← Topic 6 Exams

IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 6: Chemical Reactions -- Mock Exam 1
1 hour 15 minutes
80
7
75:00
0620

Instructions

Question Navigation

Question 1 -- Physical and Chemical Changes
Total: 12 marks
A science club in Chennai runs a stall at a school open day. At the stall the students melt candle wax, burn a candle wick, dissolve sugar in water, rust an iron nail in damp air and heat a strip of magnesium ribbon. Visitors are asked to decide which of the changes are physical and which are chemical.
(a) [4]
(i) State two features that are always shown by a chemical change but are not shown by a physical change. [2]

(ii) State two features of a physical change, referring to the substances present and to the mass of material. [2]
Model Answer -- 1(a)
(i) One or more new substances are made / new chemical bonds are formed and broken, so the products have different chemical properties from the reactants [1]
(i) The change is usually difficult to reverse, and there is an energy change (heat given out or taken in) that is large compared with a physical change [1]
(ii) No new substance is formed — the same particles are present before and after, only their arrangement, spacing or separation changes [1]
(ii) The change is easily reversed and the total mass of substance is unchanged (mass is conserved in both, but in a physical change the substance itself can be recovered unchanged) [1]
⚠ If you missed marks here: The commonest error is writing "mass changes in a chemical reaction". Mass is conserved in every reaction — a burning candle only appears to lose mass because CO₂ and H₂O escape as gases. Another loss is writing "chemical changes are irreversible" with no qualification; many are reversible (see Question 4), so say "usually difficult to reverse".
(b) [5]
The table shows the five changes carried out at the stall.
ChangeObservation at the stall
A — melting candle waxsolid wax becomes a clear liquid; it sets again on cooling
B — burning the candle wickflame, wax used up, water vapour and carbon dioxide made
C — dissolving sugar in watercolourless solution; sugar recovered by evaporating the water
D — iron nail left in damp airorange–brown solid forms on the surface
E — heating magnesium ribbonbright white flame, white powder left behind
Classify each of A to E as a physical change or a chemical change and give one reason for each classification. [5]
Model Answer -- 1(b)
A: physical — melting is a change of state only; the wax molecules are unchanged and the liquid wax solidifies back to the same wax on cooling [1]
B: chemical — the wax (a hydrocarbon) reacts with oxygen to make new substances, carbon dioxide and water; heat and light are released and the wax cannot be got back [1]
C: physical — dissolving separates the sugar molecules among the water molecules; no new substance is made and the sugar is recovered unchanged by evaporation [1]
D: chemical — iron reacts with oxygen and water to form a new orange–brown substance, hydrated iron(III) oxide (rust), with different properties from iron [1]
E: chemical — magnesium combines with oxygen to give white magnesium oxide, 2Mg + O₂ → 2MgO, releasing a lot of energy as heat and light [1]
⚠ If you missed marks here: If you called C (dissolving sugar) a chemical change because "the sugar disappears", you confused disappearing with reacting — the sugar is still sugar and is recovered by evaporation. If you gave "a colour change happened" as the reason for D, that reason on its own scores nothing; you must name the new substance (rust / hydrated iron(III) oxide).
(c) [3]
A visitor says: "The wax gave out heat when it solidified, so solidifying wax must be a chemical change." Explain why an energy change on its own does not prove that a chemical reaction has taken place. Use one physical example and one chemical example in your answer. [3]
Model Answer -- 1(c)
Physical changes also involve energy transfers, because energy is needed to overcome or to form the forces of attraction between particles as they change state [1]
Physical example: wax freezing (or steam condensing) gives out heat as the particles come closer and attractions form, yet the substance is still wax — no new substance is made [1]
The test for a chemical change is that a new substance with different properties is formed, as when the wick burns and produces carbon dioxide and water that cannot be turned back into wax by cooling [1]
⚠ If you missed marks here: Answers that simply say "an energy change can happen in both" without an example score only one mark. The examiner wants you to name the physical process that releases heat (freezing or condensing) and to state the real criterion — formation of a new substance. Saying "the wax got hot so a reaction happened" reverses the logic completely.
Question 2 -- Rate of Reaction: Surface Area and Concentration
Total: 12 marks
A class in Manchester investigates the reaction between marble chips (calcium carbonate) and dilute hydrochloric acid. The carbon dioxide produced is collected in a gas syringe and its volume is recorded every 20 seconds.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
(a) [3]
The diagram shows the apparatus used.
marble chips + dilute HCl gas syringe collecting CO₂ delivery tube
(i) State the name of one further piece of apparatus, not shown above, that is essential for obtaining the results, and say what it is used for. [1]

(ii) Give two reasons why a gas syringe gives more reliable results here than collecting the gas over water in a measuring cylinder. [2]
Model Answer -- 2(a)
(i) A stopwatch or stopclock, started the instant the acid and marble are mixed, to measure the time at which each syringe reading is taken [1]
(ii) Carbon dioxide is appreciably soluble in water, so gas collected over water gives a reading that is too low; the syringe traps all the gas made [1]
(ii) The syringe is a closed system with a clear scale read directly in cm³, so no gas escapes and there is no parallax error from reading a water level in an inverted cylinder [1]
⚠ If you missed marks here: "A balance" is not the answer to (i) — a balance measures mass loss, which is a different method, and the question specifies volume readings from a syringe. In (ii), "it is more accurate" scores nothing on its own; the marking point is the solubility of CO₂ in water and the loss of gas while the bung is fitted.
(b) [6]
Three experiments are carried out. In every experiment the same mass of calcium carbonate and the same volume of acid are used.
ExperimentForm of CaCO₃Acid concentration / mol dm⁻³Time for the first 40 cm³ of CO₂ / s
1large chips1.0100
2small chips1.050
3large chips2.052
(i) Explain, in terms of collision theory, why Experiment 2 is faster than Experiment 1. [3]

(ii) Explain, in terms of collision theory, why Experiment 3 is faster than Experiment 1. [3]
Model Answer -- 2(b)
(i) Breaking the same mass into smaller chips gives a larger total surface area, so more calcium carbonate particles are exposed at the surface where the acid can reach them [1]
(i) The frequency of collisions between H⁺ ions in the acid and the exposed carbonate surface therefore increases [1]
(i) More successful collisions occur per second (the energy of each collision is unchanged), so the rate of reaction is greater and the 40 cm³ is collected in half the time [1]
(ii) Doubling the concentration means there are more acid particles in the same volume of solution, so the solution is more crowded [1]
(ii) The acid particles collide with the marble surface more often per second [1]
(ii) The proportion of collisions with energy above the activation energy is unchanged, but because there are more collisions per second there are more successful collisions per second, so the rate increases [1]
⚠ If you missed marks here: If you wrote that smaller chips or a stronger acid make the particles "move faster" or "have more energy", that is the temperature explanation applied to the wrong variable and it earns no credit. Only temperature changes particle energy. Surface area and concentration change how often particles meet, not how hard they hit. Also avoid "more collisions" without "per second" — rate is about collision frequency.
(c) [3]
Describe three things the class must do so that the comparison between Experiment 1 and Experiment 2 is a fair test. For each, state briefly what would go wrong if it were not done. [3]
Model Answer -- 2(c)
Keep the temperature constant, for example by using a water bath and letting the acid reach room temperature first; a warmer run would be faster for a reason unconnected with surface area [1]
Use the same mass of calcium carbonate and the same volume and concentration of acid each time, with the acid in excess so that the carbonate is the limiting reactant; different amounts would change both the rate and the final volume of gas [1]
Fit the bung and start the stopwatch at the same moment in each run, and use the same syringe and flask; a delay in fitting the bung lets CO₂ escape and makes the measured rate too low [1]
⚠ If you missed marks here: A very common loss is writing "keep the surface area the same" — surface area is the independent variable here, so it is the one thing that must change. Another is naming a control without the consequence; the question asks for both. "Do it carefully" and "repeat the experiment" are not control variables.
Question 3 -- Interpreting Rate Graphs
Total: 12 marks
A student in Nairobi repeats the marble chip experiment three times and plots the volume of carbon dioxide collected against time. Curve P uses 2.0 g of small chips with 50 cm³ of 1.0 mol dm⁻³ acid, curve Q uses 2.0 g of large chips with 50 cm³ of 1.0 mol dm⁻³ acid, and curve R uses 1.0 g of small chips with 50 cm³ of 1.0 mol dm⁻³ acid. In every experiment the acid is in excess.
0 20 40 60 80 100 40 80 120 160 time / s volume of CO₂ / cm³ P Q R
(a) [4]
(i) State which curve shows the fastest reaction and justify your choice by referring to the graph. [2]

(ii) Explain, in terms of particles, why that experiment is the fastest even though the mass of calcium carbonate in P and Q is the same. [2]
Model Answer -- 3(a)
(i) Curve P is the fastest [1]
(i) P has the steepest initial gradient and reaches its plateau (100 cm³) first, at about 100 s, whereas Q needs roughly 160 s to reach the same volume [1]
(ii) P uses small chips, so the same mass of calcium carbonate presents a much larger total surface area to the acid [1]
(ii) More carbonate particles are exposed, so acid particles collide with the solid surface more frequently, giving more successful collisions per second [1]
⚠ If you missed marks here: A frequent error is choosing R because its curve levels off earliest. R stops early because it runs out of calcium carbonate (only 1.0 g), not because it is fast — finishing first is not the same as being fastest when the amounts differ. Judge speed from the steepness of the curve near t = 0, not from where the line flattens.
(b) [5]
(i) Explain why curve Q is steep at the start, becomes less steep, and finally becomes horizontal. [3]

(ii) Explain why P and Q level off at the same volume of gas but R levels off at a lower volume. [2]
Model Answer -- 3(b)
(i) At the start the concentration of acid and the exposed surface of carbonate are at their greatest, so collisions are most frequent and the rate (gradient) is highest [1]
(i) As the reaction proceeds the reactants are used up, the acid becomes more dilute and the chips get smaller, so collisions are less frequent and the gradient falls [1]
(i) The line becomes horizontal when the limiting reactant, the calcium carbonate, has been completely used up, so no more gas can be produced and the rate is zero [1]
(ii) P and Q start with the same mass (2.0 g) of calcium carbonate with acid in excess, so the same number of moles of CaCO₃ reacts and the same number of moles of CO₂ is made — surface area changes the speed but not the yield [1]
(ii) R contains only half the mass of calcium carbonate, so only half as many moles of CO₂ are produced and its plateau is at about half the volume (about 50 cm³) [1]
⚠ If you missed marks here: Writing "the reaction stops because the acid is used up" is wrong here — the stem tells you the acid is in excess, so the carbonate is the limiting reactant. Also avoid "the reaction has reached equilibrium"; this reaction goes to completion because the CO₂ escapes into the syringe. If you said R is lower "because it is slower", note that R uses small chips like P and is not slower — it simply has less carbonate.
(c) [3]
For curve P, 60 cm³ of carbon dioxide has been collected after 40 s and the reaction is complete at 100 cm³.

(i) Calculate the average rate of reaction over the first 40 s, in cm³/s, showing your working. [2]

(ii) State how the instantaneous rate at t = 80 s compares with your answer to (i), and explain your reasoning. [1]
Model Answer -- 3(c)
(i) Average rate = change in volume ÷ change in time = 60 cm³ ÷ 40 s [1]
(i) = 1.5 cm³/s (accept 1.5 cm³ s⁻¹), with the unit given [1]
(ii) The rate at 80 s is much lower than 1.5 cm³/s (close to zero), because by then most of the calcium carbonate has reacted and the curve is almost horizontal, so the gradient of the tangent is very small [1]
⚠ If you missed marks here: Dividing time by volume gives 0.67 s/cm³, which is the reciprocal and scores zero — rate is always amount per unit time. Dropping the unit also loses a mark. In (ii), saying "the rate stays the same because the average is constant" confuses the average over an interval with the instantaneous gradient at a point.
Question 4 -- Reversible Reactions
Total: 12 marks
A laboratory technician in Delhi checks whether some solvents delivered to the school are dry. She has two reagents on the shelf: blue hydrated copper(II) sulfate crystals and strips of paper soaked in pink cobalt(II) chloride solution and then dried in an oven.
(a) [5]
(i) Write an equation, using the ⇌ sign and including state symbols, for the reversible change that occurs when blue hydrated copper(II) sulfate is heated. State the colour change in both directions. [3]

(ii) Write an equation, using the ⇌ sign, for the reversible change of hydrated cobalt(II) chloride, CoCl₂.6H₂O, and state the colour change on heating. [2]
Model Answer -- 4(a)
(i) CuSO₄.5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(l) — correct formulae and balancing with 5H₂O [1]
(i) Correct use of the ⇌ sign and correct state symbols [1]
(i) On heating, blue hydrated crystals turn white (anhydrous); on adding water the white solid turns blue again and heat is released [1]
(ii) CoCl₂.6H₂O(s) ⇌ CoCl₂(s) + 6H₂O(l) [1]
(ii) On heating, the pink hydrated solid turns blue; adding water turns the blue anhydrous solid pink again [1]
⚠ If you missed marks here: Writing a single forward arrow instead of ⇌ loses the reversibility mark even if everything else is right. Swapping the cobalt colours (saying anhydrous is pink) is a classic slip — remember hydrated = pink because water is "wet and rosy", anhydrous = blue. Writing "CuSO₄.5H₂O(aq)" is also wrong; these are solids.
(b) [4]
(i) Describe how the technician would use anhydrous copper(II) sulfate to test a colourless liquid for the presence of water, giving the expected positive result. [2]

(ii) The technician tests a sample of ethanol. The paper turns pink, but the ethanol later boils at 78 °C, its correct boiling point. Explain what these two results together tell her about the sample, and describe how she could show that the sample was pure water rather than ethanol. [2]
Model Answer -- 4(b)
(i) Add a few drops of the liquid to a small amount of white anhydrous copper(II) sulfate in a dry test tube (or spotting tile) [1]
(i) If water is present the solid turns from white to blue and warms up; this shows water is present but does not show it is pure water [1]
(ii) The pink cobalt chloride paper shows that water is present in the sample, but the boiling point of 78 °C shows the bulk liquid is ethanol, so the ethanol is contaminated with a little water (it is not anhydrous) [1]
(ii) To show a liquid is pure water, measure its boiling point (must be exactly 100 °C at atmospheric pressure) and its freezing point (0 °C); a sharp boiling point at 100 °C confirms pure water [1]
⚠ If you missed marks here: The single most penalised answer is "the blue colour proves the liquid is pure water". These reagents test for the presence of water only; a solution of salt in water, or damp ethanol, gives exactly the same positive result. Purity requires a physical measurement — a sharp boiling point of 100 °C or freezing point of 0 °C.
(c) [3]
Hydrated copper(II) sulfate is heated in an open evaporating basin and all the crystals turn white. When the same reaction is carried out in a sealed tube, a mixture of blue and white solid remains no matter how long the tube is heated.

Explain this difference. In your answer state what is meant by a closed system and describe what is happening at dynamic equilibrium. [3]
Model Answer -- 4(c)
In the open basin the water vapour escapes, so the reverse reaction cannot happen and the forward dehydration goes to completion, leaving only white anhydrous solid [1]
A closed system is one in which no reactants or products can enter or leave, so in the sealed tube the water vapour is retained and can re-hydrate the solid; only in a closed system can equilibrium be reached [1]
At dynamic equilibrium the forward and reverse reactions are still occurring, but at equal rates, so the amounts (concentrations) of blue and white solid stay constant — the reaction has not stopped [1]
⚠ If you missed marks here: If you wrote that at equilibrium "the reaction has stopped" or "there are equal amounts of each substance", you have lost the meaning of dynamic equilibrium. Both reactions continue at equal rates and the amounts are constant, not equal. Also, a closed system is not the same as a sealed system at constant temperature only — the key point is that nothing enters or leaves.
Question 5 -- The Haber Process
Total: 10 marks
An ammonia plant on Jurong Island, Singapore, manufactures ammonia for fertiliser export. The graph shows how the percentage yield of ammonia at equilibrium depends on pressure at two different temperatures.
0 20 40 60 80 100 200 300 400 pressure / atm yield of ammonia / % 350 °C 550 °C
(a) [4]
(i) Write the balanced equation for the manufacture of ammonia, using the ⇌ sign. State the source of each raw material. [2]

(ii) State the temperature, the pressure and the catalyst used in a typical modern Haber process plant. [2]
Model Answer -- 5(a)
(i) N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — balanced with the reversible arrow [1]
(i) Nitrogen is obtained from the air (by fractional distillation of liquid air); hydrogen is obtained from natural gas (methane) reacted with steam, or from cracking hydrocarbons [1]
(ii) Temperature about 450 °C and pressure about 200 atm [1]
(ii) The catalyst is iron [1]
⚠ If you missed marks here: Writing "N + 3H → NH₃" loses two marks at once: nitrogen and hydrogen are diatomic molecules and the arrow must be reversible. Naming the catalyst as "platinum" or "vanadium(V) oxide" confuses the Haber process with the catalytic converter and the Contact process respectively — the Haber catalyst is iron.
(b) [4]
The forward reaction is exothermic.

(i) Use the graph to state the effect of increasing pressure on the yield of ammonia, and explain why a pressure much higher than 200 atm is not used. [2]

(ii) Explain why 450 °C is used even though the graph shows that a lower temperature would give a higher yield. [2]
Model Answer -- 5(b)
(i) Increasing the pressure increases the yield, because there are 4 molecules of gas on the left and only 2 on the right, so the equilibrium shifts to the side with fewer gas molecules [1]
(i) Very high pressures are not used because the plant would need extremely thick, expensive pipes and vessels and large amounts of energy to run the compressors, which is unsafe and uneconomic [1]
(ii) The forward reaction is exothermic, so a lower temperature moves the equilibrium to the right and gives a higher equilibrium yield, but the reaction becomes far too slow to be useful [1]
(ii) 450 °C is a compromise: it gives a somewhat lower yield but a fast enough rate, so a reasonable amount of ammonia is made in an acceptable time (and unreacted gases are recycled) [1]
⚠ If you missed marks here: If you said high pressure shifts the equilibrium left, you applied the rule backwards — count the gas molecules: 1 + 3 = 4 on the left against 2 on the right, so pressure favours ammonia. In (ii), "450 °C gives the best yield" is wrong; it gives a worse yield than a cool reactor. The mark is for recognising the yield–rate trade-off, not for claiming the yield is optimal.
(c) [2]
State what effect the iron catalyst has on the reaction, and state one thing it does not change. Explain your answer in terms of activation energy and the two rates. [2]
Model Answer -- 5(c)
The iron provides an alternative reaction pathway with a lower activation energy, so a greater proportion of collisions is successful and equilibrium is reached more quickly [1]
It does not change the position of equilibrium or the yield of ammonia, because it speeds up the forward and reverse reactions equally; it is also not used up in the process [1]
⚠ If you missed marks here: "The catalyst increases the yield of ammonia" is the classic wrong answer and scores zero — it only shortens the time taken to reach the same equilibrium position. Also avoid "the catalyst lowers the energy of the reactants"; it lowers the activation energy by offering a different pathway, leaving the energies of reactants and products unchanged.
Question 6 -- Redox
Total: 12 marks
A group of students in Leeds studies redox reactions using the extraction of iron in a blast furnace, the displacement of copper by zinc, and the reaction of chlorine with iron(II) ions.
(a) [4]
(i) Define oxidation and reduction in terms of oxygen. [2]

(ii) Define oxidation and reduction in terms of electrons, and state how each is linked to the change in oxidation number. [2]
Model Answer -- 6(a)
(i) Oxidation is the gain of oxygen by a substance [1]
(i) Reduction is the loss of oxygen from a substance [1]
(ii) Oxidation is the loss of electrons and reduction is the gain of electrons (OIL RIG) [1]
(ii) Oxidation is an increase in oxidation number and reduction is a decrease in oxidation number; the two always happen together in a redox reaction [1]
⚠ If you missed marks here: The most frequent slip is reversing the electron definitions — oxidation is LOSS of electrons even though it sounds like something is being added. If you also wrote "oxidation is a decrease in oxidation number" you have compounded the same error; losing negative electrons makes the number go up, for example Fe going to Fe²⁺ is 0 to +2.
(b) [4]
(i) Deduce the oxidation number of the underlined element in each of the following: Fe in Fe₂O₃, Mn in MnO₄⁻, S in H₂SO₄. [3]

(ii) Explain what the Roman numeral tells you in the name copper(II) sulfate, and give the formula of the copper ion present. [1]
Model Answer -- 6(b)
(i) Fe in Fe₂O₃ is +3, since 3 oxygens at −2 give −6 and the compound is neutral, so 2Fe = +6 [1]
(i) Mn in MnO₄⁻ is +7, since 4 oxygens at −2 give −8 and the overall charge is −1, so Mn = −1 + 8 = +7 [1]
(i) S in H₂SO₄ is +6, since 2H at +1 give +2 and 4 oxygens at −2 give −8, so S = +6 for a neutral molecule [1]
(ii) The Roman numeral gives the oxidation number (charge) of the metal ion in the compound; copper(II) sulfate contains Cu²⁺ [1]
⚠ If you missed marks here: Giving Fe as +6 in Fe₂O₃ means you found the total for both iron atoms and forgot to divide by 2. For MnO₄⁻ the commonest wrong answer is +8, which comes from ignoring the −1 charge on the ion. Always write out the sum: (oxidation numbers) × (number of atoms) = overall charge, then solve.
(c) [4]
Consider these two reactions.

Reaction 1: Fe₂O₃ + 3CO → 2Fe + 3CO₂
Reaction 2: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

(i) For Reaction 1, identify the oxidising agent and the reducing agent, and justify each choice. [2]

(ii) For Reaction 2, write the two ionic half equations and state which species is oxidised. [2]
Model Answer -- 6(c)
(i) The oxidising agent is Fe₂O₃, because it gives oxygen to the carbon monoxide (the iron is reduced from +3 to 0) [1]
(i) The reducing agent is CO, because it removes oxygen from the iron oxide and is itself oxidised to CO₂ (carbon goes from +2 to +4) [1]
(ii) Oxidation half equation Zn → Zn²⁺ + 2e⁻ and reduction half equation Cu²⁺ + 2e⁻ → Cu [1]
(ii) Zinc is oxidised (it loses electrons and its oxidation number rises from 0 to +2); it is therefore the reducing agent, while Cu²⁺ is the oxidising agent [1]
⚠ If you missed marks here: The classic reversal is calling CO the oxidising agent because it "contains oxygen". An oxidising agent is the species that is itself reduced — here that is Fe₂O₃. In the half equations, putting the electrons on the wrong side (Zn + 2e⁻ → Zn²⁺) makes the charges unbalanced and scores nothing; check that total charge is equal on both sides.
Question 7 -- Redox Tests and Photochemical Reactions
Total: 10 marks
A technician in a Mumbai college prepares two reagent bottles for a practical on redox: acidified potassium manganate(VII) solution and potassium iodide solution. In the same practical, students examine an old roll of black-and-white photographic film and a leaf that has been kept in the dark.
(a) [4]
(i) State the colour change seen when acidified potassium manganate(VII) is added to a reducing agent such as aqueous sulfur dioxide, and state what has happened to the manganese in terms of oxidation number. [2]

(ii) State the colour change seen when an oxidising agent such as aqueous chlorine is added to colourless potassium iodide solution, and name the substance responsible for the new colour. [2]
Model Answer -- 7(a)
(i) The solution changes from purple to colourless [1]
(i) The manganese is reduced: its oxidation number falls from +7 in MnO₄⁻ to +2 in Mn²⁺, so the manganate(VII) has acted as the oxidising agent [1]
(ii) The colourless solution turns brown (yellow–brown) [1]
(ii) The brown colour is iodine, I₂, formed as the iodide ions are oxidised: Cl₂ + 2I⁻ → 2Cl⁻ + I₂ [1]
⚠ If you missed marks here: Writing "purple to clear" is not accepted — clear means transparent, and the purple solution is already clear. The word required is colourless. In (ii), saying the brown colour is "iodide" loses the mark; iodide ions, I⁻, are colourless in solution, and it is the iodine molecules, I₂, that are brown.
(b) [3]
A student tests an unknown solution X. It turns acidified potassium manganate(VII) from purple to colourless, and it produces no colour change with potassium iodide solution.

Deduce whether X is an oxidising agent or a reducing agent, and explain your reasoning from both observations. Suggest one substance that X could be. [3]
Model Answer -- 7(b)
X is a reducing agent [1]
Decolourising the manganate(VII) shows X has reduced Mn from +7 to +2, so X itself has been oxidised (it donates electrons); the absence of any brown colour with potassium iodide shows X cannot oxidise iodide ions to iodine, confirming it is not an oxidising agent [1]
Suitable examples: aqueous sulfur dioxide (sulfite ions), potassium iodide, iron(II) sulfate solution or ethanedioic acid [1]
⚠ If you missed marks here: If you concluded X is an oxidising agent because "it changed the colour of the manganate(VII)", you have identified the wrong species as being changed — it is the manganate(VII) that is reduced, so X must be doing the reducing. The negative iodide result is not a wasted piece of information; it is the evidence that rules out the alternative.
(c) [3]
(i) State what is meant by a photochemical reaction, and write the equation for photosynthesis, naming the energy change involved. [2]

(ii) Explain, with reference to a half equation, why the reaction of silver bromide in photographic film is described as a photochemical redox reaction. [1]
Model Answer -- 7(c)
(i) A photochemical reaction is one in which light energy is absorbed and used to bring about a chemical change (light supplies the activation energy) [1]
(i) 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ in the presence of chlorophyll and light; the reaction is endothermic, converting light energy into chemical energy stored in glucose [1]
(ii) Light causes silver bromide to decompose to silver and bromine, 2AgBr → 2Ag + Br₂; the silver ions are reduced, Ag⁺ + e⁻ → Ag, while bromide ions are oxidised to bromine, so it is a redox reaction driven by light and the exposed grains turn dark grey/black [1]
⚠ If you missed marks here: Describing photosynthesis as exothermic reverses the energy flow — light energy is absorbed and stored, so it is endothermic. For the film, writing "the silver bromide turns black" without the half equation misses the redox point; you must show Ag⁺ gaining an electron to become silver metal. Also note the products are silver and bromine, not silver oxide.

Self-Assessment

Tick marks earned, then click Calculate Grade.

0
80
0%
IGCSE Chemistry 0620 -- Paper 4 Theory (Extended) -- Topic 6: Chemical Reactions -- Mock Exam 1
For examination preparation only. Not an official Cambridge Assessment paper.