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Secondary 4 Pure Chemistry Redox Electrochemistry Quiz
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Questions
Secondary 4 Pure Chemistry Quiz - Redox Electrochemistry
Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ______ / 40
Duration: 45 minutes
Total Marks: 40
Instructions:
- Answer all questions in the spaces provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- For calculations, show all working clearly.
- Use appropriate chemical notation, state symbols, and units where required.
Section A: Multiple Choice Questions (10 marks)
1. Which of the following half-equations correctly represents the oxidation of Fe²⁺ to Fe³⁺? [1]
A. Fe²⁺ + e⁻ → Fe³⁺
B. Fe²⁺ → Fe³⁺ + e⁻
C. Fe³⁺ + e⁻ → Fe²⁺
D. Fe³⁺ → Fe²⁺ + e⁻
Answer: _______
2. In the reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, what is the oxidation state change of manganese? [1]
A. +7 to +2
B. +2 to +7
C. +7 to +4
D. +4 to +2
Answer: _______
3. Which substance acts as the reducing agent in the reaction: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂? [1]
A. Cl₂
B. Br⁻
C. Cl⁻
D. Br₂
Answer: _______
4. A student sets up a simple cell using magnesium and copper electrodes in their respective sulfate solutions. Which electrode is the anode and what is the direction of electron flow? [1]
| Anode | Electron Flow | |
|---|---|---|
| A | Magnesium | Mg → Cu |
| B | Copper | Cu → Mg |
| C | Magnesium | Cu → Mg |
| D | Copper | Mg → Cu |
Answer: _______
5. Which of the following reactions is NOT a redox reaction? [1]
A. Zn + CuSO₄ → ZnSO₄ + Cu
B. 2Na + Cl₂ → 2NaCl
C. HCl + NaOH → NaCl + H₂O
D. 2Mg + O₂ → 2MgO
Answer: _______
6. In the electrolysis of concentrated aqueous sodium chloride using inert electrodes, what is the product at the anode? [1]
A. Sodium metal
B. Chlorine gas
C. Hydrogen gas
D. Oxygen gas
Answer: _______
7. The standard electrode potential for Cu²⁺/Cu is +0.34 V and for Zn²⁺/Zn is -0.76 V. What is the E°cell for a Zn/Cu electrochemical cell? [1]
A. +1.10 V
B. -1.10 V
C. +0.42 V
D. -0.42 V
Answer: _______
8. During the electrolysis of dilute sulfuric acid using platinum electrodes, what is the volume ratio of hydrogen to oxygen gas produced? [1]
A. 1:1
B. 2:1
C. 1:2
D. 4:1
Answer: _______
9. Which of the following metals CANNOT be extracted by electrolysis of its aqueous salt solution? [1]
A. Copper
B. Silver
C. Sodium
D. Lead
Answer: _______
10. In a sacrificial protection system for an iron pipeline, which metal is most suitable as the sacrificial anode? [1]
A. Copper
B. Tin
C. Magnesium
D. Silver
Answer: _______
Section B: Structured Questions (20 marks)
11. The diagram below shows a simple electrochemical cell set up in the laboratory.
<image_placeholder> id: Q11-fig1 type: experimental_setup linked_question: Q11 description: A simple electrochemical cell with zinc electrode in 1.0 mol/dm³ ZnSO₄(aq) and copper electrode in 1.0 mol/dm³ CuSO₄(aq), connected by a salt bridge (KNO₃) and a voltmeter. Wires connect each electrode to the voltmeter. Label the Zn electrode as negative terminal, Cu electrode as positive terminal. Show direction of electron flow in external circuit (Zn to Cu). Show ion movement in salt bridge (K⁺ towards cathode, NO₃⁻ towards anode). labels: Zn electrode (anode, -), Cu electrode (cathode, +), 1.0 mol/dm³ ZnSO₄(aq), 1.0 mol/dm³ CuSO₄(aq), salt bridge (KNO₃), voltmeter, electron flow direction (Zn → Cu), cation/anion movement in salt bridge values: Standard conditions: 298 K, 1.0 mol/dm³, 1 atm must_show: Two half-cells with electrodes and solutions, salt bridge connecting solutions, voltmeter in external circuit, electron flow arrow from Zn to Cu, ion migration arrows in salt bridge </image_placeholder>
(a) Write the half-equation for the reaction occurring at the zinc electrode. [1]
(b) Write the half-equation for the reaction occurring at the copper electrode. [1]
(c) Write the overall ionic equation for the cell reaction. [1]
(d) State the function of the salt bridge in this cell. [1]
12. Potassium manganate(VII), KMnO₄, is a strong oxidising agent in acidic solution.
(a) Write the balanced half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution. [2]
(b) Iron(II) sulfate reacts with acidified potassium manganate(VII). The half-equation for the oxidation of Fe²⁺ is: Fe²⁺ → Fe³⁺ + e⁻
Construct the balanced overall ionic equation for the reaction between MnO₄⁻ and Fe²⁺ in acidic solution. [2]
(c) In a titration, 25.0 cm³ of 0.0200 mol/dm³ FeSO₄ required 20.0 cm³ of KMnO₄ solution for complete reaction. Calculate the concentration of the KMnO₄ solution in mol/dm³. [3]
13. The diagram shows the electrolysis of molten lead(II) bromide using inert graphite electrodes.
<image_placeholder> id: Q13-fig1 type: experimental_setup linked_question: Q13 description: Electrolysis of molten PbBr₂ using graphite electrodes. Show PbBr₂(l) in a crucible, two graphite rods dipping into the melt, connected to a DC power supply (battery). Label anode (+) and cathode (-). Show Pb²⁺ migrating to cathode, Br⁻ migrating to anode. Show lead metal forming at cathode (silver-grey liquid), bromine vapour (brown) at anode. labels: Graphite anode (+), graphite cathode (-), molten PbBr₂, DC power supply, Pb²⁺ migration to cathode, Br⁻ migration to anode, Pb(l) at cathode, Br₂(g) at anode values: Temperature > 373°C (melting point of PbBr₂) must_show: Molten electrolyte, two inert electrodes connected to battery, ion migration arrows, products at each electrode labelled </image_placeholder>
(a) State what is meant by the term electrolysis. [1]
(b) Write the half-equation for the reaction at the cathode. [1]
(c) Write the half-equation for the reaction at the anode. [1]
(d) Explain why the electrolysis is carried out on molten lead(II) bromide rather than aqueous lead(II) bromide. [2]
14. A student investigates the electrolysis of aqueous copper(II) sulfate using copper electrodes.
(a) Write the half-equation for the reaction at the anode. [1]
(b) Write the half-equation for the reaction at the cathode. [1]
(c) Describe what happens to the mass of the anode and the mass of the cathode during electrolysis. [2]
(d) The concentration of Cu²⁺(aq) in the solution remains constant during this electrolysis. Explain why. [1]
15. The diagram shows an experimental setup for the rusting of iron.
<image_placeholder> id: Q15-fig1 type: experimental_setup linked_question: Q15 description: Three test tubes set up for rusting investigation. Tube A: Iron nail in tap water (air present). Tube B: Iron nail in boiled distilled water with layer of oil on top, stopper (no oxygen). Tube C: Iron nail in anhydrous calcium chloride, stopper (no water). All tubes left for several days. Label each tube clearly. labels: Test tube A (iron nail + tap water + air), Test tube B (iron nail + boiled distilled water + oil layer + stopper), Test tube C (iron nail + anhydrous CaCl₂ + stopper) values: Time period: several days must_show: Three distinct test tube setups with labels, conditions for rusting investigation </image_placeholder>
(a) In which test tube(s) will rusting occur? [1]
(b) Write the balanced chemical equation for the formation of rust, given that rust is hydrated iron(III) oxide, Fe₂O₃·xH₂O. [1]
(c) Explain, in terms of electron transfer, why iron rusts in the presence of both oxygen and water. [2]
(d) Galvanising is a method of preventing rusting. Explain how galvanising protects iron from rusting, even if the zinc coating is scratched. [2]
Section C: Longer Structured and Data-Based Questions (10 marks)
16. The table below shows standard electrode potentials for several half-cells.
| Half-reaction | E° / V |
|---|---|
| Ag⁺ + e⁻ ⇌ Ag | +0.80 |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 |
| Fe²⁺ + 2e⁻ ⇌ Fe | -0.44 |
| Zn²⁺ + 2e⁻ ⇌ Zn | -0.76 |
| Mg²⁺ + 2e⁻ ⇌ Mg | -2.37 |
(a) Which half-reaction shows the strongest oxidising agent? Explain your choice. [2]
(b) A cell is constructed using the Fe²⁺/Fe and Ag⁺/Ag half-cells. Calculate the E°cell for this cell. [1]
(c) Write the overall cell reaction for the Fe²⁺/Fe and Ag⁺/Ag cell. [1]
(d) Predict whether the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is spontaneous under standard conditions. Use the data in the table to justify your answer. [2]
17. A student carries out the electrolysis of 500 cm³ of 1.0 mol/dm³ copper(II) sulfate solution using inert platinum electrodes. A constant current of 0.50 A is passed for 2 hours.
(a) Calculate the total charge passed in coulombs. [1]
(b) Calculate the amount of electricity in faradays. (1 F = 96500 C) [1]
(c) The cathode reaction is: Cu²⁺ + 2e⁻ → Cu. Calculate the mass of copper deposited at the cathode. (Ar: Cu = 63.5) [2]
(d) Write the half-equation for the reaction at the anode. [1]
(e) Calculate the volume of gas produced at the anode at room temperature and pressure. (Molar gas volume at r.t.p. = 24 dm³/mol) [2]
18. Hydrogen-oxygen fuel cells are used in spacecraft to generate electricity.
(a) Write the half-equation for the oxidation of hydrogen at the anode in an alkaline fuel cell. [1]
(b) Write the half-equation for the reduction of oxygen at the cathode in an alkaline fuel cell. [1]
(c) Write the overall equation for the reaction in the fuel cell. [1]
(d) State two advantages of using a hydrogen-oxygen fuel cell over a conventional petrol engine in a vehicle. [2]
19. The diagram shows the electroplating of a steel key with silver.
<image_placeholder> id: Q19-fig1 type: experimental_setup linked_question: Q19 description: Electroplating setup: Steel key (cathode, -) and pure silver anode (+) immersed in AgNO₃(aq) electrolyte. Connected to DC power supply. Show Ag⁺ migrating to cathode, NO₃⁻ migrating to anode. Show Ag depositing on key, Ag anode dissolving. labels: Steel key (cathode, -), pure Ag anode (+), AgNO₃(aq) electrolyte, DC power supply, Ag⁺ migration to cathode, NO₃⁻ migration to anode, Ag deposit on key, Ag anode dissolving values: Electrolyte: AgNO₃(aq) must_show: Electroplating cell with labelled electrodes, electrolyte, power supply, ion migration, electrode reactions </image_placeholder>
(a) Write the half-equation for the reaction at the cathode. [1]
(b) Write the half-equation for the reaction at the anode. [1]
(c) The mass of the silver anode decreases during electroplating. Explain why. [1]
(d) A current of 0.20 A is passed for 30 minutes. Calculate the mass of silver deposited on the key. (Ar: Ag = 108; 1 F = 96500 C) [3]
20. The following data refers to the extraction of aluminium by electrolysis.
- Aluminium is extracted from purified Al₂O₃ dissolved in molten cryolite (Na₃AlF₆).
- The electrolysis uses carbon anodes and a carbon cathode (lining of the cell).
- The overall reaction is: 2Al₂O₃ → 4Al + 3O₂
(a) State why cryolite is used in the extraction of aluminium. [1]
(b) Write the half-equation for the reaction at the cathode. [1]
(c) Write the half-equation for the reaction at the anode. [1]
(d) The carbon anodes need to be replaced regularly. Explain why. [1]
(e) Calculate the volume of oxygen gas produced at r.t.p. when 270 kg of aluminium is produced. (Ar: Al = 27; Molar gas volume at r.t.p. = 24 dm³/mol) [3]
End of Quiz
Answers
Secondary 4 Pure Chemistry Quiz - Redox Electrochemistry (Answer Key)
Total Marks: 40
Section A: Multiple Choice Questions (10 marks)
1. Answer: B [1]
Explanation: Oxidation is loss of electrons. Fe²⁺ loses one electron to become Fe³⁺: Fe²⁺ → Fe³⁺ + e⁻. Option A shows reduction (gain of electron). Option C is the reduction half-equation. Option D has wrong direction and charge.
2. Answer: A [1]
Explanation: In MnO₄⁻, oxygen is -2 each (total -8). Overall charge is -1, so Mn = +7. In Mn²⁺, oxidation state is +2. Change is +7 to +2 (reduction, gain of 5 electrons).
3. Answer: B [1]
Explanation: Br⁻ is oxidised to Br₂ (loses electrons), so Br⁻ is the reducing agent. Cl₂ is reduced to Cl⁻ (gains electrons), so Cl₂ is the oxidising agent.
4. Answer: A [1]
Explanation: Mg is more reactive (more negative E°) than Cu, so Mg is the anode (oxidation: Mg → Mg²⁺ + 2e⁻). Electrons flow from anode (Mg) to cathode (Cu) in the external circuit.
5. Answer: C [1]
Explanation: HCl + NaOH → NaCl + H₂O is an acid-base neutralisation. No oxidation state changes occur (H remains +1, Cl -1, Na +1, O -2). All other options involve oxidation state changes.
6. Answer: B [1]
Explanation: Concentrated NaCl(aq): At anode, Cl⁻ is discharged preferentially over OH⁻ (higher concentration), forming Cl₂ gas. At cathode, H⁺ from water is reduced to H₂ (Na⁺ not discharged in aqueous solution).
7. Answer: A [1]
Explanation: E°cell = E°cathode - E°anode = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = +0.34 - (-0.76) = +1.10 V. Positive E°cell indicates spontaneous reaction.
8. Answer: B [1]
Explanation: Cathode: 2H⁺ + 2e⁻ → H₂. Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻. Overall: 2H₂O → 2H₂ + O₂. Volume ratio H₂:O₂ = 2:1 (Avogadro's law, same conditions).
9. Answer: C [1]
Explanation: Sodium (E° = -2.71 V) is too reactive; in aqueous solution, H⁺ from water is reduced instead (2H⁺ + 2e⁻ → H₂, E° = 0.00 V). Na⁺ is not discharged. Cu, Ag, Pb have less negative E° values and can be deposited from aqueous solution.
10. Answer: C [1]
Explanation: Sacrificial anode must be more reactive (more negative E°) than iron (E° = -0.44 V). Mg (E° = -2.37 V) will oxidise preferentially, protecting Fe. Cu, Sn, Ag are less reactive than Fe and would not protect it.
Section B: Structured Questions (20 marks)
11. (a) Zn(s) → Zn²⁺(aq) + 2e⁻ [1]
Marking: Correct species, balancing, state symbols, electron on product side.
(b) Cu²⁺(aq) + 2e⁻ → Cu(s) [1]
Marking: Correct species, balancing, state symbols, electron on reactant side.
(c) Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) [1]
Marking: Correct overall equation, state symbols, electrons cancelled.
(d) The salt bridge completes the circuit by allowing ion migration (K⁺ moves toward cathode, NO₃⁻ moves toward anode) to maintain charge neutrality in both half-cells. [1]
Marking: Mention ion migration and charge neutrality/maintaining circuit.
12. (a) MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) [2]
Marking: 1 mark for correct species and balancing; 1 mark for correct state symbols and electron count.
Method: Balance Mn → balance O with H₂O → balance H with H⁺ → balance charge with e⁻.
(b) MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq) [2]
Marking: 1 mark for multiplying Fe half-equation by 5; 1 mark for correct combined equation with state symbols.
Method: Multiply oxidation half-equation (Fe²⁺ → Fe³⁺ + e⁻) by 5 to equalise electrons, then add to reduction half-equation.
(c) Concentration of KMnO₄ = 0.00500 mol/dm³ [3]
Working:
- Moles of Fe²⁺ = 0.0200 mol/dm³ × 0.0250 dm³ = 5.00 × 10⁻⁴ mol
- Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5
- Moles of MnO₄⁻ = 5.00 × 10⁻⁴ / 5 = 1.00 × 10⁻⁴ mol
- Volume of KMnO₄ = 20.0 cm³ = 0.0200 dm³
- Concentration = 1.00 × 10⁻⁴ mol / 0.0200 dm³ = 0.00500 mol/dm³
Marking: 1 mark for moles Fe²⁺; 1 mark for mole ratio and moles MnO₄⁻; 1 mark for final concentration with units.
13. (a) Electrolysis is the decomposition of an electrolyte (molten or aqueous) by an electric current. [1]
Marking: Key terms: decomposition, electrolyte, electric current.
(b) Pb²⁺(l) + 2e⁻ → Pb(l) [1]
Marking: Correct species, state symbols, balancing.
(c) 2Br⁻(l) → Br₂(g) + 2e⁻ [1]
Marking: Correct species, state symbols, balancing.
(d) In aqueous PbBr₂, water is present. At the cathode, H⁺ from water (E° = 0.00 V) is reduced preferentially over Pb²⁺ (E° = -0.13 V), producing H₂ gas instead of Pb. At the anode, OH⁻ is oxidised to O₂ instead of Br⁻ to Br₂. Using molten PbBr₂ ensures only Pb²⁺ and Br⁻ are present, giving pure Pb and Br₂. [2]
Marking: 1 mark for explaining H⁺/H₂O reduction at cathode; 1 mark for explaining OH⁻ oxidation at anode (or general "water interferes").
14. (a) Cu(s) → Cu²⁺(aq) + 2e⁻ [1]
Marking: Copper anode oxidises (active electrode).
(b) Cu²⁺(aq) + 2e⁻ → Cu(s) [1]
Marking: Copper(II) ions reduced at cathode.
(c) Anode mass decreases (Cu dissolves as Cu²⁺); cathode mass increases (Cu deposits). [2]
Marking: 1 mark for anode mass decrease; 1 mark for cathode mass increase.
(d) For every Cu²⁺ ion reduced at the cathode, one Cu atom oxidises at the anode to form Cu²⁺. The rate of Cu²⁺ removal equals the rate of Cu²⁺ addition, so [Cu²⁺] remains constant. [1]
Marking: Mention equal rates of removal and formation of Cu²⁺.
15. (a) Test tube A only [1]
Marking: Only A has both oxygen (air) and water present.
(b) 4Fe(s) + 3O₂(g) + 2xH₂O(l) → 2Fe₂O₃·xH₂O(s) [1]
Marking: Correct reactants and product, balanced. Accept 4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃ or similar hydrated oxide forms.
(c) Iron loses electrons (oxidation): Fe → Fe²⁺ + 2e⁻. Oxygen gains electrons (reduction): O₂ + 2H₂O + 4e⁻ → 4OH⁻. Water acts as the medium for ion transport and provides OH⁻. Both O₂ and H₂O are needed for the redox reaction to proceed. [2]
Marking: 1 mark for Fe oxidation half-equation or description; 1 mark for O₂ reduction half-equation or description, and role of water.
(d) Zinc is more reactive than iron (more negative E°). Even when scratched, zinc corrodes preferentially (sacrificial protection): Zn → Zn²⁺ + 2e⁻. The electrons flow to the exposed iron, preventing its oxidation (Fe → Fe²⁺ + 2e⁻). [2]
Marking: 1 mark for Zn more reactive/sacrificial; 1 mark for electron flow preventing Fe oxidation.
Section C: Longer Structured and Data-Based Questions (10 marks)
16. (a) Ag⁺ + e⁻ ⇌ Ag (E° = +0.80 V). It has the most positive E° value, meaning Ag⁺ has the greatest tendency to gain electrons (be reduced), making it the strongest oxidising agent. [2]
Marking: 1 mark for identifying Ag⁺/Ag; 1 mark for explanation linking most positive E° to strongest oxidising agent.
(b) E°cell = E°cathode - E°anode = +0.80 - (-0.44) = +1.24 V [1]
Marking: Correct calculation, sign, and unit.
(c) Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s) [1]
Marking: Correct species, balancing, state symbols. Fe oxidised (anode), Ag⁺ reduced (cathode).
(d) Yes, spontaneous. E°cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = +0.34 - (-0.76) = +1.10 V. Positive E°cell means the reaction is thermodynamically favourable (spontaneous) under standard conditions. [2]
Marking: 1 mark for correct E°cell calculation; 1 mark for linking positive E°cell to spontaneity.
17. (a) Charge = Current × Time = 0.50 A × (2 × 60 × 60) s = 0.50 × 7200 = 3600 C [1]
Marking: Correct conversion of time to seconds, correct calculation, unit.
(b) Faradays = 3600 C / 96500 C/F = 0.0373 F [1]
Marking: Correct division, 3 significant figures.
(c) Moles of e⁻ = 0.0373 mol. Mole ratio e⁻ : Cu = 2 : 1. Moles of Cu = 0.0373 / 2 = 0.01865 mol. Mass = 0.01865 × 63.5 = 1.18 g [2]
Marking: 1 mark for moles of Cu (using 2:1 ratio); 1 mark for mass calculation with unit.
(d) 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ [1]
Marking: Correct anode reaction for inert electrode in aqueous CuSO₄ (OH⁻ discharged over SO₄²⁻).
(e) Moles of e⁻ = 0.0373 mol. Anode: 4e⁻ produce 1 O₂. Moles O₂ = 0.0373 / 4 = 0.009325 mol. Volume = 0.009325 × 24 = 0.224 dm³ (or 224 cm³) [2]
Marking: 1 mark for moles O₂ (4:1 ratio); 1 mark for volume at r.t.p. with unit.
18. (a) 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ [1]
Marking: Correct species, balancing, state symbols (alkaline conditions).
(b) O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) [1]
Marking: Correct species, balancing, state symbols.
(c) 2H₂(g) + O₂(g) → 2H₂O(l) [1]
Marking: Overall equation, balanced.
(d) Any two:
- Higher energy efficiency (direct chemical to electrical, no combustion heat loss)
- No pollutant emissions (only water produced)
- Quiet operation (no moving parts)
- Renewable if H₂ from electrolysis using renewable electricity [2]
Marking: 1 mark each for any two valid advantages.
19. (a) Ag⁺(aq) + e⁻ → Ag(s) [1]
Marking: Silver deposition at cathode.
(b) Ag(s) → Ag⁺(aq) + e⁻ [1]
Marking: Silver anode dissolution (active electrode).
(c) The silver anode is active. It oxidises (Ag → Ag⁺ + e⁻) to replenish Ag⁺ ions in solution, so its mass decreases. [1]
Marking: Mention active anode, oxidation, mass loss.
(d) Charge = 0.20 A × (30 × 60) s = 360 C. Moles e⁻ = 360 / 96500 = 0.00373 mol. Mole ratio e⁻ : Ag = 1 : 1. Moles Ag = 0.00373 mol. Mass = 0.00373 × 108 = 0.403 g [3]
Marking: 1 mark for charge; 1 mark for moles Ag (1:1 ratio); 1 mark for mass with unit.
20. (a) Cryolite lowers the melting point of Al₂O₃ from ~2050°C to ~950°C, reducing energy costs and preventing excessive cell degradation. [1]
Marking: Lowers melting point, reduces energy cost.
(b) Al³⁺(l) + 3e⁻ → Al(l) [1]
Marking: Correct species, state symbols, balancing.
(c) 2O²⁻(l) → O₂(g) + 4e⁻ [1]
Marking: Oxide ion oxidation at anode.
(d) Oxygen produced at the anode reacts with the carbon anode to form CO₂: C(s) + O₂(g) → CO₂(g). The anode is consumed and must be replaced. [1]
Marking: Reaction of O₂ with C to form CO₂, anode consumed.
(e) Moles Al = 270,000 g / 27 g/mol = 10,000 mol. From overall equation: 4 mol Al produced with 3 mol O₂. Moles O₂ = 10,000 × (3/4) = 7,500 mol. Volume O₂ = 7,500 × 24 = 180,000 dm³ [3]
Marking: 1 mark for moles Al; 1 mark for mole ratio (3:4) and moles O₂; 1 mark for volume with unit.
End of Answer Key