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Edexcel AS Chemistry
Unit 3B: Chemistry
Laboratory Skills I Alternative
Unit 1 Practical
Methods and Procedures
Edexcel AS Chemistry
A. 25cm3 of 1 mol dm–3 H2SO4
n H2SO4 = MV
1000
B. 1 mol H2SO4 reacted with 1 mol NiCO3
∴ answer from A = n NiCO3 Mass of NiCO3 = n NiCO3 × Mr NiCO3; weight it with slight excess
~3.00 g
Practical 1.1
0.00 2.00
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Edexcel AS Chemistry
C. Add little by little, stir continuously until the solid
dissolves and stop heating for a while. Then continue the same steps until all solid finishes.
D. to G. Hot mixture
Practical 1.1
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Edexcel AS Chemistry
Practical 1.1 Let the mass of the crystal obtained is 4.55 g. A green shiny crystal obtained. 1) The theoretical yield of hydrated nickel sulfate is
0.025 × [58.7 + 32.1 + 4(16) + 7(18)] = 7.02 g
2) The percentage yield of the crystal obtained from this experiment is 4.55
7.02 × 100% = 64.8 %
3) It was found that the actual yield is much lower than expected due to some reason: a. Some product left behind in the apparatus when
transferring the product from one apparatus to another. b. Some reactant may be loss during the experiment.
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Edexcel AS Chemistry
A. 25cm3 of 2 mol dm–3 H2SO4
n(H2SO4) = MV
1000 1:1 reaction; n(H2SO4) = n(Fe)
B. m(Fe) = n(Fe) × Ar Fe
Do the same step until all
solid is added
Practical 1.2 (1)
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Edexcel AS Chemistry
C. D. Addition of 10% of acid 5cm3 of 2 mol dm–3 H2SO4
Practical 1.2 (1)
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Edexcel AS Chemistry
A. 25cm3 of ammonia solution 2 mol dm–3 H2SO4
𝑀𝑎𝑉𝑎
𝑎 =
𝑀𝑏𝑉𝑏
𝑏 → V(NH3) =
𝑀𝑎𝑉𝑎
𝑎 ×
𝑏
𝑀𝑏
B. & C. Gently heat in the fume hood until it became
concentrated
Practical 1.2 (2)
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Edexcel AS Chemistry
A. B. & C. Solution from (1) Solution from (2) Allow to cool
D. to F.
Practical 1.2 (3)
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Edexcel AS Chemistry
Practical 1.2 Result and observation: A light green shiny crystal obtained.
Let the mass of the crystal obtained is ________ g.
When both the crystal and iron(II) sulphate were exposed to air overnight, it was found that the crystal does not change colour but iron(II) sulphate turns colour from light green to brown. Discussion 4) The reaction between ammonia solution and sulphuric acid is
neutralization. This is due to the fact that the oxidation number of all element remain unchanged. Whereas the reaction between iron sulphuric acid is redox. This is because there is a change of oxidation number where iron changing from 0 to +2 (oxidation) and hydrogen changing from +1 to 0 (reduction).
5) The ions present in the double salts are ammonium ion (NH4+),
iron(II) ion (Fe2+), and sulphate ion (SO42–)
Edexcel AS Chemistry
Practical 1.2 Discussion
6) Theoretical yield of the crystals which could be made is as follows:
Mr (NH4)2SO4•FeSO4•6H2O = 392
1 mole (NH4)2SO4 requires to react with 1 mole FeSO4
∴n(NH4)2SO4 = 0.05 mol
Theoretical yield = 0.05 × 392 = 19.59 g
7) Percentage yield = 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑦𝑖𝑒𝑙𝑑
19.6×100%
8) The pure iron(II) sulphate turns colour from light green to brown because Fe2+ ion was oxidized to Fe3+ ion when exposed to air.
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Edexcel AS Chemistry
Practical 1.3
Solubility and Precipitation Reactions Insoluble Dispose to sink
NaOH Soluble Result and Observation:
Name Formula Appearance Solubility Colour of Solution
Addition of NaOH
Copper(II) chloride CuCl2 Soluble
Iron(II) sulphate FeSO4 Soluble
Magnesium
carbonate MgCO3 Insoluble
Iron(III) nitrate Fe(NO3)3 Soluble
Iron(III) oxide Fe2O3 Insoluble
Potassium
carbonate K2CO3 Soluble No change
Edexcel AS Chemistry
Practical 1.3
Solubility and Precipitation Reactions Discussion:
1) There were some of the metal compounds are coloured (copper and iron compounds). Metals that give colour to the compounds are the transition metals
2) The formulae for copper(II) chloride, iron(II) sulphate and iron(III) nitrate all contain water which indicate that they are hydrated compounds. These compounds might reasonably by expected to be soluble in water.
3) All compounds provided in this experiment obeyed the solubility rules.
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Edexcel AS Chemistry
Practical 1.3
Further precipitation reactions Result and Observation:
Combination of solution
Observation Ionic equation
NaCl + Pb(NO3)2
NaCl + CuSO4
NaOH + Pb(NO3)2
NaOH + CuSO4
Pb(NO3)2 + Na2SO4
Pb(NO3)2 + CuSO4
Pb(NO3)2 + Na2CO3
CuSO4 + Na2SO4
CuSO4 + Na2CO3
Edexcel AS Chemistry
Practical 1.3
Redox Reactions Result and Observation:
Solution Ionic equations Colour of Solution
Ionic equation
after adding NaOH Before After
H2O2 2Fe2+ → 2Fe3+ + 2e- 2H+ + H2O2 + 2e- → 2H2O 2Fe2+ + 2H+ + H2O2 → 2Fe3+ + 2H2O
Green solution
Brown solution
Fe3+(aq)+ 3OH–(aq) → Fe(OH)3(s) (Brown ppt)
KMnO4
5Fe2+ → 5Fe3+ + 5e- 8H+ + MnO4
- + 5e- → Mn2+ + 4H2O 5Fe2+ + 8H+ + MnO4
- → 5Fe3+ + Mn2+ + 4H2O
Green solution
Brown solution
Fe3+(aq)+ 3OH–(aq) → Fe(OH)3(s) (Brown ppt)
Warm conc.
HNO3
6Fe2+ → 6Fe3+ + 6e- 8H+ + 2NO3
- + 6e- → 2NO + 4H2O 6Fe2+ + 8H+ + 2NO3
- → 6Fe 3+ + 4H2O + 2 NO
Green solution
Brown solution
Fe3+(aq)+ 3OH–(aq) → Fe(OH)3(s) (Brown ppt)
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Edexcel AS Chemistry
Practical 1.3
Reduction Result and Observation:
Substance Ionic Eqation
Colour of Solution Addition of NaOH
Ionic equation after adding
NaOH
Before After
Zinc, warm
2Fe3+(aq) + 2e- → 2Fe2+ (aq) Zn(s) → Zn2+(aq) + 2e- 2Fe3+(aq)+Zn(s)→ 2Fe2+(aq) + Zn2+(aq)
Brown solution
Green solution
Fe2+(aq)+ 2OH–(aq) → Fe(OH)2(s) (green ppt)
NaSO3, warm
2Fe3+ + 2e- → 2Fe2+ H2O + SO3
2– → SO32– +2H+ + 2e-
2Fe3+ + H2O + SO32–→ 2Fe2+
+ SO32– +2H+
Brown solution
Green solution
Fe2+(aq)+ 2OH–(aq) → Fe(OH)2(s) (green ppt)
Edexcel AS Chemistry
Practical 1.4 The reaction between:
1) copper(II) sulphate and zinc
m(Zn) = 0.01 × Ar Zn
2) Citric acid and sodium hydrogen carbonate
m(NaHCO3) = 0.01 × Mr NaHCO3
Thermometer
Beaker Reaction mixture
Foam polysterene
cup and lid
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Edexcel AS Chemistry
Practical 1.4 Results:
Experiment 1
(Limiting reagent: CuSO4) 2
(Limiting reagent: C6H8O7)
Moles used MV
1000
MV
1000
Initial temperature
(oC)
Final temperature
(oC)
Temperature change, ∆T (oC)
Edexcel AS Chemistry
Practical 1.4 Calculation:
Experiment 1
(Limiting reagent: CuSO4) 2
(Limiting reagent: C6H8O7)
Energy transfer, q
m(solution) × 4.18 × ∆T m(solution) × 4.18 × ∆T
Enthalpy change ∆H1 = – 𝑞
𝑛(CuSO4) ∆H2 = –
𝑞
𝑛(C6H8O7)
Thermochemical equation
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) ∆H1 = _____
C6H8O7(aq) + 3NaHCO3(s) → C3H5O7Na3(aq) + 3CO2(g) +
3H2O(l) ∆H2 = _____
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Edexcel AS Chemistry
Practical 1.4 Discussion:
4) The reaction between copper(II) sulphate and zinc is _______ whereas the reaction between citric acid and sodium hydrogencarbonate is ________.
5) As excess of solid is used in each experiment is to ensure that all solution is completely reacted.
6) The things that we should ignore when doing the calculation include the mass of solid reactant added and the specific heat capacity of the solution.
Edexcel AS Chemistry
Practical 1.4 Discussion:
7) A large container was used for the second experiment is because the carbon dioxide gas is formed which may produces bubbles or effervescences.
8) The main sources of error in this experiment include heat losses to the surrounding (environment) such as air, glass, calorimeter or even the products of the reactions, the concentration of the solutions are assumed to be correct, and slight inaccuracies in weighing the solid reactants should make little difference, as long as there is still excess.
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Edexcel AS Chemistry
Practical 1.4 Discussion:
9) The percentage error for each enthalpy change are as follows:
Experiment 1:
% error = 2×0.1
∆𝑇1 × 100%
Experiment 2:
% error = 2×0.1
∆𝑇2 × 100%
Edexcel AS Chemistry
Practical 1.5 Results:
Alcohol Propan-1-ol Butan-1-ol Octan-1-ol
Initial mass of spirit lamp + alcohol + cap (g)
Final mass of spirit lamp + alcohol + cap (g)
Mass of alcohol burned, m (g)
Initial temperature of water (oC)
Final temperature of water (oC)
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Edexcel AS Chemistry
Practical 1.5 Calculation:
1) a) n(propan-1-ol) = 𝑚(propan−1−ol)
𝑀𝑟(propan−1−ol)
b) Energy produce, q = answer from (1) × (–2021kJmol–1) c) Energy produce, q = heat capacity of apparatus, C ×
temperature rise
∴ C = 𝑎𝑛𝑠𝑤𝑒𝑟 𝑓𝑟𝑜𝑚 (2)
∆𝑇
2) d) Energy produce for butan-1-ol, q1 = answer from (3) × ∆T
Energy produce for octan-1-ol, q2 = answer from (3) × ∆T
e) n(butan-1-ol) = 𝑚(butan−1−ol)
𝑀𝑟(butan−1−ol) n(octan-1-ol) =
𝑚(octan−1−ol)
𝑀𝑟(octan−1−ol)
∆Hc butan-1-ol = – 𝑞1
𝑛(butan−1−ol) ∆Hc octan-1-ol = –
𝑞2
𝑛(octan−1−ol)
Edexcel AS Chemistry
Practical 1.5 Discussion: 1) Complete combustion may not have taken place where
carbon of carbon monoxide might have been formed instead.
2) The main sources is the heat losses to the surrounding and the precautions which should be taken is to use a ‘chimney’ or other methods of draught proofing.
3) The heat loss and incomplete combustion both will tend to give results that are less negative than the true value.
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Edexcel AS Chemistry
Practical 1.5 Discussion: 4) It was found that the enthalpy change of combustion
increase with the number of carbon atom. The difference is about 655 kJ for each extra carbon atom.
5) This is because more energy is needed to break bonds in the reactants as the number of carbon atoms increases. However, proportionally more energy is released when the new bonds form in the products. So overall reactions becomes more exothermic.
Edexcel AS Chemistry
Practical 1.6 Results:
Experiment 1
(Solid reagent: Ca) 2
(Solid reagent: CaCO3)
Moles used m
𝐴𝑟 Ca
m
𝑀𝑟 CaCO3
Initial temperature
(oC)
Maximum temperature
(oC)
Temperature change, ∆T (oC)
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Edexcel AS Chemistry
Practical 1.6 Discussion:
1) to 4)
The enthalpy change for both reactions are as follows:
Experiment 1 2
Energy transfer, q
m(solution) × 4.18 × ∆T m(solution) × 4.18 × ∆T
Enthalpy change ∆H1 = – 𝑞
𝑛(Ca) ∆H2 = –
𝑞
𝑛(CaCO3)
Thermochemical equation
Ca(a) + 2HCl(aq) → CaCl2(aq) + H2(g) ∆H1= _____
CaCO3(s) + 2HCl(s) → CaCl2(aq) + CO2(g) + H2O(l)
∆H2 = _____
Edexcel AS Chemistry
Practical 1.6 Discussion: 5) The Hess’s cycle for the formation of calcium
carbonate is shown below:
Ca(s) + 3
2O2(g) + C(graphite) CaCO3(s)
2HCl(aq) 2HCl(aq)
CaCl2(aq) + H2O(g) + CO2(g) + H2(g) 6) From the Hess’s cycle above, the ∆Hf
[CaCO3(s)] is calculated as follows: ∆Hf [CaCO3(s)] + ∆H2 = ∆H1 + ∆Hf [CO2(g)] + ∆Hf [H2O (g)]
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Edexcel AS Chemistry
Practical 1.6 Evaluation: 1) The two assumptions that must be made in this
experiment are that the specific heat capacity of HCl is same as that of water and that there is no heat loss to the surroundings.
2) The reliability of this experiment could be improved by repeating the experiment and averaging them.
3) Error could be minimised by using a lid to minimise heat loss and by a cooling curve method.
4) The precision of the experiment could be increased by using apparatus which has a low in precision error such as pipette or thermometer that measures to ±0.1oC.
Edexcel AS Chemistry
Practical 1.7 Result and Observation: Experiment Observation Inferences
1 Combustion
Combustion of cyclohexane gives a yellow (luminous) and sooty flame. The poly(ethene) softens and slowly becomes a liquid on gentle heating.
Cyclohexane has a high C:H ratio
2 Oxidation
No changes observed but slightly brown solution observe when react with poly(ethene)
Alkane does not react with the oxidising agent.
3 Action of bromine
No changes observed but bromine water colour fades when react with poly(ethene)
Alkane does not react with the bromine water at room condition
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Edexcel AS Chemistry
Practical 1.7 Result and Observation: Experiment Observation Inferences
4 Action of
bromine in sunlight
The colour of bromine water fades and white fume evolve when the test tube was irradiate under sunlight but no changes observe when test tube leave in a dark place
Alkane reacts with bromine through free radical substitution and the white fume is the hydrogen bromide.
5 Action of sulphuric
acid
No changes observed Alkane does not react with the sulphuric acid.
6 Action of
alkali No changes observed
Alkane does not react with the alkali.
Edexcel AS Chemistry
Practical 1.7 Result and Observation: Experiment Observation Inferences
7 Catalytic cracking
a) Flammability test The gas obtained is flammable.
The gas obtained is hydrocarbon.
b) Bromine test The bromine colour fades then colourless.
Alkene is present.
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Edexcel AS Chemistry
Practical 1.8 Result and Observation: Experiment Observation Inferences
1 Combustion
Combustion of cyclohexene and limonene gives a yellow (luminous) and slightly sooty flame.
Cyclohexene has a low C:H ratio.
2 Oxidation
The colour of potassium manganate(VII) decolourised. Two layers observed.
Alkene reacts with the oxidising agent due to the presents of C=C.
3 Action of bromine
The colour of bromine water decolourised. Two layers observed.
Alkene reacts with the bromine water due to the presents of C=C.
Edexcel AS Chemistry
Practical 1.8 Result and Observation: Experiment Observation Inferences
4 Action of
sulphuric acid
The solution turns dark and the bottom of the test tube is hot.
Alkene react with the sulphuric acid.
5 Polymerisation
CH3 n CH2=C CO2CH3 CH3 ~ C–CH2 ~ CO2CH3 n
The product have the same empirical formula as the starting material.
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Edexcel AS Chemistry
Unit 3B: Chemistry
Laboratory Skills I Alternative
Unit 2 Practicals
Edexcel AS Chemistry
1) The liquid is deflected towards the plastic rod.
2) Alcohols and ketones show deflection while hydrocarbons show no deflection.
3) This is because polar molecules posses a permanent dipole moment and the molecules tends to align its dipole moment along the field of the nearby charged object. Hydrocarbons do not posses a permanent dipole moment may be observed to be very slightly deflected, as a result of the induced dipole. However, hydrocarbons that show no deflection indicate that the structure is symmetry and the dipole moment are cancel out.
Practical 2.1 Finding the effect of electrostatic force on
jets of liquid
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Edexcel AS Chemistry
Water Ethanol Cyclohexane
Experiment 1
Insoluble Slightly soluble;
solution is slightly brown
Soluble; purple solution obtained
Experiment 2
Soluble Slightly soluble Insoluble
Experiment 3
Soluble; dense white precipitate obtained
Slightly soluble; slight white
precipitate observed
Insoluble; no precipitate form
Practical 2.2 Solubility of simple molecules in different
solvents
1) Iodine and sugar both have covalent bond whereas calcium chloride is ionic.
2) Iodine molecules are held together by London forces. Sugar has hydrogen bonding and calcium chloride does not have intermolecular force.
Edexcel AS Chemistry
Practical 2.2 Solubility of simple molecules in different
solvents
3) Iodine dissolves in cyclohexane due to the ability for the molecules to interact with hexane molecules through van der Waals forces. It does not dissolve in water or very well in ethanol because the molecules are both polar and therefore cannot form hydrogen bonds to the iodine molecules. Sugar is the opposite and dissolves only in polar solvents like water and ethanol (to some extent). Calcium chloride, being ionic, dissolves in polar solvents because the water molecules interact with the ionic lattice, therefore breaking the lattice and dispersing the ions.
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Edexcel AS Chemistry
Procedures: Little amount of carbonate compound are added into a test tube and heated. The gas emitted was allow to bubble through a delivery tube into another test tube containing a small quantity of limewater
Practical 2.3 Thermal decomposition of group 2 nitrates
and carbonates
Effect Ionic radius of cation/nm
MgCO3 Limewater turns cloudy 0.072
CaCO3 No change 0.100
SrCO3 No change 0.113
BaCO3 No change 0.136
Edexcel AS Chemistry
Procedures: weigh certain amount of nitrate compound with crucibles. The compound are strongly heated in crucibles for few minutes and allow it to cool then re-weighing the crucible.
Practical 2.3 Thermal decomposition of group 2 nitrates
and carbonates
Effect Ionic radius of cation/nm
Mg(NO3)2 Brown fumes observe 0.072
Ca(NO3)2 No change 0.100
Sr(NO3)2 No change 0.113
Ba(NO3)2 No change 0.136
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Edexcel AS Chemistry
Interpretation: As going down the group, the size of the cations increases, the charge density decrease. This makes the cation decrease in polarizing power and therefore less able to polarize the neighbouring anion. Hence, within the anion, less able to weaken the bonds, making the anion having less easily to be decomposed.
MgCO3 (s) → MgO (s) + CO2 (g)
2 Mg(NO3)2 (s) → 2 MgO (s) + O2 (g) + 4 NO2 (g)
Practical 2.3 Thermal decomposition of group 2 nitrates
and carbonates
Edexcel AS Chemistry
1) The coloured flame produced due to electronic excitations of electrons in the metal to a higher electronic energy levels when electrons absorb energy and then return to the lower energy levels with the emission of light energy in the visible region that corresponds to a particular colour.
2) Fireworks.
Practical 2.4 Flame tests on compounds of group 1 and 2
Compound Observation Compound Observation
Li+ Deep red Ca2+ Brick red
Na+ Yellow Sr2+ Blood red
K+ Lilac Ba2+ Apple green
Rb+ Bluish red
Cs+ Blue
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Edexcel AS Chemistry
Observation: T = 24 – 25oC; metyl orange: yellow → reddish orange
bromophenol: blue → yellow
1) Ca(OH)2 (aq) + 2 HCl (aq) → CaCl2 (aq) + 2 H2O (l)
2) 2 mol HCl reacted with 1 mol Ca(OH)2
∴ 1 mol HCl reacted with 0.5 mol Ca(OH)2
3) Number of moles HCl = 0.050×𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
1000
4) Number of moles in Ca(OH)2 in 10.0 cm3
= 0.5 × answer from (3) = x
∴ In 1000.0 cm3, number of moles Ca(OH)2
= 1000
10 × x
5) Molecular mass of Ca(OH)2 = 40 + 2(16) + 2(1) = 74
Mass of Ca(OH)2 in 1000.0 cm3 = answer from (4) × 74
6) answer from (5) g dm–3 at 24 oC
7) Percentage error = 0.1×2
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 × 100%
Practical 2.5 Simple acid-base titration
Edexcel AS Chemistry
Practical 2.6 Oxidation of metal and non-metallic
elements and ions by halogens
Chlorine Bromine Iodine
Experiment 1
Solution turns orange/brown
Solution turns orange/brown
No change
Experiment 2
The colour turns from pale yellow to
colourless
The colour turns from orange to
colourless
The colour turns from brown to
colourless
Experiment 3
Reaction with KI: solution turns from colourless to brown; Addition of hexane: Top layer: purple, bottom layer: pale
yellow to colourless
– –
Experiment 4
– –
Addition of starch: Black; addition of
sodium thiosulfate: colourless
23
Edexcel AS Chemistry
Practical 2.6 Oxidation of metal and non-metallic
elements and ions by halogens 1) Experiment 1
Ox. half equation: 2 Fe2+ (aq) → 2 Fe3+ (aq) + 2 e–
Red. Half equation: Cl2 (aq) + 2 e– → 2 Cl– (aq)
Br2 (aq) + 2 e– → 2 Br– (aq)
Overall ionic equation:
2Fe2+(aq)+Cl2(aq)→2Fe3+(aq)+2Cl–(aq)
2Fe2+(aq)+Br2(aq)→2Fe3+(aq)+2Br–(aq)
Experiment 2
Ox. half equation:
2OH– (aq) + ½Cl2 (aq) → ClO– (aq) + H2O (l) + e–
Red. Half equation: ½Cl2 (aq) + e– → Cl– (aq)
Overall ionic equation:
2OH– (aq) + Cl2 (aq) → ClO– (aq) + Cl– (aq) + H2O (l)
Edexcel AS Chemistry
Practical 2.6 Oxidation of metal and non-metallic
elements and ions by halogens 1) Experiment 3
Ox. half equation: 2I– (aq) → I 2 (aq) + 2 e–
Red. Half equation: Cl2 (aq) + 2 e– → 2 Cl– (aq)
Overall ionic equation:2I–(aq)+Cl2(aq)→I2(aq)+2Cl–(aq)
Experiment 4
Ox. half equation: 2 S2O32– (aq) → S4O6
2– (aq) + 2 e–
Red. Half equation: I2 (aq) + 2 e– → 2 I– (aq)
Overall ionic equation:
2 S2O32– (aq) + I2 (aq) → S4O6
2– (aq) + 2 I– (aq)
24
Edexcel AS Chemistry
Practical 2.6 Oxidation of metal and non-metallic
elements and ions by halogens 2) From the reaction,
2NaOH(aq)+Cl2(aq) → NaClO(aq)+NaCl(aq)+H2O(l)
0 +1 –1
The oxidation number of chlorine changes from 0 to +1 and –1 where chlorine act both as oxidizing and reducing agent simultaneously.
3) Oxidizing strength of the halogens decreases down group 7
4) The liberated iodine reacts with the starch turning it to black.
Edexcel AS Chemistry
Practical 2.7 Disproportionation reactions with cold and
hot alkalis
1) 2OH– (aq) + Cl2 (aq) → ClO– (aq) + Cl– (aq) + H2O (l)
2) The chlorine undergoes two change: part of it changes from 0 to –1 and part from 0 to +1.
3) Chlorine changes in oxidation number from 0 to +5 and from 0 to –1.
4) 6KOH(aq) + 3I2(aq) → 5KI(aq) + KIO3 (aq) + 3H2O(l)
Halogen solution
Observation on adding alkali Equation
Chlorine The colour turns from pale
yellow to colourless 2NaOH(aq)+Cl2(aq) →
NaClO(aq)+NaCl(aq)+H2O(l)
Bromine The colour turns from orange
to colourless 2NaOH(aq)+Br2(aq) →
NaBrO(aq)+NaBr(aq)+H2O(l)
Iodine The colour turns from brown
to colourless 2NaOH(aq)+I2(aq) →
NaIO(aq)+NaI(aq)+H2O(l)
25
Edexcel AS Chemistry
Practical 2.8 Iodine/thiosulfate titration and the
determination of purity of potassium iodate(V) 1) A well-mixed, accurately made solutions prepared from
good quality reagents is required in order to have an excellent result.
2) The oxidation number of sulfur in sodium thiosulfate is +2 and in sodium tetrathionate is +2½.
Part 1
1) n Na2S2O3 = 0.010×𝑣
1000; v = average titre
2) 2 mol Na2S2O3 reacted with 1 mol I2
∴ answer from step 1 divided with 2
3) 3 mol I2 produced from 1 mol KIO3
∴ answer from step 2 divided with 3 4) Molecular mass of KIO3 = 40 + 127 + 3(16) = 214
Mass of KIO3 in 10 cm3 = answer from step 3 × 214
∴ Mass of KIO3 in 1000 cm3 = 1000
10 × Mass of KIO3 in 10 cm3
5) Percentage purity = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐼𝑂3 𝑖𝑛 1 𝑑𝑚3𝑓𝑟𝑜𝑚 4
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑟𝑢𝑑𝑒 𝐾𝐼𝑂3 × 100%
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides 1) a) Observation:
Cl2 with KCl: no changes
Cl2 with KBr: the colour of the solution turns from colourless to yellow–orange
Cl2 with KI: the colour of the solution turns from colourless to brown
2KBr (aq) + Cl2(aq) → Br2 (aq) + 2KCl (aq)
2KI (aq) + Cl2(aq) → I2 (aq) + 2KCl (aq)
Addition of hydrocarbon helps in reaching a decision as the halogen dissolve well in hydrocarbons due to the formation of intermolecular interaction.
26
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides 1) b) Observation: All solutions give out two layer
Cl2 with KCl: top layer: pale green
bottom layer: light green to colourless
Cl2 with KBr: top layer: orange
bottom layer: light yellow to colourless
Cl2 with KI: top layer: purple
bottom layer: light yellow to colourless
Halogen mixed well with hydrocarbon because both halogen and hydrocarbon having London force with similar strength which allows them to interact easily between the molecules.
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides 2) A definite trend in reactivity is clearly observed in this
experiment because as the atomic radius of the halogen increases down the group, the tendency to accept electron decreases which also decreases the oxidizing power of the halogens.
2I– (aq) + Br2 (aq) → I2 (aq) + 2Br– (aq)
27
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
The reactions between halogens and halide ions
Action on
Solution added
Water Potassium chloride
Potassium bromide
Potassium iodide
Chlorine followed
by hexane
With water: soluble forming
light green solution;
Addition of hexane: Top layer: pale
green, bottom layer: light
green to colourless
No changes
With Cl: solution turns
from colourless to yellow-
orange; Addition of
hexane: Top layer:
orange, bottom layer: pale yellow to colourless
With Cl: solution turns
from colourless to brown;
Addition of hexane:
Top layer: purple, bottom
layer: pale yellow to colourless
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
The reactions between halogens and halide ions
Action on
Solution added
Water Potassium chloride
Potassium bromide
Potassium iodide
Bromine followed
by hexane
With water: soluble and
forms orange to light brown
solution Addition of hexane: Top
layer: orange, bottom layer: pale yellow to
colourless
No change No change
With Br: solution turns
from colourless to brown;
Addition of hexane:
Top layer: purple, bottom
layer: pale yellow to colourless
28
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
The reactions between halogens and halide ions
Action on
Solution added
Water Potassium chloride
Potassium bromide
Potassium iodide
Iodine followed
by hexane
With water: insoluble in
cold water but slightly soluble
in hot water and forms
brown solution Addition of hexane: Top
layer: purple, bottom layer: pale yellow to
colourless
No change No change No change
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
Some reactions of the halides
1) The silver halides
Solution
Action on
Silver nitrate Concentrated
ammonia solution Exposure to light
Potassium chloride
White precipitate obtained
The precipitate is soluble
Silver substance observed
Potassium bromide
Cream precipitate obtained
The precipitate is soluble
*Yellow substance observed
Potassium iodide
Yellow precipitate obtained
The precipitate is insoluble
No change observed
29
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
Some reactions of the halides
2) The action of concentrated sulfuric acid on potassium salts
Action on
Potassium chloride Potassium bromide
Potassium iodide
Lead(II) ethaonoate
No change No change Turns from clear
to brown
Acidified potassium
dichromate(IV) No change
Turns from orange to green
No change
Ammonia solution
White fume observe
No change No change
Edexcel AS Chemistry
Practical 2.9 Reactions between halogens and halide
ions/some reactions of the halides
Some reactions of the halides
3) The properties of the hydrogen halides
Action on
Hydrogen chloride Hydrogen bromide Hydrogen iodide
Solubility in Water
Soluble Slightly soluble Slightly soluble
Reaction with Ammonia gas
Dense white fume observed
Dense white fume observed
Dense white fume observed
Thermal stability
No change observed; the gas is
stable
Little amount of brown gas
observed; the gas is less stable
Vast amount of purple gas
observed; the gas is unstable
30
Edexcel AS Chemistry
Practical 2.10 Factors that influence the rate of chemical
reactions
1) MnO2 and PbO2
2) A catalyst works by providing an alternative reaction route of a lower activation energy
Edexcel AS Chemistry
Practical 2.11 Effect of temperature, pressure and
concentration on equilibrium
1) Initially, brown gas is observed. When a hot water bath is used to heat up the test tube containing the gas, the intensity of brown colour is increased. When a ice bath is used to cool down the test tube, the intensity of brown colour decreasing until it reaches colourless.
2) When high pressure is applied, the intensity of brown colour decreases until it reaches colourless. But when lower pressure is applied, the intensity of brown colour increases.
31
Edexcel AS Chemistry
Practical 2.12 Reactions of alcohols
1) Solubility in water
The alcohols with shorter carbon chain are more soluble than the others due to their ability to form hydrogen bonds to water molecules. As the carbon chains is getting longer, the solubility decreases. This is because the carbon chain will interrupt (break) the hydrogen bonding and replace with London forces. Thus, the trend solubility of the alcohols is decreasing as the number of carbon chain increase.
Alcohols Observation
With water pH
Ethanol Soluble 7
Methanol Soluble 7
Propan–1–ol Slightly soluble, two layer observed 7
Butan–1–ol Slightly soluble; two layer observed 7
Pentan–1–ol Less soluble, two layer observed 7
Edexcel AS Chemistry
Practical 2.12 Reactions of alcohols
2) Reaction with sodium
The sodium reacts vigorously with alcohols with shorter carbon chain. As the carbon chains is getting longer, the reaction is getting less reactive. The products produce are the alkoxide compounds. It is predicted that alcohols behave similarly with water however the reaction between alcohol and sodium is less vigorous than that between sodium and water. When the alcohols react with sodium, the O–H breaks. Thus, the trend reactivity of the alcohols is decreasing from methanol to pentan–1–ol as the number of carbon chain increase.
Alcohols Observation
Ethanol Sodium dissolve vigorously in ethanol
Methanol Sodium dissolve vigorously in methanol
Propan–1–ol Sodium dissolve slightly vigorous in propan–1–ol
Butan–1–ol Sodium dissolve slightly vigorous in butan–1–ol
Pentan–1–ol Sodium dissolve less vigorous in pentan–1–ol
32
Edexcel AS Chemistry
Practical 2.12 Reactions of alcohols
3) Reaction with sodium
The acidified dichromate(VI) solution oxidized alcohols readily. However, the reaction of the alcohols becomes progressively slower with increasing length of carbon chain but reaction with methanol is also quite slow. The products obtained produce a mild fruity smell indicate that aldehydes are form.
Alcohols Observation
Ethanol The solution turns colour from orange to green.
Methanol The solution turns colour from orange to green.
Propan–1–ol The solution turns colour from orange to green.
Butan–1–ol The solution turns colour from orange to green.
Pentan–1–ol The solution turns colour from orange to green.
Edexcel AS Chemistry
Practical 2.13 Preparation of an organic liquid
(reflux and distillation) 1) The dehydration of propan–1–ol
The product obtained from this experiment has the ability to decolourized both of the reagent indicates that the product obtained having C=C. The name of the product is ___________.
Reagent Observation
Bromine water The colour of bromine water decolourized.
Acidified potassium manganate(VII)
The colour of the solution decolourized.
33
Edexcel AS Chemistry
Practical 2.13 Preparation of an organic liquid
(reflux and distillation) 2) The oxidation of propan–1–ol
• sour smell
• neutralises a relatively large volume of sodium carbonate solution.
• It does not give any change to the Fehling solution.
• indicate carboxylic acid obtained
• The name of the product is __________,
Reagent Observation
Sodium carbonate A mild effervescence observed
Fehling solution No change
Edexcel AS Chemistry
Practical 2.14 Preparation of a halogenoalkane
from an alcohol 1) A halogenoalkane from a primary alcohol
• pungent smell
• Milky globules observed at the bottom of the conical flask
• C2H5OH (aq) + HBr (aq) → C2H5Br (aq) + H2O (aq)
• C–O bond is broken in ethanol
2) A halogenoalkane from a tertiary alcohol • ~16.5 g
• Reasons:
• The reactant used may not be pure
• The reaction may not be complete
• Product maybe left behind on the apparatus
• Product maybe evaporate
• Parallax factor
34
Edexcel AS Chemistry
Practical 2.15 Reactions of the
halogenoalkanes 1) Combustion
2) Reaction with aqueous alkali • The product obtained is alcohol
3) Comparison of halogenoalkanes • Iodo, bromo, chloro
• Tertiary, secondary, primary
Halogenoalkanes Observation
1-chlorobutane Readily burn with yellow flame
2-chloro-2-methylbutane Readily burn with yellow flame
1 – bromobutane Burn with yellow flame
1 – iodobutane Burn with yellow flame and emission of purple
fume
Edexcel AS Chemistry
Practical 2.15 Reactions of the
halogenoalkanes 4) Reaction with alcoholic alkali
• The product obtained is alkene
• Elimination reaction take place
• Due to the presents of unchanged reactant which may be caused by strong heating