri, ammeter, elektrod karbon, wayar penyambung, bikar.

Bahan : Larutan kuprum(II) sulfat 1.0 mol dm–3, kayu uji, kepingan logam kuprum, kertas pasir

Eksperimen Untuk Menentukan Pemilihan Nyahcas Ion Di Elektrod

MODULE • Chemistry FORM 4

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

The position of ions in the electrochemical series Kedudukan ion dalam siri elektrokimia When is the factor of position of ions in electrochemical series being applied? / Bilakah faktor kedudukan ion dalam siri elektrokimia digunakan?

When electrolysis is conducted on dilute solution and inert electrodes.

Apabila elektrolisis dijalankan ke atas larutan cair dan elektrod lengai.

How to choose cation and anion to be discharged based on this factor?

Bagaimana memilih kation dan anion untuk dinyahcas berdasarkan faktor ini?

The lower position of cation in the electrochemical series, or anions in the lower position of the anion discharge series will be selected to be discharged. / Lebih rendah kedudukan kation dalam siri elektrokimia atau anion yang lebih rendah kedudukan dalam siri nyahcas anion akan dinyahcas.

Cation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, and Au+ Kation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, dan Au+ Tendency to discharge increase Kecenderungan dinyahcas bertambah Anion: F–, SO42–, NO3–, Cl–, Br –, I–, and OH– Anion: F–, SO42–, NO3–, Cl–, Br –, I–, dan OH–

Choose the ion to be discharged from the pairs of ions. State the electrode where it occurs and write the half equation for the discharge of ion: Pilih ion yang dinyahcas dari pasangan ion. Nyatakan elektrod di mana ia berlaku dan tulis persamaan setengah bagi ion yang dinyahcas.

(i) Hydroxide & sulphate ions

Ion hidroksida & ion sulfat

(ii) Hydroxide & nitrate ions

2H2O + O2 + 4e

anode

: Persamaan setengah: 4OH – 2H2O + O2 + 4e : Half equation: 4OH

anod

: Half equation:

4OH–

at the 2H2O + O2 + 4e di

–

at the – 4OH 2H O + O + 4e di 2 2 Ion hidroksida & ion nitrat : Persamaan setengah: 2+ Cu + 2e Cu (iii) Hydrogen & copper(II) ions : Half equation: at the 2+ Cu + 2e Cu Ion hidrogen & ion kuprum(II) : Persamaan setengah: di 2H + 2e

(iv) Hydrogen & potassium ions : Half equation:

Ion hidrogen & ion kalium

(v) Hydrogen & silver ions

H2

+

: Persamaan setengah: : Half equation:

2H+ + 2e Ag+ + e

Ion hidrogen & ion argentum : Persamaan setengah:

H2

Ag

Ag + e +

Ag

at the di at the di

anode anod cathode katod cathode katod cathode katod

. . . . . . . . . .

(a) Electrolysis of 0.1 mol dm–3 sodium nitrate solution using carbon electrodes.

Elektrolisis larutan natrium nitrat 0.1 mol dm–3 menggunakan elektrod karbon.

Equation of electrolyte ionisation Persamaan pengionan elektrolit

NaNO3 H 2O

UNIT

Electrode / Elektrod

5

Ions that are attracted to the anode and cathode Ion yang ditarik ke anod dan katod State the ion selected to be discharged. Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas. Terangkan jawapan anda.

Half equation Persamaan setengah

Anode / Anod

Cathode / Katod

NO3–, OH–

Na+, H+

– Hydroxide ion, OH– is selected.

Ion hidroksida, OH– dipilih.

– The position of hydroxide ion is lower than nitrate ion in the discharge series of anion.

Kedudukan ion hidroksida lebih rendah daripada ion nitrat dalam siri nyahcas anion.

– Hydroxide ions release electrons to produce water and oxygen molecules.

Ion hidroksida menderma elektron untuk menghasilkan molekul air dan oksigen.

4OH–

Na+ + NO3– H+ + OH–

2H2O + O2 + 4e

– Hydrogen ion,H+ is selected.

Ion hidrogen H+ dipilih.

– The position of hydrogen ion is lower than sodium ion in the Electrochemical Series.

Kedudukan ion hidrogen lebih rendah daripada ion natrium dalam Siri Elektrokimia.

– Hydrogen ions receive electrons to produce hydrogen molecule.

Ion hidrogen menerima elektron untuk menghasilkan molekul hidrogen.

2H+ + 2e

H2

Observations Pemerhatian

Gas bubbles are released Gelembung gas dibebaskan

Gas bubbles are released Gelembung gas dibebaskan

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

Insert a glowing wooden splinter into test tube, glowing wooden splinter is lighted up Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala

When a lighted wooden splinter is placed near the mouth of the test tube, a ‘pop’ sound is produced Apabila kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan

Name the products / Inference Nama hasil / Inferens

Oxygen gas is released Gas oksigen dibebaskan

Hydrogen gas is released Gas hidrogen dibebaskan

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 100

100

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

(b) Electrolysis of 0.001 mol dm–3 sulphuric acid using carbon electrodes.

Elektrolisis larutan asid sulfurik 0.001 mol dm–3 menggunakan elektrod karbon.

Equation of electrolyte ionisation Persamaan pengionan elektrolit

H2SO4 H 2O

2H+ + SO42– H+ + OH–

Electrode / Elektrod

Anode / Anod

Cathode / Katod

Ions that are attracted to the anode and cathode Ion yang ditarik ke anod dan katod

OH–, SO42–

H+

State the ion selected to be discharged.

Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas. Terangkan jawapan anda.

Half equation Persamaan setengah

– Hydroxide ion, OH– is selected.

Ion hidroksida, OH– dipilih.

– The position of hydroxide ion is lower than sulphate ion in the discharge series of anion.

Kedudukan ion hidroksida lebih rendah daripada ion sulfat dalam siri nyahcas anion.

– Hydroxide ions release electrons to produce water and oxygen molecules.

Ion hidroksida menderma elektron untuk menghasilkan molekul air dan oksigen.

4OH–

– Hydrogen ion, H+ is selected.

Ion hidrogen H+ dipilih.

– Hydrogen ions receive electrons to produce molecule.

Ion hidrogen menerima elektron untuk menghasilkan molekul.

2H2O + O2 + 4e

2H+ + 2e

H2

Observations Pemerhatian

Gas bubbles are released Gelembung gas dibebaskan

Gas bubbles are released Gelembung gas dibebaskan

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

Insert a glowing wooden splinter into test tube, glowing wooden splinter is lighted up Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala

When a lighted wooden splinter is placed near the mouth of the test tube, a ‘pop’ sound is produced Apabila kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan

Name of the products / Inference Nama hasil / Inferens

Oxygen gas is released Gas oksigen dibebaskan

Hydrogen gas is released Gas hidrogen dibebaskan

(c) Electrolysis of 0.1 mol dm–3 copper(II) sulphate solution using carbon electrodes.

Elektrolisis 0.1 mol dm–3 larutan kuprum(II) sulfat menggunakan elektrod karbon.

Equation of electrolyte ionisation Persamaan pengionan elektrolit

CuSO4 H 2O

Cu2+ + SO42– H+ + OH–

Anode / Anod

Cathode / Katod

Ions that are attracted to the anode and cathode Ion yang ditarik ke anod dan katod

SO42–, OH–

Cu2+, H+

– Hydroxide ion, OH– is selected.

Ion hidroksida, OH– dipilih.

– The position of hydroxide ion is lower than ion sulphate in the discharge series of anion.

Kedudukan ion hidroksida lebih rendah daripada ion sulfat dalam siri nyahcas anion.

– Hydroxide ions release electrons to produce water and oxygen molecules.

Ion hidroksida menderma elektron untuk menghasilkan molekul air dan oksigen.

– Copper(II) ion, Cu2+ is selected.

Ion kuprum(II), Cu2+ dipilih.

– The position of copper(II) ion is lower than hydrogen ion in the Electrochemical Series Kedudukan ion kuprum(II) lebih rendah daripada ion hidrogen dalam Siri Elektrokimia – Copper(II) ion receive electrons to produce copper atom Ion kuprum(II) menerima elektron untuk menghasilkan atom kuprum

State the ion selected to be discharged.

Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas. Terangkan jawapan anda.

Half equation Persamaan setengah

4OH–

2H2O + O2 + 4e

Cu2+ + 2e

Observations Pemerhatian

Gas bubbles are released Gelombang gas dibebaskan

Brown solid deposited Enapan perang terbentuk

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

Insert a glowing wooden splinter into test tube, glowing wooden splinter is lighted up Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala

–

Name of the products / Inference Nama hasil / Inferens

Oxygen gas is released Gas oksigen di bebaskan

Copper is formed Kuprum terbentuk

101

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 101

UNIT

5

Electrode / Elektrod

Cu

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Concentration of electrolyte Kepekatan elektrolit When is the factor concentration of electrolyte being applied?

Bilakah faktor kepekatan elektrolit digunakan?

When electrolysis is carried out using inert electrodes and concentrated solutions.

Apabila elektrolisis dijalankan menggunakan elektrod lengai dan larutan pekat.

How to choose cation to be discharged based on this factor?

Bagaimanakah memilih kation untuk dinyahcas berdasarkan faktor ini?

Ions that are more concentrated will be discharged but this is only true for halide ions, which are Cl–, Br– and I–.

Ion yang lebih pekat akan dinyahcas tetapi ia benar untuk ion-ion halida sahaja iaitu Cl–, Br– dan I–.

State the selected ions to be discharged at the anode and cathode for the following concentrated solutions.

Nyatakan ion yang terpilih untuk dinyahcas di anod dan katod bagi larutan pekat berikut.

(i) Concentrated hydrochloric acid solution, using carbon electrodes Larutan asid hidroklorik pekat menggunakan elektrod karbon

Cl–

Anode / Anod :

Cathode / Katod :

H+

(ii) Concentrated potassium iodide solution, using carbon electrodes Larutan kalium iodida pekat menggunakan elektrod karbon

I–

Anode / Anod :

Cathode / Katod :

H+

(iii) Concentrated sodium chloride solution, using carbon electrodes Larutan natrium klorida pekat menggunakan elektrod karbon

Cl–

Anode / Anod :

Cathode / Katod :

H+

(a) Electrolysis of 0.001 mol dm–3 hydrochloric acid and 2.0 mol dm–3 hydrochloric acid, using carbon electrodes.

Elektrolisis asid hidroklorik 0.001 mol dm–3 dan asid hidroklorik 2.0 mol dm–3 menggunakan elektrod karbon.

Equation of electrolyte ionisation Persamaan pengionan elektrolit

HCI H 2O

H+ + CI– H+ + OH–

Electrolyte / Elektrolit

0.001 mol dm–3 of HCl / HCl 0.001 mol dm–3

2.0 mol dm–3 of HCl / HCl 2.0 mol dm–3

Ions that are attracted to the cathode Ion yang ditarik ke katod

H+

H+

State the ion selected to be discharged. Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas. Terangkan jawapan anda.

UNIT

Half equation at the cathode Persamaan setengah di katod

– Hydrogen ion, H+ is selected Ion hidrogen, H+ dipilih – Hydrogen ions receive electrons to produce hydrogen molecule/ Ion hidrogen menerima elektron untuk menghasilkan molekul 2H+ + 2e

H2

– Hydrogen ion, H+ is selected Ion hidrogen, H+ dipilih – Hydrogen ions receive electrons to produce molecule / Ion hidrogen menerima elektron untuk menghasilkan molekul 2H+ + 2e

H2

5

Observations at cathode Pemerhatian di katod

Gas bubbles are released Gelembung gas dibebaskan

Gas bubbles are released Gelembung gas dibebaskan

Confirmatory test at cathode (method and observations) Ujian pengesahan di katod (kaedah dan pemerhatian)

Place a burning wooden splinter into the test tube, ‘pop’ sound is produced Kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan

Place a burning wooden splinter at the mouth of the test tube, ‘pop’ sound is produced Kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan

Name the products / Inference Namakan hasil / Inferens

Hydrogen gas is released Gas hidrogen dibebaskan

Hydrogen gas is released Gas hidrogen dibebaskan

Ions that are attracted to the anode Ion bergerak ke anod State the ion selected to be discharged at the anode. Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas di anod. Jelaskan jawapan anda.

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 102

Cl–, OH– – Hydroxide ion, OH– is selected Ion hidroksida, OH– dipilih – The position of hydroxide ion is lower than chloride ion in the the discharge series of anion Kedudukan ion hidroksida lebih rendah daripada ion klorida dalam siri nyahcas anion – Hydroxide ions release electrons to produce water and oxygen molecules Ion hidroksida menderma elektron untuk menghasilkan air dan molekul oksigen

Cl–, OH– – Chloride ion, Cl– is selected Ion klorida, Cl– dipilih – The concentration of chloride ion is higher than hydroxide ion Kepekatan ion klorida lebih tinggi daripada ion hidroksida – Chloride ions release electrons to produce chlorine molecules Ion klorida menderma elektron untuk menghasilkan molekul klorin

102

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Electrolyte / Elektrolit Half equation at the anode Persamaan setengah di anod

0.001 mol dm–3 of HCl / HCl 0.001 mol dm–3 4OH–

2.0 mol dm–3 of HCl / HCl 2.0 mol dm–3

2H2O + O2 + 4e

2CI–

CI2 + 2e

Observations at anode Pemerhatian di anod

Gas bubbles are released Gelembung gas dibebaskan

Greenish yellow gas is released Gas kuning kehijauan dibebaskan

Confirmatory test at anode (method and observations) Ujian pengesahan di anod (kaedah dan pemerhatian)

Insert a glowing wooden splinter into the test tube, glowing wooden splinter is lighted up Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala

A damp blue litmus paper placed near the mouth of the test tube. The gas changed the damp blue litmus paper to red and then bleached it Sehelai kertas litmus biru lembap diletakkan di mulut tabung uji. Gas ini menukar kertas litmus biru lembap kepada merah dan kemudian melunturkannya

Name the products / Inference Nama hasil / Inferens

Oxygen gas is released Gas oksigen dibebaskan

Chlorine gas is released Gas klorin dibebaskan

The concentration of hydrochloric acid after a while. Explain.

Kepekatan asid hidroklorik selepas beberapa ketika. Terangkan.

Concentration of hydrochloric acid increases .

Hydrogen gas is released at the cathode and oxygen gas is released at the anode. Water oxygen decomposed to gas and hydrogen gas. bertambah .

Kepekatan asid hidroklorik Gas hidrogen dibebaskan di katod dan gas oksigen dibebaskan di anod. Air terurai kepada oksigen hidrogen gas dan gas .

Concentration of hydrochloric acid decreases .

Hydrogen gas is released at the cathode and chlorine gas is released at the anode.

Concentration of chloride ions decreases. berkurang Kepekatan asid hidroklorik , gas hidrogen dibebaskan di katod dan gas klorin dibebaskan di anod.

Kepekatan ion klorida berkurang.

(b) Electrolysis of 2.0 mol dm–3 sodium iodide solution using carbon electrodes.

Elektrolisis natrium iodida 2.0 mol dm–3 menggunakan elektrod karbon.

Equation of electrolyte ionisation Persamaan pengionan elektrolit

NaI H 2O

Na+ + I– H+ + OH–

Anode / Anod

Cathode / Katod

Ions that are attracted to the anode and cathode Ion yang ditarik ke anod dan katod

I–, OH–

Na+, H+

– Iodide ion, I– is selected. iodida, I– dipilih – The concentration of iodide ion is higher than hydroxide ion Kepekatan ion iodida lebih tinggi daripada ion hidroksida – Iodide ions release electrons to produce iodine molecule Ion Iodida menderma elektron untuk menghasilkan molekul iodin

Half equation Persamaan setengah

2I–

I2 + 2e

– Hydrogen ion, H+ is selected Ion hidrogen, H+ dipilih – The position of hydrogen ion is lower than sodium ion in the Electrochemical Series Kedudukan ion hidrogen lebih rendah daripada ion natrium dalam Siri Elektrokimia – Hydrogen ions receive electrons to produce hydrogen molecule Ion hidrogen menerima elektron untuk menghasilkan molekul hidrogen 2H+ + 2e

H2

Observations Pemerhatian

Brown solid deposited Enapan perang terbentuk

Gas bubbles are released Gelembung gas dibebaskan

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

– A few drops of starch solution added Beberapa titik larutan kanji ditambah – Starch solution turns to dark blue Larutan kanji menjadi biru gelap

– When a lighted wooden splinter is placed near the mouth of the test tube, a ‘pop’ sound is produced.

Apabila kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan.

Name the products / Inference Nama hasil / Inferens

Iodine is formed Iodin terbentuk

Hydrogen gas is released Gas hidrogen dibebaskan

103

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 103

UNIT

State the ion selected to be discharged.

Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas. Terangkan jawapan anda.

5

Electrode / Elektrod

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Types of electrode Jenis elektrod (a) There are two types of electrode / Terdapat dua jenis elektrod: When is the factor types of electrode being applied?

Bilakah faktor jenis elektrod digunakan?

When electrolysis is carried out using active electrode. Active electrode is an electrode that not only acts as a conductor but also undergoes chemical changes.

Apabila elektrolisis dijalankan menggunakan elektrod aktif. Elektrod aktif ialah elektrod yang bertindak bukan sahaja sebagai konduktor tetapi juga mengalami perubahan kimia.

Remark / Catatan : Inert electrode – An electrode that acts as a conductor only and does not undergo any chemical changes. Normally they are made of carbon or platinum.

Elektrod lengai – Elektrod yang bertindak sebagai konduktor sahaja dan tidak mengalami sebarang perubahan kimia.

Biasannya diperbuat daripada karbon atau platinum.

How to choose cation and kation to be discharged based on this factor?

Bagaimanakah memilih kation dan anion untuk dinyahcas berdasarkan faktor ini?

During the electrolysis, the metal atom at the anode releases electrons to form metal ion, metal anode becomes thinner, while the less electropositive cation will be selected at the cathode. They consist of metal electrodes such as copper, silver and nickel.

Semasa proses elektrolisis, atom logam di anod melepaskan elektron menjadi ion logam, anod logam menjadi nipis, manakala kation yang kurang elektropositif akan dipilih di katod. Ianya terdiri daripada elektrod logam seperti kuprum, argentum dan nikel.

(b) Explain electrolysis of 1 mol dm–3 copper(II) sulphate solution with carbon electrode and copper electrode.

Terangkan elektrolisis larutan kuprum(II) sulfat 1 mol dm–3 menggunakan elektrod karbon dan kuprum.

Equation of electrolyte ionisation Persamaan pengionan elektrolit Electrode / Elektrod

CuSO4 (aq/ak) Cu2+ + SO42– H 2O H+ + OH– Carbon electrode / Elektrod karbon

Copper electrode / Elektrod kuprum

Cu2+, H+

Cu2+, H+

– Copper(II) ion, Cu2+ is selected.

Ion kuprum(II), Cu2+ dipilih.

– The position of copper(II) ion is lower than hydrogen ion in the Electrochemical Series Kedudukan ion kuprum(II) lebih rendah daripada ion hidrogen dalam Siri Elektrokimia – Copper(II) ions receive electrons to produce copper atom Ion kuprum(II) menerima elektron untuk menghasilkan atom kuprum

– Copper(II) ion, Cu2+ is selected.

Ion kuprum(II), Cu2+ dipilih.

– The position of copper(II) ion is lower than hydrogen ion in the Electrochemical Series Kedudukan ion kuprum(II) lebih rendah daripada ion hidrogen dalam Siri Elektrokimia – Copper(II) ion receive electrons to produce copper atom Ion kuprum(II) menerima elektron untuk menghasilkan atom kuprum

The ions that move to the cathode Ion yang bergerak ke katod State the ion selected to be discharged.

Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas.

Terangkan jawapan anda.

UNIT

5

Half equation at the cathode Persamaan setengah di katod

Cu2+ + 2e

Cu

Cu2+ + 2e

Observation at cathode Pemerhatian di katod

Brown solid deposited Enapan perang terbentuk

Brown solid deposited Enapan perang terbentuk

Name of the products / Inference Nama hasil / Inferens

Copper is formed Kuprum terbentuk

Copper is formed Kuprum terbentuk

Ions that are attracted to the anode Ion bergerak ke anod State the ion selected to be discharged at the anode. Explain your answer.

Nyatakan ion yang dipilih untuk dinyahcas di anod. Terangkan jawapan anda.

Half equation at the anode Persamaan setengah di anod Observations at anode Pemerhatian di anod

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 104

Cu

SO42–, OH–

SO42–, OH–

– Hydroxide ion, OH– is selected Ion hidroksida, OH– dipilih – The position of hydroxide ion is lower than chloride ion in the the discharge series of anion Kedudukan ion hidroksida lebih rendah daripada ion klorida dalam siri nyahcas anion – Hydroxide ions release electrons to produce water and oxygen molecules Ion hidroksida membebaskan elektron untuk menghasilkan air dan molekul oksigen

– Copper atom Atom kuprum – Copper is active electrode. Copper atom releases electrons to form copper(II) ion.

Kuprum adalah elektrod aktif. Atom kuprum membebaskan elektron untuk membentuk ion kuprum(II).

4OH–

2H2O + O2 + 4e

Gas bubbles are released Gelembung gas dibebaskan

Cu

Cu2+ + 2e

Copper electrode becomes thinner Elektrod kuprum menipis

104

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Electrode / Elektrod

Anode / Anod

Cathode / Katod

Confirmatory test at anode (method and observations) Ujian pengesahan di anod (kaedah dan pemerhatian)

Insert a glowing wooden splinter into test tube, glowing wooden splinter is lighted up.

Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala.

–

Name the products / Inference Nama hasil / Inferens

Oxygen gas Gas oksigen

Copper(II) ion Ion kuprum(II)

Observation on the electrolyte. Explain.

Pemerhatian pada elektrolit. Terangkan.

– Intensity of blue colour decreases.

Keamatan warna biru berkurang.

– Concentration of copper(II) sulphate solution decreases.

Kepekatan larutan kuprum(II) sulfat berkurang.

– Copper(II) ions discharge as copper atoms and deposited at the cathode Ion kuprum(II) dinyahcas sebagai atom kuprum dan terenap di katod

– Intensity of blue colour remains unchanged.

Keamatan warna biru tidak berubah.

– Concentration of copper(II) sulphate solution remains unchanged.

Kepekatan larutan kuprum(II) sulfat tidak berubah.

– The number of copper atoms become copper(II) ions at the anode is equal to the number of copper(II) ions become copper atoms at the cathode.

Bilangan atom kuprum menjadi ion kuprum(II) di anod adalah sama dengan bilangan ion kuprum(II) menjadi atom kuprum dan terenap di katod.

Exercise / Latihan

Electrolyte Elektrolit

Dilute sulphuric acid Asid sulfurik cair Concentrated hydrochloric acid Asid hidroklorik pekat

Electrode Elektrod

Carbon Karbon

Carbon Karbon

Silver nitrate solution Larutan argentum nitrat

Carbon Karbon

Silver nitrate solution Larutan argentum nitrat

Silver Argentum

Dilute potassium iodide solution Larutan kalium iodida cair Concentrated potassium iodide solution Larutan kalium iodida pekat Dilute potassium sulphate solution Larutan kalium sulfat cair

Carbon Karbon

Carbon Karbon

Carbon Karbon

Factor that affects electrolysis Faktor yang mempengaruhi elektrolisis

Ions present Ion yang hadir

Half equation at the cathode and observation Persamaan setengah di katod dan pemerhatian

Position of ion in the Electrochemical Series Kedudukan ion dalam Siri Elektrokimia

H+, SO42–, OH–

4OH– 2H2O + O2 + 4e Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

2H+ + 2e H2 Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

Concentration of electrolyte Kepekatan elektrolit

H+, Cl–, OH–

2Cl– Cl2 + 2e Greenish yellow gas is released.

Gas kuning kehijauan dibebaskan.

2H+ + 2e H2 Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

Position of ion in the Electrochemical Series Kedudukan ion dalam Siri Elektrokimia

Ag+, NO3–, H+, OH–

4OH– 2H2O + O2 + 4e Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

Ag+ + e Ag Grey shiny solid deposited.

Pepejal kelabu berkilat terenap.

Type of electrode Jenis elektrod

Ag+, NO3–, H+, OH–

Ag Ag+ + e Anode becomes thinner.

Anod menjadi semakin nipis.

Ag+ + e Ag Grey shiny solid deposited.

Pepejal kelabu berkilat terenap

Position of ion in the Electrochemical Series Kedudukan ion dalam Siri Elektrokimia

K+, I–, H+, OH–

4OH– 2H2O + O2 + 4e Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

2H+ + 2e H2 Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

Concentration of electrolyte Kepekatan elektrolit

K+, I–, H+, OH–

2I– I2 + 2e Brown solution formed.

Larutan perang terhasil.

2H+ + 2e H2 Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

Position of ion in the Electrochemical Series Kedudukan ion dalam Siri Elektrokimia

K+, SO42–, H+, OH–

4OH– 2H2O + O2 +4e Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

2H+ + 2e H2 Colourless gas bubbles are released.

Gelembung gas tak berwarna dibebaskan.

105

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 105

Half equation at the anode and observation Persamaan setengah di anod dan pemerhatian

5

Complete the table below: / Lengkapkan jadual di bawah:

UNIT

1

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

2

The diagram below shows the set-up of the apparatus to study the electrolysis of sodium chloride solution using carbon electrodes.

Rajah di bawah menunjukkan susunan radas untuk mengkaji elektrolisis larutan natrium klorida menggunakan elektrodelektrod karbon.

Gas X Gas X

Gas Y Gas Y

Carbon electrode Elektrod karbon

Carbon electrode Elektrod karbon 2.0 mol dm–3 sodium chloride solution Larutan natrium klorida 2.0 mol dm–3

A

(a) Write all the formula of all ions present in sodium chloride solution.

Tulis formula semua ion yang hadir dalam larutan natrium klorida.

Na+ , H+, Cl–, OH– (b) State the name of gas X and gas Y.

Nyatakan nama gas X dan gas Y. Chlorine gas / Gas klorin Gas X / Gas X :

Gas Y / Gas Y :

Hydrogen gas / Gas hidrogen

(c) Explain your answer in (b) based on the factors that influence the selective discharged of ion.

Terangkan jawapan anda di (b) berdasarkan faktor yang mempengaruhi pemilihan ion untuk dinyahcas.

Answer / Jawapan: Anode / Anod

Cathode / Katod

CI–, OH–

Na+, H+

Ions attracted to each of electrodes Ion yang ditarik ke setiap elektrod

UNIT

Names of the ions selectively discharged Namakan ion yang terpilih dinyahcas

Chloride ion is selectively discharged Ion klorida terpilih dinyahcas

Hydrogen ion is selectively discharged Ion hidrogen terpilih dinyahcas

The reason why the ions are selectively discharged Alasan kenapa ion itu terpilih untuk dinyahcas

– The concentration of chloride ion is higher than hydroxide ion Kepekatan ion klorida lebih tinggi daripada ion hidroksida.

– Chloride ions release electrons to produce chlorine molecules.

Ion klorida membebaskan elektron untuk menghasilkan molekul klorin.

– The position of hydrogen ion is lower than sodium ion in Electrochemical Series.

Kedudukan ion hidrogen lebih rendah daripada ion sodium dalam Siri Elektrokimia.

– Hydrogen ions receive electrons to produce molecule.

Ion hidrogen menerima elektron untuk menghasilkan molekul hidrogen.

5

Half equation Persamaan setengah

3

2CI–

CI2 + 2e

2H+ + 2e

H2

Observation Pemerhatian

Greenish yellow gas is released.

Gas kuning kehijauan dibebaskan.

Gas bubbles are released Gelembung gas dibebaskan

Confirmatory test Ujian pengesahan

A damp blue litmus paper placed near the mouth of the test tube. The gas changed the damp blue litmus paper to red and then bleached it. Chlorine gas is released.

Sehelai kertas litmus biru lembap diletakkan di mulut tabung uji. Gas ini menukar kertas litmus biru lembap kepada merah dan kemudian melunturkannya. Gas klorin dibebaskan.

Place a burning wooden splinter at the mouth of the test tube, ‘pop’ sound is produced.

Hydrogen gas is released.

Kayu uji menyala diletakkan di mulut tabung uji, bunyi ‘pop’ dihasilkan. Gas hidrogen dibebaskan.

Describe an experiment to determine the product of electrolysis copper(II) sulphate solution with carbon electrode. Your answer should include the observation, confirmatory test for the product at the anode and half equation at the electrode.

Terangkan eksperimen untuk menentukan hasil larutan elektrolisis kuprum(II) sulfat dengan elektrod karbon. Jawapan anda hendaklah disertakan pemerhatian, ujian pengesahan untuk produk di anod dan setengah persamaan di elektrod.

Answer / Jawapan: Apparatus / Radas : Battery / power supply, carbon electrodes, wire, electrolytic cell, test tube, ammeter [from a labelled diagram] Bateri / bekalan kuasa, karbon elektrod, wayar, sel elektrolit, tabung uji, ammeter [daripada rajah

berlabel]

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 106

106

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Carbon electrode Elektrod karbon

Copper(II) sulphate Kuprum(II) sulfat

Procedure / Prosedur : –3 solution half full 1 Pour 1 mol dm copper(II) sulphate into the electrolytic cell until it is .

–3 larutan 1 mol dm separuh penuh

Tuang kuprum(II) sulfat ke dalam sel elektrolit sehingga . test tube solution 2 The apparatus is set up as shown in the diagram. Fill the with copper(II) sulphate and invert anode the test tube on the . tabung uji larutan

Radas disusun seperti yang ditunjukkan dalam rajah. Isi dengan kuprum(II) sulfat dan anod terbalikkan tabung uji di .

3 Turn on the switch.

Hidupkan suis. anode 4 Collect the gas produced at the . anod

Kumpulkan gas yang terhasil di . anode glowing wooden splinter 5 Gas produced at the is tested with a . anod kayu uji berbara

Gas yang dihasikan di diuji dengan .

Observation and confirmatory test / Pemerhatian dan ujian pengesahan : Observation / Pemerhatian

Cathode / Katod

Brown solid deposited Enapan pepejal perang

Anode / Anod

Gas bubbles are released Gelembung gas dibebaskan

– Insert a glowing wooden splinter into test tube, glowing wooden splinter is lighted up.

Masukkan kayu uji berbara ke dalam tabung uji, kayu uji berbara menyala.

Copper(II) sulphate solution is electrolysed using copper electrodes.

Larutan kuprum(II) sulfat dielektrolisis dengan menggunakan elektrod kuprum. (a) Write the formula of all the anions present in the solution.

Tuliskan formula semua anion yang terdapat dalam larutan itu.

SO42–, OH–

UNIT

4

Confirmatory Test / Ujian pengesahan

5

Electrodes / Elektrod

(b) Write the half equation for the reaction at the

Tuliskan persamaan setengah untuk tindak balas di (i)

anode / anod

: Cu

Cu2+ + 2e

2+ (ii) cathode / katod : Cu + 2e

(c) (i)

Cu

From your observations, what happen to the intensity of the blue colour of the copper(II) sulphate solution during electrolysis? / Daripada pemerhatian anda, apakah yang berlaku ke atas keamatan warna biru larutan kuprum(II) sulfat semasa proses elektrolisis?

The intensity of the blue colour of copper(II) sulphate remains unchanged.

Keamatan warna biru kuprum(II) sulfat tidak berubah.

(ii) Explain your answer. / Jelaskan jawapan anda.

The number of copper(II) ions become copper atoms at the cathode is equal to the number of copper atoms become copper(II) ions at the anode. The concentration of copper(II) sulphate solution remains unchanged.

Bilangan ion kuprum(II) menjadi atom kuprum di katod adalah sama dengan bilangan atom kuprum menjadi ion kuprum(II) di anod. Kepekatan larutan kuprum(II) sulfat tidak berubah.

107

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 107

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

(d) If the experiment is repeated with the copper electrodes being replaced by carbon electrodes, state the name of the products formed at the

Jika eksperimen diulangi dengan menggantikan elektrod kuprum dengan elektrod karbon, namakan hasil yang terbentuk di (i) anode / anod : Oxygen / Oksigen (ii) cathode / katod: Copper / Kuprum 5

The diagram below shows the set-up of apparatus of an electrolytic cell.

Rajah di bawah menunjukkan susunan radas bagi sel elektrolisis.

Carbon electrode P Elektrod karbon P

Carbon electrode Q Elektrod karbon Q Copper(II) nitrate solution Larutan kuprum(II) nitrat

(a) Write the formula of all ions present in copper(II) nitrate solution.

Tuliskan formula semua ion yang hadir dalam larutan kuprum(II) nitrat.

Cu2+, NO3–, H+ , OH– (b) Write the half equation for the reaction at the / Tuliskan persamaan setengah untuk tindak balas di 2+ – Cu 2H2O + O2 + 4e electrode P / elektrod P : Cu + 2e electrode Q / elektrod Q : 4OH (c) (i) What is the colour of copper(II) nitrate? / Apakah warna larutan kuprum(II) nitrat?

Blue / Biru (ii) What happens to the intensity of the colour of copper(II) nitrate solution? Explain your answer.

Apakah yang berlaku kepada keamatan warna larutan kuprum(II) nitrat? Jelaskan jawapan anda.

The intensity of the blue colour of copper(II) nitrate solution decreases. The concentration of Cu2+ decreases because copper(II) ions receive electrons to form copper atom at the cathode.

Keamatan warna biru larutan kuprum(II) nitrat berkurang.

Kepekatan ion Cu2+ berkurang kerana ion kuprum(II) menerima elektron untuk membentuk atom kuprum di katod.

Electrolysis in Industry / Elektrolisis dalam Industri UNIT

1

5 I

Three uses of electrolysis in industries are: / Tiga kegunaan elektrolisis dalam industri ialah: – Electroplating metal / Penyaduran logam – Purification of metal / Penulenan logam – Metal extraction / Pengekstrakan logam Electroplating metal / Penyaduran logam

What is electroplating?

Apakah penyaduran logam?

Electroplating is a process for coating of metal object with a layer of desired metal.

Penyaduran logam ialah proses menyadur objek logam dengan satu lapisan logam yang dikehendaki.

Set up of apparatus of electroplating.

Susunan rada bagi penyaduran logam.

Electroplating metal (anode) Logam penyadur logam (anod)

Metal to be electroplated (cathode) Logam yang hendak disadur (katod) Which object is made to be anode (connected to positive terminal of batteries)?

Objek yang manakah dijadikan anod (disambungkan kepada terminal positif bateri)?

The

electroplating

metal. / Logam

Electrolyte Elektrolit penyadur

.

Remark / Catatan: The electroplating metal is not an inert electrode, the electroplating metal atom ionised by releasing electrons to form metal ion. / Logam penyadur bukan elektrod lengai, atom logam penyadur mengion membebaskan elektron untuk membentuk ion logam.

X(s/p)

Xn+ (aq/ak) + ne

The anode become thinner. / Anod menipis.

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 108

108

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Which object is made to be cathode (connected to negative terminal of batteries)?

Objek yang manakah dijadikan katod (disambung kepada terminal negatif bateri)?

metal

The

logam

Objek

disadurkan

yang hendak

. .

Remark / Catatan: When cations of electroplating metal discharged at the cathode, electroplating metal deposits on the surface of the metal to be electroplated. / Apabila kation logam penyadur dinyahcas di katod, logam penyadur terenap pada permukaan logam yang hendak disadurkan.

Xn+ (aq/ak) + ne What is the electrolyte?

Apakah itu elektrolit?

electroplated

object to be

X (s/p)

The electrolyte used is an aqueous salt solution containing the ions of the electroplating metal.

Elektrolit yang digunakan ialah larutan akueus garam yang mengandungi ion logam penyadur.

Describe a laboratory experiment to electroplate the iron spoon with copper.

Huraikan satu eksperimen makmal untuk menyadur sudu besi dengan kuprum.

Copper Kuprum

Iron spoon Sudu besi

Copper(II) nitrate solution Larutan kuprum(II) nitrat

Procedure:

sandpaper – Copper plate and iron spoon are cleaned with .

Copper(II) nitrate solution beaker – is poured into a –

Iron spoon

until

is then connected to the negative terminal of battery while the

dipped

shown in the diagram. completed – The circuit is

.

– Half equation at the cathode :

– Observation of the anode:

Cu

Brown solid

– Observation of the cathode: – Half equation at the anode :

Cu2+ + 2e Cu

in the

Cu2+ + 2e

.

Iron spoon

is

copper(II) nitrate solution

as

is connected to the positive terminal of the battery using connecting wire. // copper plate made as cathode while is made as anode.

– The iron spoon and the copper plate are

half full copper plate

.

is deposited.

Copper plate becomes thinner

. .

Prosedur:

Sudu besi

–

bikar

disambungkan kepada terminal negatif bateri dan

disambungkan kepada terminal positif bateri menggunakan wayar penyambung. // dijadikan katod dan kepingan kuprum dijadikan anod.

– Sudu besi dan plat kuprum

dicelup

sehingga

5

.

kepingan kuprum Sudu besi

larutan kuprum(II) nitrat

ke dalam

UNIT

kertas pasir – Kepingan kuprum dan sudu besi dibersihkan dengan Larutan kuprum(II) nitrat – dituangkan ke dalam separuh penuh .

seperti

ditunjukkan dalam rajah. dilengkapkan .

– Litar

Cu2+ + 2e Cu – Persamaan setengah di katod : pepejal perang – Pemerhatian di katod: terenap.

– Persamaan setengah di anod : – Pemerhatian di anod:

Cu

.

Cu2+ + 2e

Kuprum menipis

. .

Observation / inference / half equation // Pemerhatian / Inferens / persamaan setengah: Observation Pemerhatian Brown solid is deposited at the cathode.

Pepejal perang terenap di katod.

Inference / half equation Inferens / Persamaan setengah Copper

is formed.

Copper(II) ion receive electrons to form copper atom Half equation : Cu2+ + 2e Cu Kuprum terbentuk.

Ion kuprum(II) menerima elektron untuk membentuk atom kuprum.

Persamaan setengah : Cu2+ + 2e Cu

109

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 109

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Observation Pemerhatian

Inference / half equation Inferens / Persamaan setengah

Copper plate become thinner Plat kuprum menipis

Copper atom

releases electrons to form copper(II) ions.

Cu

Half equation : Atom kuprum

ion kuprum(II).

melepaskan elektron untuk membentuk

Persamaan setengah : Intensity of blue colour remains unchanged Kepekatan warna biru kekal tidak berubah

The number of

Cu2+ + 2e

Cu

copper atoms

Cu2+ + 2e

form

copper(II) ions

at the anode is equal to the number of copper(II) ions form copper atoms at the cathode.

Bilangan

atom kuprum

yang membentuk ion kuprum(II)

di anod adalah sama dengan bilangan yang membentuk

atom kuprum

ion kuprum(II)

di katod.

Summary of the electroplating process Rumusan proses penyaduran.

Electroplating metal (anode) ➝ metal atom at the anode releases electrons ➝ become cation ➝ enter electrolyte ➝ cation at the cathode ➝ receive electrons ➝ discharge and deposit on the metal to be electroplated at the cathode.

Logam penyadur (anod) ➝ atom logam di anod membebaskan elektron ➝ menjadi kation➝ memasuki elektrolit ➝ kation di katod ➝ menerima elektron ➝ dinyahcas dan terenap pada logam yang hendak disadur di katod.

State two main aims of electroplating.

Nyatakan dua tujuan utama penyaduran.

corrosion – To prevent of the object through the protection of the metal layer. For example, iron object is electroplated with nickel or chromium. hakisan Untuk menghalang objek melalui perlindungan lapisan logam. Contohnya, objek besi disadurkan dengan nikel atau kromium. attractive – To make the object more with shiny appearance. For example, electroplating of metal object with gold, platinum and silver. menarik Untuk menjadikan objek lebih dengan penampilan yang berkilat. Sebagai contoh, penyaduran objek logam dengan emas, platinum dan perak.

UNIT

5

Suggest steps taken to an even and lasting layer of electroplating metal.

Cadang langkah yang diambil untuk mendapat lapisan penyaduran yang sekata dan tahan lama.

– A low

electric current Arus elektrik

is used so that electroplating is carried out slowly.

yang rendah digunakan supaya penyaduran dilakukan dengan perlahan.

– The low concentration of electrolyte is used.

Kepekatan

– The

elektrolit yang rendah digunakan.

surface of metal

Permukaan logam

to be electroplated is polished using sand paper. yang disadur digilap menggunakan kertas pasir.

II Purification of metal

Penulenan logam What is purification of metal?

Apakah penulenan logam?

It is a process removing impurities from a impure metal.

Ialah satu proses menyingkirkan bendasing daripada suatu logam tak tulen.

Remark / Catatan: The process is the same as electroplating. Its transferring the metal from anode to the cathode.

Proses ini sama seperti penyaduran. Ia memindahkan logam dari anod ke katod.

How to purify impure metal using electrolysis?

Bagaimana cara menulenkan logam tidak tulen menggunakan elektrolisis?

(a) The

(b) The

impure metal Logam tak tulen

is made to be anode. dijadikan sebagai anod.

pure metal

is made to be cathode.

Logam tulen

dijadikan sebagai katod.

(c) The electrolyte used is a Elektrolit adalah © Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 110

salt solution

larutan garam

containing the ions of the purifying metal.

yang mengandungi ion logam yang hendak ditulenkan.

110

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Exercise: Latihan: Purification of copper Penulenan kuprum

Electrode X Elektrod X

+

–

Electrode Y Elektrod Y Electrolyte Elektrolit

Impurities Tidak tulen

(a) State the name of the substance used as: Nyatakan nama bahan yang dijadikan sebagai: electrode X / elektrod X : Impure copper / Kuprum tidak tulen

electrode Y / elektrod Y : Pure copper / Kuprum tulen electrolyte Z / elektrolit Z : Copper(II) sulphate solution / Larutan kuprum(II) sulfat (b) Write the half equation that occur at the / Tuliskan persamaan setengah yang berlaku di: Cu Cu2+ + 2e electrode X / elektrod X : 2+ Cu + 2e Cu electrode Y / elektrod Y : (c) What are the observations at the / Apakah pemerhatian di electrode X / elektrod X : Electrode becomes thinner / Elektrod semakin nipis electrode Y / elektrod Y : Brown solid deposited / Logam perang terenap Summary of the purification process.

Rumusan proses penulenan.

Impure metal (anode) ➝ metal atom at the anode releases electrons ➝ become cation ➝ enter electrolyte ➝ cation at the cathode ➝ receive electrons ➝ discharge and deposit on the pure metal at the cathode Logam tak tulen (anod) ➝ atom logam di anod melepaskan elektron ➝ menjadi cation ➝ memasuki elektrolit ➝ kation di katod ➝ menerima elektron ➝ dinyahcas dan terenap pada logam tulen di katod

III Extraction of metal / Pengekstrakan logam The metal that are very reactive (placed at the top position in electrochemical series), such as sodium, calcium and aluminium are extracted from their compound by electrolysis.

Logam yang sangat reaktif (berada pada kedudukan teratas dalam siri elektrokimia), seperti natrium, kalsium dan aluminium diekstrak daripada sebatiannya melalui elektrolisis.

Example / Contoh: Extraction of aluminium from bauxite, a minerals that consists of aluminum oxide Pengekstrakan aluminium dari bauksit, satu mineral mempunyai aluminium oksida

Substance Z / Bahan Z + –

Substance W Bahan W

5

Substance Y Bahan Y

UNIT

What is extraction of metal?

Apakah pengekstrakan logam?

Substance X + cryolite Bahan X + kriolit

(a) State the name of the following substances: / Nyatakan nama bahan-bahan berikut: W : Liquid aluminium / Cecair aluminium X : Molten aluminium oxide / Leburan aluminium oksida Y : Carbon / Karbon Z : Carbon / Karbon

(b) Which substance acts as anode and cathode?

Bahan yang manakah bertindak sebagai anod dan katod?

Anode / Anod : Z Cathode / Katod : Y (c) State the name of the product at anode and cathode.

Namakan hasil yang diperoleh di anod dan katod.

Anode / Anod : Oxygen / Oksigen Cathode / Katod : Aluminium / Aluminium (d) Write the ionic equation for the reactions at Tuliskan persamaan ion bagi tindak balas yang berlaku di 2– 3+ O2 + 4e Anode / Anod : 2O Cathode / Katod : Al + 3e

Al

(e) Why is cryolite added to X? / Mengapakah kriolit ditambah ke dalam X? To reduce the melting point of aluminium oxide (from 2 045 °C to 900 °C ).

Untuk menurunkan takat lebur aluminium oksida (dari 2 045 °C ke 900 °C).

111

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 111

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Electrochemical Series / Siri Elektrokimia What is Electrochemical Series?

Apakah Siri Elektrokimia?

metals Electrochemical Series is an arrangement of according to their tendency to release positive logam electrons to form a ion. / Siri Elektrokimia ialah susunan mengikut positif kecenderungan melepaskan elektron membentuk ion bercas .

Part of Electrochemical Series of metals and their ions: Sebahagian daripada Siri Elektrokimia logam dan ionnya:

1 The position of metal in Electrochemical Series: Kedudukan logam dalam Siri Elektrokimia: K, Na, Ca, Mg, Al, Zn, Fe, Sn, Pb, Cu, Ag Tendency of metal atom to release/donate electrons increases (electropositivity increases) Kecenderungan untuk atom logam melepaskan/menderma elektron bertambah (keelektropositifan bertambah) 2 The position of metal ions (cation) in the Electrochemical Series: Kedudukan ion logam (kation) dalam Siri Elektrokimia: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, *H+, Cu2+, Ag+ Tendency of metal ion (cation) to receive/gain electrons increases Kecenderungan untuk ion logam (kation) untuk menerima elektron bertambah

What are the uses of the Electrochemical Series?

Apakah kegunaan Siri Elektrokimia?

* H+ is also in the series of ion because it is present in aqueous solution of any electrolyte (salt solution/acid/alkali) * H+ juga terdapat dalam siri ion kerana kehadiran ion H+ dalam elektrolit larutan akueus (larutan garam/asid/alkali)

UNIT

5

Four main uses of the Electrochemical Series / Empat kegunaan utama Siri Elektrokimia: (a) To predict the terminal of chemical cell / Untuk meramalkan terminal sel kimia – The more electropositive metal is the negative terminal of the cell.

Logam yang lebih elektropositif ialah terminal negatif sel.

– The less electropositive metal is the positive terminal of the cell.

Logam yang kurang elektropositif ialah terminal positif sel. (b) To predict the voltage of chemical cell / Untuk meramalkan voltan sel kimia – The further the distance between two metals in the Electrochemical Series, the higher is the voltage of the chemical cell.

Semakin jauh jarak antara dua logam dalam Siri Elektrokimia, semakin tinggi bacaan voltan sel kimia. (c) To predict the metal displacement reactions Untuk meramalkan tindak balas penyesaran logam – The more electropositive metal can displace a less electropositive metal from its salt solution.

Logam yang lebih elektropositif dapat menyesarkan logam yang kurang elektropositif daripada larutan garamnya. (d) To predict the selected ion to be discharged at the electrode in an electrolysis Untuk meramalkan pemilihan ion untuk dinyahcas di elektrod dalam proses elektrolisis

Voltaic Cell (Chemical Cell) / Sel Ringkas (Sel Kimia) What is a simple voltaic cell?

Apakah sel ringkas?

– A cell made up of two are connected by an

different external circuit

Sel yang terdiri daripada dua logam litar luar disambung dengan

metals which are dipped in an

electrolyte

and

. berlainan

elektrolit

dan

dicelup dalam

.

– It is a cell that produces electrical energy when chemical reactions occur in it.

Sel yang menghasilkan tenaga elektrik apabila berlaku tindak balas kimia di dalamnya.

What is the energy change in a simple voltaic cell?

Apakah perubahan tenaga dalam sel ringkas?

State the factor that affects the voltage reading in voltaic cell.

Nyatakan faktor yang mempengaruhi bacaan voltan dalam sel ringkas.

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 112

Energy change in voltaic cell is from chemical energy to

electrical energy

Perubahan tenaga dalam sel ringkas ialah dari tenaga kimia kepada

The voltage of chemical cell depends on the

. tenaga elektrik

.

distance

between the two metals in the higher Electrochemical Series, where the further the distance between them, the the voltage. jarak Voltan sel kimia bergantung kepada antara dua logam dalam Siri Elektrokimia di tinggi mana semakin jauh dua logam dalam Siri Elektrokimia, semakin voltannya.

112

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 How to determine the negative terminal and positive terminal of voltaic cell?

Bagaimanakah menentukan terminal negatif dan terminal positif sel voltan?

negative A more electropositive metal becomes the positive metal becomes the terminal.

Set up of apparatus of simple voltaic cell (showing that a chemical reaction in a simple voltaic cell produces electricity) Susunan radas sel ringkas (menunjukkan tindak balas kimia dalam sel ringkas menghasilkan tenaga elektrik)

Electrical current produced is detected by the voltmeter (Chemical energy ➝ Electrical energy) Arus elektrik terhasil dikesan oleh voltmeter (Tenaga kimia ➝ Tenaga elektrik)

terminal of the cell. A less electropositive

Logam yang lebih elektropositif akan menjadi terminal positif elektropositif akan menjadi terminal sel.

Negative

terminal: negatif Terminal : • More electropositive metal.

Logam lebih elektropositif.

• Metal atom will release electrons that will flow through the external circuit. Metal atom becomes metal ion (becomes thinner).

Atom logam akan melepaskan elektron yang akan mengalir di litar luar. Atom logam menjadi ion logam (semakin nipis).

*

negatif

sel. Logam yang kurang

Positive

V _

+

terminal: positif :

Terminal • Less electropositive metal.

Logam kurang elektropositif. * The electrons that flow from the external circuit are received by the positive ion in the electrolyte through this terminal.

Elektron yang akan mengalir dari litar luar diterima oleh ion positif dalam elektrolit melalui terminal ini.

Remark / Catatan: – Metal atom (solid) from negative terminal releases electrons to form metal ions (aqueous solution) – the electrode become thinner.

Atom logam (pepejal) daripada terminal negatif melepaskan elektron untuk membentuk ion logam (larutan akueus) – elektrod menipis.

– Electron flow through positive terminal, electrons received by the cations in the electrolyte: Elektron mengalir melalui terminal positif, elektron diterima oleh kation dalam elektrolit: – If hydrogen ions receive electrons, bubbles are released at the negative terminal.

Jika ion hidrogen menerima elektron, gelembung gas dibebaskan di terminal negatif.

– If metal ions receive electrons, the solid metal deposited at the negative terminal.

Jika ion logam menerima elektron, pepejal logam terenap di terminal negatif.

Exercise Latihan

V

Magnesium Magnesium

Copper Kuprum

(a) Magnesium electrode is electropositive than

negative the copper :

Elektrod magnesium adalah terminal elektropositif kuprum daripada – Magnesium atom Atom magnesium

releases

melepaskan

terminal negatif

because

magnesium

kerana

is

more

magnesium

lebih

UNIT

5

Copper(II) sulphate solution Larutan kuprum(II) sulfat

:

electrons to form magnesium ion, Mg2+.

elektron untuk membentuk ion magnesium, Mg2+.

Mg Mg2+ + 2e – Half equation / Persamaan setengah: .

– Magnesium electrode becomes

thinner

– Electron flows through external circuit to the

/ Elektrod magnesium menjadi copper electrode.

kuprum Elektron mengalir melalui litar luar ke elektrod . positive (b) Copper electrode is the terminal because copper is less than

magnesium

:

Elektrod kuprum adalah terminal magnesium daripada :

positif

kerana kuprum kurang

nipis

.

electropositive elektropositif

– *Electrons from magnesium flow through external circuit to copper electrode.

Elektron dari magnesium mengalir melalui litar luar ke elektrod kuprum.

Copper(II) receives – ion in the electrolyte electron to form copper atom.

Ion

kuprum(II)

dalam elektrolit

– Half equation / Persamaan setengah :

113

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 113

menerima

elektron untuk membentuk atom kuprum.

Cu2+ + 2e Cu . © Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 –

Brown solid

is deposited on the surface of copper electrode.

Pepejal perang

terenap di permukaan elektrod kuprum. decreases

(c) The concentration of copper(II) sulphate

intensity

form copper atom at the positive terminal. The solution decreases.

berkurang

Kepekatan larutan kuprum(II) sulfat Keamatan

atom kuprum.

because copper(II) ions discharged to of blue colour of copper(II) sulphate

kerana ion kuprum(II) dinyahcaskan kepada

warna biru larutan kuprum(II) sulfat berkurang.

decreases

(d) If the magnesium metal is replaced with a zinc metal, the voltage reading zinc is nearer to copper in the Electrochemical Series.

What is Daniel cell?

Apakah sel Daniel?

Jika logam magnesium digantikan dengan logam zink, bacaan voltan akan zink lebih dekat dengan kuprum dalam Siri Elektrokimia.

berkurang

because kerana

(a) It is an example of voltaic cell which consists of zinc electrode dipped in zinc sulphate solution, copper electrode dipped in copper(II) sulphate solution and connected by a salt bridge or porous pot.

Merupakan satu contoh sel kimia yang terdiri daripada elektrod zink yang dicelup ke dalam larutan zink sulfat, elektrod kuprum dicelupkan ke dalam larutan kuprum(II) sulfat dan dihubungkan dengan titian garam atau pasu berliang.

Zn / ZnSO4 // CuSO4 / Cu

(b) The function of porous pot or salt bridge is to allow the flow of ions through it so that the electric circuit is completed.

Fungsi pasu berliang atau titian garam adalah untuk membenarkan ion-ion mengalir melaluinya dan melengkapkan litar.

Exercise Latihan

Sulphuric acid Asid sulfurik

Copper(II) sulphate solution Larutan kuprum(II) sulfat

negative

copper

UNIT

: kuprum

5

daripada – Zinc atom Atom zink

releases

terminal because zinc is more negatif

Elektrod zink adalah terminal

– Zinc electrode becomes

electron to form zinc ion, Zn2+. elektron untuk membentuk ion zink, Zn2+.

Zn

thinner

copper

positive

:

Elektrod kuprum adalah terminal daripada

zink

.

nipis

.

electropositive

terminal because copper is less positif

.

electrode.

kuprum

Elektron mengalir melalui litar luar ke elektrod (b) Copper electrode is the

Zn2+ + 2e

/ Elektrod zink menjadi

– Electrons flow through external circuit to the

than

elektropositif

kerana zink adalah lebih

– Half equation / Persamaan setengah :

than

electropositive

:

melepaskan

zinc

Zinc sulphate solution Larutan zink sulfat Porous pot Pasu berliang Copper(II) sulphate solution Larutan kuprum(II) sulfat

Zinc sulphate solution Larutan zink sulfat

(a) Zinc electrode is the

Zinc / Zink

Copper Kuprum

Zinc Zink

Copper Kuprum

elektropositif

kerana kuprum kurang

:

– Electrons from zinc electrode flow through external circuit to copper electrode.

Elektron dari zink mengalir melalui litar luar ke elektrod kuprum.

–

Copper(II)

Ion

ion in the electrolyte

kuprum(II)

dalam elektrolit

– Half equation / Persamaan setengah : –

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 114

Brown solid

Pepejal perang

receives

menerima

electron to form copper atom. elektron untuk membentuk atom kuprum.

Cu2+ + 2e

Cu

.

is deposited on the surface of copper electrode. terenap di permukaan elektrod kuprum.

114

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 decreases

(c) The concentration of copper(II) sulphate solution discharged to copper atoms. The

intensity

because copper(II) ions are

of blue colour of copper(II) sulphate solution

decreases.

Kepekatan larutan kuprum(II) sulfat Keamatan

kepada atom kuprum.

berkurang

kerana ion kuprum(II) telah dinyahcaskan

warna biru larutan kuprum(II) sulfat berkurang.

(d) If zinc metal is replaced with a magnesium metal and zinc sulphate solution is replaced with magnesium sulphate solution, the voltage reading further

from copper in the Electrochemical Series.

Jika logam zink digantikan dengan logam magnesium dan larutan zink sulfat digantikan dengan larutan magnesium sulfat, bacaan voltan jauh

Type of cell Jenis sel

bertambah

kerana jarak antara magnesium dengan

daripada jarak antara zink dengan kuprum dalam Siri Elektrokimia.

Advantages Kelebihan

Disadvantages Kelemahan

Dry cell Sel kering

– Light and compact Ringan dan kompak – Available in different size Terdapat dalam pelbagai saiz – Cheap Murah

– Non rechargeable Tidak boleh dicas semula – Short life span Jangka hayat singkat – Current and voltage produced is low Arus dan voltan yang dihasilkan rendah

Alkaline cell Sel alkali

– Last longer than dry cell Lebih tahan lama dari sel kering – Produce higher and more stable voltage Menghasilkan voltan yang lebih tinggi dan stabil

– Non rechargeable Tidak boleh dicas semula – Expensive Mahal – Electrolyte is corrosive Elektrolitnya mengakis

Mercury cell Sel merkuri

– Small size Saiz kecil – Stable voltage Voltan stabil – Last long Tahan lama

– Non rechargeable Tidak boleh dicas semula – Expensive Mahal – Mercury compound is toxic Sebatian merkuri adalah toksik

Lead-acid accumulator Akumulator asidplumbum

– Rechargeable Boleh dicas semula – Produce high current, suitable for heavy duty work such as starting a car Menghasilkan arus yang tinggi, sesuai untuk tugas berat seperti menghidupkan kereta.

– Produce a stable voltage for a long period Menghasilkan voltan yang stabil bagi jangka masa yang panjang

– Heavy and difficult to handle Berat dan sukar untuk dipegang – Electrolyte is corrosive Elektrolit mengakis

Nickel-cadmium cell Sel nikel-kadmium

– Rechargeable Boleh dicas semula – The size is small than lead-acid accumulator Saiznya lebih kecil daripada akumulator asid plumbum – Last long Tahan lama

– Expensive Mahal – Need a transformer for recharging process Memerlukan transformer untuk proses pengecasan

5

kuprum lebih Advantages and disadvantages of various voltaic cell in daily life Kelebihan dan kelemahan pelbagai sel ringkas dalam kehidupan harian.

because magnesium is

UNIT

increases

Remark: Other types of new cells are lithium ion, nickel hydride and polymeric cell Catatan: Sel baru dari jenis lain ialah ion litium, nikel hidrida dan sel polimer

115

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 115

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Comparing Electrolytic and Voltaic Cell / Membandingkan Sel Elektrolisis dan Sel Ringkas Exercise / Latihan 1

The diagram below shows the set-up of apparatus for two types of cell.

Rajah di bawah menunjukkan susunan radas untuk dua jenis sel.

Copper Kuprum

Copper Kuprum

Zinc Zink

Copper(II) sulphate solution Larutan kuprum(II) sulfat Cell X / Sel X

Cell Y / Sel Y

Complete the following table to compare cell X and cell Y / Lengkapkan jadual berikut untuk membandingkan sel X dan sel Y : Description Perkara Type of cell Jenis sel The energy change Perubahan tenaga

Cell X Sel X

Cell Y Sel Y

Electrolytic cell / Sel elektrolisis

Chemical cell / Sel kimia

Electrical energy Tenaga elektrik

Ion presence in the electrolyte Ion hadir dalam elektrolit

Chemical energy Tenaga kimia

Cu2+, H+, SO42–, OH–

Electrical energy Tenaga elektrik

Cu2+, H+, SO42–, OH–

Electrode Elektrod

Anode / Anod: Copper / Kuprum Cathode / Katod: Copper / Kuprum

Negative terminal / Terminal negatif : Zinc / Zink Positive terminal / Terminal positif : Copper / Kuprum

Half equation Persamaan setengah

Cu2+ + 2e Anode / Anod: Cu 2+ Cu Cathode / Katod: Cu + 2e

Zn2+ + 2e Negative terminal / Terminal negatif : Zn 2+ Cu Positive terminal / Terminal positif : Cu + 2e

Observation Pemerhatian

Anode / Anod: Copper electrode becomes thinner

Negative terminal / Terminal negatif : Zinc electrode becomes thinner/Elektrod zink semakin nipis

Elektrod kuprum semakin nipis

Positive terminal / Terminal positif : Brown solid deposited / Pepejal perang terenap

UNIT

Cathode / Katod: Brown solid deposited / Pepejal perang terenap

5

Electrolyte / Elektrolit: Intensity of blue colour of copper(II) sulphate solution

2

Chemical energy Tenaga kimia

Electrolyte / Elektrolit: Intensity of blue colour of copper(II) sulphate solution decreases

remains unchanged / Keamatan warna biru larutan

Keamatan warna biru larutan kuprum(II) sulfat semakin

kuprum(II) sulfat tidak berubah

berkurang

The diagram below shows the set-up of apparatus for an experiment.

Rajah di bawah menunjukkan susunan radas bagi suatu eksperimen.

V

–

+

Zinc / Zink

Cathode / Katod Copper Kuprum

Copper Kuprum

Zinc sulphate solution Larutan zink sulfat Copper(II) sulphate solution Larutan kuprum(II) sulfat

Anode / Anod

Copper(II) sulphate solution Larutan kuprum(II) sulfat

Porous pot Pasu berliang Cell A / Sel A

Cell B / Sel B

(a) In the above diagram, label / Dalam gambar rajah di atas, label (i) the positive terminal and negative terminal in Cell A, / terminal positif dan terminal negatif bagi Sel A, (ii) anode and cathode in Cell B. / anod dan katod bagi Sel B. © Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 116

116

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

(b) What is the energy change in Cell A and Cell B? / Apakah perubahan tenaga dalam Sel A dan Sel B?

Cell A / Sel A : Chemical energy to electrical energy / Tenaga kimia kepada tenaga elektrik

Cell B / Sel B : Electrical energy to chemical energy / Tenaga elektrik kepada tenaga kimia

(c) What is the function of the porous pot in Cell A? / Apakah fungsi pasu berliang dalam Sel A? To allow the movement of ions through it. / Untuk membolehkan ion mengalir melaluinya. (d) Referring to Cell A. / Merujuk kepada Sel A. (i) What is the observation at zinc electrode?

Apakah pemerhatian di elektrod zink?

Zinc electrode becomes thinner. / Elektrod zink menipis. (ii) Write the half equation for the reaction at zinc electrode.

Tuliskan persamaan setengah untuk tindak balas di elektrod zink.

Zn Zn2+ + 2e (iii) What is the observation at copper electrode? / Apakah pemerhatian di elektrod kuprum?

Brown solid deposited. / Pepejal perang terenap. (iv) Write the half equation for the reaction at copper electrode.

Tuliskan persamaan setengah untuk tindak balas di elektrod kuprum.

Cu2+ + 2e Cu (v) After 30 minutes, what is the colour change of the copper(II) sulphate solution? Explain why.

Selepas 30 minit, apakah perubahan warna larutan kuprum(II) sulfat? Jelaskan mengapa.

The intensity of blue colour decreases. Copper(II) ions are discharged to form copper atoms. Concentration of copper(II) ions in copper(II) sulphate solution decreases.

Keamatan warna biru berkurang. Ion kuprum(II) menerima elektron untuk membentuk atom kuprum. Kepekatan ion kuprum(II) dalam larutan kuprum(II) sulfat berkurangan.

UNIT

5

(e) Referring to Cell B. / Merujuk kepada Sel B. (i) What is the observation at the anode?

Apakah pemerhatian di anod?

Copper electrode becomes thinner. / Elektrod kuprum menipis (ii) Write the half equation for the reaction at the anode.

Tuliskan persamaan setengah untuk tindak balas di anod.

Cu Cu2+ + 2e (iii) What is the observation at the cathode?

Apakah pemerhatian di katod?

Brown solid deposited. / Pepejal perang terenap (iv) Write the half equation for the reaction at cathode.

Tuliskan persamaan setengah untuk tindak balas di katod.

Cu2+ + 2e Cu (f) The intensity of blue colour of copper(II) sulphate solution in the Cell B remains unchanged during the experiment. Explain why.

Keamatan warna biru larutan kuprum(II) sulfat dalam Sel B tidak berubah semasa eksperimen. Jelaskan mengapa.

The concentration of copper(II) sulphate remain unchanged. The rate of copper(II) ions discharged to copper atom at the cathode equals to the rate of copper atoms form copper(II) ions at the anode.

Kepekatan kuprum(II) sulfat tidak berubah. Kadar ion kuprum(II) menyahcas kepada atom kuprum di katod sama dengan kadar atom kuprum mengion membentuk ion kuprum(II) di anod.

117

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 117

Š Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Constructing Electrochemical Series / Membina Siri Elektrokimia What are the basic principles used in constructing the electrochemical series?

Apakah prinsip asas yang digunakan dalam membina siri elektrokimia?

1

The electrochemical series can be constructed by comparing: Siri elektrokimia boleh dibina dengan membandingkan: (i) Potential differences between the two metals in a simple cell Beza keupayaan antara dua logam dalam sel ringkas (ii) Ability of a metal to displace another metals from its salt solution through displacement reaction Kebolehan suatu logam menyesarkan logam lain dari larutan garamnya melalui tindak balas penyesaran

Constructing the Electrochemical Series Through the Potential Difference in a Simple Cell Membina Siri Elektrokimia Melalui Beza Keupayaan dalam satu Sel Ringkas

How to construct electrochemical series based on potential difference?

Bagaimana untuk membina siri elektrokimia berdasarkan beza keupayaan?

When two different metals dipped in electrolyte and connect to external circuit through a voltmeter: Apabila dua logam berbeza dicelup dalam elektrolit dan disambung kepada litar luar melalui satu voltmeter: (i) A more electropositive metal become the negative terminal of the cell and less electropositive metal become the positive terminal.

Logam lebih elektropositif menjadi terminal negatif sel dan logam kurang elektropositif menjadi terminal positif. (ii) The bigger the potential difference (voltage value) between two metals, the further the distance between the two metals in electrochemical series.

Lebih besar beza keupayaan (nilai voltan) antara dua logam, lebih jauh jarak antara dua logam itu dalam siri elektrokimia.

Example / Contoh:

The diagram below shows the set-up of the apparatus to arrange metals W, X, Y and Z based on the potential difference of the metals.

Rajah di bawah menunjukkan susunan radas bagi eksperimen untuk menentukan kedudukan logam W, X, Y dan Z berdasarkan beza keupayaan logam.

Remark / Catatan: Voltage / Voltan

V

1.1

0.3

Y

UNIT

X

5

0.5

Electropositivity increases Keelektropositifan meningkat

Z Metal electrode Elektrod logam

Metal electrode Elektrod logam

Electrolyte Elektrolit

The table below shows the results of the experiment.

Jadual di bawah menunjukkan keputusan eksperimen.

W

Pair of metals Pasangan logam

Potential difference (V) Beza keupayaan (V)

Negative terminal Terminal negatif

W and X / W dan X

0.50

X

X and Y / X dan Y

0.30

Y

W and Z / W dan Z

1.10

Z

(a) (i) Arrange metals W, X, Y and Z in descending order of the electropositivity of metal.

Susunkan logam W, X, Y dan Z dalam tertib menurun keelektropositifan logam.

Z, Y, X, W

(ii) Explain your answer.

– Z is

above

W in the electrochemical series because Z is negative terminal.

– X is

above

W in the electrochemical series because X is negative terminal.

above

X in the electrochemical series because Y is negative terminal.

– Y is

– The potential different between Z and W is distance between Z and W is

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 118

further

– The potential different between X and Y is the

bigger

than X and W. Hence, the

than X and W. Thus Z is smallest

above

X.

, Y below Z.

118

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4

Terangkan jawapan anda.

– Z di

– X di

– Y di

– Beza keupayaan antara Z dan W Z dan W

atas

W dalam siri elektrokimia kerana Z adalah terminal negatif.

atas

W dalam siri elektrokimia kerana X adalah terminal negatif.

atas

X dalam siri elektrokimia kerana Y adalah terminal negatif.

lebih jauh

lebih besar

daripada X dan W. Oleh itu, jarak antara atas

berbanding X dan W. Jadi, Z berada di paling kecil

– Beza keupayaan antara X dan Y adalah

X.

. Y berada di bawah Z.

(b) (i) Metals X and Z are used as electrodes in the diagram. State which metal acts as positive terminal.

Logam X dan Z digunakan sebagai elektrod dalam rajah. Nyatakan logam yang manakah akan bertindak sebagai terminal positif.

Metal X / Logam X

(ii) Give reason for your answer in (b)(i).

Berikan sebab untuk jawapan anda di (b)(i).

Metal X is less electropositive than metal Z.

Logam X kurang elektropositif daripada logam Z.

(c) Predict the voltage of the cell in (b)(i).

Ramalkan nilai voltan dalam sel di (b)(i).

0.6 V ( 1.1 – 0.5)

Constructing the Electrochemical Series From Displacement Reaction Membina Siri Elektrokimia daripada Tindak Balas Penyesaran

How to predict the ability of a metal to displace another metal from its solution?

Bagaimana meramal kebolehan suatu logam menyingkir logam lain daripada larutannya?

metals

Electrochemical Series is able to displace

salt solution

below it from its

.

Logam yang berada di kedudukan atas (kecenderungan melepaskan elektron yang tinggi) dalam Siri logam

Elektrokimia dapat menyesarkan logam tersebut.

yang di bawahnya daripada

larutan garam

Example / Contoh:

Experiment / Eksperimen Silver nitrate solution Larutan argentum nitrat

Observation / Pemerhatian – Copper strip becomes thinner .

Kepingan kuprum

menipis

.

– A

grey

solid deposited.

Pepejal kelabu terenap.

– The colourless solution turns blue.

Copper Kuprum

Larutan tidak berwarna bertukar menjadi biru.

Remark / Catatan

5

3

The metal which is situated at a higher position (higher tendency to release electron) in the

Inference / Inferens: – The

grey

Pepejal

silver

solid is

kelabu

UNIT

2

.

adalah argentum .

– The blue solution is Larutan biru adalah

copper(II) nitrate kuprum(II) nitrat

. .

Explanation / Penerangan: –

Silver

ion receives electrons to form

silver

atom.

Ion argentum menerima elektron membentuk atom argentum .

– Copper atom releases electrons to form

copper(II) ion

Atom kuprum melepaskan elektron membentuk – Copper has displaced

silver

ion kuprum(II)

. .

from silver nitrate solution.

Kuprum telah menyesarkan argentum dari larutan argentum nitrat.

Cu + 2AgNO3 – Copper is

more

Cu(NO3)2 + 2Ag

.

electropositive than silver // Copper is

above

silver in the Electrochemical Series of metal.

lebih Kuprum adalah elektropositif daripada argentum // Kuprum di atas terletak argentum dalam Siri Elektrokimia logam.

119

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 119

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

MODULE • Chemistry FORM 4 Experiment / Eksperimen Copper(II) sulphate solution Larutan kuprum(II) sulfat

Observation / Pemerhatian Magnesium strip becomes thinner .

Kepingan magnesium menipis .

The

Magnesium Magnesium

brown

solid deposited.

Pepejal perang terenap.

The blue solution turns colourless.

Larutan biru bertukar menjadi tidak berwarna.

Remark / Catatan Inference / Inferens: – The

brown

Pepejal

copper

solid is

perang

adalah

– The colourless solution is

.

kuprum

.

magnesium sulphate magnesium sulfat

Larutan tidak berwarna adalah

– Copper(II) ion receives electrons to form copper atom.

Ion

kuprum(II)

menerima elektron membentuk atom kuprum.

copper

MgSO4 + Cu

– Magnesium is above

more

.

ion magnesium

.

from copper(II) sulphate solution.

kuprum

Magnesium telah menyesarkan

Mg + CuSO4

magnesium ion

Atom magnesium melepaskan elektron membentuk

– Magnesium has displaced

dari larutan kuprum(II) sulfat. .

electropositive than copper// Magnesium is

copper in the Electrochemical Series of metal. lebih

Magnesium adalah

di atas

Magnesium terletak No observable changes.

Tiada perubahan yang dapat diperhatikan.

.

Explanation / Penerangan:

– Magnesium atom releases electrons to form

Zinc sulphate solution Larutan zink sulfat

.

elektropositif daripada kuprum// kuprum dalam Siri Elektrokimia logam.

Inference / Inferens: No reaction occur.

Tiada tindak balas berlaku.

Explanation / Penerangan: – Copper

Kuprum

cannot

displace

tidak boleh

zinc

from zinc sulphate solution.

menyesarkan

zink

daripada

UNIT

larutan zink sulfat.

Copper / Kuprum

– Copper is

less

electropositive than zinc// Copper is

below

zinc

5

in the Electrochemical Series of metal.

Kuprum adalah terletak

kurang

di bawah

elektropositif daripada zink // Kuprum

zink dalam Siri Elektrokimia logam.

Soalan Tambahan Additional Question

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 120

120

11/10/2017 9:14 AM

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 121

121 Copper(II) sulphate solution

Copper

Iron and copper

Zinc and copper

Aluminium and copper

Magnesium and copper

Pair of electrodes

(f) Tabulation of data

UNIT

5

Voltmeter reading(V)

(e) Procedure: 1 Clean all the metals with sandpaper.

2 Fill a beaker with copper(II) sulphate solution until it is two-thirds full.

3 Dip the magnesium strip and copper strip into copper(II) sulphate solution.

4 Both strips are connected to a voltmeter using connecting wires.

5 Record the reading of voltmeter.

6 Repeat steps 1 to 5 using other metals as shown in the table to replace Magnesium strip.

Magnesium

V

(a) Problem statement

(a) Problem statement : How does the distance between two metals in the electrochemical series affect the voltage produced in a simple voltaic cell? (b) Manipulated variable : Metal paired with copper Responding variable : Voltage Constant variable : Electrolyte used // Copper metal (c) Hypothesis : The further the distance between two metals in the electrochemical series, the bigger the value of the voltage. (d) Materials : Magnesium strip, aluminium strip, zinc strip, iron strip copper plate, 1.0 mol dm–3 copper(II) sulphate solution Apparatus : Connecting wires with crocodile clips, beaker, sandpaper and voltmeter

Salt solution

Copper

Lead

Zinc

Magnesium

Metal strip

(f) Tabulation of data Magnesium nitrate

Zinc nitrate

Lead(II) nitrate

Copper(II) nitrate

(e) Procedure: 1 Clean 2 cm of magnesium strips with a sandpaper.

2 Pour 5 cm3 of magnesium nitrate solution, zinc nitrate solution, lead(II) nitrate solution and copper(II) nitrate solution into four separate test tubes.

3 For each test tube, place a strip of magnesium into each solution.

4 Record all the observations in the table.

5 Repeat steps 1 to 3 using strips of zinc, lead and copper to replace magnesium strip. For each repetition, use a fresh salt solution.

6 Record all observations in the table.

: How can displacement of metals from its salt solution by other metal can be used to construct electrochemical series? (b) Hypothesis : Less electropositive metal can be displaced from its salt solution by more electropositive metal. The greater the number of metals that can be displaced by a more electropositive metal from their solutions, the higher its position in the electrochemical series. (c) Manipulated variables : Metals and salt solution Responding variables : Deposition of metal Constant variables : Concentration and volume of salt solution (d) Materials : 1.0 mol dm–3 magnesium nitrate solution, 1.0 mol dm–3 zinc nitrate solution, 1.0 mol dm–3 lead(II) nitrate solution, 1.0 mol dm–3 copper(II) nitrate solution, magnesium strips, zinc strip, lead strip and copper strip Apparatus : Test tube and sandpaper

Experiment to Construct the Electrochemical Series Using the Principle of Displacement of Metals

Experiment to Construct an Electrochemical Series Through the Potential Differences (Voltage) Between Pairs of Metals

Experiment to Construct Electrochemical Series of Metals

MODULE • Chemistry FORM 4

© Nilam Publication Sdn. Bhd.

11/10/2017 9:14 AM

UNIT

5

© Nilam Publication Sdn. Bhd.

05 Chap 5 ChemF4 Bil 2017(CSY5p).indd 122

Larutan kuprum(II) sulfat

Kuprum

Ferum dan kuprum

Zink dan kuprum

Aluminium dan kuprum

Magnesium dan kuprum

Pasangan elektrod

(f) Penjadualan data Bacaan voltmeter (V)

(e) Prosedur: 1 Bersihkan semua logam dengan kertas pasir.

2 Isikan bikar dengan larutan kuprum(II) sulfat sehingga dua per tiga penuh.

3 Celupkan kepingan magnesium dan kuprum ke dalam larutan kuprum(II) sulfat.

4 Kedua-dua kepingan logam disambung kepada voltmeter menggunakan wayar penyambung.

5 Catatkan bacaan voltmeter.

6 Langkah 1 hingga 5 diulang dengan menggantikan jalur magnesium dengan logam lain seperti ditunjukkan dalam jadual.

Magnesium

V

(a) Pernyataan masalah : Bagaimanakah jarak antara dua logam dalam Siri Elektrokimia mempengaruhi voltan yang dihasilkan dalam sel ringkas? (b) Pemboleh ubah dimanipulasikan : Logam yang dipasangkan dengan kuprum Pemboleh ubah bergerak balas : Voltan Pemboleh ubah dimalarkan : Elektrolit yang digunakan // Logam kuprum (c) Hipotesis : Semakin jauh jarak antara dua logam dalam siri elektrokimia, semakin tinggi voltan. (d) Bahan : Jalur magnesium, jalur aluminium, jalur zink, jalur ferum, plat kuprum, larutan kuprum(II) sulfat 1.0 mol dm–3 Radas : Wayar penyambung dengan klip buaya, bikar, kertas pasir dan voltmeter

Eksperimen untuk Membina Siri Elektrokimia Melalui Beza Keupayaan (Voltan) antara Pasangan Logam

Larutan garam

Kuprum

Plumbum

Zink

Magnesium

Jalur logam

(f) Penjadualan data Magnesium nitrat

Zink nitrat

Plumbum(II) nitrat

Kuprum(II) nitrat

(e) Prosedur: 1 Bersihkan 2 cm jalur magnesium dengan kertas pasir.

2 Tuang 5 cm3 larutan magnesium nitrat, larutan zink nitrat, larutan plumbum(II) nitrat dan larutan kuprum(II) nitrat ke dalam empat tabung uji yang berasingan.

3 Bagi setiap tabung uji letakkan satu jalur magnesium ke dalam setiap larutan.

4 Catatkan semua pemerhatian ke dalam jadual.

5 Langkah 1 hingga 4 diulang dengan menggunakan jalur zink, plumbum dan kuprum untuk menggantikan jalur magnesium. Bagi setiap ulangan, gunakan larutan garam yang baru.

(a) Pernyataan masalah : Bagaimana penyesaran logam daripada larutan garamnya oleh logam lain boleh digunakan untuk membina Siri Elektrokimia? (b) Hipotesis : Logam yang lebih elektropositif boleh menyesarkan logam yang kurang elektropositif daripada larutan garamnya oleh logam yang lebih elektropositi.

Semakin banyak logam yang dapat disesarkan oleh logam yang lebih elektropositif daripada larutan garamnya, semakin tinggi kedudukannya dalam siri elektrokimia. (c) Pemboleh ubah dimanipulasikan : Logam dan larutan garam Pemboleh ubah bergerak balas : Enapan logam Pemboleh ubah dimalarkan : Kepekatan dan isi padu larutan garam (d) Bahan : Larutan magnesium nitrat 1.0 mol dm–3, larutan zink nitrat 1.0 mol dm–3, larutan plumbum(II) nitrat 1.0 mol dm–3, larutan kuprum(II) nitrat 1.0 mol dm–3, jalur magnesium, jalur zink, jalur plumbum dan jalur kuprum Radas : Tabung uji dan kertas pasir

Eksperimen untuk Membina Siri Elektrokimia Menggunakan Prinsip Penyesaran Logam

Eksperimen untuk Membina Siri Elektrokimia Logam

MODULE • Chemistry FORM 4

122

11/10/2017 9:14 AM