ESAT Chemistry: Electrochemistry and organic chemistry, set 2

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   Dilute sulfuric acid is electrolysed using inert platinum electrodes.

   Bubbles of gas collect steadily at both electrodes: hydrogen gas at one electrode and oxygen gas at the other, in a 2:1 ratio by volume.

   Which statement correctly identifies the electrodes involved and the type of process occurring at each?

   1. A Oxygen is produced at the cathode (the negative electrode) by reduction of ions from the water; hydrogen is produced at the anode (the positive electrode) by oxidation of ions from the water.
   2. B Hydrogen is produced at the cathode (the negative electrode) by oxidation of H+ ions; oxygen is produced at the anode (the positive electrode) by reduction of ions from the water.
   3. C Hydrogen is produced at the cathode (the negative electrode) by reduction of H+ ions; oxygen is produced at the anode (the positive electrode) by oxidation of ions from the water.
   4. D Hydrogen is produced at the cathode, which is the positive electrode here, by reduction of H+ ions; oxygen is produced at the anode, which is the negative electrode here, by oxidation of ions from the water.
2. 21 mark

   A technician sets up an electrolytic cell to electroplate a metal object, but wires it to an alternating current (ac) supply by mistake, rather than the direct current (dc) supply the process requires.

   Which outcome would be expected, and why?

   1. A No metal would be deposited overall on the object, because the polarity of each electrode reverses many times per second, so any metal deposited while the object is briefly the cathode would be redissolved when it briefly becomes the anode.
   2. B The plating would occur exactly as normal, since it is the size of the potential difference between the electrodes, not its direction, that determines whether metal ions are reduced there.
   3. C The plating would still build up, only twice as slowly, since the object is the cathode for only half of each cycle and any metal deposited during that half remains fixed in place once the current reverses.
   4. D The metal object would gradually dissolve away to nothing, since every electrode eventually acts as an anode under an ac supply, and this oxidation is not balanced by any lasting deposit.
3. 31 mark

   Concentrated aqueous sodium chloride (brine) is electrolysed using inert graphite electrodes.

   The solution contains Na+, Cl-, H+ and OH- ions (the last two from water), with Cl- present at a much higher concentration than OH-.

   Which statement correctly predicts the products formed at each electrode?

   1. A At the cathode, sodium metal is deposited, because Na+ ions are attracted to the electrode more strongly than H+ ions; at the anode, chlorine gas is produced because Cl- is present at a much higher concentration than OH-.
   2. B At the cathode, hydrogen gas is produced because H+ ions are discharged in preference to the more reactive Na+ ions; at the anode, oxygen gas is produced, because OH- ions are always discharged in preference to halide ions regardless of their relative concentrations.
   3. C At the cathode, hydrogen gas is produced because H+ ions are discharged in preference to the more reactive Na+ ions; at the anode, oxygen gas is produced, because chloride ions can never be discharged from an aqueous solution.
   4. D At the cathode, hydrogen gas is produced because H+ ions are discharged in preference to the more reactive Na+ ions; at the anode, chlorine gas is produced, because the high concentration of Cl- ions causes chlorine to be discharged there instead of oxygen.
4. 41 mark

   Molten potassium bromide, KBr, is electrolysed using inert electrodes.

   Which pair of half-equations correctly represents the reactions taking place at the cathode and at the anode?

   1. A Cathode: K+ + e- -> K. Anode: 2Br- -> Br2 + 2e-.
   2. B Cathode: K+ -> K + e-. Anode: 2Br- -> Br2 + 2e-.
   3. C Cathode: K+ + e- -> K. Anode: Br- -> Br2 + e-.
   4. D Cathode: 2K+ + 2e- -> K2. Anode: 2Br- -> Br2 + 2e-.
5. 51 mark

   In the electrolysis of concentrated aqueous sodium chloride, chlorine gas is produced at the anode.

   Which half-equation correctly represents this process?

   1. A 2Cl- + 2e- -> Cl2
   2. B Cl- -> Cl2 + e-
   3. C Cl2 + 2e- -> 2Cl-
   4. D 2Cl- -> Cl2 + 2e-
6. 61 mark

   A brass door handle is electroplated with a thin layer of nickel. The object is made the cathode, and the anode is a bar of pure nickel metal, in a solution containing nickel ions.

   Which statement correctly explains why the anode must be made of the metal being plated (nickel), rather than an inert material such as platinum?

   1. A As nickel ions are deposited at the cathode, the nickel anode dissolves, losing electrons to release fresh Ni2+ ions into the solution, replacing those removed and keeping the plating process going.
   2. B The choice of anode material makes no chemical difference to the process; a nickel anode is used only because it is cheaper to buy than a platinum one.
   3. C An inert anode such as platinum cannot conduct electricity when placed in the solution, so a reactive metal anode is required simply to allow any current to flow at all.
   4. D The nickel anode is needed only to add extra weight to the finished door handle, since the whole anode itself becomes fused to the object once plating is complete.
7. 71 mark

   Crude oil is separated by fractional distillation into fractions containing hydrocarbons of different chain lengths.

   As the average carbon chain length of the hydrocarbons in a fraction increases, which statement correctly describes how its boiling point, viscosity and flammability change?

   1. A Boiling point increases, viscosity increases, and flammability decreases.
   2. B Boiling point decreases, viscosity increases, and flammability decreases.
   3. C Boiling point increases, viscosity decreases, and flammability increases.
   4. D Boiling point increases, viscosity increases, and flammability also increases.
8. 81 mark

   Ethene, CH2=CH2, is reacted with hydrogen chloride gas, HCl.

   Which statement correctly identifies the product and the type of reaction taking place?

   1. A The product is chloroethane, CH3-CH2-Cl, formed by a substitution reaction in which a hydrogen atom on ethene is replaced by a chlorine atom, releasing hydrogen gas.
   2. B The product is chloroethane, CH3-CH2-Cl, formed by an addition reaction in which the C=C double bond opens and the H and Cl atoms of HCl each bond to one of the two carbon atoms.
   3. C The product is 1,2-dichloroethane, Cl-CH2-CH2-Cl, formed by an addition reaction in which two molecules of HCl add across the double bond.
   4. D The product is chloroethene, CH2=CH-Cl, formed by an addition reaction in which a hydrogen atom on ethene is simply replaced by a chlorine atom while the double bond remains.
9. 91 mark

   Part of an addition polymer chain is shown by its repeating unit: [-CH2-CH(CH3)-]n.

   Which monomer was used to make this polymer, and what is this type of polymerisation called?

   1. A The monomer is but-1-ene, CH2=CH-CH2-CH3; the polymer is formed by addition polymerisation.
   2. B The monomer is propan-1-ol, CH3-CH2-CH2-OH; the polymer is formed by condensation polymerisation, with water released at each new bond.
   3. C The monomer is ethene, CH2=CH2; the polymer is formed by addition polymerisation.
   4. D The monomer is propene, CH3-CH=CH2; the polymer is formed by addition polymerisation, in which the C=C double bond of each monomer opens so the monomers join together with no atoms lost.
10. 101 mark

    A polyester is formed by condensation polymerisation between molecules of a dicarboxylic acid (with a -COOH group at each end) and a diol (with an -OH group at each end), releasing a small molecule at every new ester link formed.

    Which statement correctly describes this polymer and how it differs from an addition polymer such as poly(ethene)?

    1. A Each new ester link is formed with the loss of one molecule of water; unlike poly(ethene), which is built up with no atoms lost from an unsaturated monomer, a polyester is built from two different monomers (an acid and a diol) with water released at every link.
    2. B Each new ester link is formed with the loss of one molecule of hydrogen gas; like poly(ethene), the polyester is built from a single unsaturated monomer with a C=C double bond.
    3. C Poly(ethene) is a condensation polymer formed from ethene and water, while the polyester described here is an addition polymer formed with no atoms lost.
    4. D Each new ester link is formed with no atoms lost at all, in exactly the same way that poly(ethene) is formed from ethene.
11. 111 mark

    An alcohol has the condensed structural formula CH3-CH(OH)-CH3.

    Which of the following correctly names this alcohol, and identifies its functional group?

    1. A This is propan-1-ol; its functional group is the hydroxyl group, -OH, attached to the middle (second) carbon of the three-carbon chain.
    2. B This is propan-2-ol; its functional group is an ether linkage, since an oxygen atom sits within the carbon chain.
    3. C This is propan-2-ol; its functional group is the hydroxyl group, -OH, attached to the middle (second) carbon of the three-carbon chain.
    4. D This is propanone; its functional group is a carbonyl group, C=O, on the middle carbon of the three-carbon chain.
12. 121 mark

    Propanoic acid, C2H5COOH, is reacted with an excess of sodium hydroxide solution.

    Which statement correctly describes carboxylic acids as a class of acid, and identifies the salt formed in this reaction?

    1. A Carboxylic acids are strong acids, fully ionising in water; the salt formed is sodium propanoate, C2H5COONa.
    2. B Carboxylic acids are weak acids; the salt formed is (C2H5COO)Na2.
    3. C Carboxylic acids are weak acids; the products of this reaction are an ester and water, since sodium hydroxide reacts here as if it were an alcohol.
    4. D Carboxylic acids are weak acids, only partially ionising in water; the salt formed is sodium propanoate, C2H5COONa.
13. 131 mark

    Ethene, CH2=CH2, is reacted with steam in the presence of a catalyst.

    Which equation and reaction type correctly describes this process?

    1. A C2H4 + 2H2O -> C2H5OH + H2O2; this is an addition reaction.
    2. B C2H4 + H2O -> C2H5OH; this is an addition reaction, since the C=C double bond opens and the H and OH from the water molecule each bond to one of the two carbon atoms.
    3. C C2H4 + H2O -> C2H5OH + H2O; this is a condensation reaction, since water is both a reactant and released as a by-product.
    4. D C2H4 + H2O -> C2H3OH + H2; this is a substitution reaction, in which a hydrogen atom on ethene is replaced by an -OH group and hydrogen gas is released.
14. 141 mark

    Small pieces of magnesium and zinc are separately added to iron(II) sulfate solution: both cause a grey-black deposit of iron to form, while a similar piece of copper added to the same solution causes no visible reaction.

    Iron added to copper(II) sulfate solution causes a pink-brown deposit of copper to form on its surface.

    Magnesium added to zinc sulfate solution causes a grey deposit of zinc to form, but zinc added to magnesium sulfate solution causes no reaction.

    Which row correctly ranks these four metals from most to least reactive, and correctly links this order to how easily each metal forms positive ions?

    1. A Most to least reactive: magnesium, zinc, iron, copper. The most reactive metal, magnesium, loses electrons to form its positive ions most easily, which is why it can displace each of the other three from solutions of their salts.
    2. B Most to least reactive: copper, iron, zinc, magnesium. The most reactive metal, copper, loses electrons to form its positive ions most easily, which is why the other three metals fail to displace it from solution.
    3. C Most to least reactive: zinc, magnesium, iron, copper. The most reactive metal, zinc, loses electrons to form its positive ions most easily, which is why it can displace each of the other three from solutions of their salts.
    4. D Most to least reactive: magnesium, zinc, iron, copper. The most reactive metal, magnesium, loses electrons to form its positive ions least easily, which is why it can displace each of the other three from solutions of their salts.
15. 151 mark

    Both aluminium and iron are extracted from ores that are oxides of the metal, yet aluminium is extracted by electrolysis of its molten oxide, while iron is extracted by reduction with carbon in a blast furnace.

    Iron is also known, in a different context, to be useful as a catalyst in some industrial reactions, reflecting a general property of transition metals.

    Which statement correctly explains the choice of extraction method for each metal, and correctly identifies this general property of transition metals?

    1. A Aluminium is less reactive than carbon, so carbon could reduce aluminium oxide but electrolysis is used anyway purely because it is cheaper; iron is more reactive than carbon, so it must be extracted by electrolysis instead. Transition metals are often useful as catalysts.
    2. B Aluminium is more reactive than carbon, so electrolysis must be used instead of reduction with carbon; iron is less reactive than carbon, so carbon can reduce iron oxide directly. Transition metals, including iron, are always poor catalysts, since their partially filled electron shells make them chemically unreactive.
    3. C Aluminium is more reactive than carbon, so carbon cannot reduce aluminium oxide, and electrolysis (a more expensive process) must be used instead; iron is less reactive than carbon, so carbon can reduce iron oxide directly. Transition metals, including iron, are often useful as catalysts.
    4. D The extraction method depends only on how abundant each ore is in the Earth's crust, not on the reactivity of the metal; a more abundant ore such as iron oxide is reduced with carbon simply because there is more of it available. Transition metals are often useful as catalysts.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: C

   1. An electrode's name is fixed by its charge: the cathode is always the negative electrode and the anode is always the positive electrode in an electrolytic cell.
   2. Positively charged H+ ions (from the water) are attracted to the negative cathode, where they gain electrons and are reduced: 2H+ + 2e- -> H2, producing hydrogen gas.
   3. Negatively charged ions from the water are attracted to the positive anode, where they lose electrons and are oxidised, producing oxygen gas.
   4. So the correct statement pairs cathode with hydrogen by reduction, and anode with oxygen by oxidation, matching option C.
   • Why not A: Swaps which gas forms at which electrode: positively charged H+ ions are attracted to the negative cathode, not the positive anode, so hydrogen (not oxygen) must be the gas produced there.
   • Why not B: Swaps the redox terms: gaining electrons (reduction) happens at the cathode, and losing electrons (oxidation) happens at the anode, the opposite way round from how this option describes the two processes.
   • Why not D: Swaps the polarity of the electrodes: the cathode is by definition the negative electrode and the anode the positive electrode in electrolysis, not the other way round as this option states.

2. Question 2Answer: A

   1. Electroplating relies on the object being kept permanently as the cathode, so that metal ions are continuously attracted to it and reduced, building up a lasting deposit.
   2. An alternating supply reverses which electrode is positive and negative many times a second, so the object would spend half of each cycle as the cathode, gaining a thin deposit, and the other half as the anode, where that same deposit would be oxidised and redissolved.
   3. Since the two half-cycles are equal and opposite, whatever is deposited is undone again almost immediately, so no lasting layer of metal ever builds up on the object.
   4. So the expected outcome is no net plating at all, matching option A; this is exactly why electrolysis always requires a dc, not an ac, supply.
   • Why not B: Ignores that a continuous, one-directional current is needed for the object to remain the cathode throughout; a reversing current periodically makes the object the anode instead, undoing any deposit made a moment earlier.
   • Why not C: Wrongly assumes deposited metal stays in place through the reversed half-cycle; when the object becomes the anode, the metal on its surface is itself oxidised and dissolves back into solution, rather than remaining as a slowly growing layer.
   • Why not D: Invents an unfounded asymmetry between the two half-cycles; the object spends equal time as cathode and anode, so any metal deposited while it is the cathode is undone while it is the anode, giving no net change either way, not a gradual loss.

3. Question 3Answer: D

   1. When more than one type of ion could be discharged at an electrode, the actual product depends on the reactivity of the competing ions at the cathode, and on their concentration at the anode.
   2. At the cathode, Na+ and H+ ions are both present; since sodium is far more reactive than hydrogen, the less reactive H+ ions are discharged instead, producing hydrogen gas: 2H+ + 2e- -> H2.
   3. At the anode, Cl- and OH- ions are both present; in a dilute solution OH- would normally be discharged to give oxygen, but in a concentrated solution such as brine, the much higher concentration of Cl- ions means chlorine gas is produced instead: 2Cl- -> Cl2 + 2e-.
   4. So the correct products are hydrogen at the cathode and chlorine at the anode, matching option D; this is the basis of the industrial chlor-alkali process.
   • Why not A: Wrongly assumes the more reactive metal ion is the one discharged; when more than one type of cation is present, the least reactive one is discharged preferentially, and sodium is far more reactive than hydrogen, so H+ is reduced instead of Na+.
   • Why not B: Wrongly claims OH- is always discharged in preference to a halide ion regardless of concentration; in a concentrated halide solution such as brine, the high concentration of Cl- ions means chlorine gas is produced at the anode instead of oxygen.
   • Why not C: Wrongly claims chloride ions can never be discharged from an aqueous solution at all; concentrated chloride solutions do release chlorine gas at the anode, which is exactly the basis of the industrial electrolysis of brine to manufacture chlorine.

4. Question 4Answer: A

   1. Molten potassium bromide is a binary ionic compound containing only two types of ion, K+ and Br-, since it is molten rather than dissolved in water.
   2. The K+ cations are attracted to the negative cathode, where each gains one electron and is reduced to a potassium atom: K+ + e- -> K.
   3. The Br- anions are attracted to the positive anode, where two of them each lose one electron and are oxidised to a bromine molecule: 2Br- -> Br2 + 2e-.
   4. The two half-equations balance so that the two electrons released at the anode exactly match the two electrons needed to reduce two K+ ions at the cathode, matching option A.
   • Why not B: Reverses the cathode process: it shows K+ losing an electron rather than gaining one, which is oxidation, when gaining an electron (reduction) is what actually happens at the cathode.
   • Why not C: Is not balanced: a single Br- ion cannot supply both bromine atoms needed for one Br2 molecule, and only one electron is shown released, when two Br- ions must each lose one electron for the equation to balance.
   • Why not D: Invents a non-existent diatomic potassium molecule, K2; potassium ions are reduced to individual metal atoms, K+ + e- -> K, not to a fictitious two-atom molecule.

5. Question 5Answer: D

   1. At the anode, chloride ions lose electrons and are oxidised to chlorine gas.
   2. Each chlorine atom in Cl2 comes from one Cl- ion, so two Cl- ions are needed to form one Cl2 molecule, and each Cl- ion loses one electron, so two electrons are released in total.
   3. The electrons must appear on the product side of the equation, since the ions are losing them (oxidation), not gaining them.
   4. So the correctly balanced half-equation is 2Cl- -> Cl2 + 2e-, matching option D.
   • Why not A: Places the electrons on the wrong side, as if chloride ions were gaining electrons rather than losing them; this is the direction for reduction, not the oxidation that actually occurs at the anode.
   • Why not B: Is not balanced: one chloride ion cannot supply the two chlorine atoms in Cl2, and only one electron is shown, when two electrons must be released to balance the charge lost by two Cl- ions.
   • Why not C: Writes the entire reverse reaction, the reduction of chlorine gas back into chloride ions; this is what would happen if chlorine were being reduced, not the oxidation of chloride that actually happens at the anode.

6. Question 6Answer: A

   1. In electroplating, the object to be coated is made the cathode and a bar of the plating metal is made the anode, in a solution containing ions of that metal.
   2. At the cathode, metal ions from the solution gain electrons and are reduced, depositing as a solid layer onto the object: Ni2+ + 2e- -> Ni.
   3. If nothing replaced the ions removed from solution, the ion concentration would fall and plating would slow and eventually stop; using an anode made of the plating metal solves this, because at the anode, atoms of the metal lose electrons and dissolve into solution as fresh ions: Ni -> Ni2+ + 2e-.
   4. So the correct explanation is option A: the nickel anode dissolves to replenish the ion concentration used up by deposition at the cathode, keeping the process going.
   • Why not B: Wrongly claims the anode material is chemically irrelevant; with an inert anode instead, the nickel ion concentration in the solution would steadily fall as ions are deposited at the cathode with nothing to replace them, so plating could not continue indefinitely the way it does with a reactive nickel anode.
   • Why not C: Wrongly claims an inert electrode cannot conduct electricity in solution; inert electrodes such as platinum and graphite conduct perfectly well and are used throughout electrolysis, so conductivity is not the reason a nickel anode is chosen here.
   • Why not D: Confuses the anode, a separate electrode standing in the electrolyte solution, with the object being plated; the anode gradually dissolves away and never becomes part of the door handle, which gains only the thin layer of nickel deposited onto it at the cathode.

7. Question 7Answer: A

   1. Hydrocarbon fractions are separated in fractional distillation because chain length affects several physical properties in a consistent way.
   2. As chain length increases, there is a greater surface area of contact between neighbouring molecules, so the intermolecular (van der Waals) forces between them become stronger, and more energy is needed to separate them into the gas phase, raising the boiling point.
   3. Stronger intermolecular forces also make longer molecules flow past each other less easily, increasing viscosity, and make the substance harder to vaporise, which in turn makes it harder to ignite and burn, decreasing flammability.
   4. So the correct trend is boiling point up, viscosity up, flammability down as chain length increases, matching option A.
   • Why not B: Reverses the boiling point trend; longer hydrocarbon chains have stronger intermolecular forces acting between the greater number of atoms, so more energy is needed to separate the molecules, meaning boiling point rises, not falls, with chain length.
   • Why not C: Reverses both the viscosity and flammability trends; longer, larger molecules flow less easily, so viscosity increases (not decreases), and they are harder to vaporise and ignite, so flammability decreases (not increases), with chain length.
   • Why not D: Gets boiling point and viscosity right but wrongly claims flammability also increases; a fuel must vaporise before it can ignite, and the higher boiling points of longer-chain hydrocarbons mean they vaporise less readily, so flammability actually decreases with chain length.

8. Question 8Answer: B

   1. Ethene is an unsaturated hydrocarbon (an alkene) because it contains a C=C double bond, which is the reactive site for addition reactions.
   2. Hydrogen halides such as HCl can add across the C=C double bond: the double bond opens, and the H atom bonds to one carbon while the Cl atom bonds to the other, using exactly one molecule of HCl per double bond.
   3. This converts the unsaturated ethene, C2H4, into the saturated product chloroethane, C2H5Cl, with no double bond remaining and no by-product released.
   4. So the correct description is option B: chloroethane forms by an addition reaction.
   • Why not A: Correctly identifies the product but misnames the reaction as substitution; the actual mechanism is addition, where the double bond opens and both atoms of HCl add across it, with no hydrogen gas released.
   • Why not C: Uses two molecules of HCl instead of one; a single C=C double bond reacts with exactly one molecule of a hydrogen halide, adding one H atom and one halogen atom across it, not two of each.
   • Why not D: Confuses replacing a hydrogen atom with a true addition reaction, and wrongly leaves the C=C double bond intact; addition across the double bond removes it entirely, forming a saturated product with no C=C bond remaining.

9. Question 9Answer: D

   1. In addition polymerisation, the repeating unit of the polymer has exactly the same atoms as one monomer molecule, since the C=C double bond simply opens to link monomers together with nothing added or lost.
   2. The repeating unit shown, [-CH2-CH(CH3)-]n, has a backbone of two carbon atoms per unit plus a CH3 branch, giving three carbon atoms and six hydrogen atoms in total per unit: C3H6.
   3. A monomer with formula C3H6 and a C=C double bond is propene, CH3-CH=CH2; when its double bond opens, the two carbons of the double bond become the backbone of the repeating unit, and the CH3 group becomes the branch.
   4. So the monomer is propene and the polymerisation is addition polymerisation, matching option D; the polymer formed is poly(propene).
   • Why not A: But-1-ene has four carbon atoms and would give a repeating unit with an ethyl side branch, [-CH2-CH(CH2CH3)-]n, not the shorter methyl side branch shown here, [-CH2-CH(CH3)-]n; the repeating unit shown has only three carbons per unit, pointing to propene, not but-1-ene.
   • Why not B: Proposes an alcohol monomer and condensation polymerisation, but the repeating unit shown contains only carbon and hydrogen atoms with no oxygen and no small molecule lost, both of which rule out this being a condensation polymer of an alcohol.
   • Why not C: Ethene's repeating unit is [-CH2-CH2-]n, with no branching side group at all, whereas the repeating unit shown has a CH3 branch on every other carbon, which can only come from a three-carbon monomer such as propene, not the two-carbon ethene.

10. Question 10Answer: A

    1. Condensation polymerisation joins monomers that each carry two reactive end groups, releasing a small molecule (often water) at every new bond formed between them, unlike addition polymerisation, where no atoms are lost.
    2. A polyester is formed from a dicarboxylic acid and a diol: the -COOH group of the acid reacts with the -OH group of the diol to form an ester link, releasing one molecule of water at each link.
    3. This is unlike poly(ethene), an addition polymer formed from a single monomer, ethene (CH2=CH2), whose C=C double bond simply opens so monomers join with nothing lost.
    4. So the correct description is option A: water is lost at each ester link, and the polyester is built from two different monomers, unlike the single-monomer, no-loss addition polymerisation that forms poly(ethene).
    • Why not B: Names the wrong small molecule released (hydrogen gas instead of water) and wrongly describes the polyester as built from one unsaturated monomer with a double bond, when it is in fact built from two different saturated monomers, an acid and a diol, joined with water loss.
    • Why not C: Swaps the two classifications entirely; poly(ethene) is the addition polymer (built from one unsaturated monomer with no atoms lost) and the polyester is the condensation polymer (built from two monomers with water lost at each link), the opposite way round from this option.
    • Why not D: Denies that any small molecule is released, missing the defining feature of condensation polymerisation (a molecule such as water lost at every new bond), which is exactly what distinguishes it from the addition polymerisation used to form poly(ethene).

11. Question 11Answer: C

    1. Alcohols form a homologous series containing the -OH (hydroxyl) functional group attached to a carbon chain, with general formula CnH2n+1OH.
    2. In CH3-CH(OH)-CH3, the -OH group is attached to the middle carbon of a three-carbon (propan-) chain, so the position number in the name must be 2, giving propan-2-ol.
    3. This is different from propan-1-ol, CH3-CH2-CH2-OH, where the -OH group is on an end carbon instead, and different from propanone, CH3-CO-CH3, where the middle carbon instead carries a C=O double bond with no -OH or extra hydrogen.
    4. So the correct name and functional group are propan-2-ol and the hydroxyl group, -OH, matching option C.
    • Why not A: Names the compound as propan-1-ol, which would have the -OH group on an end carbon (CH3-CH2-CH2-OH), not on the middle carbon as this structure and the -2 numbering both require.
    • Why not B: Misidentifies the functional group as an ether; there is no oxygen linking two separate carbon chains here, only a single -OH group attached to one carbon, which is the hydroxyl group of an alcohol, not an ether.
    • Why not D: Names a different compound, propanone (CH3-CO-CH3), which has a C=O double bond on the middle carbon and no hydrogen there; the structure given has an -OH group and a hydrogen on the middle carbon, which is an alcohol, not a ketone.

12. Question 12Answer: D

    1. Carboxylic acids, such as propanoic acid, are a homologous series with general formula CnH2n+1COOH, and they are weak acids, only partially ionising in water, unlike strong acids such as hydrochloric acid, which fully ionise.
    2. Like other acids, a carboxylic acid reacts with a base such as sodium hydroxide in a neutralisation reaction, producing a salt and water: C2H5COOH + NaOH -> C2H5COONa + H2O.
    3. The -COOH group has only one acidic hydrogen to lose, leaving a singly charged carboxylate ion, -COO-, so the salt formed pairs one sodium ion with one propanoate ion, giving the formula C2H5COONa, sodium propanoate.
    4. So the correct statement is option D: carboxylic acids are weak acids, and the salt formed is sodium propanoate, C2H5COONa.
    • Why not A: Correctly identifies the salt formed but wrongly calls carboxylic acids strong acids that fully ionise in water; carboxylic acids such as propanoic acid are in fact weak acids, only partially ionising.
    • Why not B: Wrongly assumes the carboxylate group carries an overall 2- charge because it contains two oxygen atoms; propanoic acid has only one acidic hydrogen, in the -COOH group, so losing it leaves a single negative charge, -COO-, which needs only one Na+ ion to balance, giving C2H5COONa.
    • Why not C: Confuses this acid-base neutralisation with esterification; sodium hydroxide is a base, not an alcohol, so reacting it with a carboxylic acid produces a salt and water, not an ester.

13. Question 13Answer: B

    1. Ethene is an alkene, an unsaturated hydrocarbon containing a C=C double bond, which reacts with steam in an addition reaction called hydration, using an acid catalyst and high temperature and pressure.
    2. The C=C double bond opens, and the H and OH from one molecule of water add across the two carbon atoms: one carbon gains the H, and the other gains the OH.
    3. This converts unsaturated ethene, C2H4, into the saturated product ethanol, C2H5OH, with every atom of both starting molecules accounted for in the single product and nothing released as a by-product.
    4. So the correct equation and reaction type are C2H4 + H2O -> C2H5OH, an addition reaction, matching option B.
    • Why not A: Uses two molecules of water and invents hydrogen peroxide as an extra by-product with no basis here; only one water molecule adds across the single C=C double bond, giving ethanol as the only product, with nothing left over.
    • Why not C: Wrongly classifies this as a condensation reaction and invents an extra water by-product; every atom of the water molecule is incorporated into the ethanol product, so no water molecule is released, and the reaction is addition, not condensation.
    • Why not D: Proposes a substitution mechanism releasing hydrogen gas and retains a double bond in the product, when in fact the C=C double bond opens completely in this addition reaction, giving the fully saturated product ethanol, C2H5OH, with no gas released.

14. Question 14Answer: A

    1. In a displacement reaction, a more reactive metal displaces a less reactive metal from a solution of its salt, because the more reactive metal loses electrons to form positive ions more readily than the less reactive one.
    2. Magnesium and zinc both displace iron from iron(II) sulfate, so both are more reactive than iron; copper causes no reaction with iron(II) sulfate, so copper is less reactive than iron, and iron itself displaces copper from copper(II) sulfate, confirming iron is more reactive than copper.
    3. Magnesium displaces zinc from zinc sulfate, but zinc cannot displace magnesium from magnesium sulfate, so magnesium is more reactive than zinc.
    4. Putting this together, the order from most to least reactive is magnesium, zinc, iron, copper, with the most reactive metal (magnesium) being the one that loses electrons to form its positive ions most easily, matching option A.
    • Why not B: Reverses the entire reactivity order; a metal that is displaced by every other metal, and displaces none of them, must be the least reactive, not the most, so copper belongs at the bottom of the list, not the top.
    • Why not C: Swaps the order of magnesium and zinc; since magnesium displaces zinc from zinc sulfate solution but zinc does not displace magnesium from magnesium sulfate solution, magnesium must be the more reactive of the two, not zinc.
    • Why not D: Gets the reactivity order right but reverses the definition of reactivity; a more reactive metal forms its positive ions more easily (loses electrons more readily), not less easily, than a less reactive one.

15. Question 15Answer: C

    1. The method used to extract a metal from its oxide ore depends on how reactive the metal is compared with carbon: a metal less reactive than carbon can have its oxide reduced directly by carbon, which is relatively cheap, while a metal more reactive than carbon cannot be reduced this way and must be extracted by electrolysis instead, which uses large amounts of electricity and is far more expensive.
    2. Aluminium is more reactive than carbon, so carbon cannot remove the oxygen from aluminium oxide, and the molten oxide must instead be electrolysed to obtain the metal.
    3. Iron is less reactive than carbon, so carbon can reduce iron oxide directly in a blast furnace, which is why this cheaper reduction method is used for iron rather than electrolysis.
    4. Iron is also a transition metal, and transition metals and their compounds are commonly useful as catalysts (for example, iron itself catalyses the Haber process); this is a general property of transition metals, alongside forming coloured compounds and multiple oxidation states, matching option C.
    • Why not A: Reverses the reactivity comparison and the extraction method for each metal; aluminium is in fact more reactive than carbon (so carbon cannot reduce its oxide, making electrolysis necessary, not merely cheaper), while iron is less reactive than carbon (so it is extracted by the far cheaper reduction with carbon, not by electrolysis).
    • Why not B: Correctly explains the extraction methods but wrongly claims transition metals are poor catalysts; transition metals and their compounds are in fact often good catalysts, linked to their ability to exist in different oxidation states, not a reason for unreactivity.
    • Why not D: Replaces the true reason (the reactivity of the metal compared with carbon) with ore abundance, which is not the deciding factor; the extraction method used for a metal is chosen according to where it sits in the reactivity series relative to carbon, not according to how common its ore is.