Electrolysis of aqueous CuSO4 with INERT electrodes — graph of O2 volume vs time, half-equations, Faraday's law calculations, electroplating modification. (25 marks)
(a)(i) Define the term 'electrolysis'. (2 marks)
Electrolysis is the breakdown (decomposition) of an ionic compound (electrolyte) — when molten or in aqueous solution — by passing a direct electric current through it.
1 mark for "decomposition / breakdown of an ionic compound"; 1 mark for "by passing a direct current / using electrical energy".
(a)(ii) Suggest ONE material that could be used as an inert electrode for the electrolysis of aqueous copper(II) sulfate. (1 mark)
Graphite (carbon) — cheap, conducts electricity, doesn't react with the electrolyte or the products. (Platinum is also accepted.)
(b)(i) Plot a graph of volume of oxygen against time using the data in Table 1. Draw a line of best fit. (5 marks)
Plot on Figure 2 (page 7): time / minutes on the x-axis; volume of O2 / cm³ on the y-axis. The points lie on a perfect straight line that, when extended back, passes through the origin.
Marks: 1 for axes labelled with units; 1 for sensible scale; 2 for plotting all six points correctly; 1 for a single ruled line of best fit.
(b)(ii) Use the graph to determine the time taken for 5.50 cm³ of oxygen to be liberated. (1 mark)
From the table, every 10 minutes liberates an extra 1.20 cm³ of O2. So 1.20 cm³ takes 10 min ⇒ 5.50 cm³ takes 5.50 ÷ 1.20 × 10 ≈ 45.8 min. Check the graph confirms ≈ 46 min.
Time ≈ 46 minutes (accept 45–47 min from the line of best fit).
(c)(i) Identify ALL the ions present in the electrolyte. (2 marks)
CuSO4(aq) is dissolved in water, so we must include the ions from water too:
Cu2+(aq), SO42−(aq), H+(aq) and OH−(aq).
1 mark for the two CuSO4 ions; 1 mark for remembering the H+ and OH− from water.
(c)(ii) Which electrode (P or Q) is the anode and which is the cathode? (1 mark)
From Figure 1, the + terminal of the battery is connected to electrode P and the − terminal to Q.
P = anode (+); Q = cathode (−).
(c)(iii) State ONE ion in the electrolyte that will drift towards the anode. (1 mark)
The anode is positively charged and attracts negative ions (anions).
SO42− (or OH−) drifts towards the anode.
(c)(iv) Write an ionic equation for the reaction taking place at the cathode. (2 marks)
At the cathode (Q), Cu2+ is preferentially discharged over H+ (lower in the electrochemical series):
Cu2+(aq) + 2e− → Cu(s)
1 mark for correct species; 1 mark for correctly balanced (the 2e−).
(d)(i) Calculate the quantity of electricity passed through the solution. (Q = I × t; 1F = 96 500 C). (2 marks)
I = 3.5 A. Time = 1 hour = 3 600 s.
Q = I × t = 3.5 × 3 600 (1 mark)
Q = 12 600 C (1 mark)
(d)(ii) Calculate the number of moles of copper deposited. (3 marks)
From the cathode half-equation: Cu2+ + 2e− → Cu. So 1 mole of Cu requires 2 moles of electrons (= 2 × 96 500 = 193 000 C).
Moles of electrons = Q ÷ F = 12 600 ÷ 96 500 = 0.1306 mol e−. (1 mark)
Moles of Cu = ½ × moles of e− = 0.1306 ÷ 2 (1 mark)
Moles of Cu deposited = 0.0653 mol (≈ 0.065 mol) (1 mark)
(d)(iii) Calculate the mass of copper deposited (Cu = 64). (1 mark)
Mass = moles × RAM = 0.0653 × 64
Mass of Cu deposited = 4.18 g (≈ 4.2 g).
(d)(iv) Suggest ONE reason for the difference between the calculated mass (4.18 g) and the measured mass (4.85 g). (1 mark)
Some H+ ions were also discharged at the cathode (releasing H2) but most current went to depositing Cu — yet the student's measurement may also include impurities or undeposited solution still clinging to the electrode.
Other accepted answers: the cathode was not fully dried before weighing (water added to the mass); current fluctuated above 3.5 A during the experiment, so more charge actually flowed than calculated; loss of solution by splashing was actually balanced by deposition continuing past the timer.
(e) State the colour change in the electrolyte after a few hours. (1 mark)
With inert electrodes, Cu2+ is steadily removed at the cathode (deposited as Cu metal) but no copper is replenished from the anode (only O2(g) is liberated).
The blue colour of the CuSO4(aq) gradually fades — the solution becomes paler and eventually colourless as the Cu2+ ions are used up.
(f) State how the electrodes can be modified to obtain pure copper from impure copper. (2 marks)
Anode: use the impure copper as the anode (it dissolves: Cu → Cu2+ + 2e−; impurities like Ag, Au fall off as anode sludge). (1 mark)
Cathode: use a thin strip of pure copper as the cathode (Cu2+ + 2e− → Cu deposits as 99.99% pure copper). (1 mark)
The CuSO4(aq) electrolyte concentration stays roughly constant because Cu2+ is replaced as fast as it is removed.
Question 2
Acids and bases — definitions; sodium bicarbonate solution; reactions of H2SO4 with iron and with NaHCO3. (15 marks)
(a)(i) Define 'acid'. (1 mark)
An acid is a substance that produces hydrogen ions (H+) as the only positive ions when dissolved in water. (Brønsted-Lowry alternative: a proton donor.)
(a)(ii) Define 'salt'. (2 marks)
A salt is an ionic compound formed when the hydrogen ion (H+) of an acid is replaced by a metal ion (or by NH4+).
1 mark for "ionic compound from an acid"; 1 mark for "H+ replaced by metal/ammonium ion".
(b)(i) Why does red litmus turn blue but blue litmus stay blue when sodium bicarbonate is dissolved in water? (1 mark)
Sodium bicarbonate solution is alkaline (basic). Bases turn red litmus blue but have no effect on blue litmus (which is already blue).
(b)(ii) State whether the pH of NaHCO3(aq) is > 7 or < 7. (1 mark)
Sodium bicarbonate solution is a weak base, typical pH about 8–9.
pH > 7 (alkaline / basic).
(c)(i) Complete Table 2 — TWO inferences from Experiment 1. (2 marks)
Experiment 1: iron + dilute H2SO4.
Observation
Inference
Solution turned pale green.
Iron(II) sulfate (FeSO4) was formed — pale green is the characteristic colour of Fe2+(aq).
Colourless gas evolved; splint goes off with a "squeaky pop".
The gas is hydrogen (H2) — the squeaky pop is the standard test for hydrogen.
1 mark per correct inference.
(c)(ii) State TWO physical properties of iron metal. (2 marks)
Silvery-grey metallic colour with a shiny lustre when freshly cut.
Magnetic at room temperature (one of the few elements that is — Fe, Co, Ni).
Other accepted: hard solid; high melting point (~1538 °C); high density; good conductor of heat and electricity; malleable and ductile.
1 mark each.
(c)(iii) Write an equation for the reaction in Experiment 1 (Fe + H2SO4). (1 mark)
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
(c)(iv) Test (with expected observation) for the CO2 given off in Experiment 2. (2 marks)
Test: Bubble the gas through limewater (calcium hydroxide solution, Ca(OH)2(aq)). (1 mark)
Observation: The colourless limewater turns milky/cloudy (a white precipitate of CaCO3 forms) — this confirms CO2. (1 mark)
(c)(v) Write a balanced equation, including state symbols, for sodium bicarbonate + sulfuric acid. (3 marks)
2 NaHCO3 reacts with 1 H2SO4 (because H2SO4 is dibasic):
1 mark for correct products; 1 mark for balancing; 1 mark for state symbols.
Question 3
Organic chemistry — Compound A (C2H4) and Compound B (C4H10); homologous-series rules; bromine and KMnO4 tests; uses. (15 marks)
(a) Write the general formula of the homologous series to which Compound A (C2H4) belongs. (1 mark)
C2H4 is ethene — an alkene.
General formula of alkenes: CnH2n
(b) List THREE general characteristics of a homologous series, other than the general formula. (3 marks)
Same functional group ⇒ similar chemical reactions.
Successive members differ by one −CH2− unit (relative molecular mass differs by 14).
Gradual change in physical properties as molar mass increases (boiling/melting points rise; density rises; volatility decreases).
1 mark each.
(c)(i) Identify the homologous series of Compound B (C4H10). (1 mark)
C4H10 fits the formula CnH2n+2 ⇒ alkane.
Alkanes (Compound B is butane, C4H10).
(c)(ii) Draw the fully displayed structures of A and B. (4 marks)
Compound A — ethene (C2H4) (2 marks)
H H
\ /
C = C
/ \
H H
Compound B — butane (C4H10) (2 marks)
H H H H
| | | |
H− C − C − C − C −H
| | | |
H H H H
2 marks each — must show every C, H and bond. The C=C must be drawn as a clear double line for ethene.
(d)(i) Compound A in test tube 1 + 1 cm³ acidified KMnO4. Compound B in test tube 2 + 1 cm³ acidified KMnO4. State observation in each. (3 marks)
Test tube 1 (ethene + KMnO4): The purple KMnO4 solution is decolorised (turns colourless). The C=C of ethene is oxidised by the manganate(VII) — manganate is reduced from purple Mn(VII) to colourless Mn(II). (1 mark)
Test tube 2 (butane + KMnO4):No reaction / no colour change — the solution remains purple, because alkanes are saturated and do not react with KMnO4 under these conditions. (2 marks: 1 for "no change", 1 for "stays purple".)
(d)(ii) Is Compound B saturated or unsaturated? (1 mark)
Saturated — butane has only single C−C bonds and the maximum number of hydrogens for 4 carbons.
(e) Give ONE use of EACH compound. (2 marks)
Compound A (ethene): raw material for making poly(ethene) plastic; also used commercially to ripen fruit (e.g. bananas, mangoes). (1 mark)
Compound B (butane): fuel — bottled gas (LPG) for camping stoves and cigarette lighters. (1 mark)
Question 4
Periodic table — electronic configuration of Si; covalent bonding in F2; isotopes of carbon; reactivity of group 2 metals with water. (15 marks)
(a)(i) State the electronic configuration of silicon (Si) and justify its placement in Period 3. (2 marks)
Silicon, atomic number 14: 14 electrons fill the shells as 2, 8, 4.
Electronic configuration of Si: 2,8,4. (1 mark)
Justification: Si has electrons in three shells, so it appears in Period 3 of the periodic table — the period number equals the number of occupied electron shells. (1 mark)
(a)(ii) Draw a dot (•) and cross (×) diagram to show the covalent bonding in F2. (2 marks)
Each fluorine atom (Group VII, 7 outer electrons) needs one more electron to complete its octet. The two F atoms share one pair of electrons (a single covalent bond) and each F retains 3 lone pairs.
•• ××
• • × ×
• F • •× ×• × F ×
• • × ×
•• ××
one shared pair shown in the middle:
F (• •)(• ×) F — each F has 6 outer
electrons of its own + 2 shared (octet OK).
1 mark for showing the single shared pair (one dot + one cross between the two F nuclei); 1 mark for showing the three lone pairs on each F atom.
(b)(i) Define 'isotope'. (1 mark)
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons (so they have the same atomic number but different mass numbers).
(b)(ii) From the list 12C, 14C, 13C, 14N — identify the isotopes of carbon. (1 mark)
Isotopes have the same proton number (Z = 6 for C) but different mass numbers.
12C, 13C and 14C are all isotopes of carbon. 14N is nitrogen, not carbon.
(b)(iii) Give TWO examples of radioisotopes and ONE use of each. (4 marks)
Radioisotope
Use
Carbon-14 (14C)
Radiocarbon dating — used by archaeologists to date ancient organic remains (e.g. wood, bone, charcoal) up to about 50 000 years old.
Cobalt-60 (60Co)
Cancer radiotherapy — its high-energy gamma rays are used to kill cancerous tumours; also used to sterilise medical equipment.
Other accepted:
Iodine-131 — diagnosis/treatment of thyroid disorders; Uranium-235 — nuclear power generation; Technetium-99m — medical imaging.
2 marks per radioisotope + use pair (1 for the isotope, 1 for the correct use).
(c)(i) Write a chemical equation, including state symbols, for the reaction of calcium with water. (3 marks)
Calcium (a Group 2 metal) reacts with cold water to form calcium hydroxide (slightly soluble) and hydrogen gas:
Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)
1 mark for products; 1 mark for balancing (the 2 in front of H2O); 1 mark for state symbols.
(c)(ii) An unknown element X is in the periodic table at Period 4 of Group 2 (below Mg and Ca). Predict the order of reactivity of X, Mg and Ca with water. (2 marks)
Reactivity of Group 2 metals increases down the group: as you go down, the outer 2 electrons are further from the nucleus and easier to lose.
X is below Ca (which is below Mg) ⇒ X is the most reactive of the three.
Order of reactivity, most → least: X > Ca > Mg.
1 mark for placing X first; 1 mark for putting Ca above Mg.
Question 5
Hydrocarbons in the Caribbean — sources; fractional distillation; cracking; substitution reactions; combustion of propane. (15 marks)
(a) State TWO natural sources of hydrocarbons. (2 marks)
Crude oil (petroleum) — Trinidad & Tobago is a major Caribbean producer.
Natural gas — primarily methane, often found alongside crude oil.
Other accepted: coal; tar sands; oil shale.
1 mark each.
(b) Name TWO of the three fractions of crude oil shown in Table 4. (2 marks)
Number of C atoms
Name of fraction (and uses)
1–4 carbons
Refinery gas / LPG — household cooking gas, lighter fuel.
1 mark per correctly named fraction. (Any TWO of the three is fine.)
(c)(i) State the name of the process illustrated in Figure 3 (long chain → shorter chains, no catalyst). (1 mark)
One long C7H16-style alkane breaks into a shorter alkane (C5H12) plus a smaller alkene (C2H4).
Cracking (specifically, thermal cracking when no catalyst is used).
(c)(ii) State the conditions for thermal cracking. (2 marks)
High temperature (~700–900 °C). (1 mark)
High pressure (~70 atm). (1 mark)
(Catalytic cracking uses lower temperature and pressure with a zeolite catalyst — but the question says "absence of a catalyst", so thermal cracking conditions apply.)
(c)(iii) State TWO uses of the C5H12 hydrocarbon (pentane) obtained from cracking. (2 marks)
Component of petrol (gasoline) for motor vehicles.
Used as a solvent in laboratories and industry (e.g. extracting oils from plant material).
Other accepted: blowing agent for polystyrene foam; in geothermal power stations as the working fluid; as a starting material for further industrial reactions.
1 mark each.
(c)(iv) Pentane reacts with bromine in sunlight. Draw the fully displayed structure of the monobromo-compound formed. (2 marks)
Free-radical substitution: one H atom on pentane is replaced by Br. The product (drawing the substitution at C-1, giving 1-bromopentane) is:
H H H H H
| | | | |
Br−C − C − C − C − C −H
| | | | |
H H H H H
1-bromopentane (C5H11Br)
1 mark for one H replaced by Br; 1 mark for fully displayed structure showing all C, H and bonds.
(c)(v) State the type of reaction in (c)(iv). (1 mark)
Free-radical substitution (one of the H atoms is substituted/replaced by Br; sunlight/UV initiates the radicals).
(d) Write a balanced equation, including state symbols, for the complete combustion of propane (C3H8). (3 marks)
Propane burns in excess O2 to form CO2 and H2O:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
1 mark for correct reactants/products; 1 mark for the balancing (5 O2, 3 CO2, 4 H2O); 1 mark for state symbols.
Question 6
Extraction of metals — aluminium vs iron; lead from PbO2; usefulness and harmfulness of metals to living systems. (15 marks)
(a)(i) Discuss the difference in extraction methods for aluminium and iron, in relation to their positions in the electrochemical series. (4 marks)
Aluminium is high in the electrochemical/reactivity series — it holds onto its electrons very tightly. Carbon is not reactive enough to displace Al from Al2O3. So Al must be extracted by electrolysis of molten aluminium oxide (in cryolite, Na3AlF6) — electricity provides the energy needed to reduce the very stable Al3+ ions to Al metal at the cathode. (2 marks)
Iron is lower in the reactivity series, below carbon. The Fe3+ ions in iron(III) oxide (Fe2O3) can be reduced by carbon (in the form of coke / carbon monoxide) — so iron is extracted by reduction in the blast furnace: Fe2O3 + 3CO → 2Fe + 3CO2. This is much cheaper than electrolysis. (2 marks)
(a)(ii) Which method (electrolysis or carbon reduction) would be more suitable for extracting lead from its ore, and why? (3 marks)
Lead is below carbon in the reactivity series — about as reactive as copper.
Carbon reduction (chemical reduction) is more suitable for extracting Pb from PbO2. (1 mark)
Reasons: Pb is below C in the electrochemical series, so carbon (as coke / CO) is reactive enough to reduce Pb4+ in PbO2 to Pb metal. (1 mark) Carbon reduction is much cheaper and uses much less energy than electrolysis — there is no need to use the expensive electrolytic process when a simple smelting reaction will do. (1 mark)
(b) Write TWO balanced equations to show the production of Pb from PbO2. (4 marks)
Step 1 — coke burns in the furnace to make carbon monoxide (the active reducing agent):
2C(s) + O2(g) → 2CO(g) (2 marks)
Step 2 — CO reduces lead(IV) oxide to lead metal:
PbO2(s) + 2CO(g) → Pb(s) + 2CO2(g) (2 marks)
(An alternative for step 2 is direct reduction with carbon: PbO2(s) + C(s) → Pb(s) + CO2(g).)
(c) Explain the usefulness of ONE metal and the harmfulness of ONE metal (from Al, Pb, Fe) to living systems. (4 marks)
Usefulness — IRON: Iron is essential to life. It is the central atom in the haemoglobin molecule of red blood cells, where it binds O2 reversibly and transports it from the lungs to every tissue. Iron is also part of cytochrome enzymes that carry out cellular respiration. A diet low in iron causes anaemia (low haemoglobin, fatigue, breathlessness). (2 marks)
Harmfulness — LEAD: Lead and its compounds are toxic to humans and animals. Lead enters the body through contaminated water (old lead pipes, lead-based paint dust, leaded petrol exhaust) and accumulates in bones and soft tissue. It interferes with the enzymes that synthesise haemoglobin, damages the nervous system (especially in children — causing reduced IQ, learning difficulties, behavioural problems) and damages the kidneys. (2 marks)
(Aluminium is generally inert to most living systems but high concentrations have been linked to neurological effects; either Al or Pb is acceptable as the harmful example.)
Solutions generated by Kairu — Student Hub's AI system, trained by The Student Hub. AI can make mistakes — always cross-check tricky answers with your teacher and class notes.