IGCSE Chemistry Paper 4 (Theory / Extended) Challenge

Topic 6: Chemical Reactions -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions -- Challenge Paper

This paper is designed to stretch A* candidates with multi-step reasoning and data analysis.
Answer all questions in the spaces provided.
Show all working for calculations -- method marks are available even if the final answer is incorrect.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Quantitative Rate Analysis

Total: 12 marks

A student in Bengaluru adds a 0.072 g strip of magnesium ribbon to 50.0 cm³ of excess dilute hydrochloric acid at 25 °C and collects the hydrogen gas in a syringe. The volume of hydrogen is recorded every 10 seconds.

Time / s	0	10	20	30	40	50	60	70	80
Volume of H₂ / cm³	0	22	39	52	61	67	70	72	72

(a) [4]

(i) Use the table to calculate the average rate of reaction between 20 s and 60 s, in cm³/s. [2]

(ii) Convert your answer to cm³/min. [2]

Model Answer -- 1(a)

Change in volume = 70 − 39 = 31 cm³ over a time interval of 60 − 20 = 40 s [1]

average rate = 31 ÷ 40 = 0.775 cm³/s [1]

There are 60 s in 1 min, so multiply the rate per second by 60 [1]

0.775 × 60 = 46.5 cm³/min [1]

⚠ If you missed marks here: The commonest slip is dividing the total volume at 60 s by the total time, 70 ÷ 60 = 1.17 cm³/s, which wrongly ignores the 39 cm³ already collected at 20 s. The second is dividing by 60 instead of multiplying when converting, giving 0.0129 cm³/min instead of 46.5 cm³/min.

1 mark -- change in volume 70 minus 39 equals 31 cm3 over 40 s 1 mark -- average rate 0.775 cm3 per second 1 mark -- multiply by 60 to convert to per minute 1 mark -- rate equals 46.5 cm3 per minute

(b) [5]

(i) A tangent has been drawn to the curve at t = 0 on the graph above. Describe how the tangent is used and calculate the initial rate of reaction in cm³/s. [3]

(ii) Explain, in terms of particles, why the gradient of the curve decreases as the reaction proceeds. [2]

Model Answer -- 1(b)

The tangent touches the curve at t = 0 and does not cross it; both points used for the gradient must lie on the straight tangent, not on the curve [1]

Points read from the tangent: (0 s, 0 cm³) and (20 s, 60 cm³) [1]

initial rate = (60 − 0) ÷ (20 − 0) = 3.0 cm³/s [1]

As the reaction proceeds the hydrochloric acid is used up so its concentration falls, and the magnesium ribbon is eaten away so its surface area decreases [1]

There are therefore fewer successful collisions per second between acid particles and the magnesium surface, so gas is produced more slowly and the gradient falls [1]

⚠ If you missed marks here: Taking one point from the tangent and one from the curve is fatal — using (0 s, 0 cm³) and the curve point (20 s, 39 cm³) gives 1.95 cm³/s, far below the true 3.0 cm³/s. Also, "the acid runs out" scores nothing: the acid is in excess, so you must say its concentration falls.

1 mark -- tangent touches curve at time zero without crossing 1 mark -- two points read from tangent 0 s and 20 s 60 cm3 1 mark -- initial rate 3.0 cm3 per second 1 mark -- acid concentration falls and magnesium surface area decreases 1 mark -- fewer successful collisions per second so gradient decreases

The experiment is repeated at 25 °C using 0.036 g of magnesium ribbon with the same 50.0 cm³ of excess hydrochloric acid.

(i) Deduce the final volume of hydrogen collected and explain your reasoning. [2]

(ii) The curve for the repeat experiment is also less steep at t = 0 than the original. Explain why. [1]

Model Answer -- 1(c)

The acid is in excess, so the magnesium is the limiting reagent and it alone fixes the maximum volume of hydrogen [1]

Half the mass of magnesium gives half the moles of magnesium and therefore half the moles of hydrogen: 72 ÷ 2 = 36 cm³ [1]

A shorter strip of ribbon presents a smaller surface area to the acid, so there are fewer collisions per second at t = 0 and the initial gradient is lower [1]

⚠ If you missed marks here: Candidates who assume the acid is limiting keep the plateau at 72 cm³ — that is the classic inversion. Equally wrong is claiming the initial rate is unchanged "because the acid concentration is the same": concentration is only half the story, the exposed magnesium surface area has also halved.

1 mark -- magnesium is limiting reagent acid in excess 1 mark -- final volume 36 cm3 half of 72 1 mark -- smaller surface area gives lower initial gradient

Question 2 -- Comparing Experiments and Anomalies

Total: 12 marks

A class in Sheffield investigates the reaction between calcium carbonate and dilute hydrochloric acid. In every experiment the calcium carbonate is in excess and the carbon dioxide produced is collected in a gas syringe.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Experiment	Volume of HCl / cm³	Concentration of HCl / mol per dm³	Temperature / °C	Form of CaCO₃
A	50.0	1.00	25	large lumps
B	50.0	1.00	25	fine powder
C	50.0	0.500	25	large lumps
D	50.0	1.00	45	large lumps

The graph below shows the results of three of these experiments, labelled P, Q and R. Experiment D is not shown. (1 mol of gas occupies 24 000 cm³ at room conditions.)

(a) [5]

(i) Identify which of experiments A, B and C produced curve P, curve Q and curve R. [3]

(ii) Explain, in terms of particles, why curve P is much steeper than curve Q during the first 20 seconds. [2]

Model Answer -- 2(a)

Curve P is experiment B: it is the steepest curve, and powdered calcium carbonate reacts fastest [1]

Curve Q is experiment A: it reaches the same final volume as P (600 cm³) but far more slowly, as expected for large lumps at the same acid concentration [1]

Curve R is experiment C: it plateaus at only 300 cm³, exactly half of 600 cm³, matching the halved acid concentration [1]

Powder has a very much larger total surface area, so many more calcium carbonate particles are exposed to the acid at any moment [1]

There is therefore a much greater frequency of collisions per second between H⁺ ions and the solid surface, giving more successful collisions per second [1]

⚠ If you missed marks here: Many candidates assign R to B "because powder is different", ignoring the plateau height. The plateau, not the steepness, is the decisive clue: only C has a different number of moles of acid, so only C can finish at 300 cm³. Saying powder gives "more energy" or "hotter particles" scores zero — surface area changes collision frequency, never collision energy.

1 mark -- curve p is experiment b powdered calcium carbonate 1 mark -- curve q is experiment a large lumps same plateau 1 mark -- curve r is experiment c half concentration plateau 300 cm3 1 mark -- powder has much larger total surface area exposed 1 mark -- greater frequency of collisions per second with the solid

(b) [4]

(i) Explain, using a calculation, why curves P and Q both level off at 600 cm³ but curve R levels off at 300 cm³. [3]

(ii) Describe, in words, exactly where the curve for experiment D would lie relative to curves P and Q. [1]

Model Answer -- 2(b)

The calcium carbonate is in excess in every experiment, so the hydrochloric acid is the limiting reagent and fixes the final volume [1]

For A and B: n(HCl) = 1.00 × 0.0500 = 0.0500 mol, so n(CO₂) = 0.0250 mol and V = 0.0250 × 24 000 = 600 cm³ [1]

For C: n(HCl) = 0.500 × 0.0500 = 0.0250 mol, so n(CO₂) = 0.0125 mol and V = 300 cm³; surface area and temperature change only the speed, never the final amount [1]

Curve D lies between P and Q: steeper than Q throughout because the higher temperature speeds the reaction, but it levels off at the same height of 600 cm³ as both P and Q [1]

⚠ If you missed marks here: The classic error is forgetting the 2:1 ratio in the equation and calculating 0.0500 × 24 000 = 1200 cm³. The other trap is drawing curve D finishing above 600 cm³: raising the temperature cannot create extra acid, so the plateau must be identical.

1 mark -- calcium carbonate in excess so acid is limiting reagent 1 mark -- 0.0500 mol hcl gives 0.0250 mol co2 equals 600 cm3 1 mark -- c uses half the moles giving 300 cm3 1 mark -- curve d steeper than q same plateau 600 cm3

Another group repeats experiment B under identical conditions but records a final volume of only 540 cm³. The gradient of the first part of their curve is also lower than for curve P.

Identify the result as anomalous and give a full explanation of what most likely went wrong. [3]

Model Answer -- 2(c)

The result is anomalous: with the same 0.0500 mol of acid it should have reached 600 cm³, so 60 cm³ of carbon dioxide is unaccounted for [1]

Carbon dioxide escaped from the flask, either through a leaking bung or in the few seconds between adding the solid and fitting the bung [1]

This loss is worst with powder because the reaction is extremely fast at the start, so a large volume of gas is generated in the first seconds and is lost before recording begins — which also flattens the early gradient [1]

⚠ If you missed marks here: Blaming "not enough calcium carbonate" contradicts the stem, which states it is in excess. Blaming "the acid was too dilute" would lower the plateau to 540 cm³ only if the concentration were 0.90 mol per dm³, which is not what the data say. The lost-gas explanation is the only one that accounts for both the low plateau and the shallow start.

1 mark -- final volume 540 cm3 anomalously low should be 600 1 mark -- carbon dioxide escaped through leak or before bung fitted 1 mark -- powder reacts very fast so gas lost in first seconds

Question 3 -- Collision Theory in Depth

Total: 12 marks

A researcher in Pune is writing safety guidance for food-processing plants. She needs to explain to engineers exactly why small changes in conditions can produce very large changes in reaction rate.

(a) [5]

Doubling the concentration of an acid roughly doubles the rate of its reaction with a metal. Raising the temperature from 20 °C to 40 °C can increase the rate by a factor of four or more, even though the average speed of the particles increases by only about 3%.

Explain fully, in terms of collisions, why a rise in temperature has such a disproportionately large effect. [5]

Model Answer -- 3(a)

Doubling the concentration doubles the number of acid particles in a given volume, so the collision frequency roughly doubles and the rate roughly doubles [1]

Raising the temperature also increases collision frequency, because the particles gain kinetic energy and move faster, but only by a few per cent [1]

This small increase in collision frequency alone cannot account for a four-fold increase in rate [1]

The dominant effect is that a much greater proportion of particles now possess energy greater than or equal to the activation energy [1]

So a far greater fraction of collisions are successful; this fraction rises very steeply with temperature and outweighs the small change in collision frequency [1]

⚠ If you missed marks here: The answer "particles move faster so they collide more often" is only worth one mark and is the reason most candidates cap at 2 out of 5. It also predicts a rate increase of about 3%, not 300%. You must explicitly separate collision frequency from the fraction of particles with energy ≥ E_a, and state that the second effect dominates.

1 mark -- doubling concentration doubles particles per unit volume and collision frequency 1 mark -- higher temperature increases speed and collision frequency slightly 1 mark -- collision frequency alone cannot explain fourfold increase 1 mark -- greater proportion of particles have energy above activation energy 1 mark -- fraction of successful collisions rises steeply and dominates

(b) [3]

Explain, in terms of particles, why grinding a solid reactant into a fine powder increases the rate of its reaction with a solution, and state clearly what does not change. [3]

Model Answer -- 3(b)

Grinding exposes many more particles at the surface of the solid; particles buried inside a lump cannot react at all until the outer layers have dissolved away [1]

More exposed particles means a greater frequency of collisions per second between the solution particles and the solid surface, so the rate increases [1]

The activation energy and the energy of the individual collisions are unchanged; only the frequency of collisions changes, and the total number of solid particles (and hence the final yield) is also unchanged [1]

⚠ If you missed marks here: Writing "more surface area so more collisions" without saying per second loses the rate mark — more total collisions over the whole reaction is not the same as a faster rate. Claiming powder "gives a bigger yield" is a straight reasoning inversion: the mass, and therefore the final volume of gas, is identical.

1 mark -- powder exposes many more particles at the surface 1 mark -- greater frequency of collisions per second with solution particles 1 mark -- activation energy and final yield unchanged only frequency changes

A large lump of flour dough will not burn easily, yet a cloud of fine flour dust suspended in the air inside a mill can explode violently if it is ignited by a spark.

Explain this difference as fully as you can. [4]

Model Answer -- 3(c)

Grinding the flour into dust gives an enormous total surface area for the same mass of material [1]

Because the dust is suspended, every tiny particle is completely surrounded by oxygen from the air, so essentially all of the flour is available to react at once [1]

The frequency of collisions between oxygen molecules and flour particles is therefore extremely high, giving an extremely fast rate of combustion [1]

Combustion is strongly exothermic, so the heat released raises the temperature, which increases the rate still further; this self-accelerating runaway releases energy faster than it can escape and produces a rapid pressure rise, i.e. an explosion [1]

⚠ If you missed marks here: Most answers stop at "bigger surface area so faster" and score 1 or 2. The two marks that separate an A* answer are the suspension point (every particle in contact with oxygen, so the whole mass reacts simultaneously) and the exothermic feedback loop. Saying the dust "contains more energy" than the dough is wrong — the energy per gram is identical; only the rate of release differs.

1 mark -- fine dust has enormous total surface area 1 mark -- suspended particles completely surrounded by oxygen in air 1 mark -- extremely high collision frequency gives very fast combustion 1 mark -- exothermic heat raises temperature causing runaway pressure rise

Question 4 -- Equilibrium and the Haber Process

Total: 12 marks

An ammonia plant on Jurong Island, Singapore, operates the Haber process. The reversible reaction reaches dynamic equilibrium inside the converter.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) the forward reaction is exothermic

(a) [4]

(i) Use the graph to state the percentage yield of ammonia at 200 atm and 450 °C, and at 200 atm and 350 °C. [2]

(ii) Explain, in terms of the position of equilibrium, why the yield falls as the temperature is raised at constant pressure. [2]

Model Answer -- 4(a)

At 200 atm and 450 °C the yield is 28% [1]

At 200 atm and 350 °C the yield is 52% [1]

The forward reaction is exothermic, so the reverse reaction is endothermic [1]

Raising the temperature shifts the position of equilibrium in the endothermic direction, i.e. to the left, to oppose the increase in temperature, so less ammonia is present at equilibrium [1]

⚠ If you missed marks here: The frequent inversion is "higher temperature means faster reaction so more ammonia", which predicts the yield rising with temperature — the exact opposite of the graph. Rate and yield are separate ideas. Also check you read the 450 °C curve at the dashed 200 atm line: reading the 350 °C curve by mistake gives 52% for both answers.

1 mark -- yield 28 percent at 200 atm and 450 degrees 1 mark -- yield 52 percent at 200 atm and 350 degrees 1 mark -- forward reaction exothermic reverse endothermic 1 mark -- equilibrium shifts left in endothermic direction lowering yield

(b) [5]

(i) Explain why the plant operates at 450 °C even though your answer to (a)(i) shows a much higher yield at 350 °C. [2]

(ii) The graph shows that raising the pressure to 1000 atm would raise the yield further. Explain why the plant operates at only about 200 atm. [3]

Model Answer -- 4(b)

At 350 °C the rate of reaction is far too slow, so equilibrium is reached only after a very long time and the plant produces too little ammonia per day to be profitable [1]

450 °C is a compromise: it gives a moderate yield of about 28% at an acceptably fast rate, maximising the mass of ammonia made per hour [1]

Producing and containing 1000 atm requires extremely thick-walled reaction vessels and pipework, so the plant would be far more expensive to build and maintain [1]

Compressing the gases to 1000 atm consumes a very large amount of energy, and very high pressures are a serious safety hazard [1]

The yield curves flatten as pressure rises, so the extra ammonia gained above about 200 atm is small and does not justify the extra capital, energy and safety costs [1]

⚠ If you missed marks here: "It is too expensive" on its own is one vague mark at best; you must name the cost — thicker vessels, compression energy — and pair it with the diminishing return visible in the flattening curve. Saying higher pressure would decrease the yield is a direct reasoning inversion: 4 moles of gas become 2, so high pressure always favours ammonia.

1 mark -- lower temperature gives unacceptably slow rate 1 mark -- 450 degrees is compromise between rate and yield 1 mark -- 1000 atm needs very thick vessels expensive plant 1 mark -- compression uses large energy and is hazardous 1 mark -- yield curve flattens so extra ammonia not worth cost

(i) State precisely what the iron catalyst does, and one important thing it does not do. [2]

(ii) The ammonia is condensed out of the mixture and the unreacted nitrogen and hydrogen are recycled. Explain, using Le Chatelier's principle, why this raises the overall conversion. [1]

Model Answer -- 4(c)

The iron catalyst provides an alternative reaction pathway of lower activation energy, speeding up the forward and reverse reactions equally, so equilibrium is reached in a much shorter time [1]

It does not change the position of equilibrium, so the percentage yield of ammonia at a given temperature and pressure is exactly the same with or without it; nor is it used up [1]

Condensing and removing ammonia lowers its concentration in the mixture, so the equilibrium shifts to the right to replace it, and recycling the unreacted nitrogen and hydrogen means almost all of the feedstock is eventually converted [1]

⚠ If you missed marks here: "The catalyst increases the yield" is the single most penalised statement in this topic — it would predict more than 28% at 450 °C and 200 atm, which the graph contradicts. Also, "it lowers the activation energy of the forward reaction" is incomplete: it lowers it for both directions, which is precisely why the position of equilibrium is unaffected.

1 mark -- iron catalyst lowers activation energy equilibrium reached faster 1 mark -- catalyst does not change position of equilibrium or yield 1 mark -- removing ammonia shifts equilibrium right recycling raises conversion

Question 5 -- The Contact Process: Quantitative Equilibrium Reasoning

Total: 10 marks

A sulfuric acid plant near Cardiff manufactures acid by the Contact process, starting from molten sulfur.

Conditions in the converter	Equilibrium yield of SO₃
450 °C, 2 atm, V₂O₅ catalyst	96%
450 °C, 10 atm, V₂O₅ catalyst	99%

(a) [4]

(i) Write the equation for the burning of sulfur in air in the first stage. [1]

(ii) Write the equation for the equilibrium between sulfur dioxide and sulfur trioxide, and state the catalyst and the temperature and pressure used. [3]

Model Answer -- 5(a)

S(l) + O₂(g) → SO₂(g) [1]

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — balanced, with the equilibrium arrow [1]

Catalyst: vanadium(V) oxide, V₂O₅ [1]

Conditions: about 450 °C and about 2 atm [1]

⚠ If you missed marks here: The commonest slip is writing SO₂ + O₂ ⇌ SO₃, which is unbalanced in oxygen (4 on the left, 3 on the right); the coefficients must be 2, 1, 2. Writing a single forward arrow also loses the mark, because the reaction is reversible and never goes to completion. Vanadium(IV) oxide or iron are wrong catalysts.

1 mark -- sulfur plus oxygen gives sulfur dioxide 1 mark -- 2so2 plus o2 equilibrium 2so3 balanced 1 mark -- vanadium five oxide catalyst 1 mark -- 450 degrees celsius and 2 atmospheres

(b) [3]

Use the yield data in the table to explain why the plant operates at only 2 atm, even though higher pressure gives a higher equilibrium yield of sulfur trioxide. [3]

Model Answer -- 5(b)

There are 3 moles of gas on the left and only 2 moles on the right, so increasing the pressure shifts the equilibrium to the right and does raise the yield of SO₃ [1]

However the yield is already 96% at only 2 atm, so the equilibrium lies far to the right without any help from high pressure [1]

Raising the pressure to 10 atm gains only a further 3 percentage points, which does not justify the cost of a high-pressure plant, the energy needed for compression or the added safety risk [1]

⚠ If you missed marks here: Answers that claim pressure has no effect "because there are the same number of moles on both sides" are simply miscounting: 2 + 1 = 3 becomes 2. The examiner wants the quantitative point — 96% to 99% is a gain of only 3% — not just the word "expensive".

1 mark -- three moles of gas become two so pressure favours so3 1 mark -- yield already 96 percent at 2 atm 1 mark -- extra 3 percent does not justify high pressure plant cost

In the final stage the sulfur trioxide is absorbed in concentrated sulfuric acid to form oleum, H₂S₂O₇, which is then diluted with water:

SO₃ + H₂SO₄ → H₂S₂O₇ then H₂S₂O₇ + H₂O → 2H₂SO₄

Explain why the sulfur trioxide is not simply added directly to water. [3]

Model Answer -- 5(c)

The direct reaction of sulfur trioxide with water is extremely, violently exothermic [1]

The heat released vaporises the product, forming a dense corrosive mist or fog of fine sulfuric acid droplets which is very difficult to condense and would escape from the plant, wasting product and polluting the air [1]

Absorbing the SO₃ in concentrated sulfuric acid instead gives oleum in a controlled, manageable reaction; the oleum can then be diluted with water safely to give twice as much sulfuric acid [1]

⚠ If you missed marks here: "Because SO₃ does not dissolve in water" is factually wrong and scores zero — it reacts far too readily, not too little. The mark is for the mist: without naming the fine acid fog that escapes uncondensed, the answer is incomplete however well the exothermic point is made.

1 mark -- reaction of so3 with water is violently exothermic 1 mark -- forms dense corrosive mist of acid droplets that escapes 1 mark -- oleum formed then diluted safely to give sulfuric acid

Question 6 -- Advanced Redox

Total: 12 marks

A student in Melbourne is working through a set of oxidation number problems taken from a university foundation course.

(a) [4]

Deduce the oxidation number of the element in bold in each species. Show your reasoning.

(i) S₂O₃²⁻ (ii) MnO₄⁻ (iii) Cr₂O₇²⁻ (iv) H₂O₂ [4]

Model Answer -- 6(a)

S₂O₃²⁻: 2x + 3(−2) = −2, so 2x = +4 and x = +2 [1]

MnO₄⁻: x + 4(−2) = −1, so x = +7 [1]

Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = +12 and x = +6 [1]

H₂O₂: 2(+1) + 2x = 0, so x = −1 (peroxide, the exception to the usual −2 for oxygen) [1]

⚠ If you missed marks here: Two errors dominate. First, forgetting to divide by the number of atoms: in Cr₂O₇²⁻ the total is +12, so quoting +12 instead of +6 per chromium is wrong. Second, assuming oxygen is always −2, which for H₂O₂ would force hydrogen to be +2 — impossible, since hydrogen has only one electron to lose.

1 mark -- sulfur plus 2 in thiosulfate 1 mark -- manganese plus 7 in manganate seven ion 1 mark -- chromium plus 6 in dichromate ion 1 mark -- oxygen minus 1 in hydrogen peroxide

(b) [4]

Chlorine reacts with cold dilute sodium hydroxide solution:

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

Use oxidation numbers to show that this is a disproportionation reaction. [4]

Model Answer -- 6(b)

In Cl₂ the oxidation number of chlorine is 0, because it is an uncombined element [1]

In NaCl the chlorine is −1; the oxidation number has decreased, so this chlorine has been reduced (it has gained an electron) [1]

In NaClO the chlorine is +1 (since Na is +1 and O is −2); the oxidation number has increased, so this chlorine has been oxidised [1]

The same element, chlorine, is simultaneously oxidised and reduced in one reaction, which is the definition of disproportionation [1]

⚠ If you missed marks here: The frequent slip is assigning chlorine −1 in NaClO out of habit, which would make the total charge on ClO −3 and leaves you unable to explain any oxidation at all. Remember: oxygen takes −2 first, so chlorine is forced to +1. Stating "chlorine is oxidised and oxygen is reduced" is a reasoning error — oxygen stays at −2 throughout.

1 mark -- chlorine is zero in cl2 uncombined element 1 mark -- chlorine falls to minus 1 in nacl reduced 1 mark -- chlorine rises to plus 1 in naclo oxidised 1 mark -- same element oxidised and reduced so disproportionation

Aqueous iodine reacts with thiosulfate ions. The thiosulfate ion S₂O₃²⁻ is converted into the tetrathionate ion S₄O₆²⁻, and the iodine is converted into iodide ions.

(i) Write the oxidation half equation, the reduction half equation, and combine them into the overall ionic equation. [3]

(ii) State which species is the oxidising agent, and justify your answer using the oxidation number of sulfur. [1]

Model Answer -- 6(c)

Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ [1]

Reduction: I₂ + 2e⁻ → 2I⁻ [1]

The electrons already balance at 2 each, so adding gives I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ [1]

Iodine is the oxidising agent: sulfur rises from +2 in S₂O₃²⁻ to +2.5 in S₄O₆²⁻ (4x − 12 = −2, so x = +2.5), so the thiosulfate is oxidised and the iodine, which is itself reduced, must be the oxidising agent [1]

⚠ If you missed marks here: Writing S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ without the coefficient 2 leaves sulfur unbalanced (2 on the left, 4 on the right) and destroys the overall equation. The other trap is rejecting +2.5 as "impossible": fractional average oxidation numbers are perfectly legitimate when the atoms of one element are not all in identical environments. Finally, the species that is reduced is the oxidising agent — naming thiosulfate here is the standard inversion.

1 mark -- oxidation half equation 2s2o3 gives s4o6 plus two electrons 1 mark -- reduction half equation iodine plus two electrons gives two iodide 1 mark -- overall equation iodine plus 2s2o3 gives 2 iodide plus s4o6 1 mark -- iodine is oxidising agent sulfur rises plus 2 to plus 2.5

Question 7 -- Redox Tests, Photochemistry and Change

Total: 10 marks

A technician in Accra finds two unlabelled bottles of colourless solution, X and Y, and tests each one with acidified potassium manganate(VII) solution and with potassium iodide solution.

Solution	With acidified KMnO₄	With KI solution
X	purple → colourless	no visible change
Y	stays purple	colourless → brown

(a) [4]

Deduce what solutions X and Y must each be, in terms of redox behaviour, and explain fully how each observation supports your deduction. [4]

Model Answer -- 7(a)

X must be a reducing agent [1]

X reduces the purple manganate(VII) ion to the almost colourless Mn²⁺ ion, manganese falling from +7 to +2; X has no effect on iodide ions because iodide is itself already a reducing agent [1]

Y must be an oxidising agent [1]

Y oxidises colourless iodide ions to iodine, which gives the brown colour; it cannot oxidise manganate(VII) any further because manganese is already at its maximum oxidation number of +7, so the purple colour remains [1]

⚠ If you missed marks here: Swapping the two labels is the classic inversion — remember that decolourising manganate(VII) means something donated electrons to it, so that reagent is the reducing agent. Saying "X bleached the solution" describes rather than explains and earns nothing; you must name the manganese change +7 to +2 and the iodide-to-iodine change.

1 mark -- solution x is a reducing agent 1 mark -- x reduces purple manganate seven to colourless manganese two plus 1 mark -- solution y is an oxidising agent 1 mark -- y oxidises colourless iodide ions to brown iodine

(b) [3]

(i) Write the equation for photosynthesis and state the energy change and the role of chlorophyll. [2]

(ii) Write the equation for the photochemical decomposition of silver bromide in traditional photographic film. [1]

Model Answer -- 7(b)

6CO₂ + 6H₂O → C₆H₂₂O₆ + 6O₂ [1]

The reaction is endothermic: light energy is absorbed and converted into chemical energy stored in the glucose. Chlorophyll is the green pigment that absorbs the light; it is a catalyst and is not used up [1]

2AgBr → 2Ag + Br₂, brought about by light; the silver formed is the dark grey deposit that darkens the exposed film [1]

⚠ If you missed marks here: Writing CO₂ + H₂O → C₆H₂₂O₆ + O₂ without the coefficient 6 is unbalanced and scores nothing. Describing photosynthesis as exothermic is a straight inversion — energy is taken in from sunlight. For the film, AgBr → Ag + Br is also wrong: bromine must be written as the diatomic molecule Br₂, which forces the coefficient 2 on both silver species.

1 mark -- 6co2 plus 6h2o gives c6h12o6 plus 6o2 1 mark -- endothermic light absorbed by chlorophyll stored as chemical energy 1 mark -- 2agbr decomposes to 2ag plus br2 in light

Grey-black iodine crystals warmed gently in a sealed tube form a purple vapour, and grey-black crystals re-form on the cool upper walls. In a separate tube, white silver bromide left on a windowsill slowly turns grey.

Classify each change as physical or chemical and justify each answer fully. [3]

Model Answer -- 7(c)

The sublimation and re-forming of iodine is a physical change [1]

No new substance is made: the purple vapour and the grey-black solid are both iodine, only the intermolecular forces between I₂ molecules are overcome, no bonds within the molecules are broken, and the original iodine is fully recovered on cooling, so the change is easily reversed [1]

The darkening of silver bromide is a chemical change: new substances, silver and bromine, are formed, the change is not reversed simply by removing the light, and the grey solid has completely different properties from white AgBr [1]

⚠ If you missed marks here: Using colour change as the test is the trap deliberately set here — iodine changes colour dramatically yet is only a physical change, so "it changed colour therefore it is chemical" gives the wrong classification for the iodine. The reliable criteria are whether a new substance is formed and whether the change is easily reversible.

1 mark -- iodine subliming and re-forming is a physical change 1 mark -- no new substance only intermolecular forces overcome easily reversed 1 mark -- silver bromide darkening chemical new silver and bromine formed

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