An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V2) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 Multiple Choice (Extended)
Choose the single correct option out of four choices.
40 Question · 40 marks
Question 1 · MCQ
1 marks
A student investigates the reaction between dilute hydrochloric acid and a large excess of calcium carbonate (marble chips). The temperature of the mixture is increased from 20 °C to 30 °C. All other conditions remain constant. Which statement explains why the rate of reaction increases?
A.The activation energy of the reaction decreases at the higher temperature.
B.The concentration of the reactant particles increases, leading to more frequent collisions.
C.A greater proportion of the colliding particles have energy equal to or greater than the activation energy.
D.The reactant particles move slower, which increases the total duration of contact between the reactants.
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Worked solution
Increasing the temperature increases the kinetic energy of the reactant particles. This means they move faster and collide more frequently, but more importantly, a significantly larger fraction of the collisions have an energy equal to or greater than the activation energy ($E \ge E_a$).
Marking scheme
[1] C is correct.
Question 2 · MCQ
1 marks
The decomposition of hydrogen peroxide, $\text{H}_2\text{O}_2$, is catalyzed by manganese(IV) oxide, $\text{MnO}_2$. Which row describes the effect of the catalyst on the activation energy and the overall enthalpy change, $\Delta H$, of this reaction?
B.activation energy: decreases | enthalpy change: no change
C.activation energy: no change | enthalpy change: decreases
D.activation energy: increases | enthalpy change: no change
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Worked solution
A catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy decreases. However, the energy levels of the reactants and products remain unchanged, so the overall enthalpy change ($\Delta H$) is unaffected.
Marking scheme
[1] B is correct.
Question 3 · MCQ
1 marks
A synthetic polymer has the repeating structure shown: $\text{---NH---(CH}_2\text{)}_6\text{---NH---CO---(CH}_2\text{)}_4\text{---CO---}$. Which type of polymerisation produces this polymer, and what is the small molecule released during the process?
A.addition polymerisation, hydrogen
B.addition polymerisation, water
C.condensation polymerisation, hydrogen chloride
D.condensation polymerisation, water
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Worked solution
This structure represents a polyamide (specifically Nylon-6,6), which contains amide linkages ($\text{---NH---CO---}$). Polyamides are formed via condensation polymerisation of diamines and dicarboxylic acids, which eliminates a small molecule of water ($\text{H}_2\text{O}$) during each link formation.
Marking scheme
[1] D is correct.
Question 4 · MCQ
1 marks
An addition polymer has the repeating unit structure: $\text{---(CH}_2\text{---CH(CH}_3\text{))}_n\text{---}$. Which monomer is used to produce this polymer?
A.propene
B.propane
C.ethene
D.but-1-ene
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Worked solution
The repeating unit contains three carbon atoms in its basic structure (two in the main chain backbone and one in a methyl side chain). Reconstructing the monomer by adding a double bond between the two chain carbons gives $\text{CH}_2\text{=CH---CH}_3$, which is propene.
Marking scheme
[1] A is correct.
Question 5 · MCQ
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes. Which products are discharged at the cathode and the anode?
A.cathode: hydrogen | anode: chlorine
B.cathode: sodium | anode: chlorine
C.cathode: hydrogen | anode: oxygen
D.cathode: sodium | anode: oxygen
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Worked solution
In concentrated aqueous sodium chloride, $\text{H}^+$ ions are discharged at the cathode because hydrogen is lower than sodium in the reactivity series, yielding hydrogen gas. $\text{Cl}^-$ ions are discharged at the anode because halide ions are in high concentration, yielding chlorine gas.
Marking scheme
[1] A is correct.
Question 6 · MCQ
1 marks
Aqueous copper(II) sulfate is electrolysed using active copper electrodes. Which row correctly describes the change in mass of each electrode?
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Worked solution
With active copper electrodes, copper atoms at the anode lose electrons to form $\text{Cu}^{2+}$ ions, causing the anode mass to decrease. At the cathode, $\text{Cu}^{2+}$ ions gain electrons and are deposited as copper metal, causing the cathode mass to increase.
Marking scheme
[1] A is correct.
Question 7 · MCQ
1 marks
A student wants to prepare a pure, dry sample of the insoluble salt, barium sulfate, $\text{BaSO}_4$. Which pair of solutions should be mixed together to achieve this?
A.barium carbonate and dilute sulfuric acid
B.barium chloride and sodium sulfate
C.barium hydroxide and copper(II) sulfate
D.barium nitrate and lead(II) sulfate
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Worked solution
Preparing an insoluble salt by precipitation requires mixing two soluble starting salts. Barium chloride is soluble, and sodium sulfate is soluble. Mixing them yields a precipitate of barium sulfate and soluble sodium chloride, which can be filtered, washed, and dried. Barium carbonate and lead(II) sulfate are insoluble, so they are not suitable.
Marking scheme
[1] B is correct.
Question 8 · MCQ
1 marks
The reaction between hydrogen and chlorine gas to form hydrogen chloride gas is exothermic: $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$. Which statement explains why this reaction is exothermic?
A.The energy released when bonds are formed in the products is greater than the energy absorbed to break the bonds in the reactants.
B.The energy released when bonds are formed in the products is less than the energy absorbed to break the bonds in the reactants.
C.The activation energy of the forward reaction is significantly greater than that of the reverse reaction.
D.The volume of the gaseous products is less than the volume of the gaseous reactants.
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Worked solution
A reaction is exothermic when bond-making (an exothermic process) releases more energy than is consumed during bond-breaking (an endothermic process).
Marking scheme
[1] A is correct.
Question 9 · multiple_choice
1 marks
A student investigates the reaction between marble chips (calcium carbonate) and excess dilute nitric acid. Which change in conditions increases the initial rate of reaction without increasing the total volume of carbon dioxide gas collected at the end of the reaction?
A.Using a larger volume of nitric acid of the same concentration.
B.Increasing the temperature of the nitric acid.
C.Using a larger mass of the same-sized marble chips.
D.Diluting the nitric acid with distilled water.
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Worked solution
Increasing the temperature increases the kinetic energy of the particles, leading to more frequent successful collisions per unit time and thus a higher initial rate. Since nitric acid is in excess, the calcium carbonate is the limiting reactant. The total volume of carbon dioxide depends only on the mass of the limiting reactant, which remains unchanged.
Marking scheme
1 mark for correct option selection.
Question 10 · multiple_choice
1 marks
Which row correctly describes the effect of adding a catalyst to a reaction mixture of iron(II) ions and acidified hydrogen peroxide at a constant temperature?
A.Activation energy: decreases | Proportion of particles with energy >= Ea: increases | Frequency of collisions: increases
B.Activation energy: decreases | Proportion of particles with energy >= Ea: increases | Frequency of collisions: remains constant
C.Activation energy: remains constant | Proportion of particles with energy >= Ea: remains constant | Frequency of collisions: increases
D.Activation energy: remains constant | Proportion of particles with energy >= Ea: increases | Frequency of collisions: remains constant
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Worked solution
A catalyst lowers the activation energy of the reaction. This increases the proportion of particles with energy greater than or equal to the activation energy. The total frequency of collisions depends on temperature and concentration, which are kept constant, so the overall frequency of collisions remains constant.
Marking scheme
1 mark for correct option selection.
Question 11 · multiple_choice
1 marks
A synthetic polymer is made by reacting a dicarboxylic acid with a diol. Which row correctly identifies the type of polymerisation and the functional group linking the monomer units?
A.Type of polymerisation: addition | Functional group link: amide
B.Type of polymerisation: addition | Functional group link: ester
C.Type of polymerisation: condensation | Functional group link: amide
D.Type of polymerisation: condensation | Functional group link: ester
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Worked solution
The reaction between a dicarboxylic acid (-COOH) and a diol (-OH) involves the elimination of water molecules, which is a condensation polymerisation. The resulting link is an ester linkage (-COO-).
Marking scheme
1 mark for correct option selection.
Question 12 · multiple_choice
1 marks
A section of a synthetic addition polymer chain is shown: -CH2-CH(C6H5)-CH2-CH(C6H5)-. Which compound is the monomer used to prepare this polymer?
A.CH2=CH2
B.CH3-CH=CH2
C.CH2=CH(C6H5)
D.CH(CH3)=CH(C6H5)
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Worked solution
The repeating unit is -CH2-CH(C6H5)-, which is derived from the unsaturated monomer phenylethene, CH2=CH(C6H5).
Marking scheme
1 mark for correct option selection.
Question 13 · multiple_choice
1 marks
Aqueous copper(II) chloride is electrolysed using inert platinum electrodes. Which ionic half-equations represent the reactions occurring at each electrode?
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Worked solution
At the cathode, copper(II) ions are discharged in preference to hydrogen ions because copper is lower in the reactivity series: Cu2+(aq) + 2e- -> Cu(s). At the anode, chloride ions are discharged in preference to hydroxide ions because halide ions are in high concentration: 2Cl-(aq) -> Cl2(g) + 2e-.
Marking scheme
1 mark for correct option selection.
Question 14 · multiple_choice
1 marks
Which statement is correct for the electrolysis of molten lead(II) bromide using inert carbon electrodes?
A.A red-brown gas is evolved at the negative electrode.
B.Lead atoms gain electrons at the positive electrode.
C.Bromide ions are oxidised at the positive electrode.
D.The mass of the positive electrode decreases.
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Worked solution
At the positive electrode (anode), bromide ions (Br-) lose electrons to form bromine gas (oxidation). Lead ions (Pb2+) are reduced at the negative electrode (cathode). Red-brown bromine gas is evolved at the anode, not the cathode.
Marking scheme
1 mark for correct option selection.
Question 15 · multiple_choice
1 marks
A student wants to prepare a pure, dry sample of barium sulfate. Which pair of aqueous solutions should be mixed together to prepare barium sulfate by precipitation?
A.Barium carbonate and dilute sulfuric acid
B.Barium chloride and aqueous sodium sulfate
C.Barium hydroxide and dilute sulfurous acid
D.Barium metal and dilute sulfuric acid
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Worked solution
Barium sulfate is an insoluble salt. To prepare an insoluble salt by precipitation, two soluble salt solutions must be mixed. Barium chloride is soluble, and sodium sulfate is soluble, making them the correct choice.
Marking scheme
1 mark for correct option selection.
Question 16 · multiple_choice
1 marks
What is the volume of hydrogen gas, measured at room temperature and pressure (r.t.p.), produced when 5.4 g of aluminium reacts completely with excess dilute hydrochloric acid? [Ar(Al) = 27; the volume of one mole of gas at r.t.p. is 24 dm3]
A.2.4 dm3
B.4.8 dm3
C.7.2 dm3
D.14.4 dm3
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Worked solution
The equation is 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g). Moles of Al = 5.4 / 27 = 0.2 mol. From the stoichiometry, 2 moles of Al produce 3 moles of H2. Therefore, moles of H2 produced = 0.2 * (3 / 2) = 0.3 mol. Volume of H2 = 0.3 * 24 = 7.2 dm3.
Marking scheme
1 mark for correct option selection.
Question 17 · MCQ
1 marks
In a series of experiments, dilute hydrochloric acid reacts with excess zinc. The temperature of the reaction mixture is increased from 20 °C to 30 °C, and the concentration of the acid is increased from 0.5 mol/dm³ to 1.0 mol/dm³. Which row describes the effect of these combined changes on the collision frequency and the percentage of collisions with energy greater than or equal to the activation energy, \(E_a\)?
A.Collision frequency: increases | Percentage of collisions with energy \(\ge E_a\): increases
B.Collision frequency: increases | Percentage of collisions with energy \(\ge E_a\): remains constant
C.Collision frequency: remains constant | Percentage of collisions with energy \(\ge E_a\): increases
D.Collision frequency: decreases | Percentage of collisions with energy \(\ge E_a\): decreases
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Worked solution
An increase in temperature increases both the average kinetic energy of the particles (hence a higher percentage of collisions have energy greater than or equal to the activation energy, \(E_a\)) and the frequency of collisions. An increase in concentration increases the number of reactant particles per unit volume, which also increases the frequency of collisions. Therefore, both parameters increase.
Marking scheme
1 mark for the correct option A.
Question 18 · MCQ
1 marks
Manganese(IV) oxide is used as a catalyst in the decomposition of hydrogen peroxide. How does the addition of this catalyst affect the activation energy and the enthalpy change, \(\Delta H\), of this reaction?
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Worked solution
A catalyst provides an alternative reaction pathway with a lower activation energy, thereby decreasing the activation energy. It does not alter the energy levels of the reactants or products, meaning the overall enthalpy change, \(\Delta H\), remains completely unchanged.
Marking scheme
1 mark for the correct option B.
Question 19 · MCQ
1 marks
A synthetic polymer is manufactured by reacting a dicarboxylic acid with a diol. Which type of polymerisation occurs and what is the type of linkage formed in the macromolecule?
A.Type of polymerisation: condensation | Linkage formed: ester
B.Type of polymerisation: condensation | Linkage formed: amide
C.Type of polymerisation: addition | Linkage formed: ester
D.Type of polymerisation: addition | Linkage formed: carbon-carbon single bond
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Worked solution
The reaction between a dicarboxylic acid and a diol involves the elimination of water molecules, which is a condensation polymerisation. The linkage formed between the monomers is an ester linkage (-COO-).
Marking scheme
1 mark for the correct option A.
Question 20 · MCQ
1 marks
Proteins are naturally occurring macromolecules. When a sample of protein is completely hydrolysed, which class of compound is formed?
A.alcohols
B.amino acids
C.carboxylic acids
D.simple sugars
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Worked solution
Proteins are condensation polymers formed from amino acid monomers. Hydrolysis of the amide (peptide) links in a protein breaks the polymer down back into its constituent amino acids.
Marking scheme
1 mark for the correct option B.
Question 21 · MCQ
1 marks
Concentrated aqueous sodium chloride (brine) is electrolysed using inert carbon electrodes. Which row correctly identifies the substances produced at the anode and at the cathode?
A.Anode (+): chlorine | Cathode (-): hydrogen
B.Anode (+): oxygen | Cathode (-): sodium
C.Anode (+): chlorine | Cathode (-): sodium
D.Anode (+): oxygen | Cathode (-): hydrogen
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Worked solution
In concentrated aqueous sodium chloride, chloride ions (\(Cl^-\)) are discharged at the positive electrode (anode) to produce chlorine gas because of their high concentration. Hydrogen ions (\(H^+\)) from water are preferentially discharged at the negative electrode (cathode) to produce hydrogen gas because hydrogen is lower in the reactivity series than sodium.
Marking scheme
1 mark for the correct option A.
Question 22 · MCQ
1 marks
Molten lead(II) bromide is electrolysed using graphite electrodes. What is the ionic half-equation for the chemical reaction occurring at the negative electrode (cathode)?
A.\(Pb^{2+} + 2e^- \rightarrow Pb\)
B.\(Pb^{2+} \rightarrow Pb + 2e^-\)
C.\(2Br^- \rightarrow Br_2 + 2e^-\)
D.\(2Br^- + 2e^- \rightarrow Br_2\)
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Worked solution
At the cathode, positive lead ions (\(Pb^{2+}\)) gain electrons (reduction) to form lead metal. The correct balanced half-equation for this process is: \(Pb^{2+} + 2e^- \rightarrow Pb\).
Marking scheme
1 mark for the correct option A.
Question 23 · MCQ
1 marks
A student wants to prepare a pure, dry sample of the insoluble salt barium sulfate, \(BaSO_4\). Which pair of aqueous solutions is most suitable to mix together?
A.barium carbonate and dilute sulfuric acid
B.barium chloride and sodium sulfate
C.barium hydroxide and copper(II) sulfate
D.barium nitrate and lead(II) sulfate
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Worked solution
To prepare an insoluble salt by precipitation, two soluble salts must be mixed. Barium chloride (\(BaCl_2\)) and sodium sulfate (\(Na_2SO_4\)) are both soluble in water, and mixing them yields insoluble barium sulfate as a precipitate and soluble sodium chloride in solution.
Marking scheme
1 mark for the correct option B.
Question 24 · MCQ
1 marks
Hydrated copper(II) sulfate crystals can be prepared by reacting excess copper(II) oxide with dilute sulfuric acid. Which sequence of experimental steps describes the correct method?
A.filter the mixture \(\rightarrow\) heat filtrate to crystallisation point \(\rightarrow\) wash and dry crystals
B.add excess copper(II) oxide to acid \(\rightarrow\) filter \(\rightarrow\) evaporate filtrate to crystallisation point \(\rightarrow\) leave to cool
C.add excess copper(II) oxide to acid \(\rightarrow\) evaporate to dryness \(\rightarrow\) filter solid \(\rightarrow\) wash residue
D.heat acid \(\rightarrow\) add copper(II) oxide until pH neutral \(\rightarrow\) evaporate to dryness \(\rightarrow\) recrystallise
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Worked solution
First, excess copper(II) oxide is added to the acid to ensure all the acid is neutralised. The unreacted solid oxide is then filtered off. The filtrate (aqueous copper(II) sulfate) is heated to evaporate some water until the crystallisation point is reached. Finally, it is cooled slowly to allow crystals to form, which are then filtered, washed, and dried.
Marking scheme
1 mark for the correct option B.
Question 25 · MCQ
1 marks
A reaction of excess dilute hydrochloric acid with a fixed mass of zinc powder is carried out at \(20^\circ\text{C}\). The volume of hydrogen gas produced is plotted over time to give Curve 1. The experiment is repeated under the same conditions but using the same mass of larger zinc pieces instead of zinc powder, to give Curve 2.
Which statement correctly describes Curve 2 compared to Curve 1?
A.The initial rate of reaction is faster and a larger final volume of hydrogen gas is collected.
B.The initial rate of reaction is faster but the same final volume of hydrogen gas is collected.
C.The initial rate of reaction is slower and a smaller final volume of hydrogen gas is collected.
D.The initial rate of reaction is slower but the same final volume of hydrogen gas is collected.
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Worked solution
Larger zinc pieces have a smaller total surface area than zinc powder, which decreases the rate of reaction, so the initial rate is slower (Curve 2 is less steep). Since the mass of the limiting reactant (zinc) is the same in both experiments and the acid is in excess, the final volume of hydrogen gas produced will be the same.
Marking scheme
1 mark: Correct option chosen (D).
Question 26 · MCQ
1 marks
Why does an increase in temperature increase the rate of a chemical reaction?
1. The particles have more kinetic energy, which increases the frequency of collisions. 2. The activation energy of the reaction is lowered. 3. A greater proportion of the colliding particles have energy equal to or greater than the activation energy.
A.1 and 2 only
B.1 and 3 only
C.2 and 3 only
D.1, 2 and 3
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Worked solution
Increasing the temperature increases the kinetic energy of the particles, meaning they move faster and collide more frequently (Statement 1 is correct). It also significantly increases the proportion of colliding particles with energy equal to or greater than the activation energy (Statement 3 is correct). Only a catalyst can lower the activation energy (Statement 2 is incorrect).
Marking scheme
1 mark: Correct option chosen (B).
Question 27 · MCQ
1 marks
A condensation polymer is formed when monomers react with the loss of small molecules. Which pair of monomers reacts to form a polyester?
A.a dicarboxylic acid and a diol
B.a dicarboxylic acid and a diamine
C.a monocarboxylic acid and a diol
D.an alkene and a diol
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Worked solution
Polyesters are condensation polymers formed by the reaction of monomers containing two carboxylic acid groups (\(-COOH\)) and two alcohol groups (\(-OH\)). Therefore, a dicarboxylic acid and a diol react to form a polyester with ester linkages, releasing water molecules.
Marking scheme
1 mark: Correct option chosen (A).
Question 28 · MCQ
1 marks
A section of an addition polymer chain is shown below.
\(-CH_2-CH(Cl)-CH_2-CH(Cl)-CH_2-CH(Cl)-\)
What is the name of the monomer used to produce this polymer?
A.chloroethane
B.chloroethene
C.1,2-dichloroethene
D.3-chloropropene
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Worked solution
The repeating unit of the polymer is \(-CH_2-CH(Cl)-\). This corresponds to the addition polymerisation of chloroethene, \(CH_2=CH(Cl)\), which is also known as vinyl chloride.
Marking scheme
1 mark: Correct option chosen (B).
Question 29 · MCQ
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly describes the products at the electrodes and the change in pH of the electrolyte near the cathode?
D.cathode product: oxygen; anode product: hydrogen; pH: no change
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Worked solution
In the electrolysis of concentrated aqueous sodium chloride (brine), hydrogen ions (\(H^+\)) are preferentially discharged at the cathode to form hydrogen gas (\(H_2\)), and chloride ions (\(Cl^-\)) are discharged at the anode to form chlorine gas (\(Cl_2\)). As \(H^+\) and \(Cl^-\) ions are removed, sodium (\(Na^+\)) and hydroxide (\(OH^-\)) ions remain in the solution, forming sodium hydroxide, which increases the pH.
Marking scheme
1 mark: Correct option chosen (C).
Question 30 · MCQ
1 marks
Which setup is correct for electroplating a steel spoon with a layer of silver?
A.Anode: pure silver; Cathode: steel spoon; Electrolyte: aqueous silver nitrate
B.Anode: steel spoon; Cathode: pure silver; Electrolyte: aqueous silver nitrate
C.Anode: pure silver; Cathode: steel spoon; Electrolyte: dilute sulfuric acid
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Worked solution
In electroplating, the object to be plated (steel spoon) is always connected as the cathode (negative electrode). The metal used for plating (pure silver) is connected as the anode (positive electrode). The electrolyte must be a soluble salt of the plating metal (aqueous silver nitrate).
Marking scheme
1 mark: Correct option chosen (A).
Question 31 · MCQ
1 marks
Which experimental method is most suitable for preparing a pure, dry sample of the insoluble salt, barium sulfate?
A.Add excess barium carbonate to dilute sulfuric acid, filter the mixture, and evaporate the filtrate to obtain crystals.
B.Mix aqueous barium chloride with dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry the residue.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid, and evaporate the resulting mixture to dryness.
D.React barium metal with dilute sulfuric acid, filter off the excess metal, and crystallise the filtrate.
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Worked solution
Barium sulfate is an insoluble salt. The standard preparation method for insoluble salts is precipitation. This involves mixing two soluble reactants (aqueous barium chloride and dilute sulfuric acid) to form the precipitate. The precipitate is then filtered off, washed with distilled water to remove soluble impurities, and dried.
Marking scheme
1 mark: Correct option chosen (B).
Question 32 · MCQ
1 marks
Copper(II) sulfate crystals are prepared by adding excess copper(II) oxide to hot dilute sulfuric acid. Why is an excess of copper(II) oxide used?
A.To ensure that all of the sulfuric acid has reacted.
B.To act as a catalyst to speed up the rate of reaction.
C.To prevent the copper(II) sulfate from dissolving in water.
D.To reduce the temperature required for crystallization.
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Worked solution
Copper(II) oxide is insoluble in water. Adding it in excess ensures that all the sulfuric acid is completely neutralised. Unreacted acid would otherwise remain in the filtrate and concentrate during crystallization, contaminating the copper(II) sulfate crystals.
Marking scheme
1 mark: Correct option chosen (A).
Question 33 · multiple_choice
1 marks
The decomposition of aqueous hydrogen peroxide is catalysed by manganese(IV) oxide. Which statement describes how the catalyst increases the rate of this reaction?
A.It increases the frequency of collisions between reactant particles.
B.It increases the kinetic energy of the reactant particles.
C.It provides an alternative reaction pathway with a lower activation energy.
D.It increases the collision energy of the colliding particles.
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Worked solution
A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows a greater proportion of colliding particles to have energy equal to or greater than the activation energy, increasing the frequency of successful collisions. It does not alter the actual kinetic energy of the particles or the frequency of overall collisions.
Marking scheme
1 mark for the correct option C.
Question 34 · multiple_choice
1 marks
An excess of calcium carbonate chunks is reacted with dilute hydrochloric acid. The volume of carbon dioxide gas produced is measured over time. The experiment is repeated under identical conditions, but using the same mass of calcium carbonate powder instead of chunks. How do the initial rate of reaction and the total volume of carbon dioxide gas produced compare to the first experiment?
A.The initial rate is higher, and the total volume of gas is greater.
B.The initial rate is higher, and the total volume of gas is the same.
C.The initial rate is the same, and the total volume of gas is greater.
D.The initial rate is lower, and the total volume of gas is the same.
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Worked solution
Using powder instead of chunks increases the surface area of the solid reactant. This increases the frequency of successful collisions, leading to a higher initial rate of reaction. However, because the mass of the limiting reactant (and excess reactant) is unchanged, the total volume of gas produced at the end of the reaction remains the same.
What is the correct name and formula of the monomer used to produce this polymer?
A.ethene, \(\text{C}_2\text{H}_4\)
B.propene, \(\text{C}_3\text{H}_6\)
C.butene, \(\text{C}_4\text{H}_8\)
D.propane, \(\text{C}_3\text{H}_8\)
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Worked solution
The repeat unit of this polymer contains a chain of two carbon atoms with a methyl group on one of them: \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\). The monomer must therefore have a carbon-carbon double bond across these two carbon atoms, which corresponds to propene, \(\text{C}_3\text{H}_6\).
Marking scheme
1 mark for the correct option B.
Question 36 · multiple_choice
1 marks
Which row correctly describes the monomers and type of linkage found in a typical polyester such as Terylene (PET)?
A.Monomers: dicarboxylic acid and diamine; Linkage: amide
B.Monomers: diol and diamine; Linkage: ester
C.Monomers: dicarboxylic acid and diol; Linkage: amide
D.Monomers: dicarboxylic acid and diol; Linkage: ester
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Worked solution
Polyesters are condensation polymers formed by the reaction of dicarboxylic acid monomers and diol monomers. During the reaction, ester linkages are formed with the elimination of water molecules.
Marking scheme
1 mark for the correct option D.
Question 37 · multiple_choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which products are formed at the anode and cathode?
A.Anode: chlorine; Cathode: hydrogen
B.Anode: hydrogen; Cathode: oxygen
C.Anode: chlorine; Cathode: sodium
D.Anode: oxygen; Cathode: sodium
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Worked solution
In concentrated aqueous sodium chloride (brine), the ions present are \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). At the anode, chloride ions are selectively discharged to form chlorine gas because they are in high concentration. At the cathode, hydrogen ions are preferentially discharged over sodium ions because hydrogen is less reactive, forming hydrogen gas.
Marking scheme
1 mark for the correct option A.
Question 38 · multiple_choice
1 marks
Molten lead(II) bromide is electrolysed using inert platinum electrodes. Which ionic half-equation correctly represents the reaction occurring at the negative electrode (cathode)?
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Worked solution
At the negative electrode (cathode), positive lead(II) ions (\(\text{Pb}^{2+}\)) migrate and gain electrons (reduction) to form lead metal. The balanced ionic half-equation is \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\).
Marking scheme
1 mark for the correct option A.
Question 39 · multiple_choice
1 marks
A student wants to prepare a pure, dry sample of the insoluble salt, barium sulfate. Which pair of aqueous solutions should be mixed together, and what is the final step to obtain the dry solid?
A.barium chloride and sodium sulfate; evaporate the filtrate to dryness
B.barium nitrate and sodium sulfate; wash the residue with distilled water and dry
C.barium carbonate and dilute sulfuric acid; filter and crystallise
D.barium hydroxide and dilute sulfuric acid; evaporate the mixture to dryness
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Worked solution
To prepare an insoluble salt, precipitation is used by mixing two soluble salts together. Barium nitrate and sodium sulfate are both soluble. Mixing them yields a precipitate of barium sulfate. The mixture is filtered, and the residue (barium sulfate) is washed with distilled water to remove soluble impurities and then dried.
A.Iron is oxidised because its oxidation state increases.
B.Carbon monoxide acts as the reducing agent, and carbon is oxidised.
C.Carbon dioxide acts as the oxidising agent.
D.The oxidation state of oxygen changes from -2 to 0.
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Worked solution
In this reaction, carbon monoxide (CO) is oxidised to carbon dioxide (\(\text{CO}_2\)) because the oxidation state of carbon increases from +2 to +4. Since CO reduces the iron(III) oxide, it acts as the reducing agent, and carbon is oxidised.
Marking scheme
1 mark for the correct option B.
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6 Question · 80 marks
Question 1 · Structured Theory
13 marks
The reaction between calcium carbonate and dilute hydrochloric acid is investigated: \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \to \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\). (a) Draw a labelled diagram of the apparatus that can be used to measure the volume of carbon dioxide gas produced over time. [3] (b) Describe how the rate of reaction changes from the start of the reaction until it stops. Explain this change in terms of particle collision theory. [4] (c) State and explain the effect of increasing the concentration of hydrochloric acid on the rate of reaction. Use collision theory in your explanation. [3] (d) Describe the expected difference in the reaction progress curve if the reaction is repeated at a higher temperature. Explain your answer. [3]
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Worked solution
(a) A diagram showing a sealed reaction flask containing calcium carbonate and acid, connected via a delivery tube to a gas syringe or an inverted measuring cylinder over water. (b) The rate is highest at the start because the concentration of acid particles is at its maximum, leading to the highest frequency of successful collisions. As the reaction proceeds, the acid particles are used up, decreasing the concentration and the collision frequency, until the reaction stops (rate becomes zero) when the limiting reactant is completely consumed. (c) Increasing the concentration increases the rate of reaction because there are more reactant particles per unit volume, which increases the frequency of successful collisions. (d) At a higher temperature, the reaction progress curve is steeper because the rate increases, but it levels off at the same volume of gas. This is because particles have more kinetic energy, so a greater fraction of collisions have energy greater than the activation energy, increasing the frequency of successful collisions.
Marking scheme
(a) M1: Delivery tube connected to a gas syringe or inverted cylinder over water [1] M2: Sealed reaction flask [1] M3: Correct labels [1] (b) M1: Rate is highest at start and decreases over time [1] M2: Rate becomes zero / reaction stops [1] M3: Concentration of acid decreases as it is used up [1] M4: Frequency of collisions decreases [1] (c) M1: Rate increases [1] M2: More particles per unit volume [1] M3: Increased frequency of collisions [1] (d) M1: Steeper curve / levels off earlier [1] M2: Levels off at same volume [1] M3: Particles have more kinetic energy / more successful collisions per second [1]
Question 2 · Structured Theory
13 marks
This question is about polymers. (a) Poly(lactic acid), PLA, is a biodegradable condensation polymer. (i) State what is meant by the term biodegradable. [1] (ii) The monomer for PLA is 2-hydroxypropanoic acid (lactic acid), \(\text{CH}_3\text{CH(OH)COOH}\). Draw the structure of poly(lactic acid) showing two repeat units. Show all the atoms and bonds in the ester linkages. [3] (b) Nylon-6,6 is a synthetic polyamide. (i) State the names of the two functional groups that react to form a polyamide. [2] (ii) Draw the structures of the two monomers that react to form the polyamide represented by the structure: \(\text{[-HN-(CH}_2\text{)}_6\text{-NH-CO-(CH}_2\text{)}_4\text{-CO-]}\). [3] (c) Compare addition polymerisation and condensation polymerisation in terms of the types of monomer used and the products formed. [4]
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Worked solution
(a)(i) Biodegradable means capable of being decomposed by bacteria or other living organisms. (a)(ii) The structure shows two repeat units linked by an ester group: \(\text{[-O-CH(CH}_3\text{)-CO-O-CH(CH}_3\text{)-CO-]}\) with continuation bonds at both ends. (b)(i) Amine (amino) group and carboxylic acid (carboxyl) group. (b)(ii) The two monomers are 1,6-diaminohexane: \(\text{H}_2\text{N-(CH}_2\text{)}_6\text{-NH}_2\) and hexanedioic acid: \(\text{HOOC-(CH}_2\text{)}_4\text{-COOH}\). (c) Addition polymerisation uses monomers with C=C double bonds (unsaturated), forming only the polymer as a product. Condensation polymerisation uses monomers with two different functional groups, forming both the polymer and a small molecule (like water) as products.
Marking scheme
(a)(i) M1: Broken down by microbes/bacteria [1] (a)(ii) M1: Ester linkage shown correctly as -O-CO- [1] M2: Correct carbon chain with methyl side groups [1] M3: Continuation bonds shown [1] (b)(i) M1: Amine / amino [1] M2: Carboxylic acid / carboxyl [1] (b)(ii) M1: Diamine structure correct [1.5] M2: Dicarboxylic acid structure correct [1.5] (c) M1: Addition uses unsaturated monomers / containing C=C [1] M2: Condensation uses monomers with bifunctional groups [1] M3: Addition produces only polymer [1] M4: Condensation produces polymer and a small molecule (e.g., water) [1]
Question 3 · Structured Theory
14 marks
The electrolysis of concentrated aqueous copper(II) chloride is carried out using carbon (graphite) electrodes. (a) State the observations and name the product formed at: (i) the anode (positive electrode) [2] (ii) the cathode (negative electrode). [2] (b) Write the ionic half-equations for the reactions occurring at: (i) the anode [1.5] (ii) the cathode. [1.5] (c) The electrolysis of dilute sulfuric acid is carried out using inert platinum electrodes. (i) Name the product formed at each electrode. [2] (ii) Write the ionic half-equation for the reaction at the anode. [2] (iii) Describe how the concentration of the acid changes during this electrolysis and explain your answer. [3]
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Worked solution
(a)(i) At the anode, bubbles of a pale green, pungent gas are observed; the product is chlorine. (a)(ii) At the cathode, a pink/brown solid is deposited on the electrode; the product is copper. (b)(i) \(2\text{Cl}^-\text{(aq)} \to \text{Cl}_2\text{(g)} + 2\text{e}^-\). (b)(ii) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \to \text{Cu(s)}\). (c)(i) Anode: Oxygen; Cathode: Hydrogen. (c)(ii) \(4\text{OH}^- \to \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (or equivalent water oxidation equation). (c)(iii) The concentration of the acid increases because water is decomposed and removed as hydrogen and oxygen gases, leaving the solute (sulfuric acid) in a smaller volume of water.
Marking scheme
(a)(i) M1: Bubbles of green gas / pungent smell [1] M2: Chlorine [1] (a)(ii) M1: Pink/brown solid deposit [1] M2: Copper [1] (b)(i) M1: Correctly balanced half-equation [1.5] (b)(ii) M1: Correctly balanced half-equation [1.5] (c)(i) M1: Oxygen at anode [1] M2: Hydrogen at cathode [1] (c)(ii) M1: Correct balanced equation [2] (c)(iii) M1: Concentration increases [1] M2: Water is decomposed / lost [1] M3: Amount of acid remains constant but volume of solvent decreases [1]
Question 4 · Structured Theory
13 marks
This question is about the preparation of salts. (a) Barium sulfate is an insoluble salt that can be prepared by a precipitation reaction. (i) Name two soluble salts that could be mixed to prepare barium sulfate. [2] (ii) Describe the steps required to obtain a pure, dry sample of barium sulfate from the reaction mixture. [4] (iii) Write the ionic equation, with state symbols, for this precipitation reaction. [2] (b) Zinc sulfate is a soluble salt. It can be prepared by reacting excess zinc oxide with dilute sulfuric acid. (i) Explain why excess zinc oxide is used rather than excess sulfuric acid. [2] (ii) State how the excess zinc oxide is removed from the reaction mixture. [1] (iii) Describe how hydrated zinc sulfate crystals are obtained from the zinc sulfate solution. [2]
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Worked solution
(a)(i) Soluble salts: Barium nitrate (or barium chloride) and sodium sulfate (or potassium sulfate). (a)(ii) Mix the two solutions to form the precipitate, then filter the mixture to obtain the barium sulfate residue. Wash the residue with distilled water to remove any soluble impurities, and finally dry the residue in a warm oven or between filter papers. (a)(iii) \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \to \text{BaSO}_4\text{(s)}\). (b)(i) Excess zinc oxide is used to ensure all the sulfuric acid is completely neutralised. Zinc oxide is insoluble, so the unreacted excess can easily be filtered off, whereas excess acid would contaminate the salt solution. (b)(ii) Remove by filtration. (b)(iii) Heat the filtrate to evaporate some water until the crystallisation point is reached, then allow it to cool so that crystals form, filter the crystals, and dry them.
Marking scheme
(a)(i) M1: Barium nitrate / barium chloride [1] M2: Sodium sulfate / potassium sulfate / sulfuric acid [1] (a)(ii) M1: Filter to get residue [1] M2: Wash residue [1] M3: Use distilled water [1] M4: Dry (oven/filter paper) [1] (a)(iii) M1: Formulae correct [1] M2: State symbols correct (aq and s) [1] (b)(i) M1: To ensure all acid is neutralised [1] M2: Excess solid is easily filtered off / prevents contamination [1] (b)(ii) M1: Filtration / filter [1] (b)(iii) M1: Evaporate to crystallization point [1] M2: Cool and filter crystals [1]
Question 5 · Structured Theory
13 marks
A student carries out an experiment to determine the formula of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\). The student heats a known mass of hydrated copper(II) sulfate in a crucible. The data obtained is: Mass of empty crucible = 15.40 g; Mass of crucible + hydrated copper(II) sulfate = 20.39 g; Mass of crucible + anhydrous copper(II) sulfate (after heating to constant mass) = 18.59 g. (a) State how the student can ensure that all the water of crystallisation has been completely removed. [1] (b) Calculate: (i) the mass of anhydrous copper(II) sulfate obtained. [1] (ii) the mass of water of crystallisation lost. [1] (iii) the number of moles of anhydrous copper(II) sulfate (Relative formula mass: \(\text{CuSO}_4 = 160\)). [2] (iv) the number of moles of water lost (Relative formula mass: \(\text{H}_2\text{O} = 18\)). [2] (v) the value of \(x\) in \(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\). [2] (c) State the colour change observed when water is added to anhydrous copper(II) sulfate. [2] (d) Write the chemical equation for the reversible reaction that occurs when hydrated copper(II) sulfate is heated. [2]
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Worked solution
(a) Heat the crucible, weigh it, and repeat this process until the mass remains constant. (b)(i) Mass of anhydrous \(\text{CuSO}_4 = 18.59 - 15.40 = 3.19\text{ g}\). (b)(ii) Mass of water lost = \(20.39 - 18.59 = 1.80\text{ g}\). (b)(iii) Moles of \(\text{CuSO}_4 = 3.19 / 160 = 0.020\text{ mol}\). (b)(iv) Moles of \(\text{H}_2\text{O} = 1.80 / 18 = 0.100\text{ mol}\). (b)(v) \(x = 0.100 / 0.020 = 5\). (c) The colour changes from white to blue. (d) \(\text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(l)} \rightleftharpoons \text{CuSO}_4 \cdot 5\text{H}_2\text{O(s)}\).
Marking scheme
(a) M1: Heat to constant mass [1] (b)(i) M1: 3.19 g [1] (b)(ii) M1: 1.80 g [1] (b)(iii) M1: Moles calculation working (3.19 / 160) [1] M2: 0.02 mol [1] (b)(iv) M1: Moles calculation working (1.80 / 18) [1] M2: 0.10 mol [1] (b)(v) M1: Ratio of moles (0.10 / 0.02) [1] M2: x = 5 [1] (c) M1: White [1] M2: to blue [1] (d) M1: Reactants and products correct [1] M2: Reversible arrow shown [1]
Question 6 · Structured Theory
14 marks
Methanol is manufactured industrially by the reversible reaction between carbon monoxide and hydrogen: \(\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}\), \(\Delta H = -91\text{ kJ/mol}\). (a) State and explain the effect of increasing the temperature on: (i) the rate of the forward reaction [2] (ii) the yield of methanol at equilibrium. [3] (b) State and explain the effect of increasing the pressure on: (i) the rate of the forward reaction [2] (ii) the yield of methanol at equilibrium. [3] (c) A catalyst is used in this process. (i) Explain why a catalyst increases the rate of reaction. [2] (ii) State the effect of the catalyst on the yield of methanol at equilibrium. [2]
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Worked solution
(a)(i) Increasing the temperature increases the rate of the reaction because particles have more kinetic energy, moving faster and colliding more frequently and with more energy. (a)(ii) The yield decreases because the forward reaction is exothermic. Increasing temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the added heat. (b)(i) Increasing the pressure increases the rate because the gas molecules are closer together, which increases the frequency of collisions. (b)(ii) The yield increases because there are fewer moles of gas on the right (1 mole) than on the left (3 moles). Increasing pressure shifts the equilibrium to the side with fewer gas molecules. (c)(i) A catalyst provides an alternative reaction pathway with a lower activation energy, meaning more particles have sufficient energy to react. (c)(ii) A catalyst has no effect on the yield of methanol because it increases the rate of both the forward and reverse reactions by the same factor.
Marking scheme
(a)(i) M1: Rate increases [1] M2: More kinetic energy / more collisions [1] (a)(ii) M1: Yield decreases [1] M2: Forward reaction is exothermic [1] M3: Shift to the left / endothermic direction [1] (b)(i) M1: Rate increases [1] M2: Particles closer together / higher collision frequency [1] (b)(ii) M1: Yield increases [1] M2: Shift to the side with fewer gas moles [1] M3: 1 mole on right vs 3 moles on left [1] (c)(i) M1: Alternative pathway [1] M2: Lower activation energy [1] (c)(ii) M1: No effect [1] M2: Increases rate of forward and reverse reactions equally [1]
Paper 62 Alternative to Practical
Answer all questions. Show planning and interpretation of laboratory data.
4 Question · 40 marks
Question 1 · Practical Theory
10 marks
A student investigates the rate of the catalytic decomposition of aqueous hydrogen peroxide, \(\text{H}_2\text{O}_2\), using manganese(IV) oxide, \(\text{MnO}_2\), as a catalyst. The equation for the reaction is:
Explain why the rate of reaction is fastest at the start of the experiment. [2]
(iv) Describe how the shape of a graph of volume of gas against time would change if the experiment were repeated at \(40\text{ }^\circ\text{C}\) with all other variables kept constant. [3]
(v) Outline an experimental procedure to prove that the manganese(IV) oxide acts as a catalyst and is not chemically consumed during the reaction. [3]
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(i) A gas syringe is the standard laboratory apparatus designed to collect and measure precise gas volumes. (ii) A volumetric pipette or a burette offers much higher accuracy and precision for measuring liquid volumes than a standard measuring cylinder. (iii) At the start of the reaction, the concentration of reactant particles (hydrogen peroxide) is at its maximum. Therefore, there is a higher frequency of successful collisions per unit time between reactant particles. (iv) Raising the temperature increases the kinetic energy of the particles, resulting in a higher rate of reaction. The curve will have a steeper initial gradient (slope) and will plateau (level off) earlier. Since the initial concentration and volume of reactant are unchanged, the final volume of gas collected remains exactly the same (\(40\text{ cm}^3\)). (v) To show that \(\text{MnO}_2\) is a catalyst: first, measure its initial dry mass before adding it to the reaction flask. After the reaction is complete, filter the mixture to recover the solid \(\text{MnO}_2\). Wash it thoroughly with distilled water to remove impurities and dry it in a warm oven. Weigh the dry solid again; the mass should be identical to the starting mass, proving it was not consumed.
Marking scheme
(i) Gas syringe [1] (ii) Use a volumetric pipette / burette / graduated pipette (reject: measuring cylinder) [1] (iii) Concentration of hydrogen peroxide molecules is highest at the start [1] leads to a higher frequency of successful collisions [1] (iv) Curve has a steeper initial gradient / slope [1] Curve levels off / plateaus earlier / in less time [1] Final volume of gas remains the same (at \(40\text{ cm}^3\)) [1] (v) Weigh the dry manganese(IV) oxide before the experiment [1] Filter, wash, and dry the solid at the end of the reaction [1] Reweigh the solid to show that the mass is unchanged [1]
Question 2 · Practical Theory
10 marks
A student investigates the electrolysis of concentrated aqueous sodium chloride (brine) using inert carbon (graphite) electrodes in a U-tube.
(i) State the observations made at each electrode during the electrolysis: Anode (positive electrode): [1] Cathode (negative electrode): [1]
(ii) A few drops of universal indicator are added to the solution around the cathode during the electrolysis. State the colour change observed and explain why this change occurs. [3]
(iii) Name the gas produced at the anode and describe a chemical test to confirm its identity. [2]
(iv) A student wants to modify the apparatus to electroplate a steel key with a thin layer of copper. State the required components for this modified setup: Material for the anode: [1] Material for the cathode: [1] Electrolyte solution: [1]
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Worked solution
(i) At the anode, chloride ions are discharged to form chlorine gas, which is seen as bubbles of a pale green / yellow-green gas. At the cathode, hydrogen ions are discharged in preference to sodium ions, producing hydrogen gas, which is seen as bubbles of a colourless gas. (ii) The indicator changes from green to blue or purple. This occurs because hydrogen ions are reduced to hydrogen gas, leaving an excess of hydroxide ions (\(\text{OH}^-\)) in the solution near the cathode. This formation of sodium hydroxide makes the local solution alkaline. (iii) The anode gas is chlorine. The chemical test for chlorine involves exposing it to damp blue litmus paper, which turns red (due to its acidic nature in water) and is then rapidly bleached white. (iv) For electroplating: the object to be plated (steel key) must be the cathode. The metal used for plating (copper) must be the anode. The electrolyte must be a soluble salt solution containing the plating metal ions (aqueous copper(II) sulfate).
Marking scheme
(i) Anode: bubbles of a pale green / yellow-green gas [1] Cathode: bubbles of a colourless / odourless gas [1] (ii) Colour change: (green to) blue / purple [1] Explanation: hydrogen ions are discharged / water is reduced to produce hydrogen [1] leaving an excess of hydroxide (\(\text{OH}^-\)) ions / forming sodium hydroxide which is alkaline [1] (iii) Gas: chlorine [1] Test: damp blue litmus paper / universal indicator paper turns red and then bleaches / turns white [1] (iv) Anode: copper / pure copper sheet [1] Cathode: steel key [1] Electrolyte: aqueous copper(II) sulfate / any soluble copper(II) salt solution [1]
Question 3 · Practical Theory
10 marks
A student prepares a pure, dry sample of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), from insoluble copper(II) oxide powder and dilute sulfuric acid.
(i) State why copper(II) oxide is added in excess to the dilute sulfuric acid. [1]
(ii) State the method used to separate the excess copper(II) oxide from the mixture. [1]
(iii) Describe how the student can obtain large, well-formed crystals of hydrated copper(II) sulfate from the filtrate. Do not describe how to dry the crystals in your answer. [3]
(iv) State how the crystals can be dried safely without losing their water of crystallisation. [1]
(v) Copper(II) carbonate can also be used instead of copper(II) oxide. - State one different observation that would be made during the reaction. [1] - Write the balanced chemical equation for the reaction between copper(II) carbonate and dilute sulfuric acid. [2] - State the chemical test used to identify the gas produced. [1]
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Worked solution
(i) Copper(II) oxide is added in excess to ensure that all of the dilute sulfuric acid has completely reacted and been neutralised, so that the final salt is not contaminated with acid. (ii) Filtration separates the insoluble excess copper(II) oxide solid (residue) from the soluble copper(II) sulfate solution (filtrate). (iii) To obtain large, well-formed crystals: heat the filtrate in an evaporating basin until the crystallization point is reached (saturated solution). Stop heating and leave the saturated solution to cool down slowly. Slow cooling allows larger crystals to grow. Once crystallization is complete, filter the mixture to recover the crystals. (iv) The crystals should be dried gently by pressing them between sheets of filter paper or placing them in a desiccator. Strong heating / using an oven must be avoided because it would dehydrate the crystals to anhydrous copper(II) sulfate. (v) - Observation: Bubbling / fizzing / effervescence will be observed due to carbon dioxide release. - Equation: \(\text{CuCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). - Test: Bubble the generated gas through limewater; it will turn cloudy/milky.
Marking scheme
(i) To ensure all the sulfuric acid is completely reacted / neutralised [1] (ii) Filtration / filtering [1] (iii) Heat the filtrate to crystallisation point / until a saturated solution forms [1] Leave the saturated solution to cool slowly [1] Filter off the crystals [1] (iv) Pat dry with filter paper / leave in a desiccator / leave in a warm place (reject: heat in an oven / dry with Bunsen flame) [1] (v) Observation: effervescence / bubbling / fizzing [1] Equation: \(\text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2\) (formulas correct [1], balancing correct [1]) Gas test: bubble through limewater AND limewater turns cloudy / milky [1]
Question 4 · Practical Theory
10 marks
A student is provided with three unlabelled bottles containing different liquid monomers used in polymerisation:
(i) Describe a chemical test to distinguish Monomer A from Monomer B, including the observations for both monomers. [3]
(ii) Monomer B and Monomer C react together to form a synthetic polymer. - Name the type of polymerisation reaction that occurs. [1] - Name the small molecule released as a byproduct during this reaction. [1] - Draw the structure of the linkage formed between Monomer B and Monomer C, showing all atoms and bonds in the linkage. [3]
(iii) Monomer A can undergo addition polymerisation to form poly(propenoic acid). Draw the structure of poly(propenoic acid) showing two repeat units. [2]
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Worked solution
(i) Monomer A contains an unsaturated carbon-carbon double bond (\(\text{C}=\text{C}\)), while Monomer B is saturated. Adding bromine water (aqueous bromine) to both is an effective test: Monomer A will decolourise the orange/brown bromine water to colourless, whereas Monomer B will show no colour change (remains orange/brown). (ii) - The reaction between a dicarboxylic acid (Monomer B) and a diamine (Monomer C) is a condensation polymerisation. - The small molecule released as a byproduct is water (\(\text{H}_2\text{O}\)). - The condensation reaction between a carboxylic acid group (\(-\text{COOH}\)) and an amine group (\(-\text{NH}_2\)) forms an amide linkage, represented as: \(-\text{C}(=\text{O})-\text{N}(\text{-H})-\). (iii) During addition polymerisation, the carbon-carbon double bond of Monomer A (\(\text{CH}_2=\text{CHCOOH}\)) opens up to form a saturated chain. Two repeat units are represented as:
(i) Test: Add aqueous bromine / bromine water [1] Observation with Monomer A: turns colourless / decolourises [1] Observation with Monomer B: remains orange / yellow / brown / no change [1] (ii) Type: Condensation (polymerisation) [1] Small molecule: Water / \(\text{H}_2\text{O}\) [1] Linkage structure: - \(-\text{C}=\text{O}\) group shown [1] - \(-\text{N}-\text{H}\) group shown [1] - single bond linking the carbon of carbonyl to the nitrogen of amine with continuation bonds shown [1] (iii) Two repeat units of addition polymer: - Correct single-bonded carbon backbone of 4 carbon atoms [1] - Correctly placed \(-\text{COOH}\) groups on alternate carbons with continuation bonds at both ends [1]
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