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Admissions tests / ESAT / Chemistry / Quantitative chemistry and redox

Demanding. 15 questions, 15 marks, about 27 minutes.

ESAT Chemistry: Quantitative chemistry and redox, set 3

Chemical equations, relative molar mass, moles, concentration, yields, limiting reagents, empirical formulae, and oxidation and reduction.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • Use the Ar values given in each question.
  • This set is pitched harder than the real test: expect multi-step routes and less signposting than a standard set.
  1. 11 mark

    Lead nitrate solution is mixed with potassium iodide solution and a bright yellow solid forms immediately: Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq). A student argues that because a solid appears where none existed before, new atoms of lead and iodine must have been created inside the precipitate at the moment of mixing. Which statement correctly identifies the flaw in this argument?

    1. A The argument is wrong because the reaction releases energy as it happens, and this energy is converted directly into a small quantity of new lead and iodine atoms, accounting for the mass of the solid.
    2. B The argument is wrong because lead ions and iodide ions are already present, dissolved separately in the two solutions, before mixing; forming the solid rearranges these existing ions into a new ionic lattice, it does not create any atom of lead or iodine.
    3. C The argument is wrong because no real chemical change has occurred at all; the yellow solid is simply the lead and iodide ions sitting next to each other as a physical mixture, with no new ionic bonds forming between them.
    4. D The argument is wrong because some of the iodide ions must first be oxidised to release the electrons needed to hold the new lattice together, and this necessary redox step is what the student has missed.
  2. 21 mark

    Ammonium phosphate is used as a fertiliser. Ammonium ions are NH4+ and phosphate ions are PO4^3-. What is the correct formula, including state symbol, for solid ammonium phosphate before it is dissolved in water?

    1. A (NH4)PO4 (s)
    2. B NH4(PO4)3 (s)
    3. C (NH4)3PO4 (aq)
    4. D (NH4)3PO4 (s)
  3. 31 mark

    Silver nitrate solution is used to test for chloride ions: AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq). Which of the following correctly gives the balanced ionic equation for this reaction, with the spectator ions removed and every species carrying its correct charge?

    1. A Ag+(aq) + Cl-(aq) -> AgCl(s)
    2. B Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)
    3. C Ag(aq) + Cl(aq) -> AgCl(s)
    4. D Ag+(aq) + Cl-(aq) -> AgCl2(s)
  4. 41 mark

    Ethanoic acid and ethanol react in a reversible esterification reaction in which every species remains liquid: CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l). A chemist adds extra ethanol to the equilibrium mixture at constant temperature, increasing its concentration. Which statement about the effect on the position of equilibrium is correct?

    1. A The equilibrium shifts to the left, producing more ethanoic acid and ethanol, in order to reduce the concentration of the ethanol that was just added.
    2. B The equilibrium is not affected at all, because every species in this reaction is a liquid rather than a gas, and concentration changes (like pressure changes) only affect equilibria involving gases.
    3. C The equilibrium shifts to the right, forming more ester, because increasing the concentration of a reactant shifts the position of equilibrium in the direction that consumes some of the added reactant.
    4. D The equilibrium shifts to the right, forming more ester, because increasing the concentration of ethanol increases the value of the equilibrium constant Kc for the forward reaction.
  5. 51 mark

    Ammonium nitrate, NH4NO3, is manufactured on an industrial scale. Using Ar: N = 14, H = 1, O = 16, how many moles of ammonium nitrate are there in 8 tonnes (8000 kg) of the pure solid?

    1. A 125,000 mol
    2. B 100,000 mol
    3. C 100 mol
    4. D 0.1 mol
  6. 61 mark

    Using L to represent Avogadro's number (6.02 x 10^23), how many oxygen atoms, in terms of L, are present in 0.5 mol of calcium carbonate, CaCO3?

    1. A 1.5 L
    2. B 0.5 L
    3. C 3 L
    4. D 0.167 L
  7. 71 mark

    50 cm3 of a solution contains 4.9 g of dissolved sulfuric acid, H2SO4. Using Ar: H = 1, S = 32, O = 16, what is the concentration of this solution, in mol dm-3?

    1. A 0.001 mol dm-3
    2. B 0.05 mol dm-3
    3. C 400 mol dm-3
    4. D 1 mol dm-3
  8. 81 mark

    In the thermite reaction, 2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s), 5.4 g of aluminium is mixed with 24 g of iron(III) oxide and allowed to react to completion. Using Ar: Al = 27, Fe = 56, O = 16, what is the maximum mass of iron that can be formed?

    1. A 16.8 g
    2. B 5.6 g
    3. C 11.2 g
    4. D 8.4 g
  9. 91 mark

    1.4 g of ethene, C2H4, burns completely in exactly 3.6 dm3 of oxygen gas, measured at room temperature and pressure where 1 mol of any gas occupies 24 dm3. Using Ar: C = 12, H = 1, which balanced equation for this combustion is consistent with the data?

    1. A C2H4 + 3O2 -> 2CO2 + H2O
    2. B C2H4 + 3O2 -> 2CO2 + 2H2O
    3. C C2H4 + 2O2 -> 2CO2 + 2H2O
    4. D C2H4 + 3O2 -> CO2 + 2H2O
  10. 101 mark

    A sample of ammonia gas, NH3, occupies 240 cm3 at room temperature and pressure, where 1 mole of any gas occupies 24 dm3. Using Ar: N = 14, H = 1, what mass of ammonia is present in this sample?

    1. A 170 g
    2. B 0.14 g
    3. C 4.08 g
    4. D 0.17 g
  11. 111 mark

    In a titration, sulfuric acid of concentration 0.25 mol dm-3 is used to neutralise 30.0 cm3 of 0.30 mol dm-3 potassium hydroxide solution: H2SO4(aq) + 2KOH(aq) -> K2SO4(aq) + 2H2O(l). What volume of the sulfuric acid is required, in cm3?

    1. A 18 cm3
    2. B 36 cm3
    3. C 72 cm3
    4. D 0.018 cm3
  12. 121 mark

    A synthesis reaction has a percentage yield of 75%. If 21.0 g of product is actually obtained, what was the theoretical (predicted) mass of product for this reaction?

    1. A 28.0 g
    2. B 15.75 g
    3. C 26.25 g
    4. D 0.28 g
  13. 131 mark

    Magnesium reacts with dilute hydrochloric acid: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). No oxygen atom is gained or lost by any species in this reaction, so the gain/loss of oxygen definition of oxidation and reduction cannot be applied directly here. Using the electron-transfer definition instead, which statement correctly identifies what is oxidised and what is reduced?

    1. A Chlorine is reduced, since the Cl- ions present in HCl are still present as Cl- ions in MgCl2, and any ion that appears unchanged in the product of a reaction must have been reduced.
    2. B Magnesium is reduced, because it changes from a solid metal into a dissolved ion, and any change from a solid element into a dissolved ion is itself a reduction.
    3. C No oxidation or reduction occurs in this reaction, because no oxygen atoms are gained or lost by any of the species involved.
    4. D Magnesium is oxidised, losing 2 electrons to form Mg^2+; hydrogen (as H+ ions) is reduced, each ion gaining 1 electron to form H2 gas; the chloride ions remain unchanged throughout and take no part in the electron transfer.
  14. 141 mark

    Hydrochloric acid is neutralised by sodium hydroxide solution: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l). By assigning an oxidation state to every element in every species, which statement correctly classifies this reaction?

    1. A This is a redox reaction, because hydrogen moves from being bonded to chlorine (in HCl) to being bonded to oxygen (in H2O), and moving to a different bonding partner always means an element's oxidation state has changed.
    2. B No oxidation or reduction occurs in this reaction; every element keeps the same oxidation state throughout: H is +1 in both HCl and H2O, Cl is -1 in both HCl and NaCl, Na is +1 in both NaOH and NaCl, and O is -2 in both NaOH and H2O.
    3. C This is reduction only, since water, H2O, has a lower total mass than the hydrochloric acid molecule it replaces, and any decrease in a molecule's mass during a reaction counts as a reduction.
    4. D This is oxidation only, since sodium changes from being part of a solid base-forming compound to being part of a dissolved salt, which represents an increase in what can be called sodium's 'reactivity state'.
  15. 151 mark

    Hydrogen peroxide decomposes: 2H2O2(l) -> 2H2O(l) + O2(g). The oxidation state of oxygen is -1 in H2O2, -2 in H2O and 0 in O2 (hydrogen remains +1 throughout). Which statement correctly classifies this reaction and identifies the species that acts as both the oxidising and reducing agent?

    1. A This is not disproportionation, because only one reactant, H2O2, is involved, and disproportionation requires at least two different reacting species.
    2. B Hydrogen is both the oxidising and reducing agent, since it appears in every species in the equation: H2O2 and H2O.
    3. C Hydrogen peroxide, H2O2, undergoes disproportionation: the oxygen within it is simultaneously reduced (oxidation state -1 to -2, in H2O) and oxidised (oxidation state -1 to 0, in O2), so this same oxygen is acting as both the oxidising and reducing agent.
    4. D This is reduction only, since the overall reaction releases oxygen gas, and releasing a gas as a product is always associated with a reduction process.

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. Before mixing, the lead nitrate solution already contains dissolved Pb^2+ and NO3- ions, and the potassium iodide solution already contains dissolved K+ and I- ions.
    2. When mixed, the Pb^2+ and I- ions come together and pack into a solid ionic lattice, PbI2, because this arrangement is more stable than remaining separately dissolved; the K+ and NO3- ions remain dissolved as spectator ions.
    3. No atom of lead or iodine is created or destroyed in this process: every lead and iodine atom present in the solid was already present, as an ion, in one of the original solutions. Only the physical arrangement and bonding of the existing ions has changed.
    4. So the student's argument is wrong because it mistakes a rearrangement of existing ions for the creation of new atoms; option B correctly locates the error.
    • Why not A: Invents a mass-to-matter conversion to explain the solid's mass, when in fact the total mass of lead and iodine present was already contributed by the two solutions before mixing; no such conversion happens in an ordinary chemical reaction.
    • Why not C: Wrongly denies that a chemical change has occurred. A genuine new ionic compound, solid PbI2, forms with its own lattice of Pb^2+ and I- ions bonded together; that is a real chemical change, just one that creates no new atoms.
    • Why not D: Assumes that forming any new bond, including an ionic lattice, requires a redox step. Precipitation is simply electrostatic attraction between ions that already carry their charges (Pb^2+ and I-); no electron is transferred and no oxidation state changes during this reaction.

  2. Question 2Answer: D

    1. Ammonium is NH4+ (charge +1) and phosphate is PO4^3- (charge -3).
    2. To balance the charges, the magnitude of each ion's charge becomes the subscript of the OTHER ion: 3 NH4+ ions (total charge +3) balance 1 PO4^3- ion (total charge -3), giving (NH4)3PO4.
    3. The question asks for the compound before it dissolves, so it must carry the state symbol for a solid, (s), not (aq) for a dissolved substance.
    4. Putting the formula and the state symbol together gives (NH4)3PO4 (s), option D.
    • Why not A: Writes a 1:1 ratio of ammonium to phosphate ions, ignoring that the charges (NH4+ and PO4^3-) must be balanced by using 3 ammonium ions for every 1 phosphate ion, not 1 of each.
    • Why not B: Crosses the charges the wrong way round: it keeps ammonium's own subscript at 1 (its own charge magnitude) and gives phosphate a subscript of 3 (its own charge magnitude), instead of crossing each ion's charge magnitude onto the OTHER ion.
    • Why not C: Gets the correct formula unit but the wrong state symbol; the question specifies the solid before it is dissolved, so it must carry (s), not (aq) for a dissolved solution.

  3. Question 3Answer: A

    1. In the full equation, silver ions and chloride ions come together to form the solid AgCl, while sodium ions and nitrate ions remain dissolved and unchanged throughout: they are spectator ions.
    2. A correctly written ionic equation removes these spectator ions and shows only the species that actually react, each carrying its correct charge: Ag+(aq) and Cl-(aq).
    3. The charges must balance: Ag+ contributes +1 and Cl- contributes -1, summing to 0, matching the neutral solid product AgCl(s), so a 1:1 ratio (not 1:2) is correct.
    4. This gives Ag+(aq) + Cl-(aq) -> AgCl(s), option A.
    • Why not B: Never cancels the spectator ions Na+ and NO3-, which appear unchanged on both sides of the equation; an ionic equation is meant to show only the species that actually react, with the unchanged spectator ions removed.
    • Why not C: Drops the ionic charges entirely, writing the species as neutral atoms (Ag, Cl) rather than the charged ions (Ag+, Cl-) that are actually present in solution.
    • Why not D: Misremembers the formula of silver chloride, using a 1:2 ratio of Ag to Cl (AgCl2) instead of the correct 1:1 pairing, which already balances the +1 charge of Ag+ against the -1 charge of Cl-.

  4. Question 4Answer: C

    1. Le Chatelier's principle states that if a system at equilibrium has one of its conditions changed, the position of equilibrium shifts to partially counteract that change.
    2. Increasing the concentration of ethanol (a reactant) means the system shifts in the direction that uses up some of the extra ethanol, which is the forward direction, producing more ester and water.
    3. This reasoning applies regardless of the physical state of the species involved; it is the pressure rule specifically, not the concentration rule, that only applies when gases are present in different amounts on each side.
    4. The equilibrium constant Kc itself is unchanged by a concentration change at constant temperature; only the position of equilibrium moves, so the equilibrium shifts right (more ester) for the reason given in option C.
    • Why not A: Misapplies Le Chatelier's principle by reversing its direction: the system responds to an increase in a reactant's concentration by favouring the reaction that CONSUMES some of that reactant (the forward reaction here), not by producing even more of it.
    • Why not B: Wrongly extends the rule that pressure changes need gas molecules to concentration changes as well. Concentration changes shift the position of equilibrium for reactions in any phase, including one where every species is a liquid; only the pressure rule is restricted to gases.
    • Why not D: Confuses shifting the position of equilibrium with changing the equilibrium constant. Kc depends only on temperature; changing a concentration shifts where the equilibrium lies, but the value of Kc itself stays the same at constant temperature.

  5. Question 5Answer: B

    1. First convert the mass to grams: 8 tonnes = 8000 kg, and 8000 kg = 8000 x 1000 g = 8,000,000 g.
    2. Find the relative molar mass: Mr(NH4NO3) = Ar(N) + 4 x Ar(H) + Ar(N) + 3 x Ar(O) = 14 + 4 + 14 + 48 = 80.
    3. Moles = mass / Mr = 8,000,000 / 80.
    4. 8,000,000 / 80 = 100,000, so there are 100,000 mol (1 x 10^5 mol) of ammonium nitrate, option B.
    • Why not A: Miscounts the oxygen atoms in the nitrate group, using only 2 oxygens (Mr = 14 + 4 + 32 = 64) instead of the correct 3 (Mr = 80), giving moles = 8,000,000 / 64 = 125,000.
    • Why not C: Uses the mass in kilograms, 8000, directly as though it were the mass in grams, giving moles = 8000 / 80 = 100 instead of first converting to 8,000,000 g.
    • Why not D: Uses the mass in tonnes, 8, directly as though it were the mass in grams, skipping the unit conversion entirely and giving moles = 8 / 80 = 0.1.

  6. Question 6Answer: A

    1. One mole of any substance contains L formula units (or particles), so 0.5 mol of CaCO3 contains 0.5 L formula units of CaCO3.
    2. Each formula unit of CaCO3 contains 3 oxygen atoms (from the carbonate ion, CO3^2-).
    3. The total number of oxygen atoms is therefore 3 x 0.5 L.
    4. 3 x 0.5 = 1.5, so there are 1.5 L oxygen atoms present, option A.
    • Why not B: Forgets that each formula unit of CaCO3 contains 3 oxygen atoms, treating it as though there were only 1 oxygen atom per formula unit, giving 0.5 L instead of 3 x 0.5 L.
    • Why not C: Correctly uses the 3 oxygen atoms per formula unit, but forgets to also scale by the 0.5 mol of CaCO3 actually present, as if a full 1 mol sample were being considered instead.
    • Why not D: Divides the number of moles by the number of oxygen atoms per formula unit (0.5 / 3) instead of multiplying, inverting the correct operation.

  7. Question 7Answer: D

    1. Find the relative molecular mass of sulfuric acid: Mr(H2SO4) = 2 x Ar(H) + Ar(S) + 4 x Ar(O) = 2 + 32 + 64 = 98.
    2. Find the number of moles: moles = mass / Mr = 4.9 / 98 = 0.05 mol.
    3. Convert the volume to dm3: 50 cm3 = 50 / 1000 dm3 = 0.05 dm3.
    4. Concentration = moles / volume (dm3) = 0.05 / 0.05 = 1 mol dm-3, option D.
    • Why not A: Correctly finds 0.05 mol of acid, but then forgets to convert the volume from cm3 to dm3, dividing by 50 instead of by 0.05, giving 0.05 / 50 = 0.001.
    • Why not B: Correctly finds the number of moles of acid present, 0.05 mol, but then reports this number directly as the concentration, without dividing by the volume of solution at all.
    • Why not C: Inverts the moles formula, calculating Mr divided by mass (98 / 4.9 = 20) instead of mass divided by Mr, and then divides this incorrect mole value by the volume in dm3 (20 / 0.05 = 400).

  8. Question 8Answer: C

    1. Find the moles of each reactant: moles Al = 5.4 / 27 = 0.2 mol, and moles Fe2O3 = 24 / (2 x 56 + 3 x 16) = 24 / 160 = 0.15 mol.
    2. The equation needs 2 mol Al for every 1 mol Fe2O3. For all 0.15 mol of Fe2O3 to react, 0.3 mol of Al would be needed, but only 0.2 mol of Al is available, so aluminium is the limiting reactant.
    3. Use the aluminium amount to find the iron produced: 0.2 mol Al x (2 mol Fe / 2 mol Al) = 0.2 mol Fe.
    4. Convert to mass: 0.2 mol Fe x 56 = 11.2 g, option C.
    • Why not A: Wrongly assumes Fe2O3 is the limiting reactant, and calculates the iron produced from all 0.15 mol of Fe2O3 reacting completely (using the 1 Fe2O3 : 2 Fe ratio), without checking whether enough aluminium is actually available.
    • Why not B: Correctly identifies aluminium as the limiting reactant, but then misapplies the mole ratio as 2 Al : 1 Fe instead of the correct 2 Al : 2 Fe (a 1:1 ratio), halving the true amount of iron produced.
    • Why not D: Wrongly assumes Fe2O3 is the limiting reactant, and additionally misapplies its ratio to iron as 1 Fe2O3 : 1 Fe instead of the correct 1 Fe2O3 : 2 Fe.

  9. Question 9Answer: B

    1. Find the moles reacting: moles C2H4 = 1.4 / 28 = 0.05 mol, and moles O2 = 3.6 / 24 = 0.15 mol.
    2. The ratio of C2H4 to O2 is 0.05 : 0.15, which simplifies to 1 : 3.
    3. Ethene has 2 carbon atoms and 4 hydrogen atoms, so complete combustion needs 2 CO2 (to balance the carbon) and 2 H2O (to balance the hydrogen, since 2 x 2 = 4).
    4. Checking oxygen: the right-hand side needs 2 x 2 + 2 x 1 = 6 oxygen atoms, which is exactly the 3 O2 molecules found from the data (3 x 2 = 6), confirming C2H4 + 3O2 -> 2CO2 + 2H2O, option B.
    • Why not A: Correctly balances the two carbon atoms into 2CO2, but only accounts for half of ethene's four hydrogen atoms when fixing the water coefficient, using 1 H2O instead of the 2 needed to balance all four H atoms.
    • Why not C: Misreads the mole ratio from the volume data: dividing the oxygen volume by 36 rather than 24 gives 3.6 / 36 = 0.1 mol O2, so a 1:2 ratio of C2H4 to O2 is used instead of the correct 1:3.
    • Why not D: Correctly balances the four hydrogen atoms into 2H2O, but drops one of ethene's two carbon atoms, using 1 CO2 instead of the 2 needed to balance both carbon atoms.

  10. Question 10Answer: D

    1. Convert the volume to dm3: 240 cm3 = 240 / 1000 dm3 = 0.24 dm3.
    2. Find the moles of gas: moles = volume (dm3) / 24 = 0.24 / 24 = 0.01 mol.
    3. Find the relative molecular mass of ammonia: Mr(NH3) = Ar(N) + 3 x Ar(H) = 14 + 3 = 17.
    4. Mass = moles x Mr = 0.01 x 17 = 0.17 g, option D.
    • Why not A: Forgets to convert the given volume from cm3 to dm3 before dividing by the molar volume, treating 240 as if it were already in dm3: 240 / 24 = 10 mol, then 10 x 17 = 170 g.
    • Why not B: Correctly finds 0.01 mol from the volume, but then uses only the Ar of nitrogen, 14, as though it were the whole molar mass of NH3, forgetting the three hydrogen atoms: 0.01 x 14 = 0.14 g.
    • Why not C: Correctly converts the volume to 0.24 dm3, but then treats this figure directly as the number of moles, skipping the division by the molar volume of 24 dm3 entirely: 0.24 x 17 = 4.08 g.

  11. Question 11Answer: A

    1. Find the moles of KOH that must be neutralised: moles = concentration x volume (dm3) = 0.30 x (30.0 / 1000) = 0.30 x 0.030 = 0.009 mol.
    2. The balanced equation shows a 1 : 2 ratio of H2SO4 to KOH, so the moles of H2SO4 needed are 0.009 / 2 = 0.0045 mol.
    3. Volume of acid (dm3) = moles / concentration = 0.0045 / 0.25 = 0.018 dm3.
    4. Convert to cm3: 0.018 dm3 x 1000 = 18 cm3, option A.
    • Why not B: Ignores the 1:2 mole ratio between H2SO4 and KOH shown in the balanced equation, treating the reaction as though it were a 1:1 neutralisation instead: moles KOH = 0.30 x 0.030 = 0.009 mol used directly as the moles of acid, giving 0.009 / 0.25 = 0.036 dm3 = 36 cm3.
    • Why not C: Inverts the mole ratio, multiplying the moles of KOH by 2 instead of dividing by 2, even though each mole of the diprotic acid H2SO4 only needs 2 moles of KOH to neutralise it: 0.009 x 2 = 0.018 mol treated as the moles of acid, giving 0.018 / 0.25 = 0.072 dm3 = 72 cm3.
    • Why not D: Correctly finds that 0.0045 mol of acid is needed and correctly divides by the concentration to get 0.018, but forgets to convert this volume from dm3 to cm3, reporting 0.018 cm3 instead of multiplying by 1000.

  12. Question 12Answer: A

    1. Percentage yield = (actual yield / predicted yield) x 100, so predicted yield = actual yield / (percentage yield / 100).
    2. Convert 75% to a decimal fraction: 75 / 100 = 0.75.
    3. Predicted yield = 21.0 / 0.75.
    4. 21.0 / 0.75 = 28.0, since 0.75 x 28.0 = 21.0, so the theoretical mass of product was 28.0 g, option A.
    • Why not B: Inverts the rearrangement of the percentage yield formula, multiplying the actual mass by the decimal yield (21.0 x 0.75) instead of dividing the actual mass by it.
    • Why not C: Treats the actual mass as though it were the full (100%) reference amount, and simply adds the 'missing' 25% of that actual mass onto it (21.0 + 0.25 x 21.0), rather than recognising the actual mass is only 75% of a larger, unknown predicted mass.
    • Why not D: Forgets to convert 75% into the decimal fraction 0.75 before dividing, dividing by 75 directly instead, which makes the answer 100 times too small.

  13. Question 13Answer: D

    1. Since no oxygen is involved, oxidation and reduction here must be identified using the electron-transfer definition: oxidation is the loss of electrons, and reduction is the gain of electrons.
    2. Magnesium starts as the neutral element, Mg (oxidation state 0), and ends up as Mg^2+ in MgCl2, so each magnesium atom loses 2 electrons: magnesium is oxidised.
    3. Hydrogen starts as H+ ions in HCl (oxidation state +1) and ends up as neutral H2 gas (oxidation state 0), so each hydrogen ion gains 1 electron: hydrogen is reduced.
    4. Chloride ions are Cl- both before (in HCl) and after (in MgCl2) the reaction, so they gain or lose no electrons and take no part in the redox change; they are spectator ions, confirming option D.
    • Why not A: Wrongly assumes that any ion appearing in a product must have been reduced. The chloride ion's charge stays at -1 throughout (in both HCl and MgCl2), so no electron is gained or lost by chlorine; it is a spectator ion, not a reduced species.
    • Why not B: Confuses a physical change of state (solid metal to dissolved ion) with the chemical definition of reduction. Magnesium actually LOSES 2 electrons to form Mg^2+, which is oxidation, not reduction.
    • Why not C: Wrongly assumes that oxidation and reduction can only be defined in terms of oxygen. This reaction involves no oxygen at all, but it is still a redox reaction because electrons are transferred: magnesium loses electrons and hydrogen ions gain them.

  14. Question 14Answer: B

    1. Assign oxidation states to every element in HCl, NaOH, NaCl and H2O, using the rule that a neutral compound's oxidation states sum to zero and known values for common ions.
    2. Hydrogen: +1 in HCl, and +1 in H2O. Chlorine: -1 in HCl, and -1 in NaCl. Sodium: +1 in NaOH, and +1 in NaCl. Oxygen: -2 in NaOH, and -2 in H2O.
    3. Every element keeps exactly the same oxidation state before and after the reaction, so no oxidation and no reduction occurs anywhere in this reaction.
    4. Acid-base neutralisation reactions like this one are a standard example of a reaction that is NOT a redox reaction, even though a chemical change clearly happens; option B correctly classifies it.
    • Why not A: Wrongly assumes that changing bonding partner always changes oxidation state. Hydrogen is +1 in HCl (paired with Cl at -1) and remains +1 in H2O (paired with O at -2), so its oxidation state does not actually change, even though its neighbouring atom does.
    • Why not C: Confuses the everyday meaning of 'reduced' (made smaller in mass) with the chemical definition, which is about a decrease in oxidation state or a gain of electrons, not about the mass of a molecule.
    • Why not D: Invents a non-existent concept, 'reactivity state', in place of oxidation state. Sodium's actual oxidation state is +1 in both NaOH and NaCl, so no genuine change occurs to sodium at all.

  15. Question 15Answer: C

    1. Disproportionation occurs when atoms of the same element, within the same reactant, are simultaneously oxidised and reduced in one reaction.
    2. Here, oxygen starts at oxidation state -1 in H2O2, and ends up at -2 in H2O (a decrease, so reduction) and at 0 in O2 (an increase, so oxidation).
    3. Both changes happen to oxygen atoms that both started out in the same reactant, H2O2, so the oxygen within H2O2 is acting as both the oxidising agent (towards the oxygen atoms being reduced) and the reducing agent (towards the oxygen atoms being oxidised).
    4. Hydrogen's oxidation state stays at +1 throughout and plays no part in the redox change, so the correct description is that H2O2 (specifically its oxygen) is both agents at once, option C.
    • Why not A: Wrongly assumes disproportionation needs two different reactant species. Disproportionation is defined by one ELEMENT within a single species being simultaneously oxidised and reduced, which is exactly the classic case of a single compound decomposing, as it does here.
    • Why not B: Wrongly attributes the redox role to hydrogen, which stays at oxidation state +1 in both H2O2 and H2O and so undergoes no change at all; the entire oxidation state change happens to the oxygen, not the hydrogen.
    • Why not D: Invents a rule that releasing a gas is always a reduction, which has no basis in oxidation-state reasoning. The correct classification requires tracking each element's oxidation state directly, not the physical state of the products formed.

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