Bab
1
Tindak Balas Redoks
Redox Reactions
Pengoksidaan dan Penurunan
1.1
Menerangkan Tindak Balas Redoks
Describe Redox Reactions
1. Pengoksidaan dan penurunan boleh dijelaskan dari segi:
IA
A
Oxidation and Reduction
Oxidation and reduction can be explained in terms of:
Pengoksidaan
Penurunan
Oxidation
Oksigen
Penambahan
Oxygen
Hidrogen
Kehilangan
Hydrogen
Loss
Elektron
Kehilangan
Electron
Loss
Nombor pengoksidaan
Loss
of oxygen
Peningkatan
Increase
Penambahan
hidrogen
Gain
of hydrogen
Menerima
elektron
Gain
of electrons
nombor pengoksidaan
Penurunan
Decrease
in the oxidation number
PA
N
Oxidation number
Kehilangan
oksigen
AS
Gain
Reduction
Reduction
oksigen
of oxygen
hidrogen
of hydrogen
elektron
of electrons
nombor pengoksidaan
in the oxidation number
2. Tindak balas redoks ialah tindak balas kimia yang melibatkan penurunan dan pengoksidaan yang berlaku secara
serentak. Contoh tindak balas redoks:
Redox reactions are chemical reactions involving reduction and oxidation occurring simultaneously. Example of redox reactions:
(a) Pemindahan elektron pada suatu jarak/ Transfer of electron at a distance
(b) Pertukaran ion ferum(II), Fe2+ kepada ion ferum(III), Fe3+ dan sebaliknya
Change of iron(II) ions, Fe2+ to iron(III) ions, Fe3+ and vice versa
(c) Penyesaran logam daripada larutan garamnya/ Displacement of metal from its salt solution
(d) Penyesaran halogen daripada larutan halidanya/ Displacement of halogen from its halide solution
3. Agen pengoksidaan ialah bahan yang menyebabkan
pengoksidaan mengalami penurunan.
oxidation
The oxidising agent is the substance that causes
reduction
. Dalam tindak balas redoks, agen
. It is reduced in the redox reaction.
penurunan
4. Agen penurunan ialah bahan yang menyebabkan
mengalami pengoksidaan.
The reducing agent is the substance that causes
pengoksidaan
. Dalam tindak balas redoks, agen penurunan
. It is oxidised in the redox reaction.
Agen penurunan/ Reducing agent
Agen pengoksidaan/ Oxidising agent
1.
Menderma/ Donates
hidrogen/ hydrogen
1.
Menerima/ Accepts
hidrogen/ hydrogen
2.
Menerima/ Accepts
oksigen/ oxygen
2.
Menderma/ Donates
oksigen/ oxygen
3.
Menderma/ Donates
elektron/ electron
3.
Menerima/ Accepts
elektron/ electron
4.
Pertambahan/ Increase
nombor pengoksidaan/
4.
Pengurangan/ Decrease
in oxidation number
1
in oxidation number
nombor pengoksidaan/
Contoh/ Examples:
1. Ferum(II) sulfat/ Iron(II) sulphate, FeSO4
2. Kalium iodida/ Potassium iodide, KI
3. Gas hidrogen/ Hydrogen gas, H2
4. Serbuk zink/ Zink powder, Zn
5. Gas sulfur dioksida/ Sulfur dioxide gas, SO2
6. Gas hidrogen sulfida/ Hydrogen sulfide gas, H2S
Contoh/ Examples:
1. Larutan kalium manganat(VII) berasid/ Acidified
potassium manganate(VII) solution, KMnO4/ H+
2. Larutan kalium dikromat(VI) berasid/ Acidified
potassium dichromate(VI) solution, K2Cr2O7 / H+
3. Air klorin/ Chlorine water, Cl2
4. Air bromin/ Bromine water, Br2
5. Hidrogen peroksida/ Hydrogen peroxide, H2O2
Contoh/Example 1
Pengoksidaan/Oxidation
Mg
+
PbO
Magnesium, Mg is
)
oksigen.
oxygen.
kehilangan
AS
kerana berlaku
loses
because it
penurunan
(c) Magnesium, Mg ialah agen
reducing
Magnesium, Mg is the
gains
penurunan
reduced
Lead(II) oxide, PbO is
Pb
penambahan
kerana berlaku
because it
(b) Plumbum(II) oksida, PbO mengalami
+
Penurunan/Reduction
pengoksidaan
oxidised
MgO

(
(a) Magnesium, Mg mengalami
)
IA
(
oxidising
Contoh/Example 2
penurunan
reduction
agent. It causes the
(d) Plumbum(II) oksida, PbO ialah agen
Lead(II) oxide, PbO is the
oxygen.
kerana menyebabkan
pengoksidaan
oksigen.
of lead(II) oxide, PbO .
pengoksidaan
kerana menyebabkan
oxidation
agent. It causes the
plumbum(II) oksida, PbO.
magnesium, Mg.
of magnesium.
PA
N
( Pengoksidaan/Oxidation )
H 2S
+
Cl2
(
(a) Hidrogen sulfida, H2S mengalami
Hydrogen sulphide, H2S is
Chlorine,Cl2 is
reduced
because it
(c) Hidrogen sulfida, H2S ialah agen
reducing
Hydrogen sulphide, H2S is the
(d) Klorin, Cl2 ialah agen
Chlorine,Cl2 is the
pengoksidaan
oxidising
loses
gains
)
kehilangan
hidrogen.
hydrogen.
penambahan
kerana berlaku
penurunan
2HCl
kerana berlaku
because it
penurunan
(b) Klorin, Cl2 mengalami
+
Penurunan/Reduction
pengoksidaan
oxidised
S

hidrogen.
hydrogen.
kerana menyebabkan
agent. It causes the
kerana menyebabkan
oxidation
agent. It causes the
reduction
penurunan
of chlorine,Cl2.
pengoksidaan
hidrogen sulfida, H2S.
of hydrogen sulphide, H2S.
Contoh/Example 3
Zn
Zn
+
Zn2+ +2e– (Persamaan setengah/ Half-equation)
CuSO4 
Cu2+ + 2e–
ZnSO4
+
Cu
Cu
(Persamaan setengah/Half-equation)
2
klorin, Cl2.
pengoksidaan
(a) Zink, Zn mengalami
oxidised
because it loses 2 electrons to form zinc ion, Zn2+.
penurunan
(b) Kuprum(II) sulfat, CuSO4 mengalami
untuk membentuk atom kuprum, Cu.
reduced
Copper(II) sulphate, CuSO4 is
(c) Zink, Zn ialah agen
Zinc, Zn is the
reducing
reduction
pengoksidaan
(d) Kuprum(II) sulfat, CuSO4 ialah agen
oxidising
CuO
+2 –2
0
pengoksidaan
oxidation
Hydrogen, H2 undergoes
+
of zinc, Zn.
increases
0
berkurang
decreases
/ The oxidation number
penurunan
0
to
PA
N
(c) Ion kuprum(II), Cu
oxidising
.
+2
+2
to
0
.
kerana mengoksidakan hidrogen. / Copper(II) ion, Cu2+ is the
agent because it oxidises hydrogen.
penurunan
(d) Hidrogen, H2 ialah agen
reducing
pengoksidaan
ialah agen
+1
kepada
kerana nombor pengoksidaan berkurang daripada
Copper(II) ion, Cu2+ undergoes reduction because the oxidation number of copper decreases from
2+
0
kerana nombor pengoksidaan bertambah daripada
because the oxidation number increases from
(b) Ion kuprum(II), Cu2+ mengalami
0
kepada
.
zink, Zn.
Cu
+1 –2
Nombor pengoksidaan
(a) Hidrogen, H2 mengalami
+1
.
oxidation
/ The oxidation number
H2O
pengoksidaan
AS
+
electrons to form copper atom, Cu.
of copper(II) sulphate, CuSO4.
agent. It causes the
bertambah
Nombor pengoksidaan
elektron
kuprum(II) sulfat, CuSO4.
kerana menyebabkan
Contoh/Example 4
H2
penurunan
kerana menyebabkan
agent. It causes the
Copper(II) sulphate, CuSO4 is the
2
because copper(II) ion, Cu2+ gains
penurunan
2
kerana ion kuprum(II), Cu2+ menerima
IA
Zinc, Zn is
kerana kehilangan 2 elektron untuk membentuk ion zink, Zn2+.
kerana menurunkan ion kuprum(II), Cu2+. / Hydrogen, H2 is the
agent because it reduces copper(II) ion, Cu2+.
Contoh/Example 5
Zn
+
0
(a) Zink, Zn mengalami
+2
.
Zinc, Zn undergoes
pengoksidaan
oxidation
CuSO4
ZnSO4 +
Cu
+2 +6 –2
+2 +6 –2
0
because the oxidation number increases from
(b) Kuprum(II) sulfat, CuSO4 mengalami
+2
0
daripada
kepada
.
0
kerana nombor pengoksidaan meningkat daripada
penurunan
0
to
+2
kepada
.
kerana nombor pengoksidaan ion kuprum(II), Cu2+ menurun
+2
Copper(II) sulphate, CuSO4 undergoes reduction because the oxidation number of copper(II) ion, Cu2+ decreases from
0
.
reducing
penurunan
(c) Zink ialah agen
kerana menurunkan ion kuprum(II), Cu2+./ Zinc is the
2+
agent because it reduces copper(II) ion, Cu .
(d) Kuprum(II) sulfat, CuSO4 ialah agen
the
oxidising
pengoksidaan
to
kerana mengoksidakan zink, Zn. / Copper(II) sulphate, CuSO4 is
agent because it oxidises zinc, Zn.
3
I
Pemindahan Elektron pada Suatu Jarak/ Transfer of Electrons at a Distance
menderma
1. Dalam tindak balas redoks, agen penurunan
elektron kepada agen pengoksidaan.
donates
In a redox reaction, a reducing agent
electrons to an oxidising agent.
2. Apabila agen pengoksidaan dan agen penurunan dipisahkan oleh elektrolit dalam suatu tiub-U, elektron
litar luar
dipindahkan melalui
.
When an oxidising agent and a reducing agent are separated by an electrolyte in a U-tube, the electrons are transferred
external circuit
through an
.
3. Pemindahan elektron boleh dikesan dengan galvanometer. Pesongan jarum galvanometer akan menunjukkan
arah
aliran elektron.
The transfer of the electrons can be detected by a galvanometer. The deflection of the galvanometer needle will show the
direction
of the flow of electrons.
negatif
IA
4. Elektrod yang diletakkan dalam larutan agen penurunan bertindak sebagai terminal
negative
The electrode placed in a reducing agent solution acts as the
terminal.
5. Elektrod yang diletakkan dalam larutan agen pengoksidaan bertindak sebagai terminal
positive
The electrode placed in an oxidising agent solution acts as the
positif
.
terminal.
1.1
Tujuan/ Aim:
Mengkaji pemindahan elektron pada suatu jarak dalam tindak balas redoks
To investigate the transfer of electrons at a distance in redox reaction
Bahan/ Materials:
Air klorin, larutan ferum(II) sulfat, FeSO4 0.5 mol dm–3 , asid sulfurik cair, H2SO4 2.0 mol dm–3, larutan kalium
dikromat(VI), K2Cr2O7 berasid larutan kalium iodida, KI 0.5 mol dm-3, larutan kalium manganat(VII), KMnO4
berasid 0.2 mol dm-3, larutan kanji 1%, larutan natrium hidroksida, NaOH
Chlorine water, 0.5 mol dm–3 iron(II) sulphate, FeSO4 solution, 2.0 mol dm–3 dilute sulphuric acid, H2SO4, acidified
potassium dichromate(VI), K2Cr2O7 solution, 0.5 mol dm-3 potassium iodide, KI solution, 0.2 mol dm-3 acidified potassium
manganate(VII), KMnO4 solution, 1% starch solution, sodium hydroxide, NaOH solution
PA
N
Eksperimen Wajib
AS
AKTIVITI
.
Radas/ Apparatus:
Tiub-U, galvanometer, elektrod karbon, wayar penyambung dengan klip buaya, elektrod karbon, kaki retort
dan pengapit, tabung uji, penitis, penutup satu lubang
U-tube, galvanometer, carbon electrodes, connecting wires with crocodile clips, carbon electrodes, retort stand with
clamp, test tube, dropper, stoppers with one hole
Prosedur/ Procedure:
G
Elektrod karbon
Carbon electrode
Elektrod karbon
Carbon electrode
Agen penurunan
Reducing agent
Agen pengoksidaan
Oxidising agent
Asid sulfurik cair H2SO4
Dilute sulphuric acid, H2SO4
Rajah/Diagram 1.1
1. Tiub-U diapit secara menegak pada kaki retort.
A U-tube is clamped upright to a retort stand.
2. Asid sulfurik, H2SO4 cair 1.0 mol dm–3 dituang ke dalam tiub-U sehingga arasnya 6 cm daripada mulut
tiub-U.
1.0 mol dm–3 dilute sulphuric acid, H2SO4 is poured into the U-tube until its levels 6 cm away from the mouth of the
U-tube.
3. Menggunakan penitis, larutan ferum(II) sulfat, FeSO4 0.5 mol dm–3 ditambahkan dengan berhati-hati pada
lengan kiri tiub-U sehingga lapisan larutan setinggi 3 cm.
Using a dropper, 0.5 mol dm-3 iron(II) sulphate, FeSO4 solution is carefully added into the left arm of the U-tube until
the layer of solution reaches the height of 3 cm.
4
4. Menggunakan cara yang sama seperti langkah 3, 0.2 mol dm-3 larutan kalium manganat(VII) berasid,
KMnO4 ditambahkan pada lengan kanan tiub-U.
In a similar manner as in step 3, 0.2 mol dm-3 acidified potassium manganate (VII), KMnO4 solution is added to the
right arm of the U-tube.
5.
Elektrod karbon diletakkan ke dalam setiap lengan tiub-U.
6.
Elektrod disambungkan kepada galvanometer dengan wayar penyambung.
7.
Pesongan jarum galvanometer diperhatikan untuk menentukan elektrod yang bertindak sebagai terminal
positif dan negatif.
A carbon electrode is placed in each arm of the U-tube.
The electrodes are connected to a galvanometer with connecting wires.
IA
The deflection of the galvanometer needle is observed to determine the electrodes that act as the positive and
negative terminal.
8.
Susunan radas dibiarkan selama 30 minit. Sebarang perubahan direkodkan.
direkodkan.
9.
Menggunakan penitis yang bersih, 1 cm3 larutan ferum(II) sulfat, FeSO4 dikeluarkan dan dimasukkan ke
dalam tabung uji. Kemudian, beberapa titis larutan natrium hidroksida, NaOH dimasukkan ke dalam
tabung uji. Sebarang perubahan direkodkan.
The set-up is left aside for 30 minutes. Any change is recorded.
Using a clean dropper, 1 cm3 of iron(II) sulphate, FeSO4 solution is draw and placed in a test tube. Then a few drops
of sodium hydroxide, NaOH solution are added to the test tube. Any change is recorded.
AS
10. Langkah 1 hingga 8 diulang dengan menggunakan larutan kalium iodida, KI 0.5 mol dm-3 dan larutan
kalium dikromat(VI), K2Cr2O7 berasid 0.2 mol dm-3.
Steps 1 to 8 are repeated using 0.5 mol dm-3 potassium iodide, KI solution and 0.2 mol dm-3 acidified potassium
dichromate(VI), K2Cr2O7 solution.
11. Menggunakan penitis yang bersih, 1 cm3 larutan kalium iodida, KI dikeluarkan dan dimasukkan ke dalam
tabung uji. Kemudian, beberapa titis larutan kanji dimasukkan ke dalam tabung uji. Sebarang perubahan
direkodkan.
Using a clean dropper, 1 cm3 of potassium iodide, KI solution is drawn and placed in a test tube. Then a few drops
of starch solution are added to the test tube. Any change is recorded.
PA
N
Mentafsir data/ Interpreting data:
1. Larutan ferum(II) sulfat, FeSO4 + larutan kalium manganat(VII) berasid, KMnO4
Iron(II) sulphate, FeSO4 solution + Acidified potassium manganate(VII), KMnO4 solution
Pemerhatian/ Observation
Inferens/ Inference
(a) Pesongan jarum galvanometer menunjukkan Elekron mengalir daripada elektrod dalam larutan
ferum(II) sulfat
elektrod dalam larutan ferum(II) sulfat bertindak
kepada elektrod dalam
negatif
sebagai terminal
manakala larutan
elektrod dalam larutan kalium manganat(VII) berasid
kalium manganat(VII) berasid
.
The electrons flow from the electrode in
iron(II) sulphate
solution to the electrode in the
The deflection of the galvanometer needle shows
acidified potassium manganate(VII)
that the electrode in iron(II) sulphate solution acts as
solution.
negative
terminal whereas the electrode in
the
acidified potassium manganate(VII) solution acts as the
positive
terminal.
bertindak sebagai terminal
positif
.
(b) Larutan hijau muda larutan ferum(II) sulfat bertukar Ion ferum(II), Fe2+
ferum(III), Fe3+.
perang
kepada
.
Pale green iron(II) sulphate solution turns
brown
.
Iron(II) ion, Fe2+ is
Fe3+.
(c) Larutan ungu kalium manganat(VII) berasid bertukar Ion manganat(VII)
tidak berwarna
mangan(II).
kepada
.
Manganate(VII)
ion
Purple acidified potassium manganate(VII) solution turns
manganese(II) ion.
colourless
.
5
dioksidakan
oxidised
to iron(III) ion,
diturunkan
is
kepada ion
kepada ion
reduced
to
perang
(d) Mendakan
terbentuk apabila
ditambah larutan natrium hidroksida.
Ion ferum(III), Fe3+
hadir.
Iron(III) ions, Fe3+
are present.
brown
precipitate is formed when sodium
A
hydroxide solution is added.
(e) Ion manganat(VII), MnO4– bertindak sebagai agen pengoksidaan . Ion ferum(II), Fe2+ bertindak sebagai
agen
penurunan
.
Manganate(VII) ions, MnO4– act as the
oxidising
agent. Iron(II) ions, Fe2+ act as the
reducing
agent.
IA
2. Larutan kalium iodida, KI + larutan kalium dikromat(VI), K2Cr2O7 berasid
Potassium iodide, KI solution + acidified potassium dichromate(VI), K2Cr2O7 solution
Pemerhatian/ Observation
Inferens/ Inference
(a) Pesongan jarum galvanometer menunjukkan Elektron mengalir daripada elektrod dalam larutan
elektrod dalam larutan kalium iodida bertindak
kalium iodida
kepada elektrod dalam larutan
negatif
sebagai terminal
manakala
elektrod dalam larutan kalium dikromat(VI) berasid
kalium dikromat(VI) berasid
.
AS
The electrons flow from the electrode in
positif
bertindak sebagai terminal
.
potassium iodide solution to the electrode in the
The deflection of the galvanometer needle shows that
acidified potassium dichromate(VI)
solution.
the electrode in potassium iodide solution acts as the
negative
terminal whereas the electrode in
acidified potassium dichromate(VI) solution acts as the
positive
terminal.
(b) Larutan tak berwarna kalium iodida bertukar kepada Ion iodida
perang
.
PA
N
Colourless potassium iodide solution turns
brown
dioksidakan
Iodide ion is
oxidised
Dichromate(VI) ion is
Orange acidified potassium dichromate(VI) solution turns
(III) ion.
green
.
biru tua
dark blue
diturunkan
kepada ion
reduced
to chromium
Iodin hadir.
apabila ditambah larutan kanji.
kanji
The brown solution turns to
solution is added.
to iodine.
.
(c) Larutan jingga kalium dikromat(VI) berasid bertukar Ion dikromat(VI)
kromium(III).
hijau
kepada
.
(d) Larutan perang bertukar kepada
kepada iodin.
Iodine is present.
when starch
(e) Ion dikromat(VI), Cr2O72- bertindak sebagai agen pengoksidaan . Ion iodida, I- bertindak sebagai agen
penurunan
.
Dichromate(VI) ions, Cr2O72- act as the
oxidising
agent. Iodide ions, I- act as the
reducing
agent.
Perbincangan/ Discussion:
1. Nyatakan bahan-bahan yang mengalami pengoksidaan dan penurunan. Terangkan jawapan anda dari segi
pemindahan elektron.
State the substance that undergoes oxidation and reduction. Explain your answer in terms of electron transfer.
(a) Pengoksidaan/Oxidation:
Ferum(II) sulfat dan kalium iodida kehilangan elektron/ Iron(II) sulphate and potassium iodide lose of electrons
6
(b) Penurunan/ Reduction:
Kalium dikromat(VI) berasid dan kalium manganat(VII) berasid menerima elektron
Acidified potassium dichromate(VI) and acidified potassium manganate(VII) accept electrons
2. Tuliskan persamaan setengah bagi pengoksidaan dan penurunan yang berikut:
Write the half-equations for the following oxidation and reduction:
(a) Tindak balas antara larutan ferum(II) sulfat, FeSO4 dengan larutan kalium manganat(VII), KMnO4
berasid
Reaction between iron(II) sulphate, FeSO4 solution with acidified potassium manganate(VII),
manganate(VII), KMnO4 solution
Pengoksidaan/ Oxidation:
IA
Fe2+  Fe3+ + e–
Penurunan/ Reduction:
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
(b) Tindak balas antara larutan kalium iodida, KI dengan larutan kalium dikromat(VI), K2Cr2O7 berasid
Reaction between potassium iodide, KI solution with acidified potassium dichromate(VI), K2Cr2O7 solution
Pengoksidaan/ Oxidation:
Penurunan/ Reduction:
AS
2I–  I2 + 2e–
Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O
3. Nyatakan bahan berikut/ State the following substance:
(a) Agen pengoksidaan/ Oxidising agents:
Larutan kalium manganat(VII) berasid/ Acidified potassium manganate(VII) solution
Larutan kalium dikromat(VI) berasid/ Acidified potassium dichromate(VI) solution
(b) Agen penurunan/ Reducing agent:
Larutan ferum(II) sulfat/ Iron(II) sulphate solution
PA
N
Larutan kalium iodida
iodida/ Potassium iodide solution
negatif
4. Elektrod dalam agen penurunan bertindak sebagai terminal
positif
agen pengoksidaan bertindak sebagai terminal
5. Elektron mengalir daripada agen
The electrons flow from
B
reducing
.
negative
The electrode in the reducing agents acts as the
positive
terminal.
agents acts as the
penurunan
manakala elektrod dalam
terminal while the electrode in the oxidising
kepada agen pengoksidaan melalui litar luar.
oxidising
agents to
agents through external circuit.
Tindak Balas Redoks Berdasarkan Perubahan Nombor Pengoksidaan
Redox Reaction Based on Changes in Oxidation Number
Unsur/ Oxidation Number of An element
I Nombor Pengoksidaan bagi Unsur
1. Peraturan umum menentukan nombor pengoksidaan
pengoksidaan/ General rules to determine oxidation number:
0
(a) Nombor pengoksidaan bagi atom suatu unsur bebas ialah sifar ,
.
The oxidation number of an atom in its elements state is
(b) Nombor pengoksidaan bagi hidrogen biasanya ialah
+1
The oxidation number of hydrogen usually is
The oxidation number of oxygen usually is
The oxidation number of Halogen usually is
.
.
–2
.
.
(d) Nombor pengoksidaan bagi halogen biasanya ialah
–1
+1
0
,
.
(c) Nombor pengoksidaan bagi oksigen biasanya ialah
–2
zero
–1
.
.
sifar
(e) Jumlah nombor pengoksidaan bagi semua unsur dalam suatu sebatian neutral mestilah
The sum of the oxidation numbers of all the elements in a neutral compound must be
7
zero
,
0
0
,
.
.
Contoh/Example 6
(a) Manakah antara persamaan berikut merupakan tindak balas redoks?
Which of the following equations is a redox reaction?
Persamaan/ Equation I: Na2SO4 + Ba(NO3)2 → BaSO4 + 2NaNO3
Persamaan/ Equation II: Mg + CuSO4 → MgSO4 + Cu
+
+1 +6 –2
Ba(NO3)2
+2 +5 –2
Nombor pengoksidaan bagi Na, S, O, Ba dan N
+
0
CuSO4 →
+2 +6 –2
BaSO4
+
+2 +6 –2
tidak berubah
did not change
The oxidation numbers of Na, S, O, Ba and N
Persamaan/ Equation II: Mg
→
MgSO4
+
2NaNO3
+1 +5 –2
bukan tindak balas redoks
. Persamaan I
not a redox reaction
. Equation I is
.
.
Cu
IA
Penyelesaian/ Solution:
Persamaan / Equation I: Na2SO4
+2 +6 –2
0
2+
Persamaan II ialah tindak balas redoks kerana terdapat perubahan nombor pengoksidaan. Mg dioksidakan kepada Mg
(0 kepada +2), manakala Cu2+ diturunkan kepada Cu (+2 kepada 0)
Equation II is a redox reaction because
AS
is reduced to Cu (+2 to 0).
.
there is a change in the oxidation numbers. Mg is oxidised to Mg2+ (0 to +2) while Cu2+
.
(b) Persamaan berikut menunjukkan perubahan Fe2+ kepada Fe3+. Namakan jenis tindak balas dan terangkan berdasarkan
perubahan nombor pengoksidaan.
The following equation shows the change of Fe2+ to Fe3+. Name the type of reaction and explain it based on the change in the oxidation
number.
Fe2+ → Fe3+ + e–
Penyelesaian/ Solution:
pengoksidaan
Proses
Oxidation
+2 kepada +3
kerana nombor pengoksidaan berubah daripada
process because the oxidation number change from
.
+2 to +3.
.
PA
N
(c) Kaji persamaan keseluruhan ion berikut dan tentukan bahan yang mengalami proses pengoksidaan dan proses penurunan.
Study the following overall ionic equation and determine the material that undergoes oxidation and reduction processes.
Penyelesaian/ Solution:
Kuprum, Cu mengalami
Cu + 2Ag+ → Cu2+ + 2Ag
pengoksidaan
penurunan
argentum, Ag+ mengalami
Copper, Cu undergoes
reduction
undergoes
0 kepada +2
kerana nombor pengoksidaan berubah daripada
kerana nombor pengoksidaan berubah daripada
oxidation
as the oxidation number change from
+1 to 0
as the oxidation number change from
0 to +2
+1 kepada 0
. Ion
.
. The argentum ion, Ag+
.
II Penamaan Sebatian Mengikut Sistem Penamaan IUPAC
Naming Compounds According to the IUPAC Nomenclature
1. Kebanyakan unsur hanya mempunyai satu nombor pengoksidaan. Sesetengah unsur seperti unsur peralihan,
karbon, nitrogen dan sulfur mempunyai lebih daripada satu nombor pengoksidaan.
Many elements have just one oxidation number, but some elements such as transition metals, carbon, nitrogen and sulphur
have more than one oxidation number.
2. Untuk mengelakkan kekeliruan, angka Roman (I, II, III...) dimasukkan dalam nama sebatian dengan unsur yang
lebih daripada satu
mempunyai
nombor pengoksidaan.
To avoid confusion, Roman numeral (I, II, III, etc) are included in naming of some compound with element which has
more than one
oxidation number.
3. Bagi sebatian ion yang ringkas, angka Roman yang menyatakan nombor pengoksidaan unsur logam ditulis di
kurungan
dalam
selepas nama logam itu.
For simple ionic compounds, the Roman numeral of the oxidation number of metal element is written in
following the name of the metal.
8
brackets
,
Contoh/Example 7
Formula kimia sebatian
Nombor pengoksidaan
FeCl2
x + 2(–1) = 0, x = +2
Ferum(II) klorida/ Iron(II) chloride
FeCl3
x + 3(–1) = 0, x = +3
Ferum( III ) klorida/ Iron( III ) chloride
Cu2O
2x + (–2) = 0, x = +1
Kuprum(
CuO
x + (–2) = 0, x = +2
Kuprum( II ) oksida/ Copper( II ) oxide
SnO
x + (–2) = 0, x = +2
SnO2
x + 2(–2) =0, x = +4
PbO
x + (–2) = 0, x = +2
PbO2
x + 2(–2) = 0, x = +4
Compound chemical formula
IUPAC name of the compound
) oksida/ Copper(
I
) oxide
IA
I
Stanum( II ) oksida/ Tin( II ) oxide
IV
Stanum( IV ) oksida/ T
Tin(
in(
) oxide
Plumbum( II ) oksida/ Lead( II ) oxide
Plumbum( IV ) oksida/ Lead( IV ) oxide
AS
Tip
Nama IUPAC bagi sebatian
Oxidation number
SPM
Bagi unsur yang mempunyai hanya satu nombor pengoksidaan, tidak perlu ditulis dalam angka Roman. Contohnya, natrium
klorida, magnesium oksida dan aluminium klorida. Unsur-unsur ini berada di Kumpulan 1, 2 dan 13 dalam Jadual Berkala Unsur.
Tip SPM
For elements with just one oxidation number, do not need to write it in Roman numerals. For example, sodium chloride, magnesium oxide and
aluminium chloride. Such elements are in Groups 1, 2 and 13 in the Periodic Table of Elements.
PA
N
III Pertukaran Ion Fe2+ kepada Ion Fe3+ dan Sebaliknya/ Change of Fe2+ Ions to Fe3+ Ions and Vice Versa
1. Logam besi (ferum) mempunyai dua nombor pengoksidaan/ Iron metal has two oxidation numbers.
(a) Dalam ion ferum(II), Fe2+, nombor pengoksidaan ialah +2/ In iron(II) ions, Fe2+, the oxidation number is +2.
(b) Dalam ion ferum(III), Fe3+, nombor pengoksidaan ialah +3/ In iron(III) ions, Fe3+ the oxidation number is +3.
2. Pertukaran ion ferum(II), Fe2+ kepada ion ferum(III), Fe3+ merupakan proses
The change of iron(II) ions, Fe2+ to iron(III) ions, Fe3+ is an
3. Agen
An
pengoksidaan
oxidising
oxidation
pengoksidaan .
process.
diperlukan untuk mengoksidakan ion ferum(II), Fe2+ kepada ion ferum(III), Fe3+
agent is required to oxidise the iron(II) ions, Fe2+ to iron(III) ions, Fe3+.
4. Pertukaran ion ferum(III), Fe3+ kepada ion ferum(II), Fe2+ merupakan proses
The change of iron(III) ions, Fe to iron (II) ions, Fe is a
3+
5. Agen
A
penurunan
reducing
2+
reduction
penurunan
.
process.
diperlukan untuk menurunkan ion ferum(III), Fe3+ kepada ion ferum(II), Fe2+.
agent is required to reduce the iron(III) ions, Fe3+ to iron(II) ions, Fe2+.
AKTIVITI
1.2
To investigate the change of Fe2+ ions to Fe3+ ions and vice versa.
Bahan/ Materials:
Larutan ferum(II) sulfat, FeSO4 0.5 mol dm–3, larutan ferum(III) klorida, FeCl3 0.5 mol dm–3, air bromin, pita
magnesium, larutan natrium hidroksida, NaOH 2.0 mol dm–3, kertas turas
0.5 mol dm–3 iron(II) sulphate, FeSO4 solution, 0.5 mol dm–3 iron(III) chloride, FeCl3 solution, bromine water, magnesium
ribbon, 2.0 mol dm–3 sodium hydroxide, NaOH solution, filter paper
Radas/ Apparatus:
Penitis, spatula, penyepit tabung uji, penunu Bunsen, corong turas, rak tabung uji, silinder penyukat, tabung
uji
Dropper, spatula, test tube holder, Bunsen burner, filter funnel, test tube rack, measuring cylinder, test tube
9
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji tindak balas pertukaran ion Fe2+ kepada ion Fe3+ dan sebaliknya.
A Pertukaran ion ferum(II), Fe2+ kepada ion ferum(III), Fe3+
Changes of iron(II) ions, Fe2+ to iron(III) ions, Fe3+
Prosedur/ Procedure:
1. 2 cm3 larutan ferum(II) sulfat, FeSO4 0.5 mol dm–3 dituang ke dalam sebuah tabung uji.
2 cm3 of 0.5 mol dm–3 iron(II) sulphate, FeSO4 solution is measured and pour into a test tube.
2. Air bromin ditambah setitis demi setitis ke dalam tabung uji yang berisi larutan dengan menggunakan
penitis sambil digoncang sehingga tiada perubahan warna dapat diperhatikan.
Bromine water is added to the solution in the test tube drop by drop using a dropper until no further changes are
observed.
3. Campuran digoncang dan dihangatkan perlahan-lahan.
The mixture is shaken well and warm gently.
4. Larutan natrium hidroksida, NaOH 0.2 mol dm–3 ditambah setitis demi setitis sehingga berlebihan.
Pemerhatian direkodkan.
IA
0.2 mol dm–3 of sodium hydroxide, NaOH solution is added drop by drop until in excess. Observations are recorded.
B Pertukaran ion ferum(III), Fe3+ kepada ion ferum(II), Fe2+
Changes of iron(III) ions, Fe3+ to iron(II) ions, Fe2+
Prosedur/ Procedure:
1. 2 cm3 larutan ferum(III) klorida, FeCl3 dituang ke dalam sebuah tabung uji.
2 cm3 of iron(III) chloride, FeCl3 solution is measured and pour into a test tube.
2. 2 cm pita magnesium ditambah ke dalam larutan.
AS
2 cm of magnesium ribbon is added into the solution.
3. Campuran digoncang dan dihangatkan perlahan-lahan sehingga tiada perubahan diperhatikan.
The mixture is shaken well and warm gently until no further changes.
4. Campuran dituras ke dalam sebuah tabung uji.
The mixture is filtered into a test tube.
5. Larutan natrium hidroksida, NaOH 0.2 mol dm–3 ditambah setitis demi setitis sehingga berlebihan.
Pemerhatian direkodkan.
0.2 mol dm–3 of sodium hydroxide, NaOH solution is added drop by drop until in excess. Observations are recorded.
Mentafsir data/ Interpreting data:
Tindak balas/
Pemerhatian/ Observation
PA
N
Reaction
A
Larutan
hijau muda
Pale green
menjadi
solution turns
Inferens/ Inference
perang
. Ion
brown
.
Fe2+
.
Fe2+
ions are oxidised to
perang
Mendakan
tak larut dalam Ion Fe
larutan natrium hidroksida berlebihan.
Fe3+
3+
precipitate insoluble in excess
sodium hydroxide solution.
Larutan
perang
Brown
menjadi
solution turns
hijau muda
pale green
. Ion
.
Fe3+
diturunkan kepada ion
Fe2+
.
Fe3+
ions are reduced to
2+
Pale green
precipitate insoluble in excess
sodium hydroxide solution.
Perbincangan/ Discussion:
1. Berdasarkan pengoksidaan ion Fe2+ kepada Fe3+/ Based on oxidation of Fe2+ ions to Fe3+ions:
(a) Tuliskan persamaan setengah bagi:
Write the half-equations for:
Fe2+  Fe3+ + e–
Br2 + 2e–  2Br-
(ii) Penurunan/ Reduction:
10
ions
hadir.
hijau muda
Mendakan
tak larut dalam Ion Fe
hadir.
larutan natrium hidroksida berlebihan.
Fe2+ ions present.
(i) Pengoksidaan/ Oxidation:
Fe3+
ions present.
Brown
B
dioksidakan kepada ion
Fe3+
Fe2+
ions
(b) Tuliskan persamaan keseluruhan ion./ Write the overall ionic equation.
2Fe2+ + Br2  2Fe3+ + 2Br–
(c) Nyatakan perubahan nombor pengoksidaan bahan berikut dan peranannya.
State the change in the oxidation number of the following substance and its role.
Bahan
Perubahan nombor pengoksidaan
Ion ferum(II), Fe2+
+2  +3
Air bromin, Br2
0  –1
Peranan
Change of the oxidation number
Iron(II) ion, Fe2+
Bromine water, Br2
2. Berdasarkan penurunan ion Fe3+ kepada ion Fe2+:
Based on reduction of Fe3+ ion to Fe2+ ion:
Role
Agen penurunan
Reducing agent
Agen pengoksidaan
IA
Substance
Oxidising agent
(a) Tuliskan persamaan setengah bagi:/ Write the half-equation for:
Mg  Mg2+ + 2e–
(i) Pengoksidaan/ Oxidation:
Fe3+ + e–  Fe2+
AS
(ii) Penurunan/ Reduction:
(b) Tuliskan persamaan keseluruhan ion./ Write the overall ionic equation.
2Fe3+ + Mg  2Fe2+ + Mg2+
(c) Nyatakan perubahan nombor pengoksidaan bahan berikut dan peranannya.
State the change in the oxidation number of the following substance and its role.
Bahan
Perubahan nombor pengoksidaan
Substance
Magnesium, Mg
0  +2
Ion ferum(III), Fe3+
+3  +2
PA
N
Magnesium, Mg
Iron(III) ion, Fe3+
C
Peranan
Change of oxidation number
Role
Agen penurunan
Reducing agent
Agen pengoksidaan
Oxidising agent
Tindak Balas Penyesaran Sebagai Tindak Balas Redoks
Displacement Reactions as Redox Reactions
I
Penyesaran Logam daripada Larutan Garamnya/ Displacement of Metal from its Salt Solution
atas
1. Logam yang berada pada kedudukan yang lebih
dalam siri elektrokimia adalah lebih elektropositif.
A metal at a
higher
2. Logam ini merupakan agen
melepaskan
tinggi untuk
It is a stronger
reducing
position in the electrochemical series is more electropositive.
penurunan
yang lebih kuat kerana mempunyai kecenderungan yang lebih
elektron.
agent because it has a higher tendency to
Kecenderungan
logam untuk
melepaskan
elektron
Tendency of metal
to lose electron
Logam
Metal
Ion positif
Positive ions
Ca
Mg
Al
Zn
Fe
Sn
Pb
Cu
Ag
Ca2+ + 2e–
Mg2+ + 2e–
Al3+ + 3e–
Zn2+ + 2e–
Fe2+ + 2e–
Sn2+ + 2e–
Pb2+ + 2e–
Cu2+ + 2e–
Ag+ + e–
Rajah/Diagram 1.2
11
lose
electrons.
3. Logam yang lebih elektropositif boleh
garamnya.
menyesarkan
displace
A more electropositive metals can
logam yang kurang elektropositif daripada larutan
a less electropositive metal from its salt solutions.
Contoh/ Example:
Kepingan zink, Zn dimasukkan ke dalam larutan kuprum(II) sulfat, CuSO4.
A zinc, Zn strip is placed in a copper(II) sulphate, CuSO4 solution.
(a) Persamaan kimia/ Chemical equation: Zn + CuSO4  Cu + ZnSO4
(b) Persamaan ion/ Ionic equation: Zn + Cu2+  Zn2+ + Cu
(c) Zink dioksidakan kerana nombor pengoksidaan bertambah daripada 0
kepada +2
Kepingan zink, Zn
Zinc, Zn strip
Zinc is oxidised because the oxidation number increases from 0 to +2.
IA
(d) Ion Cu2+ diturunkan kerana nombor pengoksidaan berkurang daripada
+2 kepada 0
Cu2+ ion is reduced because the oxidation number decreases from +2 to 0
(e) Persamaan setengah pengoksidaan/ Oxidation half-equation:
Zn  Zn2+ + 2e–
(f) Persamaan
setengah
Cu2+ + 2e–  Cu
Reduction
Rajah/Diagram 1.3
half-equation:
AS
penurunan/
Contoh/Example 8
Kepingan kuprum, Cu dimasukkan ke dalam larutan argentum nitrat, AgNO3.
A copper, Cu strip is placed in a silver nitrate, AgNO3 solution.
(a) Persamaan kimia/ Chemical equation:
Kuprum
(c)
bertambah
dioksidakan
kerana
nombor
pengoksidaan
daripada 0 kepada +2.
increases
is oxidised because the oxidation number
PA
N
Copper
0 to +2.
(d) Ion Ag+
Ag ion is
+
diturunkan
reduced
Larutan argentum nitrat,
AgNO3
Silver nitrate, AgNO3
solution
because the oxidation number decreases from
(i) Pengoksidaan/ Oxidation –
Hilang
Terima
Lose
elektron/
Cu  Cu2+ + 2e–
Gain
elektron/
Ag+ + e–  Ag
AKTIVITI
Rajah/Diagram 1.4
from
+1 kepada 0
kerana nombor pengoksidaan berkurang daripada
(e) Persamaan setengah/ Half-equation:
(ii) Penurunan/ Reduction –
Kepingan kuprum, Cu
Copper, Cu strip
Cu + 2AgNO3  2Ag + Cu(NO3)2
Cu + 2Ag+  Cu2+ + 2Ag
(b) Persamaan ion/ Ionic equation:
+1 to 0
.
electrons
electrons
1.3
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji tindak balas penyesaran logam daripada larutan garamnya.
To investigate the displacement reaction of metals from their salt solution.
Bahan/ Materials:
Kepingan magnesium, kepingan zink, larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3
Magnesium strip, zinc strip, 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution
Radas/ Apparatus:
Tabung uji, rak tabung uji, silinder penyukat, kertas pasir
Test tubes, test tube rack, measuring cylinder, sand paper
12
Larutan kuprum(II)
sulfat, CuSO4
Copper(II) sulphate,
CuSO4 solution
.
Prosedur/ Procedure:
Pita magnesium, Mg
Magnesium, Mg ribbon
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate, CuSO4 solution
Rajah/Diagram 1.5
IA
1. Sekeping pita magnesium, Mg dibersihkan dengan kertas pasir.
A piece of magnesium, Mg ribbon is cleaned with sand paper.
2. 7 cm3 larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3 dimasukkan ke dalam tabung uji.
7 cm3 of 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution is poured into a test tube.
3. Pita magnesium, Mg dimasukkan ke dalam tabung uji.
The magnesium, Mg ribbon is added into the test tube.
4. Radas dibiarkan selama 20 minit dan pemerhatian direkodkan.
The apparatus is left for 20 minutes and the observations are recorded.
AS
5. Ulang langkah 1 hingga 4 dengan menggantikan pita magnesium, Mg dengan kepingan zink, Zn.
Repeat steps 1 until 4 by replacing magnesium, Mg ribbon with a zinc, Zn strip.
Keputusan/ Result:
Tindak balas/
Pemerhatian/ Observation
Reaction
Mg + CuSO4
biru
Larutan
tak berwarna
larutan
PA
N
solution turns
perang
Pepejal
A
Zn + CuSO4
brown
Larutan
The
colourless
blue
solution turns
perang
is
copper(II) sulphate
terenap.
bertukar menjadi
disesarkan daripada
kuprum(II) sulfat
Copper
.
tak berwarna
Copper
colourless
Kuprum
.
.
displaced
from
solution.
kuprum
Logam
solid is deposited.
biru
Pepejal
brown
A
Kuprum
bertukar menjadi
.
blue
The
Inferens/ Inference
terhasil.
metal formed.
disesarkan daripada
kuprum(II) sulfat
. larutan
.
Copper
is displaced from
copper(II) sulphate
solution.
Logam
terenap.
solid is deposited.
kuprum
Copper
terhasil.
metal formed.
Mentafsir data/ Interpreting data:
1. Berdasarkan tindak balas Mg + CuSO4:/ Based on the reaction of Mg + CuSO4:
(a) Terangkan tindak balas penyesaran yang berlaku./ Explain the displacement reaction occurred.
Logam magnesium, Mg lebih elektropositif berbanding logam kuprum, Cu. Logam magnesium, Mg boleh
menyesarkan kuprum, Cu daripada larutan kuprum(II) sulfat, CuSO4
Magnesium, Mg metal is more electropositive than copper, Cu metal. Magnesium, Mg metal can displace copper,
Cu from copper(II) sulphate, CuSO4 solution.
(b) Tuliskan persamaan setengah bagi:/ Write the half-equation for:
Mg  Mg2+ + 2e–
(i) Pengoksidaan/ Oxidation:
Cu2+ + 2e–  Cu
(ii) Penurunan/ Reduction:
13
(c) Tuliskan persamaan keseluruhan ion./ Write the overall ionic equation.
Mg + Cu2+  Mg2+ + Cu
(d) Nyatakan perubahan nombor pengoksidaan bahan berikut dan peranannya.
State the change in the oxidation number of the following substance and its role.
Bahan
Perubahan nombor pengoksidaan
Magnesium, Mg
0  +2
Ion kuprum(II), Cu2+
+2  0
Substance
Peranan
Change of the oxidation number
Magnesium. Mg
Agen penurunan
Reducing agent
Agen pengoksidaan
Oxidising agent
IA
Copper(II) ion, Cu2+
Role
2. Berdasarkan tindak balas Zn + CuSO4:/Based on the reaction of Zn + CuSO4:
(a) Terangkan tindak balas penyesaran yang berlaku./ Explain the displacement reaction occurred.
Logam zink, Zn lebih elektropositif berbanding logam kuprum, Cu. Logam zink, Zn boleh menyesarkan
kuprum, Cu daripada larutan kuprum(II) sulfat, CuSO4
Zinc, Zn metal is more electropositive compare to copper, Cu metal. Zinc, Zn metal can displace copper, Cu from
AS
copper(II) sulphate, CuSO4 solution.
(b) Tuliskan persamaan setengah bagi:/ Write the half-equation for:
Zn  Zn2+ + 2e–
(i) Pengoksidaan/ Oxidation:
Cu2+ + 2e–  Cu
(ii) Penurunan/ Reduction:
(b) Tuliskan persamaan keseluruhan ion./ Write the overall ionic equation.
Zn + Cu2+  Zn2+ + Cu
PA
N
(c) Lengkapkan jadual berikut/ Complete the following table:
Bahan
Perubahan nombor pengoksidaan
Zink, Zn
0  +2
Ion kuprum(II), Cu2+
+2  0
Substance
Peranan
Change of oxidation number
Zink, Zn
Copper(II) ion, Cu2+
Role
Agen penurunan
Reducing agent
Agen pengoksidaan
Oxidising agent
Perbincangan/ Discussion:
lebih elektropositif
1. Logam yang
larutan garamnya.
A
more electropositive
boleh menyesarkan logam yang
metal can displace a
less electropositive
2. Logam yang lebih elektropositif mengalami proses
penurunan
.
The more electropositive metal undergoes
oxidation
sebagai agen
pengoksidaan
dan bertindak sebagai agen
process and acts as the
.
reduction
The less electropositive metal (metal ions) undergoes
agent.
14
daripada
metal from its salt solution.
3. Logam yang kurang elektropositif (ion logam) mengalami proses
pengoksidaan
kurang elektropositif
reducing
penurunan
process and acts as the
agent.
dan bertindak
oxidising
II Penyesaran Halogen daripada Halidanya/ Displacement of Halogen from its Halide Solution
1. Unsur-unsur Kumpulan 17 dalam Jadual Berkala Unsur dikenali sebagai halogen. Contoh-contoh halogen
ialah fluorin, klorin, bromin dan iodin.
The elements in Group 17 of the Periodic Table Elements are called the halogens.. Examples of halogens are fluorine,
chlorine, bromine and iodine.
pengoksidaan
yang kuat
2. Semua atom halogen mempunyai 7 elektron valens. Halogen merupakan agen
kerana atom halogen cenderung menerima satu elektron untuk mencapai susunan elektron oktet yang stabil.
All halogen atoms have 7 valence electrons. As a result, halogens are strong
accept an electron to achieve a stable octet electron arrangement.
oxidising
agents because they tend to
IA
lebih tinggi
dalam kumpulan adalah lebih reaktif dan boleh
3. Halogen yang berada pada kedudukan
menyesarkan
halogen yang berada pada kedudukan yang lebih rendah daripada larutan halidanya.
higher
Halogens that located
in the group are more reactive and can
lower in the group from their halide solutions.
berkurang
4. Kuasa halogen sebagai agen pengoksidaan
the halogens located
apabila menuruni kumpulan.
down the group.
AS
The strength of halogen as oxidising agents
decreases
displace
Cl2
Br2
I2
Kekuatan sebagai agen pengoksidaan berkurang
Strength as an oxidising agent decreases
Rajah/Diagram 1.6
Tip
PA
N
stra
Ek
Ekstra
Halogen
Warna dalam larutan akueus
Warna dalam larutan 1,1,1-trikloroetana
Halogen
Colour in the aqueous solution
Colour in 1,1,1-trichloroethane solution
Klorin, Cl2
Kuning kehijauan
Greenish yellow
Tanpa warna
Colourless
Bromin, Br2
Perang
Brown
Perang
Brown
Iodin, I2
Perang
Brown
Ungu
Purple
Chlorine, Cl2
Bromine, Br2
Iodine, I2
AKTIVITI
1.4
To investigate oxidation and reduction in the displacement of halogen from its halide solution.
Bahan/ Materials:
Air klorin, air bromin, larutan iodin, larutan kalium bromida, KBr 0.5 mol dm–3, larutan kalium iodida, KI 0.5
mol dm–3, larutan 1,1,1-trikloroetana
Chlorine water, bromine water, iodine solution, 0.5 mol dm–3 potassium bromide, KBr solution, 0.5 mol dm–3 potassium
iodide, KI solution, 1, 1, 1-trichloroethane solution
Radas/ Apparatus:
Tabung uji, rak tabung uji
Test tubes, test tube rack
15
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji pengoksidaan dan penurunan dalam tindak balas penyesaran halogen daripada larutan halidanya.
Prosedur/ Procedure:
Air klorin
Chlorine water
Larutan kalium bromida, KBr
Potassium bromide, KBr solution
Rajah/Diagram 1.7
1. Kira-kira 2 cm larutan kalium bromida, KBr 0.5 mol dm–3 dituang ke dalam sebuah tabung uji.
3
About 2 cm3 of 0.5 mol dm–3 potassium bromide, KBr solution is poured into a test tube.
IA
2. Dengan menggunakan penitis, kira-kira 2 cm3 air klorin ditambah setitis demi setitis ke dalam tabung uji
yang berisi larutan kalium bromida, KBr.
By using a dropper, about 2 cm3 of chlorine water is added drop by drop to potassium bromide, KBr solution in the
test tube.
3. Campuran digoncangkan.
The mixture is shaken well.
4. Kira-kira 2 cm3 larutan 1,1,1-trikloroetana ditambah kepada campuran dan campuran digoncangkan.
About 2 cm3 of 1, 1, 1-trichloroethane solution is added to the mixture and the mixture is shaken well.
5. Warna lapisan larutan akueus dan larutan 1,1,1-trikloroetana diperhatikan dan direkodkan.
The colour of the layers of aqueous solution and the 1, 1, 1-trichloroethane solution are observed and recorded.
AS
6. Langkah 1 hingga 5 diulang dengan menggunakan halogen dan larutan halida yang ditunjukkan dalam
jadual di bawah.
Steps 1 to 5 are repeated using halogens and halide solution as shown in the table below.
Pemerhatian/ Observation:
Warna/Colour of
Tindak balas
Reaction
larutan akueus
larutan 1, 1, 1-trikloroetana
Perang
Perang
aqueous solution
Air klorin +
kalium bromida
1,1,1-trichloroethane solution
Brown
PA
N
Chlorine water +
potassium bromide
Air bromin +
kalium iodida
Bromine water +
potassium iodide
Larutan iodin +
kalium bromida
lodine solution +
potassium bromide
Inferens
Inference
Perang kemerahan
Reddish brown
Perang kemerahan
Reddish brown
Brown
Bromin hadir
Bromine is present
Bromin disesarkan oleh klorin
Bromine is displaced by chlorine
Ungu
Purple
lodin hadir
lodine is present
lodin disesarkan oleh bromin
Iodine is displaced by bromine
lodin hadir
Ungu
Purple
lodine is present
Tiada tindak balas. Iodin tidak
dapat menyesarkan bromin
No reaction. Iodine cannot displace
bromine
Mentafsir data/ Interpreting data:
Pengoksidaan
Oxidation
Penurunan
Reduction
Persamaan setengah
2Br–(ak/aq)  Br2(ak/aq) + 2e–
Pemerhation
Larutan kuning kehijauan menjadi tidak
Larutan tidak berwarna menjadi perang berwarna
Half-equation
Observation
Persamaan keseluruhan ion
Overall ionic equation
The colourless solution turns brown
Cl2(ak/aq) + 2e–  2Cl–(ak/aq)
The greenish-yellow solution becomes
colourless
Cl2(ak/aq) + 2Br–(ak/aq)  2Cl–(ak/aq) + Br2(ak/aq)
16
Agen penurunan
Kalium bromida
Reducing agent
Potassium bromide
Agen pengoksidaan
Air klorin
Oxidising agent
Chlorine water
Pengoksidaan
Penurunan
2I–(ak/aq)  I2(ak/aq) + 2e–
Br2(ak/aq) + 2e–  2Br–(ak/aq)
Oxidation
Persamaan setengah
Pemerhation
IA
Half-equation
Larutan tidak berwarna menjadi perang Larutan perang menjadi tidak berwarna
Observation
The colourless solution turns brown
Persamaan keseluruhan ion
Agen penurunan
The brown solution turns colourless
Br2(ak/aq) + 2I–(ak/aq)  2Br–(ak/aq) + I2(ak/
(ak/aq)
Overall ionic equation
AS
Kalium iodida
Reducing agent
Potassium iodide
Agen pengoksidaan
Oxidising agent
Kesimpulan/ Conclussion:
bromin dan iodin
1. Air klorin menyesarkan
menyesarkan
Reduction
iodin
Air bromin
Bromine water
daripada larutan halidanya. Air bromin hanya dapat
daripada larutan halidanya.
bromine and iodine
PA
N
Chlorine water displaces
iodine
can only displace
from their respective halide solution. Bromine water
from its halide solution.
2. Halogen yang lebih reaktif dapat menyesarkan
A more reactive halogen can displace
halogen yang kurang reaktif daripada larutan halidanya.
a less reactive halogen from its halide solution
.
.
Contoh/Example 9
Air klorin boleh menyesarkan
ion iodida, I–
manakala
iodin
daripada larutan kalium iodida, KI. Klorin ialah agen
pengoksidaan
mengalami pengoksidaan.
iodine
Chlorine water can displace
iodide ions
undergo oxidation.
from potassium iodide, KI solution. Chlorine is an
(a) Persamaan kimia/Chemical equation: Cl2 + 2KI  2KCl + I2
Cl2 + 2I–  2Cl– + I2
(b) Persamaan keseluruhan ion/ Overall ionic equation:
(c) Persamaan setengah/ Half-equation:
(i) Penurunan/ Reduction –
(ii) Pengoksidaan/ Oxidation –
Terima
Hilang
Gain
elektron/
Cl2 + 2e–  2Cl–
elektron/
–
2I  I2 + 2e–
17
oxidising
Lose
electrons
electrons
agent while
Contoh/Example 10
Air klorin boleh menyesarkan bromin daripada larutan kalium bromida, KBr. Klorin ialah agen
pengoksidaan
manakala ion bromida, Br– mengalami
.
oxidising
Chlorine water can displace bromine from potassium bromide, KBr solution. Chlorine is an
oxidation
.
ions, Br– undergo
(a) Persamaan kimia/Chemical equation: Cl2 + 2KBr  2KCl + Br2
(ii)
– Terima elektron/ Gain electrons
–
Pengoksidaan
/
IA
Cl2 + 2e  2Cl
–
Oxidation
– – Hilang elektron/ Lose electrons
2Br  Br2 + 2e
–
agent while bromide
Cl2 + 2Br–  2Cl– + Br2
(b) Persamaan keseluruhan ion/ Overall ionic equation:
(c) Persamaan setengah/ Half-equation:
Penurunan
Reduction
/
(i)
pengoksidaan
–
Contoh/Example 11
AS
Air bromin boleh menyesarkan iodin daripada larutan kalium iodida, KI. Bromin ialah agen
pengoksidaan
ion iodida, I– mengalami
.
oxidising
Bromine water can displace iodine from potassium iodide, KI solution. Bromine is an
oxidation
undergo
.
pengoksidaan
manakala
agent while iodide ions, I–
(a) Persamaan kimia/ Chemical equation: Br2 + 2KI  2KBr + I2
Br2 + 2I–  2Br– + I2
(b) Persamaan keseluruhan ion/ Overall ionic equation:
(c) Persamaan setengah/ Half-equation:
Terima
Gain
elektron/
Br2 + 2e–  2Br–
PA
N
(i) Penurunan/ Reduction –
(ii) Pengoksidaan/ Oxidation –
Uji Kendiri
Hilang
elektron/
2I–  I2 + 2e
electrons
Lose
electrons
1.1
1. Nyatakan definisi bagi tindak balas redoks.
State the definition of a redox reaction.
Tindak balas kimia yang melibatkan pengoksidaan dan penurunan yang berlaku secara serentak.
Chemical reactions involving oxidation and reduction occuring simultaneously.
2. Rajah 1.8 menunjukkan susunan radas bagi mengkaji pengoksidaan dan penurunan berdasarkan pemindahan elektron
pada satu jarak.
Diagram 1.8 shows the apparatus set-up to investigate the oxidation and reduction based on the transfer of electron at a distance.
G
Karbon P
Carbon P
Karbon Q
Carbon Q
Larutan kalium
iodida
Potassium iodide
solution
Air bromin
Bromine water
Asid sulfurik
Sulphuric acid
Rajah/Diagram 1.8
18
(a) Tuliskan persamaan setengah bagi:/ Write the half-equation for:
-
-
2I  I2 + 2e
(i) Pengoksidaan/ Oxidation:
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
(ii) Penurunan/ Reduction:
(b) Tuliskan persamaan keseluruhan ion./ Write the overall ionic equation.
10I– + 2MnO4– + 16H+ → 5I2 + 2Mn2+ + 8H2O
(d) Lengkapkan jadual berikut/ Complete the following table:
Perubahan nombor pengoksidaan
Bahan
Substance
Peranan
Role
IA
Change of the oxidation number
Iodin
Agen penurunan
-1  0
Iodine
Mangan
Reduction agent
Agen pengoksidaan
+7  +2
Manganese
Oxidation
O
xidation agent
3. Hitung nombor pengoksidaan bagi mangan, Mn dan namakan sebatian mengikut tatanama IUPAC.
KMnO4
+1 + x + 4(–2) = 0
x = +7
AS
Calculate the oxidation number for manganese, Mn and name the compound according to the IUPAC nomenclature.
x + 2(–2) = 0
x = +4
Mangan(IV) oksida
PA
N
Kalium manganat(VII)
MnO
Mn
O2
Potassium manganate(VII)
Manganese(IV) oxide
Mn2O3
2x
2 + 3(–2) = 0,
2x
2 = +6
x = +3
Mangan(III) oksida
Manganese(III) oxide
4. Rajah 1.9 menunjukkan satu tindak balas penyesaran.
Diagram 1.9 shows a displacement reaction.
Serbuk logam
Metal powder
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate, CuSO4 solution
Rajah/Diagram 1.9
Namakan logam yang dapat menyebabkan keamatan warna biru larutan kuprum(II) sulfat, CuSO4 berkurang. Jelaskan
tindak balas penyesaran tersebut.
Name the metal that can cause the intensity of blue colour of copper(II) sulphate, CuSO4 solution to decrease. Explain the
displacement reaction.
Magnesium atau zink. Magnesium atau zink lebih elektropositif berbanding kuprum. Magnesium atau zink akan
menyesarkan kuprum daripada larutan kuprum(II) sulfat, CuSO4.
Magnesium or zinc. Magnesium or zinc are more electropositive than copper. Magnesium or zinc will displace copper from
copper(II) sulphate, CuSO4 solution.
19
1.2
A
Keupayaan Elektrod Piawai
Standard Electrode Potential
Keupayaan Elektrod Piawai
Standard Electrode Potential
1. Beza keupayaan merentasi terminal sel ketika arus sifar (tiada beban) akan bertindak sebagai penggerak atau
litar luar
. Beza keupayaan ini disebut sebagai daya
“tekanan elektrik” yang menolak elektron melalui
gerak elektrik, d.g.e. atau daya keupayaan sel, Esel.
The potential difference across the cell terminals when zero current (no load) will act as a mover or “electric pressure” that
pushes electrons through the external circuit . This potential difference is referred as electromotive force, emf or cell potential
force, Ecell.
keupayaan elektrod piawai
IA
2. Nilai d.g.e sel yang diukur pada keadaan piawai disebut sebagai
The emf value of a cell measured at standard conditions is referred as
standard electrode potential
, E0sel
, E0cell
3. Keadaan piawai bagi sel/ Standard condition for cells:
(a) kepekatan akueus ion 1.0 mol dm-3/ aqueous concentration of ions 1.0 mol dm-3
(b) suhu 25 °C atau 298 K/ temperature 25 °C or 298 K
(c) tekanan 1 atm atau 101 kPa/ pressure of 1 atm or 101 kPa
(d) platinum digunakan sebagai elektrod lengai/ platinum is used as inert electrode
AS
4. Dalam elektrokimia, keupayaan elektrod piawai, E0 ditakrifkan sebagai ukuran keupayaan elektrod , berbanding
sel hidrogen
platinum
dengan elektrod
pada keadaan piawai dengan kepekatan akueus ion
–3
1.0 mol dm
1 atm
25
°C
, suhu
.
dan tekanan
electrode potential
In electrochemistry, standard electrode potential, E0 is defined as the measure of
, compared
hydrogen cell
platinum
electrode at standard conditions with aqueous concentration of ions at
with
to
–3
1.0 mol dm
1 atm
25
°C
°
C
and pressure of
, temperature
.
5. Tenaga keupayaan elektron dalam setiap setengah-sel adalah berbeza.
The electron potential energy in each half-cell is different.
PA
N
Contoh/Example:
Dalam sel Daniell:/ In Daniell’s cell:
(a) Elektron dalam zink mempunyai tenaga keupayaan yang lebih tinggi berbanding kuprum. Maka, elektron akan
bergerak dari zink ke kuprum.
Electrons in zinc have a higher potential energy than copper. Thus, electrons will move from zinc to copper.
(b) Perbezaan tenaga keupayaan di antara dua elektrod menghasilkan d.g.e sel yang boleh diukur dalam unit Volt.
The potential energy difference between the two electrodes produces a measurable value of a cell emf in Volt units.
H2 pada 298K dan 1 atm
H2 at 298K and 1 atm
Voltmeter
Voltmeter
V
Titian garam
Salt bridge
Elektrod platinum
Platinum electrode
Jalur zink
Zink strip
Larutan asid, H+ 1.0 mol dm–3
1.0 mol dm–3 H+, acid solution
Larutan Zn2+ 1.0 mol dm–3
1.0 mol dm–3 Zn2+ solution
Rajah/Diagram 1.10
B
Agen Pengoksidaan dan Agen Penurunan Berdasarkan Nilai Keupayaan Elektrod Piawai
Oxidising Agent and Reducing Agent Based on Standard Electrode Potential Values
1. Siri elektrokimia ialah satu siri unsur yang disusun mengikut nilai
Electrochemical series is a series of elements arranged according to
20
keupayaan elektrod piawai
standard electrode potential
.
values.
2. Elektrod hidrogen dianggap mempunyai nilai keupayaan elektrod piawai
penurunan
ditakrif dalam bentuk
.
zero
Hydrogen electrode is assumed to have
reduction
.
the form of
Agen Pengoksidaan
Oxidising agent
Agen penurunan
Reducing agent
S2O8 (ak/aq) + 2e
–
MnO4–(ak) + 8H+(ak/aq) + 5e–
Cl2(g/g) + 2e–
Cr2O7 (ak/aq) + 14H (ak/aq) + 6e
+
–
MnO2(p/s) + 4H+(ak/aq) + 2e–
O2(g/g) + 4H+(ak/aq) + 4e–
Br2(ce/l) + 2e
–
2F–(ak/aq)
Agen penurunan lemah
Weak reducing agent
Nilai E0 (V)
E0 value (V)
+2.87
2SO42–(ak/aq)
+2.01
Mn2+(ak/aq) + 4H2O(ce/l)
l
l)
+1.51
2Cl–(ak/aq)
+1.36
2Cr3+(ak/aq)) + 7H2O(ce/
O(ce/l)
+1.33
Mn2+(ak/aq)) + 2H2O(ce/l)
+1.23
2H2O(ce/l)
l
l)
+1.23
2Br (ce/l)
l
l)
+1.07
–
IA
F2(g/g) + 2e–
2–
2NO2(g/g) + 2H2O(ce/
O(ce/l)
O(ce/l
l)
l)
+0.80
Ag+(ak/aq) + e–
Ag(p/s)
Ag(p/
+0.80
Fe3+(ak/aq) + e–
Fe2+(ak/aq)
aq)
aq)
+0.77
2I–(ak/
(ak/aq)
(ak/aq
aq)
aq)
+0.53
4OH–(ak/
(ak/aq)
(ak/aq
aq)
aq)
+0.40
Cu(p/s)
Cu(p/s
Cu(p/
s))
+0.34
AS
2NO3–(ak/aq) + 4H+(ak/aq) + 2e–
I2(p/s) + 2e
–
O2(g/g) + 2H2O(ce/l)) + 4e–
Cu (ak/aq)) + 2e
2+
–
aq) + 2e–
aq
SO42–(ak/aq) + 4H+(ak/aq)
SO2(g/
(g/g)
(g/g
g)) + 2H2O(ce/
g
O(ce/l)
O(ce/l
l)
l)
+0.20
(ak/aq) + e–
Cu2+(ak/
Cu+(ak/
(ak/aq)
(ak/aq
aq)
aq)
+0.15
2H+(ak/aq)) + 2e–
H2(g/
(g/g)
(g/g
g)
g)
0.00
Pb2+(ak/aq)) + 2e–
Pb(p/s)
Pb(p/s
Pb(p/
s))
–0.13
Sn2+(ak/aq)
aq)) + 2e–
aq
Sn(p/s)
Sn(p/s
Sn(p/
s))
–0.14
(ak/aq)
(ak/aq
aq)) + 2e
aq
Ni (ak/
Ni(p/s)
Ni(p/s
Ni(p/
s))
–0.26
(ak/aq)
(ak/aq
aq)) + 2e–
aq
Fe2+(ak/
Fe(p/s)
Fe(p/s
Fe(p/
s))
–0.44
(ak/aq)
(ak/aq
aq)) + 2e–
aq
Zn2+(ak/
Zn(p/s)
Zn(p/s
Zn(p/
s))
–0.76
H2(g/
(g/g) + 2OH–(ak/aq
(ak/aq)
–0.83
–
PA
N
2+
O(ce/l)
O(ce/l
ll)) + 2e–
2H2O(ce/
Mn (ak/
(ak/aq)
(ak/aq
aq)) + 2e
aq
Mn(p/s)
Mn(p/
–1.19
(ak/aq)
(ak/aq
aq)) + 3e–
aq
Al3+(ak/
Al(p/s)
–1.66
(ak/aq)
(ak/aq
aq)) + 2e–
aq
Mg2+(ak/
2+
–
Mg(p/s)
s)
–2.37
–
(ak/aq)
(ak/aq
aq)) + e
aq
Na (ak/
Na(p/s)
Na(p/
–2.71
(ak/aq)
(ak/aq
aq)) + e–
aq
Ca2+(ak/
Ca(p/s)
Ca(p/
–2.87
(ak/aq) + e–
K+(ak/
K(p/s)
K(p/
–2.93
Li (ak/aq) + e
Li(p/s)
–3.05
+
+
–
. Keupayaan elektrod
standard electrode potential value. Electrode potential is defined in
Jadual Keupayaan Elektrod Piawai, E0
Table of Standard Electrode Potential, E0
Agen pengoksidaan kuat
Strong oxidising agent
2–
sifar
Semakin positif
nilai E0, semakin
mudah untuk spesies
kimia mengalami
penurunan.
penurunan
More positif E0
value, easier for the
chemical species to
undergo reduction
reduction.
Semakin negatif
nilai E0, semakin
mudah untuk spesies
kimia mengalami
pengoksidaan.
More negative E0
value, easier for the
chemical species to
undergo oxidation.
Agen penurunan lemah
Weak reducing agent
Agen pengoksidaan kuat
Strong oxidising agent
Rajah/Diagram 1.11
3. Semakin positif nilai keupayaan elektrod piawai, E0, semakin mudah untuk spesies kimia mengalami
penurunan
pengoksidaan
, iaitu sebagai agen
yang kuat.
The more positive the standard electrode potential, E0 value, the easier the chemical species to undergo
oxidising
agent.
is as a strong
reduction
, that
4. Semakin negatif nilai keupayaan elektrod piawai, E0, semakin mudah untuk spesies kimia mengalami
pengoksidaan
penurunan
, iaitu sebagai agen
yang kuat.
The more negative the standard electrode potential, E0 value, the easier the chemical species to undergo
reducing
agent.
is as a strong
5. Rumus nilai daya keupayaan sel ialah:
The formula for the value of cell's potential force:
Voltan sel/ Cell voltage, E0sel/cell = E0katod/cathode – E0anod/anode
21
oxidation
, that
Contoh/Example 1
Lengkapkan jadual bagi pasangan elektrod berikut.
Complete the table for the following pair of electrodes.
Nilai Eo (V)
Pair of electrodes
E0 value (V)
Elektrod magnesium, Mg dalam larutan
magnesium sulfat, MgSO4 dan elektrod
kuprum, Cu dalam larutan kuprum(II)
sulfat, CuSO4.
Magnesium, Mg electrode in magnesium
sulphate, MgSO4 solution and copper, Cu
electrode in copper(II) sulphate, CuSO4
solution.
Copper, Cu electrode in copper(II) sulphate,
CuSO4 solution and argentum, Ag electrode in
argentum nitrate, AgNO3 solution.
Uji Kendiri
1.2
Negative
terminal
Terminal
positif
Positive
terminal
Mg2+ │ Mg
= –2.37 V
Cu
Cu
= +0.34 V
2+ │
Cu2+ │ Cu
= +0.34 V
Mg
Cu
AS
Elektrod kuprum, Cu dalam larutan
kuprum(II) sulfat, CuSO4 dan elektrod
argentum, Ag dalam larutan argentum
nitrat, AgNO3.
Terminal
negatif
Agen
pengoksidaan
Oxidising agent
Ion Cu2+
IA
Pasangan elektrod
Cu
Ag Ag
= +0.80 V
+│
Ag
Cu2+ ion
Ion Ag+
Ag+ ion
Agen
penurunan
Reducting
agent
Mg
Cu
PA
N
1. Merujuk pada Rajah 1.11, hitungkan nilai keupayaan sel, E0 bagi elektrolisis zink, Zn dan kuprum, Cu menggunakan
larutan kuprum(II) sulfat, CuSO4.
Referring to Diagram 1.11, calculate the cell potential value, E0 for the electrolysis of zinc, Zn and copper, Cu using copper(II)
sulphate, CuSO4 solution.
Zn2+ (ak/aq) + 2e–  Zn (p/s)
E0 = –0.76 V
Cu2+ (ak/aq) + 2e–  Cu (p/s)
E0 = +0.34 V
E0sel/cell = E0 katod/cathode – E0 anod/anode
= +0.34 – (–0.76)
= 1.10 V
2. Merujuk nilai E0 pada Rajah 1.11, kirakan nilai keupayaan sel, E0 bagi tindak balas berikut.
Referring to Diagram 1.11, calculate the cell potential value, E0 of the following reaction.
2Ag+ + Cd  2Ag + Cd2+
Ag+(ak/aq) + 2e–  Ag(p/s)
E0 = +0.80 V
Cd2+(ak/aq) + 2e– → Cd(p/s)
E0 = –0.40 V
E0sel/cell = E0 katod/cathode – E0 anod/anode
= +0.80 – (–0.40)
= 1.20 V
22
1.3
A
Sel Kimia
Voltaic Cell
Tindak Balas Redoks dalam Sel Kimia
Redox Reaction in Voltaic Cells
logam
berlainan yang dicelup ke dalam suatu elektrolit
1. Sebuah sel kimia ringkas terdiri daripada dua
galvanometer/
voltmeter
dan disambung kepada
yang mencatatkan bacaan.
which are immersed into an electrolyte and connected to the
IA
metals
A simple voltaic cell consists of two different
galvanometer/ voltmeter
that records a reading.
2. Beza keupayaan antara dua elektrod menyebabkan pengerakan elektron yang menghasilkan arus elektrik.
Potential difterence between two electrodes causes the movement of electrons that produces electric current.
Contoh/Example 1
Elektrod zink dan elektrod kuprum dalam asid
sulfurik cair, H2SO41.0 mol dm–3.
AS
V
Zinc and copper electrodes in 1.0 mol dm–3 dilute
sulphuric acid, H2SO4.
Elektrod kuprum
Copper electrode
Elektrod zink
Zinc electrode
Rajah/Diagram 1.12
Asid sulfurik cair,
H2SO4 1.0 mol dm–3
1.0 mol dm–3 dilute
sulphuric acid, H2SO4
PA
N
(a) Tentukan terminal negatif (anod) dan Terminal negatif/ Negative terminal:
terminal positif (katod). Nyatakan alasan.
Zink. Agen penurunan yang lebih kuat sebagai sumber elektron.
Determine the negative terminal (anode) and Zinc. Stronger reducing agent as the source of electrons.
positive terminal (cathode). State the reason.
Terminal positif/ Positive terminal:
Kuprum/ Copper
(b) Persamaan setengah bagi tindak balas di Terminal negatif/ Negative terminal:
terminal negatif dan terminal positif
Zn  Zn2+ + 2e–
(Pengoksidaan/ Oxidation)
Half-equations for reactions at the negative and
Terminal
positif/
Positive terminal:
positive terminals
2H+ + 2e–  H2
(Penurunan/ Reduction)
(c) Hasil yang terbentuk di terminal negatif Terminal negatif/ Negative terminal:
dan terminal positif
Ion zink/ Zinc ions
Products formed at the negative and positive
Terminal positif/ Positive terminal:
terminals
Gas hidrogen/ Hydrogen gas
(d) Pemerhatian di
terminal positif
terminal
negatif
dan Terminal negatif/ Negative terminal:
Elektrod zink menipis/ Zinc electrode becomes thinner
Observations at the negative and positive Terminal positif/ Positive terminal:
terminals
Gelembung gas tidak berwarna/ Colourless gas bubbles
(e) Arah pengaliran elektron dalam litar luar
Direction of electron flow in the outer circuit
(f) E0sel/cell = E0 katod/cathode – E0 anod/anode
zink
Daripada
zinc
From
kepada
to
Nilai voltan/ Voltage value:
0.34 – (-0.76) = 1.1 V
23
copper
kuprum
Contoh/Example 2
Elektrod magnesium dan elektrod ferum dalam asid
sulfurik cair, H2SO4 1.0 mol dm–3.
V
Magnesium and iron electrodes in 1.0 mol dm–3 dilute sulphuric
acid, H2SO4.
Elektrod ferum
Iron electrode
Elektrod magnesium
Magnesium electrode
Asid sulfurik cair,
H2SO4 1.0 mol dm–3
1.0 mol dm–3 dilute
acid, H2SO4
sulphuric acid
IA
Rajah/Diagram 1.13
(a) Tentukan terminal negatif (anod) dan positif Terminal negatif/ Negative terminal:
(katod). Nyatakan alasan.
Magnesium. Agen penurunan yang lebih kuat
Determine the negative terminal (anode) and the
Magnesium. Stronger reducing agent
positive terminal (cathode). State the reason.
Terminal positif/ Positive terminal:
Ferum/ Iron
AS
(b) Persamaan setengah bagi tindak balas di Terminal negatif/ Negative terminal:
terminal negatif dan terminal positif
Mg  Mg2+ + 2e–
(Pengoksidaan/ Oxidation)
Half-equations for reactions at the negative and
positive terminals
Terminal positif/ Positive terminal:
2H+ + 2e–  H2
(Penurunan/ Reduction)
(c) Hasil yang terbentuk di terminal negatif Terminal negatif/ Negative terminal:
dan terminal positif
Ion magnesium/ Magnesium ion
Products formed at the negative and positive
Terminal positif/ Positive terminal:
terminals
PA
N
Gas hidrogen/ Hydrogen gas
(d) Pemerhatian di
terminal positif
terminal
negatif
dan Terminal negatif/ Negative terminal:
Elektrod magnesium menipis
Observations at the negative and positive
Magnesium electrode becomes thinner
terminals
Positif/ Positif:
Gelembung gas tidak berwarna
Colourless gas bubbles
(e) Arah pengaliran elektron dalam litar luar
Direction of electron flow in the outer circuit
magnesium
Daripada
From
magnesium
to
kepada
ferum
iron
Nilai voltan / Voltage value:
-0.44 – (-2.37) = 1.93 V
(f) E0sel/cell = E0 katod/cathode – E0 anod/anode
zink
zink sulfat
3. Dalam sel Daniell, elektrod
direndam ke dalam larutan
, manakala
kuprum
kuprum
(II)
sulfat
elektrod
direndam ke dalam larutan
. Dua larutan itu
pasu
berliang
dipisahkan dengan
atau disambung dengan titian garam.
In the Daniell cell,
zinc
electrode is immersed into
or connected by a salt bridge.
zinc sulphate
electrode is immersed into
solution while
copper (II) sulphate
solution. The two solutions are separated by a
24
copper
porous pot
Voltmeter
Voltmeter
Suis
Switch
E0 = +0.34 V
Tip
V
Elektrod kuprum
Copper electrode
Elektrod zink
Zinc electrode
Fungsi pasu berliang ialah untuk
melengkapkan
litar
dengan
membenarkan ion-ion melaluinya.
Tip SPM
E0 = –0.76 V
The function of the porous pot is to
complete the circuit by allowing ions
to pass through it.
Larutan zink sulfat, ZnSO4
Zinc sulphate, ZnSO4
solution
Larutan kuprum(II)
sulfat, CuSO4
Copper(II) sulphate,
CuSO4 solution
SPM
Pasu berliang
Porous pot
IA
Rajah/Diagram 1.14
(a) Tentukan terminal negatif (anod) dan Terminal negatif/ Negative terminal:
terminal positif (katod). Nyatakan alasan.
Elektrod zink. Agen penurunan yang lebih kuat.
Determine the negative terminal (anode) and Zinc electrode. Stronger reducing agent
positive (cathode). State the reason.
Terminal positif/ Positive terminal:
Elektrod kuprum/ Copper electrode
PA
N
AS
(b) Persamaan setengah bagi tindak balas di Terminal negatif/ Negative terminal:
terminal negatif dan terminal positif
Zn  Zn2+ + 2e–
(Pengoksidaan/ Oxidation)
Half-equations for reactions at the negative and Terminal positif/ Positive terminal:
positive terminals
Cu2+ + 2e–  Cu
(Penurunan/ Reduction)
(c) Hasil yang terbentuk di terminal negatif Terminal negatif/ Negative terminal:
dan terminal positif
Ion zink/ Zinc ion
Products formed at the negative and positive Terminal positif/ Positive terminal:
terminals
Logam kuprum/ Copper metal
(d) Pemerhatian di terminal negatif dan Terminal negatif/ Negative terminal:
terminal positif
Elektrod zink menjadi semakin nipis/ Zinc electrode becomes thinner.
Observations at the negative and positive Terminal positif/ Positive terminal:
terminals.
Elektrod kuprum menebal/ Copper electrode becomes thicker.
(e) Arah pengaliran elektron dalam litar luar
zink
kuprum
kepada
Daripada
Direction of electron flow in the outer circuit
From
(f) Perubahan warna elektrolit
Change in the colour of electrolyte
zinc
copper
to
Keamatan warna biru larutan kuprum(II) sulfat, CuSO4 berkurang
kerana kepekatan ion Cu2+ berkurang.
Intensity of blue colour of copper(II) sulphate, CuSO4 solution decreases
because the concentration of Cu2+ ions decreases.
(g) E0sel/cell = E0 katod/cathode – E0 anod/anode
stra
Ek
Ekstra
Nilai voltan/ Voltage value:
0.34 – (-0.76) = 1.1 V
Tip
Sel Daniell hanya menggunakan zink dan kuprum sebagai elektrod.
Daniell cell only use copper and zinc as electrodes.
4. Sel kimia boleh dibina dengan menggunakan pasangan elektrod logam berlainan jenis dalam larutan akueus
masing-masing dan disambungkan dengan titian garam.
Voltaic cell (chemical cell) can be constructed using two different types of metal electrodes in their respective aqueous solutions
and connected by a salt bridge.
Tip
V
SPM
Tatacara menulis notasi sel/ The procedure for
Titian garam
Salt bridge
writing cell notation:
Logam A
Metal A
Logam B
Metal B
Larutan akueus
garam A
Aqueous
solution of salt A
Larutan akueus
garam B
Aqueous solution
of salt B
Tip SPM
Zn(p/s) I Zn2+ (ak/aq) II Cu2+ (ak/aq) I Cu(p/s)
Anod/Anode
Katod/Cathode
II – mewakili titian garam/ represents the salt
bridge
I – mewakili sempadan elektrod-elektrolit
atau fasa berbeza/ represents the boundary
of electrode-electrolytes or different phase
Rajah/Diagram 1.15
25
Contoh/Example 3
Elektrod ferum dalam larutan ferum(II) sulfat, FeSO4 dan elektrod kuprum dalam larutan kuprum(II) sulfat, CuSO4.
Iron and copper electrodes in iron(II) sulphate, FeSO4 and copper(II) sulphate, CuSO4 solutions respectively
Fe (p/s) l Fe2+ (ak/aq, 1.0 M) ll Cu2+ (ak/aq, 1.0 M) l Cu (p/s)
Fe2+ (ak/aq) + 2e–  Fe (p/s)
E0 = –0.44 V
Cu2+ (ak/aq) + 2e–  Cu (p/s)
E0 = +0.34 V
(a) Tentukan terminal negatif (anod) dan Terminal negatif/ Negative terminal:
terminal positif (katod). Nyatakan alasan.
Ferum/ Iron
IA
Determine the negative terminal (anode) and the
Terminal positif/ Positive terminal:
positive terminal (cathode). State the reason.
Kuprum/ Copper
Alasan/ Reason:
Semakin negatif nilai E0, semakin mudah suatu atom atau ion mengalami
pengoksidaan
AS
The more negative the value of E0, the easier for the atom or ion to undergo
oxidation
(b) Persamaan setengah bagi tindak balas di Terminal negatif/ Negative terminal:
terminal negatif dan terminal positif
Fe  Fe2+ + 2e–
(Pengoksidaan/ Oxidation)
Half-equations for the reactions at the negative
and positive terminals
Terminal positif/ Positive terminal:
Cu2+ + 2e–  Cu
(Penurunan/ Reduction)
(c) Hasil yang terbentuk di terminal negatif Terminal negatif/ Negative terminal:
dan terminal positif
Ion ferum(II)/ Iron(II) ions, Fe2+
PA
N
Products formed at the negative and positive
terminals
Terminal positif/ Positive terminal:
Logam kuprum/ Copper metal, Cu
(d) Pemerhatian di
terminal positif
terminal
negatif
dan Terminal negatif/ Negative terminal:
Elektrod ferum menipis/ Iron electrode becomes thinner
Observations at the negative and positive
terminals
Terminal positif/ Positive terminal:
Elektrod kuprum menebal/ Copper electrode becomes thicker
(e) Arah pengaliran elektron dalam litar luar
Direction of electron flow in the outer circuit
(f) Perubahan warna elektrolit
Change in the colour of electrolyte
ferum
Daripada
From
iron
berkurang
copper
to
Keamatan warna
berkurang
kuprum
kepada
biru
larutan
kerana
kuprum(II)
kepekatan
ion
sulfat,
CuSO4
kuprum(II),
.
decreases
Intensity of blue colour of copper(II) sulphate, CuSO4 solution
decreases
.
because the concentration of copper(II) ions, Cu2+
Nilai voltan/ Voltage value:
(g) E0sel/cell = E0 katod/cathode – E0 anod/anode
0.34 – (-0.44)
= 0.78 V
26
Cu2+
AKTIVITI
1.5
To identify the anode and cathode and to determine the voltage value by using a Daniell and a simple voltaic cells.
Bahan/ Materials:
Kepingan magnesium, zink dan kuprum, larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3, larutan magnesium
nitrat, Mg(NO3)2 1.0 mol dm–3, larutan zink sulfat, ZnSO4 1.0 mol dm–3, larutan natrium nitrat, NaNO3
1.0 mol dm–3, kertas turas, kertas pasir
IA
Strips of magnesium, zinc and copper, 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution, 1.0 mol dm–3 magnesium nitrate,
Mg(NO3)2 solution, 1.0 mol dm–3 zinc sulphate, ZnSO4 solution, 1.0 mol dm–3 sodium nitrate, NaNO3 solution, filter paper,
sand paper
Radas/ Apparatus:
Bikar, voltmeter, wayar penyambung dengan klip buaya, gunting, kertas turas, pasu berliang
Beaker, voltmeter, connecting wires with crocodile clips, scissors, filter paper, porous pot
A Sel Daniell/ Daniell cell
Prosedur/ Procedure:
AS
V
Jalur kertas turas
Filter paper strip
Elektrod kuprum, Cu
Copper, Cu electrode
Elektrod zink, Zn
Zinc, Zn electrode
Larutan zink sulfat, ZnSO4
Zinc sulphate, ZnSO4 solution
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate,, CuSO4 solution
Rajah/Diagram 1.16
1. Kepingan zink dan kepingan kuprum dibersihkan dengan kertas pasir.
Zinc and copper strips are cleaned by using the sand paper.
PA
N
2. Larutan zink sulfat, ZnSO4 dan larutan kuprum(II) sulfat, CuSO4 dituang ke dalam dua buah bikar berasingan
sehingga separuh penuh.
Zinc sulphate, ZnSO4 and copper(II) sulphate, CuSO4 solutions are poured into two separate beakers until half full.
3. Hujung kedua-dua jalur kertas turas yang telah direndam dalam larutan natrium nitrat, NaNO3 dicelup ke
dalam larutan zink sulfat, ZnSO4 dan kuprum(II) sulfat, CuSO4 masing-masing di dalam bikar.
Both ends of the filter paper strip which is damped with sodium nitrate, NaNO4 solution is immersed into zinc
sulphate, ZnSO4 and copper(II) sulphate, CuSO4 solutions respectively in the beakers.
4. Kepingan zink dan kepingan kuprum disambungkan kepada voltmeter dengan wayar penyambung.
Zinc and copper strips are connected to the voltmeter with connecting wires.
5. Kepingan zink dan kepingan kuprum dicelupkan ke dalam larutan garam masing-masing.
Zinc and copper strips are immersed into their salt solutions respectively.
6. Bacaan voltmeter dan pemerhatian direkodkan./ The voltmeter reading and observations are recorded.
Keputusan/ Result:
Elektrod/ Electrode
Nilai keupayaan elektrod piawai, E (V)
0
Standard electrode potential, E0 value (V)
Logam disambung pada terminal positif dan negatif voltmeter
Metals connected at the positive and negative terminals of voltmeter
Persamaan setengah bagi tindak balas
Half-equations for the reactions
Pengoksidaan atau penurunan
Cell notation
Kuprum/ Copper
-0.76
+0.34
Negatif
Positif
Negative
Positive
Zn  Zn2+ + 2e–
Cu2+ + 2e–  Cu
Pengoksidaan
Penurunan
Oxidation
Oxidation or reduction
Notasi sel
Zink/ Zinc
Reduction
Zn (p/s) l Zn2+ (ak/aq, 1.0 M) ll Cu2+ (ak/aq, 1.0 M) l Cu (p/s)
Logam yang bertindak sebagai anod dan katod
Anod
Anode
Metals that act as anode and cathode
Keupayaan sel, E0sel
Katod
Cathode
Jawapan murid
Student answer
Cell potential, E0 cell
27
Eksperimen Wajib
Tujuan/ Aim:
Mengenal pasti anod dan katod serta menentukan nilai voltan menggunakan sel Daniell dan sel kimia ringkas.
B Sel Kimia Ringkas/ Simple Voltaic Cell
Prosedur/ Procedure:
V
Jalur kertas turas
Filter paper strip
Elektrod kuprum, Cu
Copper, Cu electrode
Elektrod magnesium, Mg
Magnesium, Mg electrode
Larutan magnesium nitrat, Mg(NO3)2
Magnesium nitrate, Mg(NO3)2 solution
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate, CuSO4 solution
Rajah/Diagram 1.17
IA
1. Kepingan magnesium dan kepingan kuprum dibersihkan dengan kertas pasir.
Magnesium and copper strips are cleaned by using the sand paper.
2. Larutan magnesium nitrat, Mg(NO3)2 dan larutan kuprum(II) sulfat, CuSO4 dituang ke dalam dua bikar
berasingan sehingga separuh penuh.
Magnesium nitrate, Mg(NO3)2 and copper(II) sulphate, CuSO4 solutions are poured into two separate beakers until
half full.
3. Hujung kedua-dua jalur kertas turas yang telah direndam dalam larutan natrium nitrat, NaNO3 dicelup ke
dalam larutan magnesium nitrat, Mg(NO3)2 dan kuprum(II) sulfat, CuSO4 masing- masing di dalam bikar.
AS
Both ends of the filter paper strip which is damped with sodium nitrate, NaNO3 solution is immersed into magnesium
nitrate, Mg(NO3)2 and copper(II) sulphate, CuSO4 solutions respectively in the beakers.
4. Kepingan magnesium dan kepingan kuprum disambungkan kepada voltmeter dengan wayar penyambung.
Magnesium and copper strips are connected to the voltmeter with connecting wires.
5. Kepingan magnesium dan kepingan kuprum dicelupkan ke dalam larutan garam masing-masing.
Magnesium and copper strips are immersed into their salt solutions respectively.
6. Bacaan voltmeter dan pemerhatian direkodkan.
Voltmeter readings and observations are recorded.
Keputusan/ Result:
Elektrod
Electrode
Magnesium
Kuprum
-2.37
+0.34
Magnesium
Nilai keupayaan elektrod piawai, E (V)
PA
N
0
Standard electrode potential, E0 value (V)
Logam disambung pada terminal positif dan negatif voltmeter
Negatif
Persamaan setengah bagi tindak balas
Half-equation for the reactions
Pengoksidaan atau penurunan
Positif
Negative
Positive
Mg  Mg2+ + 2e–
Cu2+ + 2e–  Cu
Pengoksidaan
Penurunan
Metals connected at the positive and negative terminals of voltmeter
Oxidation
Oxidation or reduction
Notasi sel
Copper
Reduction
Mg (p/s) l Mg2+ (ak/aq, 1.0 M) ll Cu2+ (ak/aq, 1.0 M) l Cu (p/s)
Cell notation
Logam yang bertindak sebagai anod dan katod
Anod
Metals that act as anode and cathode
Keupayaan sel, E0sel
Cell potential, E
0
Katod
Anode
Cathode
Jawapan murid
Student answer
cell
Mentafsir data/ Interpreting data:
1. Anod ialah elektrod di mana
penurunan
berlaku.
Anode is the electrode where
takes place.
pengoksidaan
oxidation
berlaku dan katod ialah elektrod di mana
takes place and cathode is the electrode where
2. Semakin negatif nilai keupayaan elektrod piawai, E0, semakin
ion mengalami
easier
pengoksidaan
mudah
reduction
suatu atom atau
. The more negative the value of standard electrode potential, E0, the
for the atom or ion to undergo
oxidation
28
.
Uji Kendiri
1.3
Lengkapkan jadual bagi sel kimia yang menggunakan elektrod magnesium dan elektrod kuprum masing-masing di dalam
larutan magnesium nitrat, Mg(NO3)2 1.0 mol dm–3 dan larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3 yang disambungkan
dengan titian garam.
Complete the table for chemical cell that used magnesium and copper electrodes in magnesium nitrate, Mg(NO3)2 1.0 mol dm–3 solution
and copper(II) sulphate, CuSO4 1.0 mol dm–3 solution respectively connected by a salt bridge.
E0 Mg2+ │ Mg = –2.37 V
IA
E0 Cu2+ │ Cu = +0.34 V
(a) Tentukan terminal negatif (anod) dan Terminal negatif/ Negative terminal:
terminal positif (katod). Nyatakan alasan. Magnesium. Agen penurunan yang lebih kuat
Determine the negative terminal (anode) Magnesium. Stronger reducing agent
and the positive terminal (cathode). State the
Terminal positif/ Positive terminal:
reason.
Kuprum/ Copper
AS
Alasan/Reason:
E0 Mg lebih negatif berbanding Cu
E0 of Mg is more positive compare to Cu
(b) Persamaan setengah bagi tindak balas di Terminal negatif/ Negative terminal:
terminal negatif dan terminal positif
Mg  Mg2+ + 2e–
(Pengoksidaan/ Oxidation)
Half-equations for reactions at the negative
and positive terminals
Terminal positif/ Positive terminal:
Cu2+ + 2e–  Cu
(Penurunan/ Reduction)
PA
N
(c) Hasil yang terbentuk di terminal negatif Terminal negatif/ Negative terminal:
dan terminal positif
Ion magnesium/ Magnesium ions, Mg2+
Products formed at the negative and positive
terminals
Terminal positif/ Positive terminal:
Atom kuprum/ Copper atoms, Cu
(d) Pemerhatian di terminal negatif dan Terminal negatif/ Negative terminal:
terminal positif
Elektrod magnesium menipis/ Magnesium electrode becomes thinner
Observations at the negative and positive
terminals
Terminal positif/ Positive terminal:
Elektrod kuprum menebal/ Copper electrode becomes thicker
(e) Arah pengaliran elektron dalam litar luar
Direction of electron flow in the outer circuit
(f) Perubahan warna elektrolit
Change in the colour of electrolyte
magnesium
Daripada
From
magnesium
Keamatan warna
berkurang
kepada
to
biru
larutan
copper
kuprum
.
.
kuprum(II)
sulfat,
CuSO4
kerana kepekatan ion kuprum(II), Cu2+
berkurang
.
Intensity of the blue colour of copper(II) sulphate, CuSO4 solution
decreases
because the concentration of copper(II) ion, Cu2+
decreases
.
(g) E0sel/cell = E0 katod/cathode – E0 anod/anode
(h) Notasi sel
Cell notation
Nilai voltan/ Voltage value:
0.34 – (-2.37) = 2.71 V
Mg (p/s) l Mg2+ (ak/aq, 1.0 M) ll Cu2+ (ak/aq, 1.0 M) l Cu (p/s)
29
1.4
A
Sel Elektrolisis
Electrolytic Cell
Elektrolisis
Electrolysis
1. Elektrolisis ialah proses
unsur juzuknya
Electrolysis is a process of
it.
penguraian
sebatian
suatu
dalam keadaan lebur atau akueus kepada
apabila arus elektrik dialirkan melaluinya.
breaking down a compound
constituent elements
into its
by passing electricity through
leburan
2. Elektrolit ialah bahan yang boleh mengkonduksikan arus elektrik dalam keadaan
akueus
perubahan kimia
dan mengalami
.
molten
free moving ions
Electrical conductivity of electrolyte is due to the presence of
cation
and
dan
anion
.
.
anion
elektrolit sahaja yang hadir.
of the electrolyte are present.
AS
In a molten state, only
kation
states and undergo
ion-ion yang bergerak bebas
3. Kekonduksian elektrik bagi elektrolit adalah disebabkan kehadiran
(a) Dalam keadaan leburan, hanya
aqueous
or
IA
Electrolytes are substances that can conduct electricity in
chemical changes
.
atau
(b) Dalam larutan akueus, selain daripada kation dan anion elektrolit, ion
turut hadir.
H+
In an aqueous solution, apart from the cation and anion of the electrolyte,
H+
OH–
dan
and
OH–
daripada air
ions are also present.
molekul-molekul
yang neutral.
4. Bukan elektrolit terdiri daripada
neutral Bukan elektrolit tidak
mengalami sebarang perubahan kimia dan tidak mengkonduksi elektrik kerana tiada ion-ion yang bergerak bebas.
molecules
Non-electrolytes are composed of neutral
. Hence, non-electrolytes will not undergo any chemical
changes and do not conduct electricity because there are no free-moving ions.
pepejal
PA
N
5. Konduktor ialah bahan yang boleh mengkonduksikan arus elektrik dalam keadaan
leburan
perubahan kimia
tetapi tidak mengalami sebarang
.
solid
Conductors are substances that can conduct electricity in
chemical changes
.
any
6. Kekonduksian elektrik bagi konduktor adalah disebabkan oleh
Electrical conductivity of conductor is due to the
flow of electrons
AKTIVITI
or
molten
pengaliran elektron
atau
states but do not undergo
.
.
1.6
Eksperimen Wajib
Tujuan/ Aim:
Mengelaskan bahan kepada elektrolit dan bukan elektrolit
To classify substances into electrolytes and non-electrolytes
Bahan/ Materials:
Serbuk plumbum(II) bromida, PbBr2, asetamida, CH3CONH2 dan sulfur, larutan natrium hidroksida, NaOH,
larutan glukosa, larutan kuprum(II) sulfat, CuSO4
Lead(II) bromide, PbBr2, acetamide, CH3CONH2 and sulphur powder, sodium hydroxide, NaOH solution, glucose solution,
copper(II) sulphate, CuSO4 solution
Radas/Apparatus:
Elektrod karbon, wayar penyambung dengan klip buaya, tungku kaki tiga, penunu Bunsen, alas segi tiga
tanah liat, mangkuk pijar, bikar 100 cm3, bateri, mentol, suis
Carbon electrodes, connecting wires with crocodile clips, tripod stand, Bunsen burner, pipe-clay triangle, crucible, 100
cm3 beaker, batteries, bulb, switch
30
A Bahan leburan/ Molten substances
Prosedur/ Procedure:
Bateri
Battery
Suis
Switch
Mentol
Bulb
Elektrod karbon
Carbon electrodes
Mangkuk
pijar
Crucible
Serbuk
plumbum(II)
bromida, PbBr2
Lead(II) bromide,
PbBr2 powder
Panaskan
Heat
Rajah/Diagram 1.18
Carry out the this activity in the
fume chamber or in an open
space.
• Jangan hidu sebarang gas
yang terbebas.
IA
Alas segi tiga
tanah liat
Pipe-clay
triangle
Awas/ Caution
• Jalankan aktiviti ini dalam
kebuk wasap atau tempat
yang terbuka.
Do not inhale any gas liberated.
1. Serbuk plumbum(II) bromida, PbBr2 diisi ke dalam sebuah mangkuk pijar sehingga separuh penuh.
Lead(II) bromide, PbBr2 powder is filled in a crucible until it is half full.
2. Dua elektrod karbon dimasukkan ke dalam serbuk plumbum(II) bromida, PbBr2.
AS
Two carbon electrodes are put into the lead(II) bromide, PbBr2 powder.
3. Litar dilengkapkan dengan menyambung kedua-dua elektrod kepada suis, mentol dan bateri seperti
ditunjukkan dalam Rajah 1.18.
The circuit is completed by connecting the electrodes to a switch, a bulb and batteries as shown in Diagram 1.18.
4. Serbuk plumbum(II) bromida, PbBr2 dipanaskan sehingga lebur sepenuhnya.
The lead(II) bromide, PbBr2 powder is heated until it completely melts.
5. Suis dihidupkan. Pemerhatian pada mentol direkodkan.
The switch is turned on. Observation at the bulb is recorded.
6. Langkah 1 hingga 5 diulang dengan menggantikan serbuk plumbum(II) bromida, PbBr2 dengan serbuk
asetamida, CH3CONH2 dan sulfur, S8.
PA
N
Steps 1 to 5 are repeated by replacing the lead(II) bromide, PbBr2 powder with acetamide, CH3CONH2 powder and
sulphur, S8 powder respectively.
B Larutan akueus/ Aqueous solution
Prosedur/ Procedure:
Bateri
Battery
Suis
Switch
Bikar
Beaker
Awas/ Caution
• Jalankan aktiviti ini dalam
kebuk wasap atau tempat
yang terbuka.
Mentol
Bulb
Elektrod karbon
Carbon electrodes
Larutan natrium
hidroksida, NaOH
Sodium hydroxide, NaOH
solution
Carry out the this activity in the
fume chamber or in an open
space.
• Jangan hidu sebarang gas
yang terbebas.
Do not inhale any gas liberated.
Rajah/Diagram 1.19
1. 20 cm3 larutan natrium hidroksida, NaOH dituangkan ke dalam sebuah bikar.
20 cm3 of sodium hydroxide, NaOH solution is poured into a beaker.
2. Dua elektrod karbon dimasukkan ke dalam larutan itu.
Two carbon electrodes are placed into the solution.
3. Litar dilengkapkan dengan menyambung kedua-dua elektrod kepada suis, mentol dan bateri seperti
ditunjukkan dalam Rajah 1.19.
The circuit is completed by connecting the electrodes to a switch, a bulb and batteries as shown in Diagram 1.19.
4. Suis dihidupkan. Pemerhatan pada mentol direkodkan.
The switch is turned on. Observation at the bulb is recorded.
5. Langkah 1 hingga 4 diulang dengan menggantikan larutan natrium hidroksida, NaOH dengan larutan
glukosa, C6H12O6 dan larutan kuprum(II) sulfat, CuSO4.
Steps 1 to 4 are repeated by replacing sodium hydroxide, NaOH solution with glucose, C6H12O6 solution and copper(II)
sulphate, CuSO4 solution respectively.
31
Keputusan/ Result:
Adakah mentol
menyala?
Bahan
Substance
Does the bulb light up?
Are there free–moving ions?
Ya/ Yes
Ya/ Yes
Leburan asetamida/ Molten acetamide
Tidak/ No
Tidak/ No
Leburan sulfur/ Molten sulphur
Tidak/ No
Tidak/ No
Ya/ Yes
Ya/ Yes
Tidak/ No
Tidak/ No
Ya/ Yes
Ya/ Yes
Leburan plumbum(II) bromida/ Molten lead(II) bromide
Larutan glukosa/ Glucose solution
Larutan kuprum(II) sulfat/ Copper(II) sulphate solution
Mentafsir data/ Interpreting data:
Leburan plumbum(II) bromida, PbBr2
larutan kuprum(II) sulfat, CuSO4
dan
adalah elektrolit dan mengalami perubahan kimia
apabila arus elektrik mengalir melaluinya.
Molten lead(II) bromide, PbBr2
sodium hydroxide, NaOH solution
,
copper(II) sulphate, CuSO4 solution
and
are electrolytes. They undergo chemical changes when
electric current is passed through them.
2.
larutan natrium hidroksida, NaOH
,
AS
1.
IA
Larutan natrium hidroksida/ Sodium hydroxide solution
Leburan asetamida, CH3CONH2
leburan sulfur
,
dan
adalah bukan elektrolit dan tidak mengalami sebarang perubahan kimia.
molten sulphur
,
non-electrolytes. They do not undergo any chemical changes.
PA
N
Molten acetamide, CH3CONH2
B
Adakah terdapat ion
bergerak bebas?
larutan glukosa
glucose solution
and
are
Elektrolisis Sebatian Lebur
Electrolysis of Molten Compounds
AKTIVITI
1.7
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji elektrolisis leburan plumbum(II) bromida, PbBr2 dengan elektrod karbon
To investigate the electrolysis of molten lead(II) bromide, PbBr2 with carbon electrodes
Prosedur/ Procedure:
Bateri
Battery
Suis
Switch
Mentol
Bulb
Elektrod karbon
Carbon electrodes
Mangkuk
pijar
Crucible
Alas segi tiga
tanah liat
Pipe-clay
triangle
Panaskan
Heat
Serbuk
plumbum(II)
bromida, PbBr2
Lead(II) bromide,
PbBr2 powder
Rajah/Diagram 1.20
32
Awas/ Caution
• Jalankan aktiviti ini dalam
kebuk wasap atau tempat
yang terbuka.
Carry out the this activity in the
fume chamber or in an open
space.
• Jangan hidu sebarang gas
yang terbebas.
Do not inhale any gas liberated.
1. Susunan radas seperti ditunjukkan dalam Rajah 1.20 disediakan.
The apparatus set-up as shown in Diagram 1.20 is prepared.
2. Serbuk plumbum(II) bromida, PbBr2 dipanaskan dengan perlahan sehingga melebur.
Lead (II) bromide, PbBr2 powder is heated gently until all melted.
3. Pemanasan dihentikan selepas 10 minit dan bahan leburan dituang ke dalam mangkuk pijar.
The heating is stop after 10 minutes and the molten substance is poured into a crucible.
Pemerhatian/ Observation:
Elektrod/ Electrodes
Pemerhatian/ Observation
Anod
Gas perang berbau sengit terbebas
Katod
Pepejal kelabu terbentuk
IA
Anode
A brown gas with a pungent smell is released
A grey solid is formed
Cathode
Mentafsir data/ Interpreting data:
(a) Ion yang hadir dalam elektrolit
Pb2+, Br-
AS
Ions present in the electrolyte
(b) Ion yang tertarik ke anod dan katod
Anod/Anode (+):
Ions attracted to the anode and cathode
Br-
Katod/Cathode (–):
Pb2+
(c) Persamaan setengah tindak balas di anod Anod/ Anode:
dan katod
2Br–  Br2 + 2e–
Half-equations for reactions at the anode and
cathode
(Pengoksidaan/ Oxidation)
Katod/ Cathode:
PA
N
Pb2+ + 2e–  Pb
(d) Hasil yang terbentuk di anod dan katod
(Penurunan/ Reduction)
Anod/ Anode:
Products formed at the anode and cathode
Gas bromin
Bromine gas
Katod/ Cathode:
Logam plumbum
Lead metal
Ya. Pengoksidaan dan penurunan yang berlaku serentak adalah
tindak balas redoks.
(e) Adakah elektrolisis sebatian lebur satu
tindak balas redoks? Jelaskan.
Is the electrolysis of molten compound a redox Yes. Oxidation and reduction that occur simultaneously is a redox
reaction? Explain.
reaction.
Uji Kendiri
1.4
Lengkapkan yang berikut bagi elektrolisis leburan plumbum(II) bromida, PbBr2 dengan menggunakan elektrod karbon.
Complete the following for the electrolysis of molten lead(II) bromide, PbBr2 with carbon electrodes.
anod
1. Ion bromida, Br– mengalami pengoksidaan di
Bromide ion, Br undergo oxidation at the
–
Anod/ Anode:
anode
.
2Br  Br2 + 2e–
–
katod
2. Ion plumbum(II), Pb2+ mengalami penurunan di
Lead(II) ion, Pb undergo reduction at the
cathode
Katod/ Cathode:
Pb
2+
.
2+
(Pengoksidaan/ Oxidation)
.
.
+ 2e–  Pb
33
(Penurunan/ Reduction)
C
Faktor-faktor yang Mempengaruhi Elektrolisis Larutan Akueus
Factors Affecting Electrolysis of Aqueous Solutions
1. Nilai keupayaan elektrod piawai, E0/ Standard electrode potential, E0 value
(a) Pada katod, jika terdapat lebih daripada satu spesies yang boleh diturunkan di dalam larutan, spesies dengan E0
yang lebih positif dipilih untuk diturunkan.
At cathode, if there is more than one reducible species in the solution, the species with more positive E0 is preferentially
reduced.
Na+(ak/aq) + e–  Na(p/s)
E0 = –2.71 V
2H2O(cec/l) + 2e–  H2(g/g) + 2OH–(ak/aq)
E0 = –0.83 V
0
Oleh kerana nilai E air, H2O lebih positif, maka air, H2O dipilih untuk diturunkan berbanding ion natrium, Na+.
Since the E0 value is more positive for water, H2O, thus water, H2O is preferentially reduced than sodium ion, Na+.
IA
(b) Pada anod, jika terdapat lebih daripada satu spesies yang boleh dioksidakan di dalam larutan, spesies dengan E0
yang lebih negatif dipilih untuk dioksidakan.
At anode, if there is more than one oxidisable species in the solution, the species with more negative E0 is preferentially
oxidised.
I2(p/s) + 2e–  2I–(ak/aq)
E0 = +0.53 V
–
–
Br2(cec/l) + 2e  2Br (ak/aq)
E0 = +1.07 V
Oleh kerana nilai E0 ion iodida, I– lebih negatif, maka ion iodida, I– dipilih untuk dioksidakan berbanding ion
bromida, Br–.
Since the E0 value is more negative for iodide ion, I–, thus iodide ion, I– is preferentially oxidised than bromide ion, Br–.
2. Kepekatan ion dalam elektrolit/ Concentration of ions in the electrolyte
AS
Ion yang mempunyai kepekatan lebih tinggi dalam elektrolit akan dipilih untuk dinyahcas.
Ions with higher concentration in the electrolyte will be selected for discharge.
3. Jenis elektrod/ Types of electrode
Apabila elektrod reaktif (argentum, kuprum) digunakan, elektrod anod akan mengion.
When reactive electrodes (silver, copper) are used, the anode electrode will ionise.
Tip
SPM
O2 + 4H+ + 4e–  2H2O
E0 = +1.23 V
Cl2 + 2e–  2Cl–
E0 = +1.36 V
Tip SPM
Na+ + e–  Na
2H2O + 2e–  H2+ 2OH–
(Rujukan untuk ion OH /Reference
Reference for OH ion
ion))
–
PA
N
–
∴ OH dipilih untuk dinyahcaskan kerana
nilai E0 lebih negatif.
–
OH–
Na+
Cl–
H+
∴ OH– is selected to be discharged because the E0
value more negative.
A
E0 = –2.71 V
E0 = –0.83 V
(Rujukan untuk ion H+/Reference f or H+ ion)
∴ H+ dipilih untuk dinyahcaskan kerana
nilai E0 lebih positif
∴ H+ is selected to be discharged because the E0
value more positive.
Rajah/Diagram 1.21 Elektrolisis berdasarkan nilai keupayaan elektrod/ Electrolysis based on electrode potential value
Eksperimen
1.1
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji Eksperimen
kesan nilai keupayaan elektrod piawai, E0 terhadap pemilihan ion untuk dinyahcas pada elektrod.
To investigate the effect of the standard electrode, E0 value on selective discharge of ions at the electrodes.
Pernyataan masalah/ Problem statement:
Bagaimanakah nilai keupayaan elektrod piawai, E0 mempengaruhi pemilihan ion untuk dinyahcas di elektrod?
How does the standard electrode potential, E0 value affect the selective discharge of ions at the electrodes?
Hipotesis/ Hypothesis:
Di anod, semakin negatif nilai E0 semakin mudah spesies dinyahcas, manakala di katod semakin positif nilai E0 semakin
mudah spesies dinyahcaskan.
At the anode, the more negative the value of E0 the easier the species to be discharged, while at the cathode, the more positive the value
of E0 the easiar the species to be discharged.
34
Pemboleh ubah/ Variables:
(a) Dimanipulasikan/ Manipulated: Nilai keupayaan elektrod piawai, E°/ Standard electrode potential, E0 value
(b) Bergerak balas/ Responding: Hasil elektrolisis/ Products of electrolysis
(c) Dimalarkan/ Fixed: Jenis elektrod, kepekatan larutan elektrolit/ Types of electrode, concentration of electrolyte solution
Bahan/ Materials:
Larutan asid sulfurik cair, H2SO4 1.0 mol dm–3, larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3, kayu uji
1.0 mol dm–3 dilute sulphuric acid, H2SO4, 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution, wooden splint
IA
Radas/ Apparatus:
Sel elektrolitik, tabung uji, bateri, ammeter, elektrod karbon, wayar penyambung
Electrolytic cell, test tubes, batteries, ammeter, carbon electrodes, connecting wires
Prosedur/ Procedure:
Asid sulfurik cair
cair,, H2SO4
Dilute sulphuric acid
acid,, H2SO4
AS
Karbon elektrod
Carbon electrodes
Tabung uji
Test tube
Suis
Switch
A Ammeter
Ammeter
Bateri
Battery
Rajah/Diagram 1.22
1. Larutan asid sulfurik cair, H2SO4 1.0 mol dm–3 dituang ke dalam sel elektrolitik hingga menutupi elektrod
karbon.
1.0 mol dm–3 dilute sulphuric acid, H2SO4 is poured into an electrolytic cell until it covers the carbon electrodes.
PA
N
2. Tabung uji diisi dengan larutan asid sulfurik cair, H2SO4 1.0 mol dm–3 dan ditengkupkan ke atas elektrod
karbon.
The test tubes are filled with 1.0 mol dm–3 dilute sulphuric acid, H2SO4 and overturn onto the carbon electrodes.
3. Elektrod karbon disambung kepada bateri dengan menggunakan wayar penyambung.
Carbon electrodes are connected to the batteries using the connecting wires.
4. Arus elektrik dialirkan melalui elektrolit selama 15 minit.
The electric current is passed through the electrolyte for 15 minutes.
5. Pemerhatian di anod dan katod direkodkan.
Observations at the anode and cathode is recorded.
6. Gas yang terkumpul di anod diuji dengan kayu uji berbara manakala gas yang terkumpul di katod diuji dengan
kayu uji menyala.
Gas collected at the anode is tested with a glowing wooden splint while gas collected at the cathode is tested with a burning wooden
splint.
7. Langkah 1 hingga 6 diulang dengan menggunakan larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3.
Steps 1 to 6 are repeated by using 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution.
Pemerhatian/ Observation:
Elektrolit
Electrolyte
Anod/Anode (+)
Asid sulfurik cair, H2SO4
Dilute sulphuric acid, H2SO4
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate solution, CuSO4
Gelembung gas dibebaskan. Gas tidak Gelembung gas dibebaskan. Gas tidak
berwarna menyebabkan kayu uji berbara berwarna menyebabkan kayu uji berbara
menyala.
menyala
Gas bubbles are released. Colourless gas ignites the Gas bubbles are released. A colourless gas ignites the
glowing wooden splint.
glowing wooden splint.
Katod/Cathode (–) Gelembung gas dibebaskan. Gas tidak Pepejal perang terenap.
berwarna menghasilkan bunyi “pop” dengan Brown solid is deposited
kayu uji menyala.
Gas bubbles released. Colourless gas produces a “pop”
sound with the glowing wooden splint.
35
Mentafsir data/ Interpreting data:
Elektrolit
Asid sulfurik cair, H2SO4
Electrolytes
(a) Ion yang
elektrolit
hadir
Dilute sulphuric acid, H2SO4
dalam H2SO4  2H+ + SO42–
H2O H+ + OH–
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate solution, CuSO4
CuSO4  Cu2+ + SO42–
H2O H+ + OH–
Ions present in the electrolytes
Anod/Anode (+):
SO42–, OH–
Katod/ Cathode (–):
H+, Cu2+
(c) Ion yang dinyahcas di anod Anod/Anode:
dan katod. Nyatakan alasan. OH–(H2O)
Ions discharged at the anode and Alasan/Reason:
cathode. State the reason.
Ion SO42– tidak dipilih kerana
S telah mencapai nombor
pengoksidaan yang tinggi.
Anod/Anode:
OH–(H2O)
Alasan/Reason:
Ion SO42– tidak dipilih kerana S telah
mencapai nombor pengoksidaan yang
tinggi.
IA
(b) Ion yang tertarik ke anod Anod/Anode (+):
dan katod
SO42–, OH–
Ions attracted to the anode and Katod/ Cathode (–):
cathode
H+
SO42– ions were not selected because S had SO42– ions were not selected because S had reached
reached higher oxidation number.
higher oxidation number.
Katod/Cathode:
Katod/Cathode:
Cu2+
Alasan/Reason:
E0 bagi ion Cu2+ lebih positif
AS
OH–(H2O)
Alasan/Reason:
–
E0 for Cu2+ ion more positive
(d) Persamaan setengah tindak
balas di anod dan katod
Anod/Anode:
2H2O  O2 + 4H++ 4e–
Half-equations for reaction at the
Katod/Cathode:
anode and cathode
–
Anod/Anode:
2H2O  O2 + 4H++ 4e–
2H2O + 2e  H2 + 2OH
–
Katod/Cathode:
Cu2+ + 2e–  Cu
PA
N
Kesimpulan/ Conclusion:
Di anod, semakin negatif nilai E0 semakai mudah spesies dinyahcaskan, manakala di katod, semakin positif nilai E0,
semakin mudah spesies dinyahcaskan.
At the anode, the more negative the value of E0 the easier the species to be discharged, while at the cathode, the more positive the value
of E0 the easier the species to be discharged.
Eksperimen
1.2
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji Eksperimen
kesan kepekatan ion dalam elektrolit terhadap pemilihan ion untuk dinyahcas pada elektrod
To investigate the effect of the concentration of ions on the selective discharge of ions at the electrodes
Pernyataan masalah/ Problem statement:
Bagaimanakah kepekatan ion dalam elektrolit mempengaruhi pemilihan ion untuk dinyahcas di anod?
How does the concentration of ions in the electrolyte affect the discharge of ions at the anode?
Hipotesis/ Hypothesis:
Ion halida yang mempunyai kepekatan lebih tinggi akan dinyahcas secara pilihan di anod
Halide ions of higher concentrations will be selectively discharged at the anode
Pemboleh ubah/ Variables:
(a) Dimanipulasikan/ Manipulated: Kepekatan elektrolit/ Concentration of electrolyte
(b) Bergerak balas/ Responding: Hasil elektrolisis di anod/ Products of electrolysis at the anode
(c) Dimalarkan/ Fixed: Jenis elektrod, jenis elektrolit/ Types of electrode, type of electrolyte
36
Bahan/ Materials:
Asid hidroklorik, HCl 2.0 mol dm–3, asid hidroklorik, HCl 0.001 mol dm–3, kertas litmus biru, kayu uji
2.0 mol dm–3 hydrochloric acid, HCl, 0.001 mol dm–3 hydrochloric acid, HCl, blue litmus paper, wooden splint
Radas/ Apparatus:
Sel elektrolisis, tabung uji, bateri, ammeter, elektrod karbon, wayar penyambung
Electrolytic cell, test tubes, batteries, ammeter, carbon electrodes, connecting wires
Prosedur/ Procedure:
Tabung uji
Test tube
IA
Asid hidroklorik
hidroklorik, HCl
Hydrochloric
HCl
acid, H
acid
Karbon elektrod
Carbon electrodes
Ammeter A
Ammeter
Bateri
Battery
Suis
Switch
AS
Rajah/Diagram 1.23
1. Sebuah sel elektrolisis diisi dengan asid hidroklorik, HCl 2.0 mol dm–3 sehingga separuh penuh.
An electrolytic cell is filled with 2.0 mol dm–3 hydrochloric acid, HCl until it is half full.
2. Litar dilengkapkan dengan menyambungkan elektrod karbon, suis, bateri dan ammeter dengan wayar penyambung
seperti dalam Rajah 1.23.
The circuit is completed by connecting the electrodes to the switch, ammeter and batteries as shown in Diagram 1.23.
3. Suis dihidupkan.
The switch is turned on.
4. Gas di anod dan di katod dikumpulkan dan diuji dengan kertas litmus biru lembap, kayu uji berbara dan kayu uji
menyala. Pemerhatian direkodkan.
PA
N
The gases produced at the anode and the cathode are collected and tested with a moist blue litmus paper, a glowing wooden splint and
a lighted wooden splint. The observation is recorded.
5. Langkah 1 hingga 4 diulang dengan menggunakan asid hidroklorik, HCl 0.001 mol dm–3.
Steps 1 to 4 are repeated using 0.001 mol dm–3 hydrochloric acid, HCl.
Pemerhatian/ Observation:
Elektrolit
Electrolyte
Anod/
Anode (+)
Asid hidroklorik, HCl 2.0 mol dm–3
2.0 mol dm–3 hydrochloric acid, HCl
kuning
tidak
Gelembung gas dibebaskan. Gas
Gelembung gas dibebaskan. Gas
kehijauan
berbau sengit
berwarna
yang
menyebabkan kayu uji
menyala
dibebaskan. Kertas litmus biru lembap menjadi berbara
.
merah
dilunturkan
dan kemudian
.
Gas bubbles are released.
pungent smell
with a
A greenish-yellow gas
blue litmus paper turns
bleached
then
Katod/
Cathode (–)
Asid hidroklorik, HCl 0.001 mol dm–3
0.001 mol dm-3 hydrochloric acid, HCl
is released. Moist splint.
red
and
wooden splint.
colourless
the glowing wooden
.
tidak
Gelembung gas dibebaskan. Gas
berwarna
bunyi
menghasilkan
“pop”
dengan kayu uji menyala.
Gas bubbles are released. A
“pop” sound
gas produces a
Gas bubbles are released. A
ignites
gas
colourless
tidak
Gelembung gas dibebaskan. Gas
berwarna
bunyi
menghasilkan
“pop”
dengan kayu uji menyala.
Colourless
Gas bubbles released.
gas
“pop”
sound
with the lighted produces a
with the lighted
wooden splint.
37
Kesimpulan/ Conclusion:
1. Elektrolisis asid hidroklorik, HCl 0.001 mol dm–3 menghasilkan gas
elektrolisis asid hidroklorik, HCl 2.0 mol dm–3 menghasilkan gas
Electrolysis of 0.001 mol dm hydrochloric acid, HCl produces
chlorine
gas at the anode.
hydrochloric acid, HCl produces
–3
tinggi
2. Ion halida yang berkepekatan
oxygen
oksigen
klorin
di anod manakala
di anod.
gas at the anode. Electrolysis of 2.0 mol dm–3
di dalam elektrolit akan dipilih untuk dinyahcaskan berbanding
ion hidroksida, OH .
–
Halide ion with a
higher
concentration in the electrolyte will be selectively
ectively discharged compare to hydroxide ion, OH–.
0.001 mol dm-3 hydrochloric acid, HCl with carbon electrodes
IA
Perbincangan/ Discussion:
1. Asid hidroklorik, HCl 0.001 mol dm-3 dengan elektrod karbon
(a) Ion yang hadir dalam elektrolit
HCl  H+ + Cl–
H2O H+ + OH–
(b) Ion yang tertarik ke anod dan katod.
Anod/ Anode (+):
Cl–, OH–
Ions present in the electrolyte
Ions attracted to the anode and cathode
AS
Katod/ Cathode (–):
H+
(c) Ion yang dinyahcas di anod dan katod. Anod/ Anode: OH–(H2O)
Alasan/ Reason:
Nyatakan alasan.
Ions discharged at the anode and cathode. State the Kerana E0 lebih negatif
reason.
Because the E0 more negative
Katod/ Cathode: H+(H2O)
PA
N
(d) Persamaan setengah bagi tindak balas di Anod/ Anode:
anod dan katod
2H2O  O2 + 4H+ + 4e–
Half-equations for reactions at the anode and
Katod/ Cathode:
cathode
2H2O + 2e–  H2 + 2OH– (Penurunan/ Reduction)
(e) Hasil yang terbentuk di anod dan katod
Anod/ Anode:
Gas oksigen/ Oxygen gas
Katod/ Cathode:
Gas hidrogen/ Hydrogen gas
(f ) Pemerhatian di anod dan katod
Anod/ Anode:
Gelembung gas tidak berwarna/ Colourless gas bubbles
Katod/ Cathode:
Gelembung gas tidak berwarna/ Colourless gas bubbles
Products formed at the anode and cathode
Observations at the anode and cathode
2
(Pengoksidaan/ Oxidation)
Asid hidroklorik, HCl 2.0 mol dm-3 dengan elektrod karbon
2.0 mol dm-3 hydrochloric acid, HCl with carbon electrodes
(a) Ion yang hadir dalam elektrolit
Ions present in the electrolyte
(b) Ion yang tertarik ke anod dan katod
Ions attracted to the anode and cathode
HCl  H+ + Cl–
H2O
H+ + OH–
Anod/ Anode:
Cl–, OH–
Katod/ Cathode:
H+
38
(c) Ion yang dinyahcas di anod dan katod. Anod/ Anode: Cl–
Nyatakan alasan.
Alasan/ Reason:
Ions discharged at anode and cathode. State the Kerana ion Cl– lebih pekat
reason.
Because Cl– ion more concentrated
Katod/ Cathode: H+(H2O)
(d) Persamaan setengah bagi tindak balas di Anod/ Anode:
anod dan katod.
2Cl–  Cl2 + 2e–
2H2O + 2e–  H2 + 2OH–
(e) Hasil yang terbentuk di anod dan katod
Anod/ Anode:
Gas klorin/ Chlorine gas
Katod/ Cathode:
Gas hidrogen/ Hydrogen gas
(f ) Pemerhatian di anod dan katod
Anod/ Anode:
Gelembung gas kuning kehijauan
(Penurunan/Reduction)
AS
Products formed at the anode and cathode
(Pengoksidaan/ Oxidation)
IA
Half-equations for reactions at the anode and
Katod/ Cathode:
cathode
Observations at the anode and cathode
Greenish-yellow gas bubbles
Katod/ Cathode:
Gelembung gas tidak berwarna
Colourless gas bubbles
1.3
Tujuan/ Aim:
Mengkaji Eksperimen
kesan jenis elektrod terhadap pemilihan ion untuk dinyahcas di elektrod
To investigate the effect of the types of electrode on the selective discharge of ions at the electrodes
Pernyataan masalah/ Problem statement:
Bagaimanakah jenis elektrod mempengaruhi hasil yang terbentuk semasa elektrolisis?
How does the types of electrode affect the types of products formed during the electrolysis?
Hipotesis/ Hypothesis:
Apabila elektrod kuprum digunakan untuk menggantikan elektrod karbon, hasil yang terbentuk pada anod dan katod
adalah berbeza.
When copper electrodes are used instead of carbon electrodes, the types of products formed at the anode and cathode are different.
Bahan/ Materials:
Larutan kuprum(II) sulfat, CuSO4 1.0 mol dm–3, kayu uji, kepingan logam kuprum, kertas pasir
1.0 mol dm–3 copper(II) sulphate solution, CuSO4, wooden splint, copper metal plates, sand paper
Radas/ Apparatus:
Sel elektrolitik, tabung uji, bateri, ammeter, elektrod karbon, wayar penyambung, bikar
Electrolytic cell, test tubes, batteries, ammeter, carbon electrodes, connecting wires, beakers
Pemboleh ubah/Variables:
(a) Dimanipulasikan/ Manipulated: Jenis elektrod/ Types of electrode
(b) Bergerak balas/ Responding: Hasil elektrolisis/ Products of electrolysis
(c) Dimalarkan/ Fixed: Kepekatan elektrolit, jenis elektrolit/ Concentration of electrolyte, types of electrolyte
39
Eksperimen Wajib
PA
N
Eksperimen
Prosedur/ Procedure:
Tabung uji
Test tube
Larutan kuprum(II) sulfat, CuSO4
Copper(II) sulphate, CuSO4 solution
Karbon elektrod
Carbon electrodes
Bateri
Battery
Ammeter
Ammeter A
Suis
Switch
Rajah/Diagram 1.24
1. Larutan kuprum(II) sulfat, CuSO4 dituangkan ke dalam bikar hingga separuh penuh.
Copper(II) sulphate solution, CuSO4 is poured into a beaker until half full.
IA
2. Elektrod karbon disambungkan kepada bateri serta ammeter dengan menggunakan wayar penyambung seperti
yang ditunjukkan dalam Rajah 1.24.
Carbon electrodes are connected to the batteries and ammeter using the connecting wires as shown in Diagram 1.24.
3. Elektrod karbon dimasukkan ke dalam larutan kuprum(II) sulfat, CuSO4.
Carbon electrodes are immersed into the copper(II) sulphate solution, CuSO4.
4. Arus elektrik dialirkan melalui elektrolit selama 15 minit.
Electric current is passed through the electrolyte for 15 minutes.
5. Pemerhatian pada anod, katod dan elektrolit direkodkan.
AS
Observations at the anode, cathode and electrolyte are recorded.
6. Langkah 1 hingga 5 diulang dengan menggantikan elektrod karbon dengan elektrod kuprum dan menggunakan
susunan radas ditunjukkan dalam Rajah 1.25.
Steps 1 to 5 are repeated by replacing carbon electrodes with copper electrodes and using the apparatus set-up as shown in Diagram
1.25.
Bateri
Battery
Suis
Switch
A
Ammeter
Ammeter
Elektrod kuprum
Copper electrodes
PA
N
Bikar
Beaker
Larutan kuprum(II) sulfat
sulfat, CuSO4
Copper(II) sulphate, CuSO4 solution
Rajah/Diagram 1.25
Pemerhatian/ Observations:
Anod
Elektrod
Anode (+)
Electrode
Karbon
Carbon
Katod
Elektrolit
Cathode (–)
Gelembung gas dibebaskan.
tidak berwarna
Gas
Electrolyte
perang
Pepejal
terenap.
berkurang
Keamatan warna biru
kerana kepekatan ion kuprum(II), Cu2+
berkurang
solid is
.
Brown
menyebabkan kayu uji berbara
menyala
. deposited.
decreases
The intensity of blue colour
because the concentration of copper(II) ions,
decreases
Cu2+
.
Gas bubbles are released. A
colourless gas
ignites
the glowing wooden splint.
Kuprum
Copper
Elektrod
Electrode becomes
menipis
. Elektrod
thinner
.
menebal
Electrode becomes
thicker
40
. Keamatan warna biru tidak berubah
kerana kepekatan ion kuprum(II), Cu2+
tidak berubah
.
.
unchanged
The intensity of blue colour
because the concentration of copper(II) ions,
remains unchanged
Cu2+
.
Kesimpulan/ Conclusion:
oksigen
1. Elektrolisis larutan kuprum(II) sulfat, CuSO4 dengan elektrod karbon menghasilkan
air
di anod serta logam kuprum di katod.
Electrolysis of copper(II) sulphate, CuSO4 solution using carbon electrodes produces
copper metal
at the cathode.
at the anode and
oxygen
Perbincangan/ Discussion:
ion kuprum(II)
copper(II) ions
di
at the anode and
IA
Electrolysis of copper(II) sulphate, CuSO4 solution using copper electrodes produces
copper metal
at the cathode.
water
and
2. Elektrolisis larutan kuprum(II) sulfat, CuSO4 dengan elektrod kuprum menghasilkan
anod dan logam kuprum di katod.
dan
1. Larutan kuprum(II) sulfat, CuSO4 dengan elektrod karbon
Copper(II) sulphate, CuSO4 solution with carbon electrodes
Ions present in the electrolyte
CuSO4  Cu2+ + SO42–
AS
(a) Ion yang hadir dalam elektrolit
(b) Ion yang tertarik ke anod dan katod
Ions attracted to the anode and cathode
H2O
H+ + OH–
Anod/ Anode (+): SO42–, OH–
Katod/ Cathode (–) : H+, Cu2+
PA
N
(c) Ion yang dinyahcas di anod dan katod. Anod/ Anode: OH–(H2O)
Alasan/ Reason:
Nyatakan alasan.
Ions discharged at the anode and cathode. State the Ion SO42– tidak dipilih kerana S telah mencapai nombor
reason.
pengoksidaan yang tinggi.
SO42– ions were not selected because S had reached higher oxidation
number.
Katod/ Cathode: Cu2+
Alasan/ Reason:
E0 ion Cu2+ lebih positif
E0 of Cu2+ ion more positive
(d) Persamaan setengah bagi tindak balas di Anod/ Anode: 2H2O  O2 + 4H+ + 4e
anod dan katod
Half-equations for reactions at the anode and Katod/ Cathode: Cu2+ + 2e–  Cu
cathode
(e) Perubahan warna elektrolit
Change in the colour of the electrolyte
Keamatan warna biru
berkurang
disebabkan kepekatan
berkurang
.
decreases
because the concentration
decreases
.
ion kuprum(II), Cu2+
The intensity of blue colour
of copper(II) ions Cu2+
2. Larutan kuprum(II) sulfat, CuSO4 dengan elektrod kuprum
Copper(II) sulphate, CuSO4 solution with copper electrodes
(a) Ion yang hadir dalam elektrolit
Ions present in the electrolyte
(b) Ion yang tertarik ke anod dan katod
Ions attracted to the anode and cathode
CuSO4  Cu2+ + SO42–
H2O
H+ + OH–
Anod/ Anode (+): Cu
Katod/ Cathode (–): H+, Cu2+
41
(c) Ion yang dinyahcas di anod dan katod. Anod/ Anode: Cu
Nyatakan sebab.
Alasan/ Reason:
Ions discharged at the anode and cathode. State the Logam kuprum, Cu lebih reaktif
reason.
Copper, Cu metal more reactive
Katod/ Cathode: Cu2+
Alasan/ Reason:
E0 ion Cu2+ lebih positif
E0 of Cu2+ ion more positive
Cu  Cu2+ + 2e–
(d) Persamaan setengah bagi tindak balas di Anod/ Anode:
anod dan katod
Half-equations for reactions at the anode and Katod/ Cathode:
IA
cathode
(e) Hasil yang terbentuk di anod dan katod
Products formed at the anode and cathode
Cu2+ + 2e–  Cu
Anod/ Anode:
Ion kuprum(II), Cu2+/ Copper (II) ions, Cu2+
Katod/ Cathode:
Logam kuprum, Cu/ Copper, Cu metal
(f ) Perubahan warna elektrolit
tidak berubah
kerana kepekatan
tidak
berubah
ion kuprum(II), Cu
. Ini disebabkan
kadar pengionan atom kuprum di anod adalah sama dengan
AS
Keamatan warna biru
Change in the colour of the electrolyte
2+
kadar nyahcas ion kuprum(II), Cu2+ di katod.
.
unchanged
because the concentration of
remains
unchanged
. This is because the
copper(II) ions, Cu
rate of ionisation of copper atom at the anode is equal to the rate of
Intensity of blue colour
2+
PA
N
discharge of copper(II) ions, Cu2+ at the cathode.
D
.
Membanding Sel Kimia dan Sel Elektrolisis
Comparing Voltaic Cell and Electrolytic Cell
I
Tindak Balas Redoks dalam Sel Elektrolisis/ Redox Reactions in Electrolytic Cells
1. Sebuah sel elektrolisis terdiri daripada bateri, elektrolit dan dua elektrod yang disambung kepada bateri.
An electrolytic cell consists of a battery, an electrolyte and two electrodes which are connected to the battery.
e
e–
Anod
Anode
–
Bateri
Battery
–
–
+
e–
Katod
Cathode
–
+
+
Rajah/Diagram 1.26
2. Elektrod yang disambung kepada terminal positif bateri dikenali sebagai
yang disambung kepada terminal negatif bateri dikenali sebagai
anode
The electrode connected to the positive terminal of the battery is known as
cathode
to the negative terminal of the battery is known as
42
anod
katod
manakala elektrod
.
while the electrode connected
.
litar luar
3. Dalam sel elektrolisis, elektron mengalir melalui
dari anod ke katod.
external circuit .
In an electrolytic cell, electrons are flowed from anode to cathode through the
anod
4. Semasa elektrolisis, anion (ion negatif) tertarik ke
katod
ke
.
anode
During electrolysis, anions (negative ions) are attracted to
cathode
.
to
menderma
5. Di anod, anion dinyahcas dengan
pengoksidaan
berlaku di anod.
donating
elektron kepada anod. Oleh itu, proses
oxidation
electrons to anode. Thus,
menerima
6. Di katod, kation dinyahcas dengan
penurunan
berlaku di katod.
accepting
At the cathode, cations are discharge by
cathode.
whereas cations (positive ions) are attracted
IA
At the anode, anions are discharge by
manakala kation (ion positif) tertarik
occurs at anode.
elektron daripada katod. Oleh itu, proses
reduction
electrons from cathode. Thus,
occurs at
II Tindak Balas Redoks dalam Sel Kimia/ Redox Reactions in Voltaic Cells
AS
1. Sebuah sel kimia ringkas terdiri daripada dua logam berlainan yang dicelup ke dalam suatu elektrolit dan
disambung dengan wayar penyambung
A simple voltaic cell consists of two different metals which are immersed into an electrolyte and connected by wire.
Voltmeter
Voltmeter
V
e–
e–
PA
N
Terminal
negatif (anod)
Negative
terminal (anode)
e
–
–
e–
+
e – e–
e–
Terminal
positif (katod)
Positive
terminal (cathode)
+
Rajah/Diagram 1.27
2. Semakin negatif nilai keupayaan elektrod, logam tersebut akan bertindak sebagai terminal
The more negative the electrode potential value, the metal will act as a
negative
At the negative terminal (anode), the more electropositive metal is corroded by
oxidation
occurs at the negative terminal (anode).
4. Di terminal positif (katod), kation dalam elektrolit dinyahcas dengan
penurunan
berlaku di terminal positif (katod).
katod. Oleh itu,
At the positive terminal (cathode), cations in the electrolyte are discharge by
reduction
occurs at the positive terminal (cathode).
Thus,
5. Di terminal positif (katod), kation (ion positif) daripada elektrolit akan
terminal positif (katod).
At the positive terminal (cathode), cations (positive ions) from the electrolyte will
the positive terminal (cathode).
donating
menerima
accepting
menerima
accept
6. Logam yang lebih elektropositif akan menderma elektron. Elektron akan mengalir melalui
dari terminal negatif (anod) ke terminal positif (katod).
The more electropositive metal
donates
positive terminal (cathode) through the
.
terminal.
menderma
3. Di terminal negatif (anod), logam yang lebih elektropositif terkakis dengan
pengoksidaan
itu,
berlaku di terminal negatif (anod).
negatif
elektron. Oleh
electrons. Thus,
elektron daripada
electrons from cathode.
elektron daripada
the electrons from
litar luar
electrons. The electrons are flowed from negative terminal (anode) to
external circuit .
43
Contoh/Example 1
Rajah 1.28 menunjukkan satu sel kimia yang dibina dengan menggunakan elektrod zink dan elektrod kuprum. Kedua-dua
elektrod disambung kepada voltmeter dengan wayar penyambung dan litar dilengkapkan dengan titian garam.
Diagram 1.28 shows a voltaic cell constructed using zinc and copper electrodes. Both electrodes are connected to the voltmeter using the
connecting wires and the circuit is completed with a salt bridge.
V
E0 Zn2+Zn = –0.76 V
E0 Cu2+Cu = +0.34 V
e–
e–
Titian garam
Salt bridge
–
+
Elektrod zink
Zinc electrode
Larutan zink sulfat, ZnSO4
1.0 mol dm–3
1.0 mol dm–3 zinc sulphate,
ZnSO4 solution
Elektrod kuprum
Copper electrode
IA
Larutan kuprum(II)
sulfat, CuSO4 1.0 mol dm–3
1.0 mol dm–3 copper(II) sulphate,
CuSO4 solution
Rajah/Diagram 1.28
negatif
1. Berdasarkan nilai E0, zink, Zn bertindak sebagai terminal
positif
terminal
.
negative
Based on the E0 value, zinc, Zn acts as a
ppositivc
ositivc
terminal while copper, Cu acts as a
menderma
2. Zink, Zn mengalami proses pengoksidaan dengan
terminal.
elektron.
AS
donating
Zinc, Zn undergo an oxidation process by
electrons.
Zn → Zn2+ + 2e-
(Pengoksidaan/ Oxidation)
menerima
elektron.
3. Ion kuprum(II), Cu2+ mengalami proses penurunan dengan
accepting
Copper(II) ion, Cu2+ undergo a reduction process by
Cu2+ + 2e– → Cu
penurunan
4. Zink, Zn bertindak sebagai agen
reducing
electrons.
(Penurunan/ Reduction)
manakala ion kuprum(II), Cu2+ bertindak sebagai agen pengoksidaan .
agent while copper(II) ions, Cu2+ act as an
PA
N
Zinc, Zn acts as a
manakala kuprum, Cu bertindak sebagai
oxidising
agent.
III Perbezaan antara Sel Kimia dan Sel Elektrolisis/ Differences Between a Voltaic Cell and an Electrolytic Cell
Sel elektrolisis/ Electrolytic cell
+
Sel kimia/ Voltaic cell
–
V
e–
Anod
Anode
+
–
Katod
Cathode
+
Terminal positif
Positive terminal
e–
–
Terminal negatif
Negative terminal
Elektrolit
Electrolyte
Elektrolit
Electrolyte
Rajah/Diagram 1.29
Elektrod daripada bahan yang
The electrodes are from the
materials.
Tenaga elektrik ditukar kepada
Electrical energy is converted to
Elektron
kepada
sama
same
Rajah/Diagram 1.30
atau berbeza. Elektrod mesti daripada jenis logam
or different The electrodes must be from
metal.
tenaga kimia
chemical energy
.
.
anod
mengalir daripada
katod
melalui litar luar.
anode
Electrons flow from
through external circuit.
to
Tenaga kimia
Chemical energy
berbeza
different
.
types of
ditukar kepada tenaga elektrik.
is converted to electrical energy.
Elektron mengalir daripada logam lebih elektropositif
kepada logam kurang elektropositif melalui litar luar.
cathode
Electrons flow from
less electropositive
44
more electropositive
metal to
metal through external circuit.
Penurunan
menerima elektron.
Reduction
electrons.
berlaku pada katod (–). Ion positif
occurs at cathode (-). Positive ions accept
Pengoksidaan
menderma elektron.
Oxidation
electrons.
Penurunan
berlaku pada terminal positif (+).
Ion positif menerima elektron.
Reduction
occurs at the positive terminal (+).
Positive ions accept electrons.
Pengoksidaan
berlaku di anod (+). Ion negatif
berlaku pada terminal negatif (–)
menderma
elektron untuk membentuk
dengan
occurs at anode (+). Negative ions donate kation.
Oxidation
electrons to form cations.
Penyaduran dan Penulenan Logam secara Elektrolisis
Electroplating and Purification of Metals by Electrolysis
I
Kepentingan Penyaduran Logam/ The Importance of Electroplating of Metals
katod
1. Dalam proses penyaduran, objek yang hendak disadur dijadikan
yang tulen dijadikan
anod
.
manakala logam penyadur
AS
E
occurs at the negative terminal (–) by
IA
donating
In electroplating process, object to be electroplated is used as the
anode
.
as the
cathode
2. Elektrolit yang digunakan ialah larutan akueus yang mengandungi ion
plating metal
The electrolyte used is an aqueous solution that contains the
while the pure plating metal is used
logam penyadur
.
ions.
mengelakkan logam daripada kakisan
3. Penyaduran logam adalah penting untuk
prevent the metal from corrosion
Electroplating of metal is important to
.
.
PA
N
4. Menyadur sudu besi dengan logam kuprum:/To electroplate an iron spoon with copper:
A
Kuprum sebagai logam penyadur
Copper metal as the plating metal
Larutan kuprum(II) sulfat, CuSO4 sebagai elektrolit
Copper(II) sulphate, CuSO4 solution as the electrolyte
Sudu besi yang akan disadur
Iron spoon to be electroplated
Rajah/Diagram 1.31
katod
(a) Sudu besi digunakan sebagai
The iron spoon is used as the
(b) Larutan
cathode
kuprum(II) sulfat
Copper(II) sulphate
, while copper metal is used as the
.
solution is used as the electrolyte.
katod
cathode
Copper(II) ions, Cu2+ will be discharged at the
Katod/ Cathode:
anode
digunakan sebagai elektrolit.
(c) Ion kuprum(II), Cu2+ akan dinyahcas di
Anod/ Anode:
anod
, manakala logam kuprum digunakan sebagai
Cu  Cu + 2e
2+
–
dan terenap pada permukaan sudu besi.
and deposited on the surface of the iron spoon.
(Pengoksidaan/ Oxidation)
Cu2+ + 2e–  Cu
(Penurunan/ Reduction)
II Kepentingan Penulenan Logam/ The Importance of Purification of Metals
1. Kuprum dan perak tulen boleh diperoleh melalui elektrolisis.
Pure copper and silver can be obtained through electrolysis.
mendapatkan logam yang tulen
2. Penulenan logam adalah penting untuk
Purification of metal is important to
obtain a pure metal
3. Menulenkan logam kuprum:/ To purify copper metal:
anod
(a) Logam kuprum tidak tulen digunakan sebagai
The impure copper metal is used as the
anode
45
.
.
.
.
.
katod
(b) Logam tulen kuprum digunakan sebagai
cathode
The pure copper metal is used as the
kuprum(II) sulfat
(c) Larutan
Copper(II) sulphate
.
.
digunakan sebagai elektrolit.
solution is used as the electrolyte.
A
+
–
Logam kuprum tidak tulen
Impure copper metal
Logam kuprum tulen
Pure copper metal
IA
Larutan kuprum(II) sulfat, CuSO4
sebagai elektrolit
Copper(II) sulphate, CuSO4 solution
as the electrolyte
Rajah/Diagram 1.32
mengion
(d) Anod akan
dan kekotoran akan jatuh ke dasar bikar.
ionise
The anode will
and the impurities will settle to the bottom of the beaker.
Cu  Cu2+ + 2e–
Anod/ Anode:
(Pengoksidaan/ Oxidation
Oxidation)
AS
(e) Ion kuprum(II), Cu2+ dan ion hidrogen, H+ akan tertarik ke katod. Ion kuprum(II), Cu2+ akan
dinyahcas
secara pilihan di katod. Lapisan kuprum dienapkan pada kuprum tulen.
Copper(II) ions, Cu2+ and hydrogen ions, H+ will attracted to the cathode. Copper(II) ions, Cu2+ are selectively
discharged
at the cathode. A layer of copper is formed on the pure copper.
Cu2+ + 2e–  Cu
Katod/ Cathode:
1.5
Pengestrakan Logam daripada Bijihnya
Extraction of Metals from Their Ores
Pengekstrakan Logam Reaktif daripada Bijihnya melalui Proses Elektrolisis
Extraction of Reactive Metals from Their Ores through Electrolysis
PA
N
A
(Penurunan/ Reduction)
1. Logam reaktif seperti aluminium boleh diekstrak daripada bauksit yang mengandungi aluminium oksida, Al2O3
melalui elektrolisis.
Reactive metal such as aluminium can be extracted from bauxite, which contains aluminium oxide, Al2O3 by electrolysis.
Sisa gas
Waste gas
+
Aluminium diekstrak keluar daripada tangki keluli
Aluminium extracted out of stainless steel tank
V = 5V
I = 200 kA
Elektrod grafit
Graphite electrodes
Campuran kriolit dan aluminium oksida
Mixture of cryolite and aluminium oxide
Aluminium lebur
Molten aluminium
–
Tangki keluli
Stainless steel tank
Elektrod
Electrode
Lapisan karbon sebagai katod
Carbon lining as a cathode
Rajah/ Diagram 1.33 Pengekstrakan aluminium oksida, Al2O3/ Extraction of aluminium oxide, Al2O3
2. Elektrolisis aluminium oksida lebur menggunakan elektrod
Electrolysis of molten aluminium oxide used
3. Kriolit ditambah untuk
Cryolite is added to
menurunkan
lower
carbon
karbon
.
electrodes.
takat lebur aluminium oksida, Al2O3 kepada 980 °C.
the melting point of aluminium oxide, Al2O3 to 980 °C.
2Al3+ + 3O2–
Δ
Katod/ Cathode: Al3+ + 3e–  Al (Penurunan/ Reduction)
Anod/ Anode: 2O2–  O2 + 4e– (Pengoksidaan/ Oxidation)
Al2O3
46
4. Pengekstrakan aluminium daripada bauksit boleh mendatangkan kesan buruk terhadap alam sekitar.
Extraction of aluminium from bauxite can have a harmful effect on the environment.
karbon dioksida , karbon monoksida ,
(a) Proses peleburan dan pemprosesan membebaskan gas seperti
sulfur dioksida
dan perfluorokarbon yang boleh mencemarkan udara serta menyebabkan hujan asid dan
pemanasan global.
The smelting and processing processes release gases such as carbon dioxide , carbon monoxide ,
and perfluorocarbons that can pollute the air, as well as cause acid rain and global warming.
hakisan tanah
(b) Perlombongan bauksit akan mengakibatkan
Bauxite mining will cause
soil erosion
sulfur dioxide
dan memusnahkan habitat hidupan liar.
and destroy wildlife habitats.
IA
(c) Lumpur dan sisa toksik yang termendap di lombong yang digali akan meresap ke dalam tanah dan mencemarkan
sumber air
.
Mud and toxic waste deposited in excavated mines will seep into the soil and pollute the
(d) Tanah yang telah dilombong akan menjadi
infertile
Land that has been mined will be
.
dan tidak sesuai untuk pertanian.
and unsuitable for agriculture.
Pengekstrakan Logam daripada Bijihnya melalui Proses Penurunan oleh Karbon
AS
B
tidak subur
water sources
Extraction of Metals from Their Ores through Reduction by Carbon
I
Pengekstrakan Besi/ Extraction of Iron
1. Logam besi diekstrak daripada bijihnya, iaitu hematit dan magnetit melalui proses
karbon di dalam relau bagas.
Iron is extracted from its ores, hematite and magnetite through a
reduction
penurunan
oleh
process in a blast furnace.
PA
N
Bijih besi + arang kok
+ batu kapur
Iron ore + coke
+ limestone
Gas buangan
yang panas
Hot waste
gases
800°C
Gas buangan yang panas
(disalur semula ke
bahagian bawah relau)
Hot waste gases
(recycled to heat
the furnace)
1 500°C
2 000°C
Udara panas
Hot air
Udara panas
Hot air
Leburan besi
Molten iron
Leburan sanga
Molten slag
Rajah/ Diagram 1.34 Pengekstrakan besi di dalam relau bagas/ Extraction of iron in a blast furnace
2. Hematit mengandungi ferum(III) oksida, Fe2O3 manakala magnetit mengandungi triferum tetraoksida, Fe3O4.
Hematite contains iron(III) oxide, Fe2O3 while magnetite contains triiron tetraoxide, Fe3O4.
3. Tindak balas kimia dalam relau bagas./ Chemical reactions in the blast furnace.
(a) Karbon bertindak balas dengan oksigen dalam udara panas untuk menghasilkan
The carbon reacts with oxygen in the hot air to form
(b)
Karbon dioksida
karbon dioksida
.
carbon dioxide .
C + O2 → CO2
yang terhasil bertindak balas dengan karbon selebihnya untuk membentuk
karbon monoksida (agen penurunan yang kuat).
Carbon dioxide
formed reacts with more hot carbon to form carbon monoxide (strong reducing agent)
CO2 + C → 2CO
47
menurunkan
(c) Karbon monoksida dan karbon
reduce
Carbon monoxide and carbon
2Fe2O3
+
2Fe2O3
+
Hematit/Hematite
Hematit/Hematite
3C
4Fe
+
3CO2
3CO
2Fe
Karbon monoksida/Carbon monoxide
+
Fe3O4
+
Magnetit/ Magnetite
the iron oxides to iron.
Karbon/Carbon
Fe3O4
Magnetit/ Magnetite
oksida besi kepada besi.
2C
3Fe
Karbon/Carbon
+
+
3CO2
2CO2
4CO
Karbon monoksida/Carbon monoxide
3Fe
+
4CO2
leburan
.
IA
4. Pada suhu yang tinggi dalam relau bagas, besi yang terbentuk wujud dalam keadaan
Leburan besi mengalir ke bahagian bawah relau.
molten
At high temperature in the blast furnace, the iron formed is in
of the blast furnace.
state. The molten iron flows to the bottom
state.
5. Besi lebur dituang ke dalam acuan dan dibiarkan menyejuk. Besi yang diperoleh dikenali sebagai besi tuangan
dan mengandungi kira-kira 4% karbon dan benda asing lain.
The molten iron is poured into moulds and is allowed to cool. The iron obtained is known as cast iron and contains about 4%
of carbon and other impurities.
AS
6. Pada suhu yang tinggi, batu kapur terurai untuk membentuk kalsium oksida dan karbon dioksida. Kalsium oksida
bertindak balas dengan bendasing berasid seperti pasir (silikon dioksida) untuk membe
membentuk
ntuk leburan sanga.
At high temperatures, limestone is decomposed to produce calcium oxide and carbon dioxide. Calcium oxide reacts with
acidic impurities, for example sand (silicon dioxide) to produce a molten slag.
7. Di bahagian bawah relau leburan sanga terapung di atas lapisan leburan besi. Leburan sanga dikeluarkan mengikut
sela masa yang ditetapkan.
At the bottom of the furnace, slag floats on top of the molten iron. The slag is tapped off at regular intervals.
kedudukan
8. Kaedah yang digunakan dalam pengekstrakan logam daripada bijihnya bergantung kepada
logam dalam siri kereaktifan.
The method used to extract the metal from its ore depends on the
position
of the metal in the reactivity series.
PA
N
9. Siri kereaktifan merupakan satu senarai logam yang disusun berdasarkan kereaktifan logam terhadap
oksigen
.
The reactivity series is a list of metal which are arranged according to their reactivity with
penurunan
10. Logam-logam yang berada di atas dalam siri kereaktifan ialah agen
logam ini dapat menurunkan oksida bagi logam yang kurang reaktif.
Metals at higher position in the reactivity series are strong
reactive metals.
reducing
oxygen
.
yang kuat. Logam-
agent. They can reduce the oxides of less
Logam
Kaedah pengekstrakan
K
Na
Ca
Mg
Elektrolisis leburan klorida/ Electrolysis of molten chloride
Al
Elektrolisis leburan oksida/ Electrolysis of molten oxide
Zn
Fe
Sn
Pb
Penurunan oksida oleh karbon/ Reduction of oxide by carbon
Cu
Ag
Hg
Memanaskan bijih secara terus dalam udara/ Heating ores directly in air
Au
Wujud sebagai logam bebas/ Exist as free metals
Metals
Extraction method
48
Uji Kendiri
1.5
Tuliskan persamaan setengah tindak balas penurunan dan pengoksidaan dalam elektrolisis leburan natrium klorida, NaCl.
Write the half-equation reactions of reduction and oxidation in the electrolysis of molten sodium chloride, NaCl.
NaCl
Anod/ Anode:
1.6
2Cl–  Cl2 + 2e–
(Penurunan/Reduction)
(Pengoksidaan/Oxidation)
Pengaratan
Rusting
Proses Pengaratan sebagai Tindak Balas Redoks
Rusting Process as a Redox Reaction
AS
A
Na+ + e–  Na
Na+ + Cl-
IA
Katod/ Cathode:
Δ
1. Pengaratan ialah kakisan besi. Pengaratan besi memerlukan kehadiran kedua-dua
air
.
Rusting is the corrosion of iron. The rusting of iron requires both
oxygen
water
and
oksigen
dan
.
2. Mekanisme pengaratan besi/ Iron rusting mechanism:
Karat/ Rust
(Fe2)3.xH2O
PA
N
Titisan air
Water droplet
OH–
O2
O2
Katod (+)
Cathode
Fe2+
e–
Anod(–)
Anode
Katod (+)
Cathode
Besi/ Iron
Rajah/ Diagram 1.35 Mekanisme pengaratan besi/ Iron rusting mechanism
(a) Permukaan besi, di tengah titisan air bertindak sebagai anod. Di anod, atom besi
dengan
kehilangan
dua elektron dan membentuk ion ferum(II), Fe2+.
Iron surface, in the centre of a water droplet acts as the anode. At the anode iron atoms are
lose
two electrons and form iron(II) ion, Fe2+.
dioksidakan
oxidised
by
Anod/ Anode: Fe  Fe2+ + 2e– (Pengoksidaan/ Oxidation)
(b) Permukaan besi di pinggir titisan air bertindak sebagai katod. Elektron bergerak ke
titisan air. Elektron diterima oleh oksigen dan molekul air membentuk ion
penurunan
berlaku.
hidroksida
edge
Iron surface at the edge of the water droplet serves as the cathode. Electrons flow to the
hydroxide
ions OH–. A
droplet. The electrons are received by oxygen and water molecules to form
process occurs.
Katod/ Cathode: 2H2O + O2 + 4e–  4OH– (Penurunan / Reduction)
49
pinggir
, OH–. Proses
of the water
reduction
(c) Ion ferum(II), Fe2+ bergabung dengan ion hidroksida, OH– membentuk pepejal berwarna
ferum(II) hidroksida, Fe(OH)2.
green
Iron(II) ions, Fe2+ combine with hydroxide ions, OH– to form
hijau
,
substance, iron(II) hydroxide, Fe(OH)2.
Fe2+ + 2OH–  Fe(OH)2
(d) Dalam kehadiran udara, ferum(II) hidroksida, Fe(OH)2
oksida terhidrat, Fe2O3.xH2O. Bahan berwarna
dioksidakan
perang
In the presence of air, iron(II) hydroxide, Fe(OH)2 is
brown
substance known as rust.
Fe2O3.xH2O. This
oleh oksigen membentuk ferum(III)
ini dikenali sebagai karat.
oxidised
by oxygen to form hydrated iron(III) oxide,
4Fe(OH)2(p/s) + O2(g/g) + 2H2O(ce/l)  4Fe(OH)3(p/s
(p/s)
(p/
s)
B
IA
O(p/s)) + H2O(ce/
O(p/
O(ce/l)
O(ce/l
l)
l)
2Fe(OH)3(p/s)  Fe2O3.2H2O(p/s
Mencegah Pengaratan Besi
To Prevent Rusting of Iron
1. Pengaratan besi dapat dicegah atau sekurang-kurangnya diminimumkan dengan melindungi permukaan besi
daripada terdedah kepada oksigen dan kelembapan.
Rusting of iron can be prevented, or at least minimised, by shielding the iron surface from oxygen and moisture.
2. Tiga kaedah untuk mencegah pengaratan:
AS
The three methods to prevent rusting:
(a) Pelindungan fizikal – melindungi permukaan besi dengan mengecat, menyapu gris, menyalut dengan plastik
dan penyaduran dengan logam kurang reaktif seperti kromium, kuprum dan stanum.
Physical protection – covers the surfaces of iron from water and air by painting, applying grease, coating with plastic and
electroplating with metals that are less reactive such as chromium, copper and tin.
(b) Perlindungan elektrokimia – menyadurkan permukaan besi dengan logam yang lebih elektropositif seperti
zink.
Electrochemical protection – electroplating the surface of iron with metals that are more electropositive such as zinc.
(c) Penggalvanian – menyalut permukaan ferum dengan lapisan logam
zink
. Zink adalah lebih
dioksidakan
berbanding ferum. Oleh itu, atom zink lebih mudah
berbanding
ferum. Maka, pengoksidaan atom ferum yang menyebabkan pengaratan besi dapat dielakkan.
PA
N
elektropositif
zinc
Galvanising – a process of covering the surface of iron with a layer of
metal. Zinc is more electropositive
compared to iron. So, atom zinc is easier to be oxidised compared to iron. Thus, the oxidation of iron atoms which caused
rusting is prevented.
Zn  Zn2+ + 2e–
Eksperimen
1.4
Eksperimen Wajib
Tujuan/ Aim:
Mengkaji Eksperimen
kesan logam lain yang bersentuhan dengan besi terhadap pengaratan besi
To investigate the effect of other metals in contact with iron on the rusting of iron
Penyataan masalah/ Problem statement:
Bagaimana logam berlainan jenis yang bersentuhan dengan besi mempengaruhi pengaratan besi?
How do different types of metal in contact with iron affect the rusting of iron?
Hipotesis/ Hypothesis:
Logam yang lebih elektropositif akan menghalang pengaratan besi. Logam yang kurang elektropositif akan
mempercepatkan pengaratan besi.
A more electropositive metal will prevent iron from rusting. A less electropositive metal will speed up rusting.
Pemboleh ubah/ Variables:
(a) Dimanipulasikan/ Manipulated: Jenis logam yang bersentuhan dengan paku besi
Types of metal in contact with iron nail
50
(b) Bergerak balas/ Responding: Pengaratan paku besi/ Kehadiran warna biru
Rusting of iron nails/ Presence of blue colouration
(c) Dimalarkan/ Fixed: Paku besi, suhu persekitaran/ Iron nails, temperature of the surroundings
Bahan/ Materials:
Paku besi, pita magnesium, jalur kuprum, larutan agar-agar panas, larutan kalium heksasianoferat(III), K3[Fe(CN)6],
penunjuk fenolftalein
IA
Iron nails, magnesium ribbon, copper strip, hot agar-agar solution, potassium hexacyanoferrate(III), K3[Fe(CN)6] solution, phenolphthalein
indicator
Radas/ Apparatus:
Tabung uji, rak tabung uji, kertas pasir
Test tubes, test tube rack, sand paper
Prosedur/ Procedure:
AS
Agar-agar + kalium heksasianoferat(III) + fenolftalein
Agar-agar + potassium hexacyanoferrate(III) + phenolphthalein
A
B
Paku besi
Iron nail
Pita magnesium
Magnesium ribbon
C
Paku besi
Iron nail
Jalur kuprum
Copper strip
Rak tabung uji
Test tube rack
Paku besi
Iron nail
PA
N
Rajah/Diagram 1.36
1. Tiga batang paku besi, pita magnesium dan jalur kuprum dibersihkan dengan kertas pasir.
Three iron nails, magnesium ribbon and copper strip are cleaned using a sand paper.
2. Pita magnesium dililitkan pada paku besi pertama dan jalur kuprum dililitkan pada paku besi kedua.
The magnesium ribbon is wrapped onto the first iron nail and the copper strip is wrapped onto the second iron nail.
3. Paku besi kemudian dimasukkan ke dalam tiga tabung uji dan dilabelkan dengan A, B dan C seperti ditunjukkan
dalam Rajah 1.36.
The iron nails are then inserted into three test tubes labelled as A, B and C as shown in Diagram 1.36.
4. Larutan agar-agar panas yang ditambahkan dengan kalium heksasianoferat(III) dan penunjuk fenolftalein dituang
ke dalam setiap tabung uji sehingga menutupi seluruh paku besi.
Hot agar-agar solution is added with potassium hexacyanoferrate (III) and phenolphthalein indicator is poured into each test tube
until it covers the whole iron nail.
5. Tabung uji dibiarkan pada rak tabung uji selama tiga hari. Pemerhatian direkodkan.
The test tubes are kept in a test tube rack for three days. The observations are recorded.
Pemerhatian/ Observations:
Tabung uji
Test tube
A
(Mg + Fe)
Keamatan warna biru
Intensity of the blue
colouration
Tiada warna biru terbentuk
No blue colour formed
Keamatan warna merah
jambu/ Intensity of the
Inferens
Inference
pink colouration
Warna merah jambu Tiada ion
OH–
dengan keamatan tinggi ion
terbentuk
mencegah
High intensity pink colour
formed
Fe2+
hadir. Banyak
hadir.
Magnesium
pengaratan.
ions not presence. A lot of
present. Magnesium
51
Fe2+
prevents
OH–
ions
rusting.
B
(Cu + Fe)
Fe2+
Warna biru dengan keamatan Tiada warna merah jambu Banyak ion
hadir. Kuprum
menggalakkan
tinggi terbentuk
terbentuk
pengaratan.
High intensity blue colour formed
No pink colour formed
Fe2+
A lot of
promotes
Warna biru dengan keamatan
C
(Fe sahaja/ rendah terbentuk
Low intensity blue colour formed
only)
Ion
Fe2+
Fe2+
ions present. Copper
rusting
hadir. Besi berkarat.
ions presence. Iron undergoes rusting.
Kesimpulan/ Conclusion:
mencegah
elektropositif daripada besi akan
pengaratan manakala logam
menggalakkan
elektropositif daripada besi akan
pengaratan.
yang
kurang
Metals that are
more
electropositive than iron will
speed up
electropositive than iron will
rusting.
IA
lebih
Logam yang
prevent
rusting while metals that are
less
Perbincangan/ Discussion:
1. Larutan kalium heksasianoferat(III), K3[Fe(CN)6], digunakan untuk mengesan kehadiran ion Fe2+. Apabila ion
keamatan warna
biru tua
terhasil. Semakin banyak ion Fe2+terhasil, semakin
tinggi
AS
Fe2+. hadir, warna
biru tua
.
Potassium hexacyanoferrate(III), K3[Fe(CN)6] solution is used to detect the presence of Fe2+ ions. When Fe2+ ions presence, a
dark blue
colour produced. The more Fe2+ ions formed, the
higher
the intensity of the dark blue colour formed.
2. Penunjuk fenolftalein digunakan untuk mengesan kehadiran ion OH−. Kehadiran ion OH− meningkatkan
kealkalian larutan dan memberi warna
merah jambu
kepada larutan.
Phenolphthalein indicator is used to detect the presence of OH ions. The presence of OH– ions increases the alkalinity of the solution
–
pink
colour to the solution.
PA
N
and gives
3. Tuliskan persamaan setengah bagi proses pengoksidaan yang berikut.
Write the half-equations for the following oxidation processes.
Tabung uji
Persamaan setengah
A (Mg + Fe)
Mg  Mg2+ + 2e–
B (Cu + Fe)
Fe  Fe2+ + 2e–
C (Fe sahaja/
sahaja/only)
only
Fe  Fe2+ + 2e–
Test tube
Half-equation
4. Eksperimen dalam tabung uji B telah diulang dengan menggantikan jalur kuprum, Cu dengan jalur zink, Zn.
Experiment in test tube B is repeated by replacing copper strip, Cu with zinc strip, Zn.
(a) Ramalkan pemerhatian.
Predict the observations.
Warna merah jambu terbentuk.
Pink colour formed.
(b) Nyatakan inferens.
State the inference.
Zink lebih elektropositif daripada besi. Zink mengalami pengoksidaan. Zink mencegah pengaratan.
Zinc is more electropositive than iron. Zinc will undergoes oxidation. Zinc prevent rusting.
52
Contoh/Example 1
Rajah 1.37 menunjukkan aktiviti yang dijalankan untuk mengkaji kesan logam yang berlainan terhadap pengaratan besi.
Diagram 1.37 shows an activity carried out to investigate the effect of other metals on the rusting of iron.
Agar-agar + kalium
heksasianoferat(III)
+ fenolftalein
Agar-agar + potassium
hexacyanoferrate(III)
+ phenolphthalein
Paku besi
Iron nail
Paku besi
Iron nail
Pita magnesium
Magnesium ribbon
I
IA
Agar-agar + kalium
heksasianoferat(III)
+ fenolftalein
Agar-agar + potassium
hexacyanoferrate(III)
+ phenolphthalein
Jalur kuprum
Copper strip
II
AS
Rajah/Diagram 1.37
Jadual di bawah menunjukkan keputusan aktiviti.
Table below shows the results of the activity.
Set
Pasangan logam
Pair of metals
I
Besi + Magnesium
II
Besi + Kuprum
Pemerhatian
Observation
Warna merah jambu dengan keamatan tinggi terbentuk
Iron + Magnesium
High intensity pink colour formed.
PA
N
Warna biru dengan keamatan tinggi terbentuk.
Iron + Copper
High intensity blue colour formed.
menghalang
1. Logam magnesium
Magnesium metal
prevents
pengaratan besi manakala logam kuprum
the rusting of iron while copper metal
penurunan
2. Logam magnesium ialah agen
besi
. Maka, besi
sebelum
Magnesium metal is a stronger
iron
. Thus, iron is
3. Logam kuprum ialah agen
kuprum
sebelum
reducing
prevented
speed up
yang lebih kuat berbanding besi.
dihalang
daripada berkarat.
agent compared to iron.
yang kurang kuat berbanding besi.
dipercepatkan .
. Maka, pengaratan besi
reducing
Thus, the rusting of iron is
speed up
pengaratan besi.
the rusting of iron.
Magnesium
Magnesium
akan mengion
will ionise before
from rusting.
penurunan
Copper metal is a weaker
mempercepatkan
agent compared to iron.
.
53
Iron
Besi
will ionise before
akan mengion
copper
.
Uji Kendiri
1.6
Rajah 1.38 menunjukkan suatu eksperimen untuk mengkaji kesan zink, Zn dan logam X terhadap pengaratan paku besi.
Keputusan eksperimen direkodkan dalam Jadual 1.
Diagram 1.38 shows an experiment to study the effects of zinc, Zn and metal X on the corrosion of iron nails. The results of the
experiment are recorded in Table 1.
Paku besi
Iron nail
Logam zink
Zinc metal
IA
Paku besi
Iron nail
Logam X
Metal X
A
B
Larutan agar-agar + Larutan kalium heksasianoferat(III) + fenolftalein
Agar-agar solution + potassium hexacyanoferrate(III) solution + phenolphthalein
Rajah/Diagram 1.38
Test tube
A
B
Pemerhatian
AS
Tabung uji
Observation
Warna biru tua yang banyak terbentuk
Lots of dark blue colour formed.
Warna merah jambu terbentuk. Gelembung gas terbentuk.
A pink colour formed. Gas bubbles are formed.
Jadual/Table 1
1. (a) Nyatakan nama bagi ion yang memberikan warna biru tua dalam tabung uji A.
State the name of the ion that give the dark blue colour in test tube A.
PA
N
Ion ferum(II), Fe2+/ Ferum(II) ion, Fe2+
(b) Tuliskan persamaan setengah untuk mewakili pembentukan ion di (a).
Write the half-equation to represent the formation of ions in (a).
Fe(p/s)
Fe2+(ak/aq) + 2e–
2. (a) Nyatakan nama bagi ion yang memberikan warna merah jambu dalam tabung uji B.
State the name of the ion that give the pink colour in test tube B.
Ion hidroksida, OH–/ Hydroxide ion, OH-
(b) Tuliskan persamaan setengah untuk mewakili pembentukan ion di (a).
Write the half-equation to represent the formation of ions in (a).
O2(ak/aq) + 2H2O(ce/l) + 4e–
4OH–(ak/aq)
3. (a) Paku besi di dalam tabung uji yang manakah tidak berkarat? Jelaskan sebabnya.
Iron nail in which test tube does not rust? Explain the reason.
Tabung uji B.. Zink lebih elektropositif daripada besi.
Test tube B. Zinc is more electropositive than iron.
4. (a) Namakan dua logam yang mungkin bagi X.
Name two possible metals for X.
Plumbum dan kuprum/ Lead and copper
54