ESAT Chemistry: Electrochemistry and organic chemistry, set 3

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

1. 11 mark

   Molten magnesium chloride, MgCl2, is electrolysed using inert electrodes and a constant electric current.

   The half-equations are: cathode, Mg2+ + 2e- -> Mg; anode, 2Cl- -> Cl2 + 2e-.

   For a given quantity of charge passed through the cell, which statement correctly compares the number of moles of magnesium produced at the cathode with the number of moles of chlorine gas produced at the anode?

   1. A Twice as many moles of chlorine gas are produced as moles of magnesium, because the formula MgCl2 contains two chloride ions for every magnesium ion.
   2. B Equal numbers of moles of magnesium and chlorine gas are produced, because each half-equation transfers exactly two electrons per particle formed (per Mg atom, and per Cl2 molecule), so the same quantity of charge produces the same number of moles of each.
   3. C Twice as many moles of magnesium are produced as moles of chlorine gas, because reduction at the cathode always proceeds twice as fast as oxidation at the anode.
   4. D The comparison cannot be made without knowing the size of the current and the time for which it flows, since a quantity of charge only fixes the mass of a product, not the number of moles.
2. 21 mark

   Impure copper is purified by electrolytic refining. A block of impure copper is connected as one electrode and a thin sheet of pure copper as the other, both dipped into copper(II) sulfate solution and connected to a dc supply, with the impure copper wired as the anode and the pure copper wired as the cathode.

   Which statement correctly describes what happens at each electrode during this process?

   1. A At the anode, the impure copper dissolves into solution as Cu2+ ions by oxidation; at the cathode, Cu2+ ions from the solution are deposited as pure copper by reduction, so copper transfers from the impure block onto the pure copper cathode.
   2. B At the anode, Cu2+ ions from the solution are deposited as pure copper by reduction; at the cathode, the impure copper dissolves into solution as Cu2+ ions by oxidation.
   3. C Oxygen gas is produced at the anode and hydrogen gas is produced at the cathode, exactly as would happen with inert graphite electrodes, since the electrolyte is still copper(II) sulfate solution.
   4. D The impure copper anode stays completely unreacted throughout the process, while pure copper is deposited at the cathode using only the Cu2+ ions that were already dissolved in the solution before the current was switched on.
3. 31 mark

   Aluminium is extracted by electrolysis of molten aluminium oxide (dissolved in molten cryolite to lower its melting point), using inert electrodes.

   The two half-equations are: cathode, Al3+ + 3e- -> Al; anode, 2O2- -> O2 + 4e-.

   Which overall equation is obtained when these two half-equations are combined so that the electrons cancel exactly?

   1. A 4Al3+ + 6O2- -> 4Al + 3O2
   2. B Al3+ + O2- -> Al + O2
   3. C 4Al3+ + 6O2- -> 4Al + 6O2
   4. D 3Al3+ + 8O2- -> 3Al + 4O2
4. 41 mark

   A copper medal is electroplated with gold using a pure gold anode, a solution containing gold ions, the copper medal as the cathode, and a dc supply.

   A technician then wires the cell up the wrong way round, so that the copper medal is connected as the anode and the pure gold bar is connected as the cathode.

   Which statement correctly describes what would happen, and why?

   1. A The copper medal would still be plated with gold, because it is easier to reduce Au3+ ions to gold than to oxidise the copper medal, regardless of which electrode it is connected to.
   2. B The copper medal would not be plated: as the anode, it would lose electrons and dissolve into the solution as copper ions, while gold would instead be deposited onto the pure gold bar, which is now acting as the cathode.
   3. C The copper medal would be plated with gold twice as fast, because both electrodes now attract gold ions.
   4. D Nothing would change, because swapping the electrodes of a dc supply has no effect on which reactions occur at each electrode.
5. 51 mark

   The molecular formula C5H12 has exactly three structural isomers.

   Which of the following correctly lists and names all three?

   1. A Pentane, 3-methylbutane (a methyl branch on the third carbon of a four-carbon chain) and 2,2-dimethylpropane.
   2. B Pentane, 2-methylbutane and cyclopentane.
   3. C Pentane, 2-methylbutane, 3-methylbutane and 2,2-dimethylpropane.
   4. D Pentane (straight-chain), 2-methylbutane (a methyl branch on the second carbon of a four-carbon chain) and 2,2-dimethylpropane (two methyl branches on the central carbon of a three-carbon chain).
6. 61 mark

   An unbranched (straight-chain) alkane contains 18 hydrogen atoms per molecule.

   Using the general formula for alkanes, which of the following correctly gives its molecular formula and its name?

   1. A C8H18, octane
   2. B C9H18, nonane
   3. C C8H18, heptane
   4. D C8H16, octane
7. 71 mark

   But-1-ene, CH2=CHCH2CH3, reacts with hydrogen bromide gas in an addition reaction.

   Which equation correctly represents this reaction and its product?

   1. A C4H8 + 2HBr -> C4H10Br2
   2. B C4H8 + HBr -> C4H7Br + H2
   3. C C4H8 + HBr -> C4H9Br
   4. D C4H8 + HBr -> C4H7Br + H2O
8. 81 mark

   Phenylethene (styrene), C6H5-CH=CH2, is used as the monomer to make poly(phenylethene), commonly known as polystyrene, by addition polymerisation.

   Which structure correctly represents the repeating unit of this polymer?

   1. A -[-CH=CH(C6H5)-]-
   2. B -[-CH2-CH2-]-
   3. C -[-CH(C6H5)-CH(C6H5)-]-
   4. D -[-CH2-CH(C6H5)-]-
9. 91 mark

   A polyester is formed by condensation polymerisation between a molecule with two -OH groups (a diol) and a molecule with two -COOH groups (a dicarboxylic acid).

   Which statement correctly describes this polymerisation and the resulting polymer?

   1. A The diol and dicarboxylic acid join together directly with no small molecule released, unlike protein formation from amino acids.
   2. B Each new ester link formed between an -OH group and a -COOH group releases one small molecule of water; because the monomers each have two reactive end groups, the chain can keep growing at both ends, and polyesters such as this one are generally non-biodegradable, unlike naturally occurring condensation polymers such as proteins.
   3. C Each new ester link releases a molecule of hydrogen gas rather than water.
   4. D Because it is a condensation polymer, this polyester is automatically biodegradable, just like proteins.
10. 101 mark

    When ethanol, C2H5OH, reacts completely with excess sodium metal, hydrogen gas is released.

    If 0.4 mol of ethanol reacts completely, how many moles of hydrogen gas, H2, are produced?

    1. A 0.2 mol
    2. B 0.4 mol
    3. C 0.8 mol
    4. D 0.1 mol
11. 111 mark

    Butanoic acid, CH3CH2CH2COOH, is reacted with methanol, CH3OH, in the presence of a small amount of concentrated sulfuric acid as a catalyst, producing an ester and water.

    What is the correct name of the ester formed?

    1. A butyl methanoate
    2. B methyl butanoic acid
    3. C propyl butanoate
    4. D methyl butanoate
12. 121 mark

    Small, equal-sized pieces of four metals - magnesium, aluminium, iron and silver - are each added to separate test tubes of dilute hydrochloric acid.

    Magnesium reacts immediately, producing a rapid, vigorous stream of hydrogen gas bubbles. Aluminium shows almost no reaction at first, because its surface is coated in a thin, unreactive oxide layer; once this layer has been broken down, aluminium reacts steadily, producing bubbles more slowly than magnesium but faster than iron. Iron reacts steadily but slowly throughout. Silver shows no reaction at all, even after standing for a long time.

    Which row correctly ranks these four metals from most to least reactive?

    1. A Magnesium, aluminium, iron, silver
    2. B Silver, iron, aluminium, magnesium
    3. C Aluminium, magnesium, iron, silver
    4. D Magnesium, iron, aluminium, silver
13. 131 mark

    Titanium is widely used in aircraft components and in artificial hip joint replacements, often alloyed with small amounts of other elements.

    Which statement correctly explains why titanium (and its alloys) are suited to these particular uses?

    1. A Titanium is used in these applications mainly because it is an excellent conductor of electricity and heat, better than copper, which reduces the risk of the aircraft overheating.
    2. B Titanium has a low density combined with high strength and excellent resistance to corrosion, giving a high strength-to-weight ratio that is valuable in aircraft, and it is also non-toxic and does not react with body fluids, which suits it to surgical implants.
    3. C Titanium is used because it is the most reactive of all metals, which allows it to bond strongly and permanently with living bone tissue in a hip joint.
    4. D Titanium alloys are used because alloying always makes a metal both lighter and cheaper than the pure metal, which is why alloys are preferred over pure metals in every application.
14. 141 mark

    Iron is extracted from iron(III) oxide, Fe2O3, in a blast furnace, where the ore is reduced by carbon monoxide gas.

    Which equation correctly and completely balances this reduction reaction?

    1. A Fe2O3 + 3CO -> 2Fe + 3CO2
    2. B Fe2O3 + CO -> 2Fe + 3CO2
    3. C Fe2O3 + 3CO -> 2Fe + 3CO
    4. D 2Fe2O3 + 3CO -> 4Fe + 3CO2
15. 151 mark

    Iron, copper and manganese(IV) oxide each illustrate one of the characteristic properties of transition metals: forming ions in more than one oxidation state, forming coloured compounds, and acting as a catalyst.

    Which statement correctly matches an example to the property it demonstrates?

    1. A Copper shows the property of forming coloured compounds because pure copper metal itself is a deep blue colour.
    2. B Manganese(IV) oxide is a catalyst that gets permanently used up in the reaction it speeds up, which is why more must be added for each new batch of reactant.
    3. C Iron acts as a catalyst in the Haber process for making ammonia, illustrating that transition metals (or their compounds) can act as catalysts.
    4. D Iron shows the property of forming ions in different oxidation states because pure iron metal can exist as both a solid and a liquid depending on temperature.

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: B

   1. At the cathode, each magnesium atom deposited requires two electrons: Mg2+ + 2e- -> Mg.
   2. At the anode, each chlorine molecule produced also involves the transfer of two electrons, since two Cl- ions each lose one electron: 2Cl- -> Cl2 + 2e-.
   3. Because the same current passes through the whole circuit, the same total number of electrons is transferred at both electrodes in any given time; since two electrons are needed per Mg atom and two electrons are released per Cl2 molecule, an equal number of moles of magnesium and chlorine gas must form for a given quantity of charge passed.
   4. So the correct statement is B: the moles of magnesium and chlorine gas produced are equal, in a 1:1 ratio, for the same quantity of charge, regardless of the 2:1 ratio of ions in the formula MgCl2.
   • Why not A: Confuses the 2:1 ratio of ions in the formula MgCl2 with the ratio of products formed during electrolysis; what fixes the mole ratio of products is the number of electrons transferred per particle formed, which is two for both Mg and Cl2 here, giving equal (1:1), not 2:1, molar amounts for a given charge.
   • Why not C: Invents a rule with no basis: the same current flows through every part of a series circuit, so the rate of electron transfer at the cathode and at the anode is necessarily the same; the mole ratio of products depends only on how many electrons each half-equation needs per particle formed, not on one electrode process being arbitrarily faster than the other.
   • Why not D: While the actual number of moles produced does depend on the current and the time (via Faraday's law), the RATIO between the two products for a given quantity of charge is fixed by the electron count in each half-equation alone, so the comparison can be made without knowing the specific current or time.

2. Question 2Answer: A

   1. With reactive (active) electrodes made of the same metal as the electrolyte's ions, such as copper electrodes in copper(II) sulfate solution, the electrode reactions involve the electrode metal itself rather than water.
   2. At the anode (the impure copper), oxidation occurs: copper atoms lose electrons and dissolve into solution as Cu2+ ions: Cu -> Cu2+ + 2e-.
   3. At the cathode (the pure copper), reduction occurs: Cu2+ ions from solution gain electrons and are deposited as pure copper: Cu2+ + 2e- -> Cu.
   4. Overall, copper effectively transfers from the impure anode to the pure cathode, leaving behind any insoluble impurities, which is how this process purifies copper industrially.
   • Why not B: Swaps oxidation and reduction between the two electrodes: reduction (electron gain, depositing metal) always happens at the cathode and oxidation (electron loss, dissolving metal) always happens at the anode, whichever object is wired as which electrode; this option has the two processes the wrong way round.
   • Why not C: Assumes the electrodes are inert, as graphite electrodes would be; but copper electrodes are reactive (active) electrodes here, so instead of water being oxidised to oxygen at the anode, the copper metal of the anode itself is oxidised and dissolves, and instead of hydrogen being produced at the cathode, Cu2+ ions from solution are reduced and deposited as copper.
   • Why not D: Denies that the anode reacts at all; the whole point of electrorefining is that the impure anode continuously dissolves as Cu2+ ions to replace the Cu2+ ions being removed from solution at the cathode, keeping the solution's concentration roughly constant, rather than the process relying only on a fixed initial supply of dissolved ions.

3. Question 3Answer: A

   1. To combine two half-equations, each is scaled so that the number of electrons is the same in both, so that the electrons cancel when they are added.
   2. The cathode equation has 3 electrons and the anode equation has 4, so the cathode equation is multiplied by 4 (giving 4Al3+ + 12e- -> 4Al) and the anode equation is multiplied by 3 (giving 6O2- -> 3O2 + 12e-).
   3. Adding these and cancelling the 12 electrons on each side gives 4Al3+ + 6O2- -> 4Al + 3O2, which is balanced for aluminium (4 = 4), oxygen (6 = 6) and charge (12+ from Al3+ cancels 12- from O2-, giving 0 charge on the right too).
   4. This is the ionic form of the overall equation 2Al2O3 -> 4Al + 3O2 for extracting aluminium by electrolysis.
   • Why not B: Simply adds the two half-equations as given without first scaling them so the electrons cancel: 3e- from the cathode equation and 4e- from the anode equation are not equal, so they cannot be added directly like this.
   • Why not C: Correctly scales the cathode and anode half-equations so the electrons cancel (12 on each side), but keeps the anode's O2 coefficient equal to the O2- coefficient (6) instead of the correctly balanced 3, so the oxygen atoms no longer balance (6 on the left, 12 on the right).
   • Why not D: Multiplies each half-equation by its own number of electrons instead of by the other half-equation's number of electrons, so the electrons still do not cancel (9e- against 16e-); the two multipliers must be swapped.

4. Question 4Answer: B

   1. In electroplating, the object to be plated is made the cathode, so metal ions from solution are reduced (gain electrons) and deposit onto its surface; the plating metal is made the anode, so it is oxidised and dissolves into solution to replace the ions removed at the cathode.
   2. If the copper medal is wired as the anode instead, it is the medal that is oxidised: Cu -> Cu2+ + 2e-, so the medal dissolves rather than being coated.
   3. Meanwhile the pure gold bar, now the cathode, is where Au3+ ions are reduced and deposited: Au3+ + 3e- -> Au, so gold builds up on the gold bar instead of the medal.
   4. So reversing the connections stops the medal being plated at all, and instead causes it to corrode.
   • Why not A: Assumes the metal that gets deposited is decided purely by relative reactivity rather than by which electrode (cathode or anode) each object is connected to; in electrolysis, reduction always happens at the cathode, whichever object that is.
   • Why not C: Invents a nonsensical doubling effect and ignores that only the cathode is where reduction (deposition) occurs; the anode is where oxidation happens instead.
   • Why not D: Assumes the connection to the supply is irrelevant; reversing which terminal an object is connected to swaps whether it undergoes oxidation (as the anode) or reduction (as the cathode), so the reactions at the medal and the gold bar are reversed.

5. Question 5Answer: D

   1. A structural isomer of C5H12 must have the same molecular formula but a different arrangement of atoms.
   2. The straight chain gives pentane. Moving one carbon off the main chain to form a single methyl branch gives 2-methylbutane, with the branch on carbon 2, the lowest locant available, not carbon 3.
   3. Moving two carbons off the main chain, both attached to the same central carbon, gives 2,2-dimethylpropane, a three-carbon chain with two methyl branches on the central carbon.
   4. No other arrangement gives a different compound with the same molecular formula, so there are exactly three structural isomers; a ring compound such as cyclopentane is not one of them, because forming a ring changes the molecular formula to C5H10.
   • Why not A: Uses the higher locant (3) instead of the lower locant (2) required by IUPAC naming rules; numbered from the other end of the chain, the same branch sits on carbon 2, so the compound must be named 2-methylbutane, not 3-methylbutane.
   • Why not B: Cyclopentane has the molecular formula C5H10, not C5H12, because forming a ring removes two hydrogen atoms compared with the equivalent open-chain alkane, so it cannot be a structural isomer of C5H12.
   • Why not C: Treats 2-methylbutane and 3-methylbutane as two different compounds, when they are in fact the same molecule (numbered from opposite ends); C5H12 has only three distinct structural isomers, not four.

6. Question 6Answer: A

   1. Alkanes have the general formula CnH2n+2, where n is the number of carbon atoms.
   2. Setting 2n + 2 = 18 and solving gives 2n = 16, so n = 8.
   3. Eight carbon atoms in a straight-chain alkane is octane, so the molecule is C8H18.
   • Why not B: Uses the alkene general formula CnH2n instead of the alkane formula CnH2n+2, giving n = 9 instead of correctly solving 2n + 2 = 18 for n = 8.
   • Why not C: Correctly finds n = 8 carbon atoms but then applies the name for a seven-carbon chain (heptane) instead of the correct eight-carbon name, octane.
   • Why not D: Finds the correct number of carbon atoms (8) but then uses the alkene general formula CnH2n to calculate the hydrogens (16) instead of the alkane general formula CnH2n+2 (18), giving a formula that does not actually have 18 hydrogen atoms as the question states.

7. Question 7Answer: C

   1. But-1-ene is an unsaturated hydrocarbon containing one C=C double bond; alkenes undergo addition reactions in which a small molecule adds across the double bond.
   2. In addition, the double bond becomes a single bond and the two atoms of the added molecule, here H and Br, each bond to one of the two carbon atoms that were double-bonded.
   3. Every atom in the reactants ends up in the single product, so C4H8 + HBr -> C4H9Br, giving bromobutane, with no other product formed.
   • Why not A: Adds two molecules of HBr across the one C=C double bond, as though each of the two double-bonded carbon atoms needed a separate HBr molecule; in fact one HBr molecule provides exactly one new C-H bond and one new C-Br bond, using up the double bond completely.
   • Why not B: Treats the reaction as a substitution that removes a hydrogen atom from but-1-ene and releases it as hydrogen gas, rather than as an addition reaction in which the double bond opens up and both atoms of HBr join the carbon chain with nothing given off.
   • Why not D: Mistakes this addition reaction for a condensation reaction and assumes water is lost as a by-product, as happens in esterification; but there is no oxygen atom present in either but-1-ene or hydrogen bromide, so water could not be formed.

8. Question 8Answer: D

   1. In addition polymerisation, the C=C double bond of each monomer opens up and forms new single covalent bonds directly to the next monomer unit, joining many monomers into one long chain with no atoms lost.
   2. Phenylethene, C6H5-CH=CH2, has two carbon atoms in its C=C double bond, one of which also carries a phenyl (C6H5) side group.
   3. So the repeating unit keeps both backbone carbons and the phenyl branch, and swaps the double bond for two single bonds to neighbouring units: -[-CH2-CH(C6H5)-]-, with the phenyl group on alternate backbone carbons and no C=C bond remaining.
   • Why not A: Leaves the C=C double bond in the repeating unit, but addition polymerisation uses up the double bond of each monomer to form new single C-C bonds joining the monomers, so no double bond remains in the polymer backbone.
   • Why not B: This is the repeating unit of poly(ethene), formed from ethene, CH2=CH2. It omits the phenyl (C6H5) side branch that comes from styrene's benzene ring, so it does not represent poly(phenylethene) at all.
   • Why not C: Places a phenyl branch on both backbone carbons instead of just one, effectively inventing an extra benzene ring that is not present in the styrene monomer, which has only one phenyl group attached to one of the two double-bonded carbons.

9. Question 9Answer: B

   1. Condensation polymerisation joins monomers by forming a new bond between two functional groups with the loss of a small molecule at each link; here, an -OH group and a -COOH group react to form an ester link, releasing one molecule of water each time.
   2. Because both the diol and the dicarboxylic acid have two reactive end groups, further monomers can keep adding at both ends of the growing chain, building up a long polyester molecule.
   3. This is the same type of polymerisation (condensation) as protein formation from amino acids, but not every condensation polymer is biodegradable: synthetic polyesters are generally non-biodegradable, whereas proteins are broken down naturally.
   • Why not A: Confuses condensation polymerisation with addition polymerisation: in condensation polymerisation, forming each new bond (here, an ester link) releases a small molecule, water, which is exactly what also happens when amino acids join to form proteins.
   • Why not C: Names the wrong by-product: esterification releases water, not hydrogen gas, which is instead released when an alcohol reacts with reactive sodium metal, a different reaction entirely.
   • Why not D: Assumes being a condensation polymer guarantees biodegradability; most synthetic polyesters are in fact non-biodegradable even though they are condensation polymers, because they lack the natural pathways that break proteins down.

10. Question 10Answer: A

    1. The balanced equation for the reaction between ethanol and sodium is 2C2H5OH + 2Na -> 2C2H5ONa + H2, showing that 2 mol of ethanol reacts with 2 mol of sodium to release 1 mol of hydrogen gas.
    2. So the mole ratio of ethanol to hydrogen gas is 2 to 1.
    3. 0.4 mol of ethanol therefore produces 0.4 divided by 2, which is 0.2 mol, of hydrogen gas.
    • Why not B: Assumes a 1:1 mole ratio between ethanol and hydrogen gas, as if each ethanol molecule alone released one whole H2 molecule, instead of the correct 2:1 ratio.
    • Why not C: Doubles the amount of ethanol instead of halving it, as if multiplying by the '2' in the balanced equation's ethanol coefficient rather than dividing by it.
    • Why not D: Halves the correct answer again, as if using a 4:1 mole ratio of ethanol to hydrogen instead of the correct 2:1 ratio.

11. Question 11Answer: D

    1. An ester's name has two parts: the first part names the alkyl group from the ALCOHOL, and the second part names the acid's carbon chain with its ending changed from '-oic acid' to '-oate'.
    2. Methanol, CH3OH, has one carbon atom, giving the alkyl group 'methyl'. Butanoic acid, CH3CH2CH2COOH, has four carbon atoms, giving the acid-derived part 'butanoate' once its -COOH has reacted.
    3. Combining these in the correct order gives the name methyl butanoate, CH3CH2CH2COOCH3, with water, H2O, released as the other product.
    • Why not A: Swaps the alcohol and acid parts of the name around: the alkyl group named first must come from the ALCOHOL (methanol, giving 'methyl'), and the '-oate' stem must come from the ACID (butanoic acid, giving 'butanoate'); this name in fact describes the ester of methanoic acid and butan-1-ol, the opposite pairing.
    • Why not B: Keeps the 'acid' ending instead of changing it to the '-oate' ending that signals an ester has formed; once the acid's -OH has been replaced by the alkoxy group from the alcohol, the compound is no longer an acid.
    • Why not C: Miscounts the carbon atoms in methanol, CH3OH, which has only one carbon (giving the 'methyl' group), as though it had three carbons (giving 'propyl'); the alkyl part of an ester's name must match the actual carbon count of the alcohol used.

12. Question 12Answer: A

    1. The rate and vigour of a metal's reaction with dilute acid reflects its position in the reactivity series: a more reactive metal reacts faster and produces gas more vigorously.
    2. Magnesium reacts immediately and vigorously, so it is the most reactive metal here.
    3. Aluminium is a genuinely reactive metal, but it is naturally coated with a thin, tough layer of aluminium oxide that must be broken down before the underlying metal can react with the acid; once this happens, aluminium reacts faster than iron, confirming it sits above iron in the reactivity series despite its slow initial appearance.
    4. Silver does not react with dilute hydrochloric acid at all, showing it is one of the least reactive metals, below hydrogen in the reactivity series. So the correct order from most to least reactive is magnesium, aluminium, iron, silver, matching option A.
    • Why not B: Reverses the entire order: a metal that reacts fastest and most vigorously with acid is the most reactive, not the least, so silver (which does not react at all) must be placed last, not first, and magnesium (the fastest reaction) must be placed first, not last.
    • Why not C: Swaps magnesium and aluminium; aluminium's reaction only becomes noticeable once its protective oxide layer has broken down, whereas magnesium reacts immediately and vigorously from the very start, showing magnesium is the more reactive of the two, not aluminium.
    • Why not D: Swaps aluminium and iron; aluminium's apparent delay is caused only by its oxide layer, not by low reactivity of the metal itself, and once that layer is removed aluminium reacts faster than iron, so aluminium sits above iron in the true reactivity order, not below it.

13. Question 13Answer: B

    1. A metal's or alloy's suitability for a use depends on matching its actual physical and chemical properties to what that use requires.
    2. Titanium has a notably low density for a strong metal, giving components a high strength-to-weight ratio, which matters for aircraft where reducing weight saves fuel.
    3. Titanium also forms a stable, unreactive oxide layer that resists corrosion and does not react with body fluids, so it is not rejected or corroded when used in hip joint replacements; alloying with small amounts of other elements can further improve its strength without losing these properties.
    • Why not A: Attributes titanium's suitability to high electrical and thermal conductivity, a property associated with metals like copper; titanium is not chosen for either application because of its conductivity.
    • Why not C: Titanium is valued for being relatively unreactive (chemically inert and corrosion-resistant) in body fluids, which is why the body does not reject it, not for being highly reactive; a highly reactive metal implant would corrode and could poison the surrounding tissue.
    • Why not D: Makes a false blanket generalisation: alloying changes a metal's properties, often making it harder, but it does not always make a metal lighter, and titanium alloys are in fact more expensive than many pure metals, not cheaper.

14. Question 14Answer: A

    1. In the blast furnace, carbon monoxide reduces iron(III) oxide to iron, and is itself oxidised to carbon dioxide.
    2. Fe2O3 contains 2 iron atoms and 3 oxygen atoms, so 2Fe must appear on the product side to balance iron, and the three oxygen atoms from the oxide must end up somewhere on the product side too.
    3. Balancing oxygen and carbon together requires 3CO reacting with the three oxygen atoms already in the oxide to form 3CO2, giving Fe2O3 + 3CO -> 2Fe + 3CO2: 2 iron atoms, 3 + 3 = 6 oxygen atoms, and 3 carbon atoms balance on both sides.
    • Why not B: Forgets to also multiply the carbon monoxide reactant by 3, even though three CO2 molecules are correctly produced on the product side; carbon is not conserved (1 carbon atom on the left, 3 on the right).
    • Why not C: Forgets to add the extra oxygen atom that turns each CO into CO2 during the reduction, so oxygen is not conserved (3 oxygen atoms from Fe2O3 plus 3 from CO equals 6 on the left, but only 3 on the right).
    • Why not D: Doubles the iron oxide and iron on both sides without doubling everything else consistently, so oxygen no longer balances (2 x 3 + 3 = 9 oxygen atoms on the left, but only 3 x 2 = 6 on the right).

15. Question 15Answer: C

    1. Transition metals and their compounds characteristically show several properties: they can form stable ions in more than one oxidation state, they often form coloured compounds, and they (or their compounds) often act as catalysts.
    2. Iron is the catalyst used in the Haber process, where nitrogen and hydrogen react together to form ammonia; the iron itself is not used up overall and could in principle be reused.
    3. This illustrates the catalytic property specifically; the coloured-compounds property is shown by, for example, blue copper(II) sulfate solution (not by the colour of copper metal itself), and the variable-oxidation-states property is shown by, for example, iron forming both Fe2+ and Fe3+ compounds (not by iron changing between solid and liquid).
    • Why not A: Confuses the colour of copper compounds in solution, such as blue copper(II) sulfate solution, with the colour of the pure metal itself, which is a reddish-brown metallic colour, not blue.
    • Why not B: Misunderstands the definition of a catalyst: a catalyst speeds up a reaction without itself being permanently used up or changed overall, so the same manganese(IV) oxide can be reused repeatedly, not consumed batch by batch.
    • Why not D: Confuses 'different oxidation states', a chemical property referring to ions such as Fe2+ and Fe3+ forming different compounds, with a physical change of state on melting, which is not what oxidation state means at all.