October/November 2025 Paper 32 Worked Answers (IGCSE Chemistry 0620 Core)

57 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksElement used as a lubricant
   Step 1: Carbon exists in different forms, including graphite. Step 2: Graphite has layers that can slide over each other, so solid carbon (graphite) is used as a lubricant. Step 3: Chlorine, calcium and neon are not used as lubricants, so the symbol is C.
   Method:

   Recall which element has a slippery layered form used as a lubricant.
   Examiner tips

   • Graphite layers slide, which is why carbon is used as a lubricant.

2. Question 1(b)

   1 marksIon present in every dilute acid
   Step 1: An acid in water produces hydrogen ions, $\text{H}^+$. Step 2: Every dilute acid contains these hydrogen ions, which give the acid its acidic properties. Step 3: Oxide, chloride and sodium ions are not present in every acid, so the element is hydrogen, H.
   Method:

   Recall the ion that all acids have in common.
   Examiner tips

   • The $\text{H}^+$ ion is what makes a solution acidic.

3. Question 1(c)

   1 marksMetal that does not react with steam
   Step 1: Reactive metals such as potassium, sodium and calcium react with water or steam. Step 2: Silver is low in the reactivity series and does not react with steam. Step 3: So the metal that does not react with steam is silver, Ag.
   Method:

   Pick the least reactive metal, which will not react with steam.
   Examiner tips

   • Unreactive metals like silver do not react with water or steam.

4. Question 1(d)

   1 marksPeriod 3 element forming a 1- ion
   Step 1: Period 3 elements are in the third row of the Periodic Table. Step 2: Chlorine is in Period 3 and Group VII, so it gains one electron to form a 1- ion. Step 3: Sodium forms a 1+ ion, oxygen is in Period 2, and calcium forms a 2+ ion, so the element is chlorine, Cl.
   Method:

   Find the Period 3 element whose group means it gains one electron.
   Examiner tips

   • Group VII elements gain one electron to form a 1- ion; Period 3 is the third row.

5. Question 2(a)

   2 marksElectrons and neutrons in a krypton atom
   Step 1: In a neutral atom the number of electrons equals the proton number, so there are 36 electrons. Step 2: The number of neutrons = nucleon number - proton number = 84 - 36 = 48. Step 3: So one atom has 36 electrons and 48 neutrons.
   Method:

   Use the proton number for electrons and subtract it from the nucleon number for neutrons.
   Examiner tips

   • Neutron number = nucleon number minus proton number.

6. Question 2(b)

   2 marksEffect of temperature and pressure on gas volume
   Step 1: Cooling a gas makes its particles move more slowly and take up less space, so the volume decreases. Step 2: Increasing the pressure squeezes the gas into a smaller space, so the volume also decreases. Step 3: So both decreasing the temperature and increasing the pressure cause the volume to decrease.
   Method:

   Apply how cooling and increased pressure each change a gas volume.
   Examiner tips

   • Gas volume falls when it is cooled and when the pressure on it is increased.

7. Question 2(c)

   2 marksArrangement and motion of particles in a liquid
   Step 1: In a liquid the particles are still close together, usually touching, but they are not in a fixed pattern. Step 2: The particles have enough energy to slide over each other and move around in random directions. Step 3: So in liquid krypton the particles are close together and slide over each other.
   Method:

   Describe both the spacing and the motion of particles in a liquid.
   Examiner tips

   • Liquid = close together + sliding/moving randomly (not fixed like a solid).

8. Question 3(a)(i)

   1 marksNegative ion with the highest concentration
   Step 1: All six ions listed are negative ions, so compare their masses in the same 500 cm³ sample. Step 2: Nitrate has the largest mass at 8.5 mg. Step 3: Hydrogencarbonate (6.0 mg), chloride (4.2 mg) and iodide (0.4 mg) are all smaller, so the negative ion with the highest concentration is nitrate.
   Method:

   Compare the masses of the negative ions and select the largest.
   Examiner tips

   • Highest concentration in the same volume means the largest mass.

9. Question 3(a)(ii)

   1 marksFormula of the ammonium ion
   Step 1: The ammonium ion is formed from one nitrogen and four hydrogen atoms and carries a single positive charge. Step 2: Its formula is therefore $\text{NH}_4^+$. Step 3: $\text{NH}_3$ is ammonia (a molecule), and $\text{NO}_3^-$ is the nitrate ion, so the ammonium ion is $\text{NH}_4^+$.
   Method:

   Recall the make-up and charge of the ammonium ion.
   Examiner tips

   • Ammonium = $\text{NH}_4^+$ (positive); ammonia = $\text{NH}_3$ (neutral).

10. Question 3(a)(iii)

    2 marksTest for chloride ions
    Step 1: Halide ions are tested by acidifying with dilute nitric acid and adding aqueous silver nitrate. Step 2: With chloride ions a white precipitate of silver chloride forms. Step 3: A yellow precipitate indicates iodide ions, limewater tests for carbon dioxide and the warming with sodium hydroxide tests for ammonium ions, so chloride gives a white precipitate.
    Method:

    Recall the acidified silver nitrate halide test and match the colour to chloride.
    Examiner tips

    • Chloride → white, bromide → cream, iodide → yellow precipitate with silver nitrate.

11. Question 3(a)(iv)

    1 marksMass of ions in a smaller volume
    Step 1: The mass of ions is proportional to the volume of water. Step 2: Mass in 200 cm³ = 6.0 × (200 ÷ 500) = 6.0 × 0.40. Step 3: This gives 2.4 mg of hydrogencarbonate ions in 200 cm³.
    Method:

    Set up the proportion of mass to volume and solve for the smaller volume.
    Examiner tips

    • Mass is proportional to volume: multiply by (new volume ÷ original volume).

12. Question 3(b)

    1 marksProcess to make water suitable for practical chemistry
    Step 1: Water for practical chemistry must be pure, with dissolved substances removed. Step 2: Distillation boils the water and condenses the pure water vapour, leaving dissolved substances behind. Step 3: Chlorination only kills microbes, filtration only removes insoluble solids and crystallisation makes solid crystals, so the process is distillation.
    Method:

    Recall the method that produces pure water by boiling and condensing.
    Examiner tips

    • Distillation removes dissolved substances to give pure water.

13. Question 3(c)(i)

    1 marksUse of carbon in water treatment
    Step 1: Carbon (often as activated charcoal) absorbs dissolved substances that cause unpleasant tastes and smells. Step 2: Passing water through carbon therefore removes bad tastes and odours. Step 3: Killing microbes is done by chlorination and removing large solids is done by filtration, so carbon is used to remove tastes and odours.
    Method:

    Recall the specific job carbon does in water treatment.
    Examiner tips

    • Carbon = removes tastes and odours; chlorine = kills microbes.

14. Question 3(c)(ii)

    1 marksReason for chlorinating water
    Step 1: Untreated water can contain harmful microbes such as bacteria. Step 2: Chlorine kills these microbes, making the water safe to drink. Step 3: Removing tastes is done by carbon and removing solids is done by filtration, so chlorination is used to kill microbes.
    Method:

    Recall what adding chlorine achieves in water treatment.
    Examiner tips

    • Chlorination kills bacteria and other microbes in water.

15. Question 3(d)(i)

    1 marksSeparation method for coloured substances
    Step 1: Separating coloured substances by letting a solvent carry them up a piece of paper is paper chromatography. Step 2: The substances travel different distances, so they separate into spots. Step 3: Filtration, crystallisation and distillation do not separate dissolved colours up paper, so the method is paper chromatography.
    Method:

    Recognise the technique that separates colours as they move up paper.
    Examiner tips

    • Coloured substances separating up paper = chromatography.

16. Question 3(d)(ii)

    1 marksThe solvent front on a chromatogram
    Step 1: As the solvent soaks up the paper it reaches a highest level before the paper is removed. Step 2: This highest level reached by the solvent is called the solvent front. Step 3: The base line is where the mixture was placed, and the spots and the solvent surface in the beaker are different, so the solvent front is the highest level reached by the solvent.
    Method:

    Identify the line marking how far the solvent travelled up the paper.
    Examiner tips

    • Solvent front = the highest point the solvent reaches on the paper.

17. Question 4(a)

    1 marksMolecular formula from a displayed formula
    Step 1: The molecular formula shows the number of each type of atom in one molecule. Step 2: Compound A has 10 carbon, 18 hydrogen and 1 oxygen atom. Step 3: Writing these counts gives $\text{C}_{10}\text{H}_{18}\text{O}$.
    Method:

    Count each type of atom in the structure and write the molecular formula.
    Examiner tips

    • Molecular formula = total count of each atom: C then H then O.

18. Question 4(b)

    1 marksPart of a molecule that decolourises bromine
    Step 1: Aqueous bromine is decolourised by a carbon-carbon double bond, which is the feature of an unsaturated molecule. Step 2: Bromine adds across the C=C double bond, so the orange colour disappears. Step 3: Single C-H and C-C bonds and the oxygen atom do not decolourise bromine, so the responsible part is the C=C double bond.
    Method:

    Recall the bond that reacts with bromine to remove its colour.
    Examiner tips

    • Bromine adds across C=C double bonds, so the colour disappears.

19. Question 4(c)

    2 marksRelative molecular mass calculation
    Step 1: Multiply each atom count by its relative atomic mass: carbon 10 × 12 = 120, hydrogen 16 × 1 = 16, oxygen 2 × 16 = 32. Step 2: Add the totals: 120 + 16 + 32 = 168. Step 3: So the relative molecular mass of $\text{C}_{10}\text{H}_{16}\text{O}_2$ is 168.
    Method:

    Multiply each atom count by its mass and add the results.
    Examiner tips

    • Relative molecular mass = sum of (number of atoms × relative atomic mass).

20. Question 4(d)

    1 marksNaming a fossil fuel
    Step 1: Fossil fuels are formed from the remains of living things over millions of years. Step 2: Coal (and natural gas) are fossil fuels, like petroleum. Step 3: Hydrogen and ethanol are not fossil fuels and uranium is a nuclear fuel, so coal is the other fossil fuel.
    Method:

    Recall which fuels are fossil fuels and choose one other than petroleum.
    Examiner tips

    • Remember the three fossil fuels: coal, oil (petroleum) and natural gas.

21. Question 4(e)(i)

    1 marksFraction with the longest chain length
    Step 1: Fractions collected lower in the column contain larger molecules with longer carbon chains. Step 2: Lubricating oil is collected near the bottom, so it has the longest chains. Step 3: Gasoline, naphtha and kerosene are collected higher up and have shorter chains, so the longest chains are in lubricating oil.
    Method:

    Pick the fraction collected lowest, which has the longest chains.
    Examiner tips

    • Bottom of the column = longest chains, highest boiling point.

22. Question 4(e)(ii)

    1 marksFraction with the highest volatility
    Step 1: Volatility describes how easily a substance evaporates; small molecules with low boiling points are the most volatile. Step 2: Gasoline is collected near the top of the column and has the smallest molecules, so it is the most volatile. Step 3: Lubricating oil, fuel oil and diesel oil have larger molecules and lower volatility, so the most volatile fraction is gasoline.
    Method:

    Pick the fraction with the smallest molecules and lowest boiling point.
    Examiner tips

    • Smaller molecules = more volatile = collected near the top of the column.

23. Question 4(e)(iii)

    1 marksFraction used for home heating
    Step 1: Fuel oil is a heavy fraction burned to provide heat. Step 2: It is used in home heating systems and to fuel large furnaces and ships. Step 3: Gasoline fuels cars, naphtha is a chemical feedstock and lubricating oil is used for lubrication, so the home-heating fraction is fuel oil.
    Method:

    Match the heavy heating fuel to the correct fraction.
    Examiner tips

    • Fuel oil = heating and large furnaces/ships.

24. Question 4(f)(i)

    1 marksIdentifying an unsaturated hydrocarbon
    Step 1: Unsaturated hydrocarbons are alkenes, which contain a carbon-carbon double bond and whose names end in -ene. Step 2: Nonene ends in -ene, so it is the alkene and is unsaturated. Step 3: Undecane, dodecane and tridecane all end in -ane and are saturated alkanes, so the unsaturated hydrocarbon is nonene.
    Method:

    Pick the hydrocarbon whose name ends in -ene.
    Examiner tips

    • -ene ending = alkene = unsaturated; -ane ending = alkane = saturated.

25. Question 4(f)(ii)

    1 marksExplaining why a hydrocarbon is unsaturated
    Step 1: Alkenes are unsaturated and have names ending in -ene and the general formula $\text{C}_n\text{H}_{2n}$. Step 2: Nonene ends in -ene and fits $\text{C}_n\text{H}_{2n}$, so it is an alkene and unsaturated. Step 3: Ending in -ane or having the largest molecules or highest boiling point does not show unsaturation, so the reason is the -ene name and the $\text{C}_n\text{H}_{2n}$ formula.
    Method:

    Use the alkene naming and general formula to justify unsaturation.
    Examiner tips

    • Alkene test: -ene name and $\text{C}_n\text{H}_{2n}$ formula.

26. Question 4(g)(i)

    1 marksOther product of complete combustion
    Step 1: Complete combustion of a hydrocarbon in excess oxygen produces carbon dioxide and water. Step 2: The other product, besides carbon dioxide, is therefore water. Step 3: Carbon monoxide and carbon (soot) are products of incomplete combustion, and hydrogen is not a combustion product, so the answer is water.
    Method:

    Recall the products of complete combustion of a hydrocarbon.
    Examiner tips

    • Complete combustion: hydrocarbon + oxygen → carbon dioxide + water.

27. Question 4(g)(ii)

    1 marksEnvironmental effect of carbon dioxide
    Step 1: Carbon dioxide is a greenhouse gas that traps heat in the atmosphere. Step 2: Increasing its amount increases global warming, which leads to climate change. Step 3: Acid rain is caused mainly by sulfur dioxide, ozone damage is caused by other gases and blood poisoning is caused by carbon monoxide, so the effect of carbon dioxide is increased global warming.
    Method:

    Recall the environmental problem caused by extra carbon dioxide.
    Examiner tips

    • Carbon dioxide → greenhouse effect → global warming/climate change.

28. Question 4(h)(i)

    2 marksDisplayed formula of ethanol
    Step 1: Ethanol has two carbon atoms and one -OH (hydroxyl) group. Step 2: The displayed formula shows a CH3 group bonded to a CH2 group, and the CH2 group bonded to an O-H group, with every bond drawn as a line. Step 3: A single carbon, a C=C double bond with two OH groups, or no oxygen are all wrong, so the correct structure is CH3-CH2-O-H with all bonds shown.
    Method:

    Build the two-carbon chain, then add the O-H group and all bonds.
    Examiner tips

    • Draw 2 carbons, fill C-H bonds, then add one O-H group.

29. Question 4(h)(ii)

    3 marksManufacture of ethanol from ethene
    Step 1: Ethanol is made by the catalytic addition of steam to ethene (hydration). Step 2: The conditions used are a temperature of about 300 °C and a pressure of about 60 atm, with a catalyst. Step 3: Oxygen or hydrogen are wrong reactants and the swapped values are wrong conditions, so the answer is steam at 300 °C and 60 atm.
    Method:

    Recall the reactant and the conditions for the hydration of ethene.
    Examiner tips

    • Hydration of ethene: steam, ~300 °C, ~60 atm, catalyst.

30. Question 5(a)

    1 marksWhy an element is in Group V
    Step 1: The group number of a main-group element equals the number of electrons in its outer shell. Step 2: Nitrogen has 5 outer electrons (configuration 2,5), so it is in Group V. Step 3: The number of shells gives the period, and proton or neutron counts do not set the group, so the reason is 5 outer electrons.
    Method:

    Link the group number to the number of outer-shell electrons.
    Examiner tips

    • Outer-shell electrons = group number for main-group elements.

31. Question 5(b)

    2 marksPhysical state from melting and boiling points
    Step 1: A substance is liquid between its melting point and its boiling point. Step 2: At 650 °C, tellurium is above its melting point (450 °C) but below its boiling point (988 °C). Step 3: So at 650 °C tellurium is a liquid.
    Method:

    Compare the temperature with the melting and boiling points to deduce the state.
    Examiner tips

    • Below mp = solid; between mp and bp = liquid; above bp = gas.

32. Question 5(c)

    2 marksColour and state of bromine
    Step 1: Going down Group VII the halogens get darker and change state. Step 2: Bromine is a red-brown liquid at room temperature and pressure. Step 3: Chlorine is a green gas and iodine is a grey-black solid, so bromine is a red-brown liquid.
    Method:

    Recall the colour and state of bromine at room temperature.
    Examiner tips

    • Group VII at rtp: chlorine green gas, bromine red-brown liquid, iodine grey solid.

33. Question 5(d)(i)

    3 marksObservations when sodium reacts with water
    Step 1: Sodium is a reactive Group I metal and is less dense than water, so it floats and moves on the surface. Step 2: The reaction gives off hydrogen gas, so it fizzes, and the heat released melts the sodium into a ball. Step 3: Sinking with no reaction, turning blue with a brown solid, or no change are all wrong, so the correct observations are floating/moving, fizzing and melting into a ball.
    Method:

    Recall the typical observations of a Group I metal reacting with water.
    Examiner tips

    • Group I + water: float, move, fizz, melt into a ball, may catch fire.

34. Question 5(d)(ii)

    1 marksWord equation for sodium and water
    Step 1: A reactive metal reacting with water forms a metal hydroxide and hydrogen gas. Step 2: Sodium therefore forms sodium hydroxide and hydrogen. Step 3: Sodium oxide, chloride and carbonate are not formed in this reaction, so the missing product is sodium hydroxide.
    Method:

    Apply the metal + water → hydroxide + hydrogen pattern.
    Examiner tips

    • Group I metal + water → hydroxide + hydrogen.

35. Question 5(e)

    2 marksSymbol equation for a carbonate and an acid
    Step 1: A carbonate reacting with an acid forms a salt, carbon dioxide and water. Step 2: To make sodium sulfate the acid must be sulfuric acid, $\text{H}_2\text{SO}_4$, and the other product is water, $\text{H}_2\text{O}$. Step 3: Hydrochloric acid would not give sulfate, and hydrogen or sulfur dioxide are not products, so the gaps are $\text{H}_2\text{SO}_4$ and $\text{H}_2\text{O}$.
    Method:

    Use the carbonate-plus-acid pattern and match the acid to the sulfate salt.
    Examiner tips

    • Carbonate + acid → salt + $\text{CO}_2$ + water; match the acid to the salt.

36. Question 5(f)(i)

    2 marksColour change of thymolphthalein
    Step 1: Thymolphthalein is colourless in neutral and acidic solution. Step 2: Sodium hydroxide dissolves to form an alkali, which turns thymolphthalein blue. Step 3: So the colour change is from colourless to blue.
    Method:

    Recall the thymolphthalein colours in neutral and alkaline solution.
    Examiner tips

    • Thymolphthalein: colourless (neutral/acid) → blue (alkali).

37. Question 5(f)(ii)

    1 marksDefinition of an alkali
    Step 1: A base is a substance that neutralises an acid. Step 2: An alkali is simply a base that is soluble in water. Step 3: An acid, a salt and an indicator are different things, so an alkali is a soluble base.
    Method:

    Recall the definition linking alkalis to soluble bases.
    Examiner tips

    • Alkali = soluble base.

38. Question 5(g)

    3 marksDot-and-cross diagram for sodium fluoride
    Step 1: Sodium (2,8,1) loses its single outer electron to form $\text{Na}^+$ with configuration 2,8 and a 1+ charge. Step 2: Fluorine (2,7) gains that electron to form $\text{F}^-$ with configuration 2,8 and a 1- charge. Step 3: Both ions end up with the configuration 2,8, the sodium ion 1+ and the fluoride ion 1-.
    Method:

    Transfer one electron from sodium to fluorine and assign the resulting configurations and charges.
    Examiner tips

    • Group I → 1+ ion; Group VII → 1- ion; both reach the stable 2,8 configuration.

39. Question 6(a)(i)

    1 marksMain ore of aluminium
    Step 1: Aluminium is found mainly as aluminium oxide in an ore. Step 2: This main ore of aluminium is called bauxite. Step 3: Haematite is an iron ore, limestone is calcium carbonate and cryolite is added to lower the melting point, so the main ore is bauxite.
    Method:

    Recall the name of the aluminium ore.
    Examiner tips

    • Aluminium ore = bauxite; iron ore = haematite.

40. Question 6(a)(ii)

    1 marksWhy aluminium is not extracted by carbon
    Step 1: Carbon can only remove (reduce) a metal from its ore if the metal is less reactive than carbon. Step 2: Aluminium is above carbon in the reactivity series, so it is more reactive than carbon and carbon cannot reduce it. Step 3: This is why aluminium must be extracted by electrolysis, so the reason is that aluminium is more reactive than carbon.
    Method:

    Compare the reactivity of aluminium and carbon to explain the extraction method.
    Examiner tips

    • Metals above carbon (e.g. aluminium) must be extracted by electrolysis.

41. Question 6(a)(iii)

    1 marksShowing that an oxide is reduced
    Step 1: Reduction is the loss of oxygen. Step 2: In the equation, $\text{Fe}_2\text{O}_3$ becomes Fe, so the iron(III) oxide has lost its oxygen. Step 3: Because it has lost oxygen, the iron(III) oxide has been reduced.
    Method:

    Use the oxygen-loss definition of reduction applied to the oxide.
    Examiner tips

    • Lost oxygen = reduced; gained oxygen = oxidised.

42. Question 6(b)(i)

    2 marksMeaning of the term alloy
    Step 1: An alloy is made by mixing a metal with other elements. Step 2: So an alloy is a mixture of a metal with one or more other elements. Step 3: A pure metal, a metal oxide compound and a dissolved non-metal are not alloys, so an alloy is a mixture of a metal with other elements.
    Method:

    Recall the definition of an alloy as a metal mixture.
    Examiner tips

    • Alloy = mixture of a metal + other element(s).

43. Question 6(b)(ii)

    2 marksSubstances needed for iron to rust
    Step 1: Rusting is the corrosion of iron in the presence of two substances from the air and surroundings. Step 2: Iron rusts only when both water and oxygen are present. Step 3: Carbon dioxide, nitrogen and hydrogen are not both required, so the two substances are water and oxygen.
    Method:

    Recall the two substances needed for iron to rust.
    Examiner tips

    • Rusting requires water AND oxygen together.

44. Question 6(b)(iii)

    2 marksChemical name for rust
    Step 1: Rust is the product of iron corroding in water and oxygen. Step 2: Its chemical name is hydrated iron(III) oxide. Step 3: Iron(II) chloride, iron(II) sulfate and other compounds are not rust, so the chemical name is hydrated iron(III) oxide.
    Method:

    Recall the chemical name given to rust.
    Examiner tips

    • Rust is hydrated iron(III) oxide.

45. Question 6(c)(i)

    2 marksOrdering metals by reactivity from observations
    Step 1: The more reactive a metal, the faster the bubbles and the greater the temperature rise with acid. Step 2: Nb shows no reaction (least reactive), Fe reacts very slowly, Mn reacts quickly and Sr reacts most vigorously (most reactive). Step 3: So from least to most reactive the order is Nb, Fe, Mn, Sr.
    Method:

    Order the metals by how vigorously each reacts with the acid.
    Examiner tips

    • Rank by reaction vigour: no reaction = least reactive, most vigorous = most reactive.

46. Question 6(c)(ii)

    2 marksWord equation for a metal and an acid
    Step 1: A metal reacting with an acid produces a salt and hydrogen gas. Step 2: Strontium with hydrochloric acid gives the salt strontium chloride and hydrogen gas. Step 3: Oxide and water, hydroxide and oxygen, or chlorine are wrong products, so the products are strontium chloride and hydrogen.
    Method:

    Apply the metal + acid → salt + hydrogen pattern with hydrochloric acid.
    Examiner tips

    • Metal + hydrochloric acid → metal chloride + hydrogen.

47. Question 7(a)

    1 marksNaming the process of passing electricity through a molten compound
    Step 1: Passing electricity through a molten ionic compound to break it down is called electrolysis. Step 2: The molten magnesium chloride is decomposed into magnesium and chlorine. Step 3: Electroplating coats an object, neutralisation is an acid-base reaction and distillation separates liquids, so the process is electrolysis.
    Method:

    Name the decomposition process that uses electricity.
    Examiner tips

    • Electrolysis = breaking down a molten/aqueous ionic compound using electricity.

48. Question 7(b)

    1 marksIdentifying the electrolyte
    Step 1: The electrolyte is the molten or dissolved ionic compound that conducts electricity and is broken down. Step 2: Here that is the molten magnesium chloride. Step 3: The platinum rods are the electrodes, the power supply provides the current and the crucible is the container, so the electrolyte is the molten magnesium chloride.
    Method:

    Identify which part conducts the current and is decomposed.
    Examiner tips

    • Electrolyte = the molten/aqueous ionic substance conducting the current.

49. Question 7(c)

    1 marksProduct formed at the anode
    Step 1: At the anode (positive electrode), negative ions are discharged. Step 2: In molten magnesium chloride the negative ions are chloride ions, which form chlorine gas at the anode. Step 3: Magnesium forms at the cathode, and hydrogen and oxygen are not present, so the anode product is chlorine.
    Method:

    Decide which ions move to the anode and what they form.
    Examiner tips

    • Anode = positive electrode; negative ions (chloride) discharge there.

50. Question 7(d)(i)

    1 marksProperty of platinum electrodes
    Step 1: An electrode must carry the current and not take part in the reaction. Step 2: Platinum conducts electricity and is inert (unreactive), so it carries the current without reacting. Step 3: Dissolving, conducting poorly or reacting with the products would make it unsuitable, so the property is that platinum conducts electricity and is inert.
    Method:

    Recall the properties an electrode material needs.
    Examiner tips

    • Electrode needs: good conductor + inert (unreactive).

51. Question 7(d)(ii)

    1 marksNon-metal used as electrodes
    Step 1: Electrodes need a material that conducts electricity. Step 2: Carbon, in the form of graphite, is a non-metal that conducts electricity, so it is used as electrodes. Step 3: Sulfur, phosphorus and iodine do not conduct electricity, so the non-metal used is carbon (graphite).
    Method:

    Pick the conducting non-metal used for electrodes.
    Examiner tips

    • Graphite (carbon) is the conducting non-metal used as electrodes.

52. Question 7(e)

    1 marksPhysical or chemical change in electrolysis
    Step 1: A chemical change makes new substances, while a physical change does not. Step 2: Electrolysis breaks magnesium chloride into the new substances magnesium and chlorine. Step 3: Because new substances are made, this is a chemical change.
    Method:

    Decide the type of change by whether new substances are formed.
    Examiner tips

    • If new substances are made, the change is chemical.

53. Question 7(f)

    2 marksForms of magnesium chloride
    Step 1: Aqueous means dissolved in water, so aqueous $\text{MgCl}_2$ is dissolved in water. Step 2: Anhydrous means containing no water, so it is the solid with no water; hydrated means chemically combined with water, so it is the solid that contains water of crystallisation. Step 3: Only the first set matches all three correctly.
    Method:

    Match each term (aqueous, anhydrous, hydrated) to its meaning.
    Examiner tips

    • anhydrous = no water; hydrated = water of crystallisation; aqueous = dissolved.

54. Question 8(a)

    1 marksReading the final volume from a graph
    Step 1: When the reaction is complete, no more gas is made and the total volume stays constant (the curve becomes horizontal). Step 2: Reading the height of the flat (plateau) part of the curve gives the final total volume. Step 3: The curve levels off at 132.5 cm³, so that is the total volume of $\text{H}_2\text{S}$ produced.
    Method:

    Read off the constant value where the curve has levelled off.
    Examiner tips

    • The final volume is the height of the flat (plateau) part of the curve.

55. Question 8(b)(i)

    1 marksEffect of higher concentration on reaction time
    Step 1: Increasing the concentration of the acid means there are more acid particles in the same volume. Step 2: More frequent collisions make the reaction faster, so it finishes in less time. Step 3: So the time taken for the reaction to finish decreases.
    Method:

    Use collision theory to predict the effect of higher concentration on time.
    Examiner tips

    • More concentrated acid → more collisions → faster → less time.

56. Question 8(b)(ii)

    1 marksEffect of higher concentration on total gas volume
    Step 1: The acid is in excess, so the amount of sodium sulfide controls how much $\text{H}_2\text{S}$ can be made. Step 2: Changing the concentration of the excess acid changes the rate but not the total amount of product. Step 3: So the total volume of $\text{H}_2\text{S}$ produced does not change.
    Method:

    Separate the effect on rate from the effect on total amount of product.
    Examiner tips

    • Higher concentration → faster, but same total product when a reactant amount is fixed.

57. Question 8(c)

    2 marksType of energy change from a reaction pathway diagram
    Step 1: If the products are at a lower energy level than the reactants, energy is released to the surroundings. Step 2: A reaction that releases energy is exothermic. Step 3: Because the products are below the reactants in energy, the reaction is exothermic.
    Method:

    Compare the energy levels of products and reactants to decide the energy change.
    Examiner tips

    • Exothermic: products lower in energy; endothermic: products higher.