KSSM Form 5 Experiments
Activity 1.1 To investigate oxidation and reduction in the transfer of electrons at a distance.
Aim: To investigate oxidation and reduction in the transfer of electrons at a distance.
Materials: 2 mol dm-3 sulphuric acid, 0.5 mol dm-3 freshly prepared iron(II) sulphate solution,
0.2 mol dm-3 acidified potassium mangante(VII) solution, 0.5 mol dm-3 potassium iodide
solution, 0.2 mol dm-3 acidified potassium dichromate(VI) solution, 0.2 mol dm-3
potassium thiocynate solution, 1% starch solution.
Apparatus: U-tube, galvanometer, connecting wires with crocodile clips, carbon electrodes, retort
stand and clamp, test tube, dropper, stoppers with one hole.
Procedure:
1. A U-tube is clamped to a retort stand.
2. 20 cm3 of 1 mol dm-3 dilute sulphuric acid is poured into the U-tube.
3. 10 cm3 of 0.5 mol dm-3 iron(II) sulphate solution is added slowly drop by drop into the left arm
of U-tube using a dropper.
4. 10 cm3 of 0.2 mol dm-3 acidified potassium manganate(VII) solution is added slowly drop by
drop into the right arm of U-tube using a dropper.
5. Carbon electrodes are immersed/ placed into the solutions and connected to a galvanometer.
6. The electrodes are connected to a galvanometer as shown in diagram. Based on the deflection
of the galvanometer, the electrodes that act as the positive terminal and negative terminal are
determined.
7. The solutions are left for 20 minutes to react. Any change is observed.
8. Using a clean dropper, 1 cm3 of iron(II) sulphate is drawn out and placed in a test tube, Then, a
few drops of 0.2 mol dm-3 potassium thiocyanate solution are added to the test tube. Any
change is observed.
9. Steps 1 to 7 are repeated using 0.5 mol dm-3 potassium iodide solution and 0.2 mol dm-3
acidified potassium dichromate(VI) solution to replace the iron(II) sulphate solution and
acidified potassium manganate(VII) solution. Step 8 is repeated to test the potassium iodide
solution with 1 % starch solution.
Results:
1. Solution used: Iron(II) sulphate solution and acidified potassium manganate(VII)
solution:
Observation
Inference
(a) The electrode in the iron(II) sulphate
Electron flow from iron(II) sulphate solution
solution acts as the negative terminal
to acidified potassium mangante(VII)
while the electrode in the acidified
solution through external wire.
potassium manganate(VII) solution acts
as the positive terminal.
(b) Iron(II) sulphate solution changes from
At the end of the reaction, iron(III) ions are
pale green to yellow. It gives blood-red
present. Iron(II) ions have changed to
colouration with potassium thiocyanate
iron(III) ions.
solution.
(c) The purple acidified potassium
Mangnate(VII) ions that give the solution its
manganate(VII) solution decolourises
purple colour are used up in the reaction.
1
KSSM Form 5 Experiments
2. Solution used: potassium iodide solution and acidified potassium dichromate (VI)
solution:
Observation
Inference
(a) The electrode in the potassium iodide
Electrons flow from potassium iodide
solution acts as the negative terminal,
solution to acidified potassium
whereas the electrodes in the acidified
dichromate(VI) solution through external
potassium dichromate(VI) solution acts wire.
as the positive terminal.
(b) The colourless potassium iodide
At the end of the reaction, iodine is
solution turns brown. It gives a dark
present. Iodide ions have changed to
blue colouration with starch solution.
iodine.
(c) Potassium dichromate(VI) solution
Dichromate(VI) ions have changed to
changes colour from orange to green.
chromate(III) ions.
Discussion:
1. Iron(II) sulphate solution and acidified potassium manganate(VII) solution
(a) Iron(II) ions act as the reducing agent, releasing electrons to become iron(III) ions. Thus,
iron(II) sulphate solution changes colour from pale green to yellow.
Fe2+  Fe3+ + e(b) Manganate(VII) ions act as the oxidizing agent, accepting the electrons and therefore,
undergoing reduction to become colourless manganese(II) ions.
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
(c) Thus, there is a transfer of electrons from electrode in iron(II) sulphate solution to acidified
potassium manganate(VII) solution through wire.
(d) The overall ionic equation is as follows:
5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+ + 4H2O
2. Potassium iodide solution and acidified potassium dichromate(VI) solution
(a) iodide ions act as the reducing agent, releasing electrons to become iodine molecules.
Thus,the colourless potassium iodide solution turns brown.
2I-  I2 +2 e(b) Dichromate(VI) ions act as the oxidizing agent, accepting the electrons and therefore,
undergoing reduction to become chromate(III) ions.
Cr2O72- + 14 H+ + 6e-  2Cr3+ + 7H2O
(c) Thus, there is a transfer of electrons from electrode in potassium iodide solution to
acidified potassium dichromate(VI) solution through wire.
(d) The overall ionic equation is as follows:
6I- + Cr2O72- + 14H+  3I2 + 2Cr3+ + 7H2O
3. Sulphuric acid has two functions:
(a) To separate the reducing agent from the oxidizing agent
(b) To complete the circuit by allowing the transfer of ions to occur
Conclusion:
The transfer of electrons occurs from the reducing agent to the oxidizing agent through the connecting
wires.
2
KSSM Form 5 Experiments
Activity 1.2
To investigate oxidation and reduction in the change of iron(II) ions to iron(III) ions and vice
versa.
Aim: To investigate oxidation and reduction in the change of iron(II) ions to iron(III) ions and vice
versa.
Material: 0.5 mol dm-3 freshly prepared iron(II) sulphate solution, 0.5 mol dm-3 iron(III) sulphate
solution, bromine water, zinc powder, 2.0 mol dm-3 sodium hydroxide solution, filter paper.
Apparatus: Dropper, spatula, test tubes, test tube holder, Bunsen burner, filter funnel, test tube rack.
A. Changing of iron(II) ions to iron(III) ions
1. 2 cm3 of 0.5 mol dm-3 iron(II) sulphate solution
is poured into a test tube.
2. Using a dropper, bromine water is added to
the solution drop by drop.
3. The test tube is warmed gently.
4. 2.0 mol dm-3 sodium hydroxide solution
is added slowly to the mixture until in excess.
B. Changing of iron(III) ions to iron(II) ions
1. 2 cm3 of 0.5 mol dm-3 iron(III) sulphate solution
is poured into a test tube.
2. Half a spatula of zinc powder is added to the solution.
3. The mixture is filtered.
4. 2.0 mol dm-3 sodium hydroxide solution is added
slowly to the filtrate until in excess.
Observation:
Activity
Reagent
Bromine water
A
B
Sodium hydroxide solution
Zinc powder
Sodium hydroxide solution
Observation
Bromine water decolourises. The solution changes colour
from pale green to yellow.
Brown precipitate is formed. It is insoluble in excess alkali.
Some of the zinc powder dissolves. The solution changes
colour from brown to pale green.
Green precipitate is formed. It is insoluble in excess alkali.
Discussion:
A. Changing of iron(II) ions to iron(III) ions
1. Bromine water oxidises iron(II) ions, Fe2+ to iron(III) ions, Fe3+. The presence of Fe3+
ions is confirmed by the formation of brown precipitate with sodium hydroxide solution.
2. Fe2+ ions lose their electrons and are oxidized to Fe3+ ions.
3. Bromine molecules, which give the bromine water its brown colour, gain the electrons and
are reduced to colourless bromide ions, Br -. This explains why the bromine water is
decolourised.
4. In this reaction, bromine water acts as the oxidizing agent, whereas Fe2+ ions act as the
reducing agent.
5. Oxidation half-equation: Fe2+ Fe3+ + eReduction half-equation: Br2 + 2e 2Br –
Overall ionic equation: 2Fe2+ + Br2  2Fe3+ + 2Br 3
KSSM Form 5 Experiments
B. Changing of iron(III) ions to Iron(II) ions
1. Zinc powder reduces iron(III) ions, Fe2+ to iron(II) ions, Fe2+. The presence of Fe2+
ions is confirmed by the formation of green precipitate with sodium hydroxide solution.
2. Zinc atoms lose their electrons and are oxidized to zinc ions,Zn2+. The explains why zinc
powder dissolves in iron(III) sulphate solution.
3. Fe3+ ions accept these electrons and are reduced to Fe2+ ions.
4. In this reaction, Fe3+ ions act as the oxidizing agent, whereas zinc acts as the reducing
agent.
5. Oxidation half-equation: Zn  Zn2+ + 2eReduction half-equation: Fe3+ + e-  Fe2+
Overall ionic equation: Zn + 2Fe3+  Zn2+ + 2Fe2+
Conclusion:
1. Bromine water acts as an oxidizing agent, changing iron(II) ions to iron(III) ions.
2. Zinc acts as a reducing agent, changing iron(III) ions to iron(II) ions.
4
KSSM Form 5 Experiments
Activity 1.1: To construct electrochemical series based on the potential difference between two
metals.
Aim: To construct Electrochemical Series based on potential difference between two metals.
Problem statement: How can the Electrochemical Series be constructed based on the potential difference
between two metals?
Hypothesis: The further apart the pair of metals in the Electrochemical Series, the greater is their potential
difference.
Variables:
(a) Manipulated variable: Pair of metals.
(b) Responding variable: Potential difference
(c) Controlled variables: Type of electrolyte, concentration of electrolyte, copper electrode
Material: 0.1 mol dm-3 copper (II) sulphate solution, copper strip, lead strip, iron nail, zinc strip, magnesium
ribbon, aluminium strip, sand paper.
Apparatus: Voltmeter, 250 cm3 beaker and connecting wires with crocodile clips.
Procedure:
1.
2.
3.
4.
Two-thirds of a beaker is filled with copper (II) sulphate solution.
A magnesium ribbon and copper strip are cleaned with sandpaper.
The magnesium and copper electrodes are placed into the copper (II) sulphate solution.
The electrodes are connected to the voltmeter as shown in figure below.
5.
6.
7.
The reading of the voltmeter is recorded.
The metal strip which becomes the negative terminal is determined and recorded.
Steps 1 to 6 are repeated using other metals to replace the magnesium ribbon as electrode A.
Results:
Pair of metals
Magnesium and copper
Aluminium and copper
Zinc and copper
Iron and copper
Lead and copper
Copper and copper
Potential difference (V)
Negative terminal of the cell
-
5
KSSM Form 5 Experiments
Discussion:
1.
Figure below shows the potential differences of different voltaic cells using different metals as electrodes.
2.
In all the cells, the copper plate acts as the positive terminal of the cell. Therefore, copper is situated at the
lowest position among the metals in the Electrochemical Series.
When the magnesium ribbon is connected to the copper plate, the voltmeter shoes the highest reading. This
shows that magnesium is situated furthest from copper in the Electrochemical Series.
No current will flow if both the electrodes are made of copper metal.
3.
4.
Conclusion:
The descending order of the electropositivity of the metals in the Electrochemical Series is magnesium,
aluminium, zinc, iron, lead, copper. The hypothesis is accepted.
6
KSSM Form 5 Experiments
Experiment 1.2 Study of electrolysis of sulphuric acid and copper (II) sulphate solution
using carbon electrodes
Aim: To study the E° value on the selection of chemical species to be reduced or oxidised at the
electrodes.
Materials: 0.1 mol dm-3 sulphuric acid, 0.1 mol dm-3 copper (II) sulphate solution
Apparatus: Batteries, Carbon electrodes, Electrolytic cell, connecting wires with crocodile
clips, ammeter, test tubes, switch
Procedures:
1. An electrolytic cell is filled with dilute sulphuric acid, H2SO4 until it is half full.
2. The apparatus is set up as shown in the figure below. The test tubes must be fully filled with
sulphuric acid at the beginning of the activity.
3.
4.
5.
6.
7.
The switch is turned on to allow electricity to pass through the electrolyte for 15 minutes.
The observations at the anode, cathode and electrolyte are recorded.
The gas collected at the cathode is tested using lighted wooden splint.
The gas collected at the anode is tested using a glowing wooden splint.
Steps 1 to 4 are repeated using 0.1 mol dm-3 copper (II) sulphate solution to replace 0.1 mol
dm-3 sulphuric acid.
8. The gas collected at anode is tested using a glowing wooden splint.
Observations:
Electrolyte
Observation
Cathode
Anode
Change in solution
Dilute
sulphuric acid
Copper (II)
sulphate
solution
7
KSSM Form 5 Experiments
Discussion:
1. Dilute sulphuric acid consists of hydrogen ions, H+, sulphate ions, SO42-, and hydroxide ions,
OH- that move freely.
(a) During the electrolysis, the H+ ions move to the cathode. The H+ ions are discharged by
accepting electrons to form hydrogen gas.
2H+ + 2e- H2
(b) The SO42-, and OH- ions move to the anode. The OH- ions are selectively discharged by
donating electrons to from oxygen and water.
4OH- 2H2O + O2 + 4e(c) The concentration of sulphuric acid decreases gradually as water is decomposed into
hydrogen gas and oxygen gas. The volume of hydrogen gas formed is twice the volume of
oxygen gas.
2. The aqueous solution of copper (II) sulphate consists of copper (II) ions, Cu2+, sulphate ions,
SO42-, hydrogen ions, H+ and hydroxide ions, OH- that move freely.
(a) During the electrolysis, the Cu2+ ions and H+ ions move to the cathode. The Cu2+ ions are
selectively discharged whereby each Cu2+ ion accepts two electrons to form copper metal.
Cu2+ + 2e- Cu
(b) The SO42-, and OH- ions move to the anode. The OH- ions are selectively discharged by
donating electrons to from oxygen and water.
4OH- 2H2O + O2 + 4e(c) The intensity of blue colour of the electrolyte decreases as the concentration of Cu2+ ions
decreases when more copper is deposited on the cathode.
(d) The electrolyte becomes more acidic because of the H+ ions and SO42- ions left.
Conclusion:
During the electrolysis of dilute sulphuric acid using carbon electrodes, _____________ is given off at
the cathode and ________ is produced at the anode.
During the electrolysis of copper (II) sulphate solution using carbon electrodes, ____________ is
deposited at the cathode and __________ is produced at the anode.
8
KSSM Form 5 Experiments
Experiment 1.4 Study of electrolysis of copper (II) sulphate solution using copper electrodes
Aim: To investigate the effects of the type of electrode on the selection of chemical species to be
reduced or oxidised at the electrodes.
Materials: 0.1 mol dm-3 copper (II) sulphate solution
Apparatus: Batteries, carbon electrodes, Electrolytic cell, connecting wires with crocodile clips,
ammeter, test tubes, switch, sandpapers, copper electrodes
Procedures:
1. Two carbon electrodes are cleaned with sandpapers.
2. An electrolytic cell is filled with copper (II) sulphate solution until it is half full.
3. The apparatus is set up as shown in the figure below.
4. The switch is turned on to allow electricity to pass through the electrolyte for 15 minutes.
5. The observations at the anode, cathode and electrolyte are recorded.
6. Step 1 to 5 are repeated using copper electrodes to replace the carbon electrodes.
Observations:
Electrodes
Observation
Anode
Cathode
Electrolyte
Carbon
Copper
Discussion:
1. The aqueous solution of copper (II) sulphate consists of copper (II) ions, Cu2+, sulphate ions,
SO42-, hydrogen ions, H+ and hydroxide ions, OH- that move freely.
(a) During the electrolysis using carbon electrodes, the Cu2+ ions and H+ ions move to the
cathode. The Cu2+ ions are selectively discharged whereby each Cu2+ ion accepts two
electrons to form copper metal.
Cu2+ + 2e- Cu
9
KSSM Form 5 Experiments
(b) The SO42-, and OH- ions move to the anode. The OH- ions are selectively discharged by
donating electrons to from oxygen and water.
4OH- 2H2O + O2 + 4e(c) The intensity of blue colour of the electrolyte decreases as the concentration of Cu2+ ions
decreases when more copper is deposited on the cathode.
2. During the electrolysis using copper electrodes, the Cu2+ ions and H+ ions move to the
cathode. The Cu2+ ions is lower than the H+ ions in the electrochemical series. Hence the Cu2+
ions are selectively discharged to form copper metal.
Cu2+ + 2e- Cu
The intensity of blue colour of electrolyte remains unchanged. This is because the
concentration of the blue Cu2+ ions remains unchanged. For one Cu2+ ions discharged to form
copper atom at the cathode, one copper atom from the anode will dissolve to form Cu2+ ion.
Conclusion:
During the electrolysis of copper (II) sulphate solution, _________________________ are formed at
the anode when carbon electrodes are used, while the _________ dissolves to form copper (II) ions
when copper electrodes are used.
10
KSSM Form 5 Experiments
Activity 2.2 To compare hexane, C6H14 and hexene, C6H12 for sootiness of flames during
combustion
Aim:
To compare the chemical properties of alkanes and alkenes.
Apparatus:
Porcelain dishes, dropper, test tubes, Bunsen burner
Materials:
Hexane, hexene, bromine in 1, 1, 1 – trichloroethane, 0.1 mol dm– 3 potassium
manganate (VII) solution, dilute sulphuric acid, wooden splinter, filter paper
A. Reaction with oxygen
Procedure:
1. About 1 cm3 of hexane and 1 cm3 of hexene are placed into two separate porcelain dishes.
2. A lighted wooden splinter is used to ignite the two liquids.
3. The sootiness of the flame produced from the two burning hydrocarbons is observed.
4. A piece of filter paper is held above each flame in both dishes as shown in above.
5. The amount of soot collected on the two pieces of filter paper is noted.
B. Reaction with bromine
Procedure:
1. About 2 cm3 of hexane is poured into a test tube.
2. 2 to 3 drops of bromine in 1, 1, 1 – trichloroethane are added to the hexane. The mixture is
shaken.
3. Any change that occurs is noted.
4. Steps 1 to 3 are repeated using hexene to replace hexane.
C. Reaction with acidified potassium manganate (VII) solution
Procedure:
1. About 2 cm3 of dilute potassium manganate (VII) solution is poured into a test tube.
2. About 2 cm3 of dilute sulphuric acid is added. The mixture is shaken.
3. About 2 cm3 of hexane is added to the acidified potassium manganate (VII) solution. The
mixture is shaken.
4. Any change that occurs is noted.
5. Steps 1 to 4 are repeated using hexene to replace hexane
11
KSSM Form 5 Experiments
Observations:
Observation
Reaction
Hexane
Hexene
Reaction with oxygen
Hexane burns with a
yellow sooty flame.
Hexene burns with a yellow and very
sooty flame
Reaction with bromine
No visible change.
Hexene decolourises reddish – brown
bromine.
Reaction with acidified
potassium manganate (VII)
solution
No visible change.
Hexene decolourises purple acidified
potassium manganate (VII) solution.
Discussion:
1. Hexane, C6H14 is a member of the alkane family, whereas hexene, C6H12 belongs to the alkene
family.
2. Both hexane and hexene burn incompletely in air with a sooty flame. This is because both
hydrocarbons have a high percentage by mass of carbon. Furthermore, air has only 21%
oxygen.
Percentage of carbon in hexane
Percentage of carbon in hexene
Hexene burns with a more sooty flame compared to hexane as its carbon content is much
higher.
3. Hexane is an unreactive hydrocarbon which does not react with bromine water or acidified
potassium manganate (VII) solution.
4. Hexene is a reactive hydrocarbon. It decolourises reddish – brown bromine 1, 1, 1 –
trichloroethane in an addition reaction.
C6H12 (l) + Br2 (l) ⟶ C6H12Br2 (l)
hexene
It also decolourises purple acidified potassium manganate (VII) to form a diol.
C6H12 (l) + H2O (aq) + [O] ⟶ C6H12(OH)2 (l)
hexene
hexane – 1, 2 – diol
Conclusion:
Alkanes and alkenes have different chemical properties.
12
KSSM Form 5 Experiments
Activity 2.6 Reaction of Ethanoic Acid with Ethanol
Aim:
To prepare a sample of ethyl ethanoate in the laboratory.
Materials:
Absolute ethanol, butan-1-ol, glacial ethanoic acid, concentrated sulphuric acid, water.
Apparatus:
Boiling tube, beaker, dropper, measuring cylinder, Bunsen burner, test tube
holder, test tube rack
Procedure:
1. About 2 cm3 of glacial ethanoic acid is poured into a boiling tube.
2. About 2 cm3 of absolute ethanol is added to the acid. The boiling tube is shaken to mix the
liquids well.
3. A dropper is used to add about 1 cm3 of concentrated sulphuric acid to the mixture. The
boiling tube is shaken well.
4. The mixture is carefully heated over a small flame. The mixture is allowed to boil gently for
about 2 to 3 minutes.
5. A beaker is half-filled with some water.
6. The contents of the boiling tube are poured into the beaker.
7. Any change that occurs is observed.
8. Steps 1 to 7 are repeated using butan-1-ol to replace absolute ethanol.
Observations:
Alcohol
Ethanol
Butan-1-ol
●
●
●
●
Observation
A colourless oily layer floats on the water.
The liquid has a fruity smell.
A colourless oily layer floats on the water.
The liquid has fruity smell.
Discussion:
1. Esterification occurs when ethanoic acid reacts with an alcohol to form an ester and water.
2. Ethanoic acid reacts with ethanol to produce ethyl ethanoate.
CH3COOH (l) + C2H5OH (l) → CH3COOC2H5 (l) + H2O (l)
Ethyl ethanoate
3. Ethanoic acid reacts with butan-1-ol to produce butyl ethanoate.
CH3COOH (l) + C4H9OH (l) → CH3COOC4H9 (l) + H2O (l)
Butyl ethanoate
4. Concentrated sulphuric acid functions as a catalyst.
Conclusion:
An ester is produced from an esterification reaction between a carboxylic acid an alcohol.
13
KSSM Form 5 Experiments
Experiment 3.1 To investigate the heat of displacement of copper by magnesium is
higher than the heat of displacement of copper by zinc.
Aim: To compare the heat of displacement of copper by zinc with the heat of displacement of copper
by iron.
Problem statement: Are the heats of displacement of a metal by two different metals different?
Hypothesis: The heat of displacement of copper by zinc is higher than the heat of displacement of
copper by iron.
Variables:
(a) Manipulated variable: Different metals used to displace copper
(b) Responding variable: Heat of displacement of copper
(c) Controlled variables: Volume and concentration of copper (II) sulphate solution, plastic cup,
mass of metals
Materials: 0.2 moldm-3 copper (II) sulphate solution, iron powder, zinc powder
Apparatus: Thermometer, plastic cup with cover, 50 cm3 measuring cylinders, electronic balance,
weighing bottle
Procedure:
zinc
1. 50 cm3 of 0.2 moldm-3 copper (II) sulphate solution is measured using a measuring cylinder and
poured into a plastic cup.
2. The initial temperature of the solution is measured and recorded after a few minutes.
3. About 2 g of zinc powder is weighed in a weighing bottle.
4. The zinc powder is then added quickly and carefully into the copper (II) sulphate solution.
5. The mixture in the plastic cup is stirred using the thermometer and the highest temperature reached
is recorded.
6. Step 1 to 5 are repeated using iron powder to replace the zinc powder.
Results:
Metal
Initial temperature of copper (II) sulphate
solution (oC)
Highest temperature of the mixture (oC)
Increase in temperature (oC)
Zinc
数据 1 decimal
Iron
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
14
KSSM Form 5 Experiments
Interpreting data:
1. Heat of displacement of copper by zinc
Number of moles of CuSO4 used = MV
= 0.2 moldm-3 x 50/1000 dm3
= 0.01 mol
Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
From the equation:
1 mole of CuSO4 produces 1 mole of copper,
0.01 mole of CuSO4 produces 0.01 mole of copper.
Mass of solution used = volume of solution x density of solution
= 50 cm3 x 1 gcm-3
= 50 g
Heat produced during reaction, H = mcѲ
= 50 g x 4.2 Jg-1oC-1 x 温差 oC
= ___________ J
Heat produced during the displacement of 0.01 mole of copper = __________ J
Heat produced during the displacement of 1 mole of copper = _______ J = _______ kJ
The heat of displacement of copper by zinc, ∆H = -__________ kJ mol-1
The thermochemical equation is:
Zn (s) + Cu2+ (aq) → Zn 2+ (aq) + Cu (s)
∆H = -_________ kJ mol-1
2. Heat of displacement of copper by iron
Number of moles of CuSO4 used = MV
= 0.2 moldm-3 x 50/1000 dm3
= 0.01 mol
Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
From the equation:
1 mole of CuSO4 produces 1 mole of copper,
0.01 mole of CuSO4 produces 0.01 mole of copper.
Mass of solution used = volume of solution x density of solution
= 50 cm3 x 1 gcm-3
= 50 g
Heat produced during reaction, H = mcѲ
= 50 g x 4.2 Jg-1oC-1 x 温差 oC
= ___________ J
Heat produced during the displacement of 0.01 mole of copper = ___________ J
Heat produced during the displacement of 1 mole of copper = _______ J = _______ kJ
The heat of displacement of copper by iron, ∆H = -__________ kJ mol-1
The thermochemical equation is:
Fe (s) + Cu2+ (aq) → Fe 2+ (aq) + Cu (s)
∆H = -__________ kJ mol-1
15
KSSM Form 5 Experiments
Discussion:
1. Iron and zinc are more electropositive than copper, therefore both iron and zinc can displace
copper from copper (II) sulphate solution. The heat of displacement of copper by zinc is higher
than iron because zinc is more electropositive than iron.
2. In both the reactions, brown solids (copper metal) are formed. The intensity of the blue
solutions decreases until they become colourless.
3. The equations for the reactions are shown below.
(a) Displacement of copper by zinc:
Chemical equation: Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Ionic equation: Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
(b) Displacement of copper by iron:
Chemical equation: Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
Ionic equation: Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
4. The practical values of heat of displacement of copper for both the reactions are shown below.
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
∆H = -_____ kJ mol-1
Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
∆H = -_____ kJ mol-1
5. The energy level diagrams for both reactions are shown below.
Fe + Cu2+
-
kJmol-1
-
kJmol-1
Fe2+ + Cu
6. Excess zinc and iron are used to make sure all the copper (II) ions are displaced to form copper.
7. The amount of heat absorbed by the remaining unreacted zinc and iron is very little and can be
neglected in the calculation of the heat of displacement.
8. The volume and concentration of copper (II) sulphate solution used in both reactions are the same
so that the same number of moles of copper is formed.
9. The following precautions need to be taken during the experiment to get more accurate results.
(a) Metals in the powder form are used so that the reactions will take a shorter time to complete,
and thus less heat is lost to the surroundings.
(b) The initial temperature of the copper (II) sulphate solution is taken after a few minutes to let
the solution achieve a uniform temperature.
(c) The metals are added quickly to the solution to reduce heat loss to the surroundings.
(d) The mixture is stirred slowly throughout the experiment to make sure the temperature of the
solution is uniform.
(e) The reading of the thermometer is observed carefully so that the highest temperature of the
solution can be recorded. This is done to ensure that the reactions are completed and all the
heat has been given out.
Conclusion:
The heat of displacement of copper by zinc is higher than the heat of displacement of copper by iron.
Hence, the hypothesis is accepted.
16
KSSM Form 5 Experiments
Experiment 3.2 To determine the heat of neutralization between acids and alkalis
of different strengths.
Aim: To determine and compare the heat of neutralization between acids and alkalis of different
strengths.
Problem statement: How are heats of neutralization can be determined and compared?
Hypothesis: The heat of neutralization between hydrochloric acid and sodium hydroxide solution is
higher than the heat of neutralization between ethanoic acid and sodium hydroxide
solution.
Manipulated variable: Different types of acids
Responding variable: Heat of neutralization
Fixed variables: Concentrations and volumes of acid and alkali used and the plastic cup
Materials: 1.0 moldm-3 hydrochloric acid, 1.0 moldm-3 sodium hydroxide solution,
1.0 moldm-3 ethanoic acid
Apparatus: 50 cm3 measuring cylinders, thermometer, plastic cups with covers
Procedure:
50 cm3 of 1.0 moldm-3 sodium
hydroxide solution
50 cm3 of 1.0 moldm-3
hydrochloric acid
1. 50 cm3 of 1.0 moldm-3 sodium hydroxide solution is measured using a measuring cylinder and
poured into a plastic cup. The initial temperature of the solution is measured after a few minutes.
2. 50 cm3 of 1.0 moldm-3 hydrochloric acid is measured using another measuring cylinder and poured
into a plastic cup. The initial temperature of the solution is measured after a few minutes.
3. The hydrochloric acid is then poured quickly and carefully into the sodium hydroxide solution.
4. The mixture is stirred using a thermometer and the highest temperature reached is recorded.
5. Step 1 to 4 are repeated using ethanoic acid to replace hydrochloric acid.
17
KSSM Form 5 Experiments
Results:
Reacting mixture
Initial temperature of alkali (oC)
Initial temperature of acid (oC)
Average temperature of acid and
alkali (oC)
Highest temperature of mixture
(oC)
Increase in temperature (oC)
Sodium hydroxide
solution and
hydrochloric acid
数据 1 decimal
数据 1 decimal
数据 1 decimal
Sodium hydroxide
solution and ethanoic
acid
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
数据 1 decimal
Interpreting data:
1. Reaction between sodium hydroxide solution and hydrochloric acid
Number of moles of OH- ions used
Number of moles of H+ ions used
= number of moles of NaOH
= number of moles of HCl
-3
3
= 1.0 moldm x 50/1000 dm
= 1.0 moldm-3 x 50/1000 dm3
= 0.05 mol
= 0.05 mol
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
From the equation:
1 mole of H+ ions react with 1 mole of OH- ions to form 1 mole of water
So 0.05 mol of H+ ions react with 0.05 mol of OH- ions to form 0.05 mol of water
Mass of solution used = (50 + 50) cm3 x 1 g cm-3 = 100 g
Heat given out in the reaction = heat absorbed by the solution
= 100 x 4.2 x 温差
= ____________ J
Formation of 0.05 mol of water gives out __________ J of heat.
So formation of 1 mole of water gives out ____________ J = ___________ kJ of heat
The heat of neutralization between sodium hydroxide solution and hydrochloric acid
is - ______kJmol-1
2. Reaction between sodium hydroxide solution and ethanoic acid
Number of moles of OH- ions used
Number of moles of H+ ions used
= number of moles of NaOH
= number of moles of CH3COOH
-3
3
= 1.0 moldm x 50/1000 dm
= 1.0 moldm-3 x 50/1000 dm3
= 0.05 mol
= 0.05 mol
CH3COOH (aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
From the equation:
1 mole of H+ ions react with 1 mole of OH- ions to form 1 mole of water
So 0.05 mol of H+ ions react with 0.05 mol of OH- ions to form 0.05 mol of water
Mass of solution used = (50 + 50) cm3 x 1 g cm-3 = 100 g
18
KSSM Form 5 Experiments
Heat given out in the reaction = heat absorbed by the solution
= 100 x 4.2 x 温差
= ____________ J
Formation of 0.05 mol of water gives out __________ J of heat.
So formation of 1 mole of water gives out ____________ J = ___________ kJ of heat
The heat of neutralization between sodium hydroxide solution and ethanoic acid
is - ______kJmol-1
Discussion:
1. It is found that the heat of neutralization between sodium hydroxide solution and hydrochloric acid
is higher than the heat of neutralization between sodium hydroxide solution and ethanoic acid.
2. Ethanoic acid is a weak acid. It dissociates partially in water. Most of the ethanoic acid still exist as
molecules. Some of the heat given out during the neutralization reaction is used to dissociate the
weak acid completely in water.
3. The equation for the neutralization reactions are as follows:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
CH3COOH (aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
4. The energy level diagram for both reactions are shown below:
CH3COOH + NaOH
HCl + NaOH
-
-
kJmol-1
NaCl + H2O
kJmol-1
CH3COONa + H2O
5. It is necessary to mix the acid and the alkali quickly to reduce heat loss to the surrounding.
6. A plastic cup is used in the experiment to reduce heat loss to the surrounding.
Conclusion:
The heat of neutralization between hydrochloric acid and sodium hydroxide solution is higher than the
heat of neutralization between ethanoic acid and sodium hydroxide solution. Hence, the hypothesis
can be accepted.
19
KSSM Form 5 Experiments
Activity 5.1 : Preparation of Soap by Saponification
Aim
: To prepare soap through the saponification process.
Problem statement: How is soap prepared in the laboratory?
Hypothesis : The vegetable oil will react with the alkali to form soap during saponification process.
Material: Sodium hydroxide 5 mol dm-3, Sodium Chloride, Corn oil, Soy oil, Palm oil, Olive oil
Apparatus: Beaker, Glass rod, Measuring cylinder, Tripod stand, Wire gauze, Spatula, Bunsen burner
Procedure:
1.
2.
3.
4.
5.
6.
7.
8.
9.
5 cm3 of palm oil are measured by measuring cylinder and poured into a beaker.
30 cm3 of sodium hydroxide are measured and added into the beaker.
The mixture is stirred constantly and heated until boiling for about 10 minutes.
50 cm3 of water and one spatula of sodium chloride are added into the beaker.
The mixture is boiled for another 5 minutes and then cool down to room temperature.
Product formed are filtered and dried with pieces of filter paper.
The residue is then touched using fingers.
The residue is then mixed and stirred together with water
The experiment is repeated with different vegetable oils.
Observation:
Discussion:
1.
2.
The white semi-solid is called soap as it will form bubbles with water.
The chemical equations for the formation of soap are shown below:
Palm oil + alkali → sodium salt of fatty acid + glycerol
Corn oil + alkali → sodium salt of fatty acid + glycerol
Soy oil + alkali → sodium salt of fatty acid + glycerol
Olive oil + alkali → sodium salt of fatty acid + glycerol
3.
4.
The presence of sodium chloride helps the soap to precipitate. Sodium chloride makes the soap
less soluble by helping the solid soap to form from the liquid soap solution.
The flame should be controlled during the boiling process to avoid bubbling of mixture.
Conclusion:
The boiling of different vegetables oil with concentrated alkaline solution will produce soap.
20