Megan (2026)_Chemistry_Recap_13 October 2026

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HKDSE Chemistry Recap: Electrolysis | Megan (2026)

HKDSE Chemistry — Electrolysis

Subject: Chemistry

Megan (2026)

Issue date: Tuesday, 13 October 2026

Total: 50 marks  |  Suggested time: 60–70 minutes  |  Data: relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3, Cu = 63.5, Zn = 65.4, Ag = 108, Pb = 207; molar volume of gas at room temperature and pressure (r.t.p.) = 24.0 dm3 mol−1; ionic product of water Kw = 1.0 × 10−14 mol2 dm−6 at 25 °C. Electrodes are inert (graphite) unless stated otherwise. When cells are connected in series, the same number of moles of electrons passes through each cell.

01

Comprehensive Structured Question 1

Electrolysis of Molten Electrolytes: Lead(II) Bromide and Magnesium Chloride

Fig. 1 shows the set-up used to electrolyse lead(II) bromide, PbBr2, in a fume cupboard. Both electrodes X and Y are graphite rods, and a bulb is connected in series with the d.c. supply. At the beginning, the PbBr2 is solid and the bulb does not light up. When the crucible is heated strongly until the PbBr2 melts, the bulb lights up and changes are seen at both electrodes.

X Y + − d.c. supply bulb molten PbBr2 crucible graphite electrodes Bunsen flame
Fig. 1 — Electrolysis of molten lead(II) bromide (not drawn to scale)

(a) Explain why the bulb lights up only after the PbBr2 has melted. State the type of particles that carry the charge in the graphite electrodes and in the molten PbBr2 respectively. (2 marks)

(b) X is connected to the positive terminal of the d.c. supply. Write the half equation for the reaction at X and that at Y, and state one observation at each electrode. (3 marks)

(c) The cell in Fig. 1 is now connected in series with a second cell containing molten magnesium chloride, MgCl2, with graphite electrodes. In the first cell, 2.07 g of lead is formed. For the second cell, calculate

  1. the mass of magnesium formed; (2 marks)
  2. the volume of chlorine gas formed, measured at r.t.p. (1 mark)

(d) A student tries to obtain magnesium by electrolysing aqueous magnesium chloride with graphite electrodes. No magnesium is formed. Explain why, and name the product at the cathode. (2 marks)

10 marks
02

Comprehensive Structured Question 2

Electrolysis of Copper(II) Sulphate Solution: Inert and Copper Electrodes

Fig. 2 shows two cells connected in series to a d.c. supply. Cell I contains 1.0 mol dm−3 copper(II) sulphate solution with graphite electrodes P and Q. Cell II contains copper(II) sulphate solution of the same concentration with copper electrodes R and S. Electrode P is connected to the positive terminal of the d.c. supply, and electrode S is connected to the negative terminal.

P Q R S + − d.c. supply Cell I CuSO4(aq), graphite electrodes Cell II CuSO4(aq), copper electrodes
Fig. 2 — Two electrolytic cells connected in series (not drawn to scale)

(a) For Cell I, write the half equation for the reaction at electrode P and at electrode Q, and state one observation at each electrode. (4 marks)

(b) As the electrolysis in Cell I continues, the colour of the electrolyte changes and its pH changes. State these two changes and explain each of them in terms of the ions involved. (2 marks)

(c) Write the half equation for the reaction at electrode R and at electrode S in Cell II. Hence explain why the colour of the electrolyte in Cell II does not change. (2 marks)

(d) After some time, the mass of electrode Q has increased by 0.635 g. Calculate

  1. the volume of gas formed at electrode P, measured at r.t.p.; (1 mark)
  2. the change in mass of electrode R. (1 mark)
10 marks
03

Comprehensive Structured Question 3

Electrolysis of Sodium Chloride Solution: Effect of Concentration

Fig. 3 shows the electrolysis of 200 cm3 of concentrated sodium chloride solution using graphite electrodes P and Q. A few drops of universal indicator are added to the solution, which is initially green. Electrode P is connected to the negative terminal of the d.c. supply. Gas X and gas Y are collected in the two inverted test tubes.

− + gas X gas Y P Q concentrated NaCl(aq) with a few drops of universal indicator graphite electrodes
Fig. 3 — Electrolysis of concentrated sodium chloride solution (not drawn to scale)

(a) State the names of all the ions present in the concentrated sodium chloride solution before electrolysis. (1 mark)

(b) Write the half equation for the reaction at electrode P and at electrode Q. (2 marks)

(c) State the colour change of the universal indicator near electrode P and explain it. Describe a chemical test for gas Y. (2 marks)

(d) The experiment is repeated using a very dilute sodium chloride solution. A different gas is now the main product at electrode Q. Name this gas and explain why it is formed instead of gas Y. (2 marks)

(e) In the experiment with concentrated sodium chloride solution, 240 cm3 of gas X (measured at r.t.p.) was collected, but the volume of gas Y collected was clearly less than 240 cm3.

  1. Suggest why the volume of gas Y collected was less than that of gas X. (1 mark)
  2. Calculate the pH of the solution at the end of the electrolysis. Assume that the volume of the solution remains 200 cm3 and that the hydroxide ions formed do not react with any other substance. (2 marks)
10 marks
04

Comprehensive Structured Question 4

Electro-refining of Copper

Fig. 4 shows the purification of copper by electrolysis. Electrode X is a block of impure copper which contains small amounts of zinc and silver. Electrode Y is a strip of pure copper. X is connected to the positive terminal and Y to the negative terminal of the d.c. power supply. The electrolyte is an aqueous solution of a suitable copper(II) salt. As the electrolysis proceeds, a small amount of solid collects at the bottom of the cell below X.

X Y solid collected at the bottom + − d.c. power supply electrolyte X: impure copper (contains Zn and Ag) Y: pure copper
Fig. 4 — Purification of copper by electrolysis (not drawn to scale)

(a) Suggest a suitable electrolyte for the cell. (1 mark)

(b) Write the half equations for the reactions of copper and of zinc at X, and for the reaction at Y. (3 marks)

(c) Explain why silver in X does not dissolve but is found in the solid collected at the bottom of the cell, whereas zinc in X does dissolve. Explain why zinc is not deposited at Y. (2 marks)

(d) In one operation, the mass of Y increased by 6.35 g. At the same time, 1.31 g of zinc dissolved from X. Assume that, apart from silver, the only metals which dissolve from X are copper and zinc, and that the volume of the electrolyte remains 1.00 dm3. Calculate the decrease in the concentration of copper(II) ions in the electrolyte. (3 marks)

(e) A student suggests purifying impure zinc in the same way, using impure zinc as the positive electrode, pure zinc as the negative electrode and aqueous zinc sulphate as the electrolyte. Explain why this method does not work. (1 mark)

10 marks
05

Comprehensive Structured Question 5

Electrolysis of Dilute Sodium Sulphate Solution and Electroplating

Fig. 5 shows a Hofmann voltameter containing dilute sodium sulphate solution with a few drops of universal indicator. The solution is initially green. The electrodes P and Q are graphite rods. P is connected to the negative terminal of the d.c. supply. Gas X is collected in the left-hand tube and gas Y in the right-hand tube.

− + gas X gas Y P Q dilute Na2SO4(aq) with a few drops of universal indicator
Fig. 5 — Hofmann voltameter (not drawn to scale)

(a) State all the ions present in the electrolyte. Explain why sodium sulphate is added to the water. (2 marks)

(b) Write the half equation for the reaction at electrode P and at electrode Q. (2 marks)

(c) Describe how gas X and gas Y can each be identified. (1 mark)

(d) State the colour of the universal indicator near P and near Q shortly after the electrolysis starts. After the electrolysis, the solution is stirred thoroughly and the green colour returns. Explain why. (2 marks)

(e) The Hofmann voltameter is connected in series with an electroplating cell. In this cell, a metal spoon is the cathode, a silver rod is the anode, and the electrolyte is silver nitrate solution. 48.0 cm3 of gas X (measured at r.t.p.) is collected in the voltameter.

  1. Write the half equation for the reaction at the spoon. (1 mark)
  2. Calculate the mass of silver deposited on the spoon. (2 marks)
10 marks

Question 1 — Electrolysis of Molten Electrolytes: Lead(II) Bromide and Magnesium Chloride

(a) Conduction

In solid PbBr2, the Pb2+ and Br− ions are held in fixed positions in the lattice and are not free to move, so the solid does not conduct. When molten, the ions become mobile and can carry charge, so the bulb lights up. (1)
In graphite (and the metal wires), the charge is carried by delocalised (mobile) electrons. In molten PbBr2, it is carried by mobile ions. (1)

(b) Half equations and observations

At X (anode): 2Br−(l) → Br2(g) + 2e− (1)
At Y (cathode): Pb2+(l) + 2e− → Pb(l) (1)
Observations: reddish-brown fumes (bromine vapour) at X; a silvery/grey bead of molten lead collects at the bottom near Y. (1, both correct)

(c) Cells in series

(i) Amount of Pb = 2.07 ÷ 207 = 0.0100 mol. Since Pb2+ + 2e− → Pb, amount of electrons = 2 × 0.0100 = 0.0200 mol (1)
Mg2+ + 2e− → Mg, so amount of Mg = 0.0100 mol; mass of Mg = 0.0100 × 24.3 = 0.243 g (1)
(ii) 2Cl− → Cl2 + 2e−, so amount of Cl2 = 0.0200 ÷ 2 = 0.0100 mol; volume = 0.0100 × 24.0 = 0.240 dm3 (240 cm3) (1)

(d) Aqueous magnesium chloride

The solution contains Mg2+, H+ (from water), Cl− and OH− ions. At the cathode, H+ is preferentially discharged because it is lower than Mg2+ in the electrochemical series (more easily reduced), so Mg2+ stays in solution. (1)
The product at the cathode is hydrogen gas: 2H+ + 2e− → H2 (or 2H2O + 2e− → H2 + 2OH−). (1)
Common errors: (1) saying that “electrons flow through the molten salt”; (2) writing Pb2+ + 2e− → Pb(s) at the cathode when lead is molten at that temperature; (3) forgetting that 2 mol of electrons give only 1 mol of Cl2.

Question 2 — Electrolysis of Copper(II) Sulphate Solution: Inert and Copper Electrodes

(a) Cell I

Ions present: Cu2+, H+, SO42−, OH−.
At P (anode): 4OH−(aq) → 2H2O(l) + O2(g) + 4e− (accept 2H2O(l) → O2(g) + 4H+(aq) + 4e−) (1)
Observation at P: colourless gas bubbles (1)
At Q (cathode): Cu2+(aq) + 2e− → Cu(s) (1)
Observation at Q: a reddish-brown (pink) solid is deposited (1)

(b) Colour and pH of the electrolyte in Cell I

The blue colour becomes paler, because blue Cu2+(aq) ions are removed from the solution at Q. (1)
The pH decreases, because OH− ions are discharged at P, leaving an excess of H+ ions (sulphuric acid is formed). (1)
Overall equation: 2Cu2+(aq) + 2H2O(l) → 2Cu(s) + O2(g) + 4H+(aq).

(c) Cell II

At R (anode): Cu(s) → Cu2+(aq) + 2e−. At S (cathode): Cu2+(aq) + 2e− → Cu(s). (1)
The copper anode is oxidised in preference to OH− ions. For each Cu2+ ion discharged at S, one Cu2+ ion is formed at R, so the concentration of Cu2+(aq) (and the blue colour) stays the same. (1)

(d) Quantitative link between the cells

Amount of Cu at Q = 0.635 ÷ 63.5 = 0.0100 mol, so amount of electrons = 0.0200 mol.
(i) 4e− per O2: amount of O2 = 0.0200 ÷ 4 = 0.00500 mol; volume = 0.00500 × 24.0 = 0.120 dm3 (120 cm3) (1)
(ii) The same amount of electrons passes through Cell II, so 0.0100 mol of Cu dissolves from R: mass of R decreases by 0.635 g (1)
Common errors: (1) saying that Cu2+ is discharged at the anode in Cell I; (2) saying that the pH does not change in Cell I because “water is only split”; (3) using 2e− per O2 in (d)(i).

Question 3 — Electrolysis of Sodium Chloride Solution: Effect of Concentration

(a) Ions present

Sodium ions, chloride ions, hydrogen ions and hydroxide ions (the last two come from water). (1)

(b) Half equations

At P (cathode): 2H+(aq) + 2e− → H2(g) (accept 2H2O(l) + 2e− → H2(g) + 2OH−(aq)) (1)
At Q (anode): 2Cl−(aq) → Cl2(g) + 2e− (1)

(c) Indicator and test for gas Y

Near P, the indicator turns from green to blue/purple, because H+ ions are removed (water is reduced), so OH− ions accumulate and the solution becomes alkaline. (1)
Gas Y is chlorine: damp blue litmus paper turns red and is then bleached (accept: moist starch–iodide paper turns blue-black). (1)

(d) Very dilute sodium chloride solution

The gas is oxygen: 4OH−(aq) → 2H2O(l) + O2(g) + 4e−. (1)
In the very dilute solution the concentration of Cl− is low, so the effect of concentration no longer outweighs the position in the electrochemical series. OH− is lower than Cl− in the electrochemical series and is preferentially discharged. (1)

(e) Volume of gas Y and pH

(i) Chlorine is fairly soluble in water (and partly reacts with it), whereas hydrogen is almost insoluble, so less chlorine gas is collected. (1)
(ii) Amount of H2 = 0.240 ÷ 24.0 = 0.0100 mol. From 2H2O + 2e− → H2 + 2OH−, amount of OH− formed = 0.0200 mol.
[OH−] = 0.0200 ÷ 0.200 = 0.100 mol dm−3; [H+] = 1.0 × 10−14 ÷ 0.100 = 1.0 × 10−13 mol dm−3 (1)
pH = −log(1.0 × 10−13) = 13.0 (1)
Common errors: (1) saying that Na+ is discharged at the cathode; (2) explaining the dilute case by “OH− is more concentrated” (wrong; it is the position in the electrochemical series that decides when Cl− is dilute); (3) using the volume of the gas, not the volume of the solution, as the volume in the [OH−] calculation; (4) using pH = −log[OH−].

Question 4 — Electro-refining of Copper

(a) Electrolyte

Copper(II) sulphate solution (accept acidified CuSO4(aq), or copper(II) nitrate solution). (1)

(b) Half equations

At X: Cu(s) → Cu2+(aq) + 2e− (1)
At X: Zn(s) → Zn2+(aq) + 2e− (1)
At Y: Cu2+(aq) + 2e− → Cu(s) (1)

(c) Silver, zinc and the products at Y

Silver is less reactive than copper (it is more difficult to oxidise), so it is not oxidised. It falls off X as the copper around it dissolves and collects as anode sludge. Zinc is more reactive than copper and is oxidised to Zn2+, which dissolves. (1)
At Y, Cu2+ is lower than Zn2+ in the electrochemical series and is more easily reduced, so Cu2+ is discharged in preference to Zn2+. Zn2+ ions stay in solution. (1)

(d) Decrease in [Cu2+]

Amount of Cu deposited at Y = 6.35 ÷ 63.5 = 0.100 mol, so amount of electrons = 0.200 mol (1)
Amount of Zn dissolved = 1.31 ÷ 65.4 = 0.0200 mol, which supplies 0.0400 mol of electrons.
The remaining 0.200 − 0.0400 = 0.160 mol of electrons come from Cu, so amount of Cu dissolved = 0.0800 mol (1)
Net change in Cu2+ = 0.0800 − 0.100 = −0.0200 mol, so [Cu2+] decreases by 0.0200 mol dm−3 (1)
Remark: Unlike a plating cell with a pure anode, [Cu2+] does not stay constant here. Some of the current is used to dissolve zinc, which does not replace the Cu2+ removed at Y. In industry, the electrolyte therefore has to be replaced or purified from time to time.

(e) Refining zinc

In aqueous ZnSO4, H+ (from water) is preferentially discharged at the cathode because it is lower than Zn2+ in the electrochemical series, so hydrogen gas is formed instead of zinc. (1)
Common errors: (1) saying that silver is “oxidised but stays as the solid”; (2) forgetting that zinc also dissolves, so that Cu dissolved ≠ Cu deposited in (d); (3) writing an equation that shows Zn2+ being discharged at Y.

Question 5 — Electrolysis of Dilute Sodium Sulphate Solution and Electroplating

(a) Ions and the role of Na2SO4

Ions present: Na+, SO42−, H+ and OH−. (1)
Pure water contains very few ions and is a very poor conductor. Adding Na2SO4 provides many mobile ions to increase the conductivity. Na+ and SO42− are not discharged, so only water is electrolysed. (1)

(b) Half equations

At P (cathode): 2H+(aq) + 2e− → H2(g) (accept 2H2O(l) + 2e− → H2(g) + 2OH−(aq)) (1)
At Q (anode): 4OH−(aq) → 2H2O(l) + O2(g) + 4e− (accept 2H2O(l) → O2(g) + 4H+(aq) + 4e−) (1)

(c) Identification of the gases

Gas X is hydrogen: it burns with a “pop” sound with a burning splint. Gas Y is oxygen: it relights a glowing splint. (1, both correct)

(d) Universal indicator

Near P the indicator turns blue/purple (OH− accumulates), and near Q it turns red/orange (H+ accumulates, because OH− is discharged there). (1)
The overall change is 2H2O → 2H2 + O2. Equal amounts of OH− (at P) and H+ (at Q) are produced, so after stirring they neutralise each other (H+ + OH− → H2O) and the solution is neutral, with the same green colour. (1)

(e) Electroplating cell in series

(i) Ag+(aq) + e− → Ag(s) (1)
(ii) Amount of H2 = 0.0480 ÷ 24.0 = 0.00200 mol. 2H+ + 2e− → H2, so amount of electrons = 2 × 0.00200 = 0.00400 mol (1)
Amount of Ag = 0.00400 mol; mass of Ag = 0.00400 × 108 = 0.432 g (1)
Common errors: (1) stating that the solution becomes acidic or alkaline overall; (2) forgetting that H2 needs 2 mol of electrons per mole, whereas Ag needs only 1; (3) using the volume of the gas in cm3 with 24.0 dm3 mol−1 without converting.
Mark weighting: Question 1 (10) + Question 2 (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) = 50 marks.
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