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Secondary 3 Chemistry Redox Electrochemistry Quiz
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Questions
Secondary 3 Chemistry Quiz - Redox Electrochemistry
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Use appropriate chemical terminology and state symbols where required.
- The number of marks is given in brackets [ ] at the end of each question or part question.
Section A: Short Answer and Structured Response (10 marks)
Answer all questions in this section. Questions 1-5.
1. Define oxidation in terms of electron transfer. [1]
2. State the oxidation state of sulfur in SO₂. [1]
3. Identify the oxidising agent in the following reaction: [1]
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
4. A student adds aqueous potassium iodide to a solution of iron(III) chloride. A brown solution forms.
(a) Write the ionic equation for the reaction that occurs. [2]
(b) Explain, in terms of electron transfer, which species is reduced. [1]
5. Describe a chemical test to distinguish between an oxidising agent and a reducing agent. Include the reagent used and the expected observations. [3]
Section B: Electrolysis Principles (10 marks)
Answer all questions in this section. Questions 6-10.
6. The diagram below shows the electrolysis of molten lead(II) bromide using inert electrodes.
(a) State the product formed at the anode. [1]
(b) Write the half-equation for the reaction occurring at the cathode. [2]
(c) Explain why the lead(II) bromide must be molten for electrolysis to occur. [2]
7. Aqueous copper(II) sulfate is electrolysed using copper electrodes.
(a) State what is observed at the anode during electrolysis. [1]
(b) Explain why the concentration of copper(II) sulfate in the solution remains constant during this electrolysis. [2]
8. A student sets up a simple cell using a strip of zinc and a strip of copper dipped into a beaker of dilute sulfuric acid.
(a) State which metal acts as the negative electrode. [1]
(b) Write the half-equation for the reaction occurring at the positive electrode. [1]
9. A student investigates the electrolysis of aqueous sodium sulfate using inert electrodes. A gas is collected at each electrode.
(a) Name the gas collected at the cathode. [1]
(b) The volume of gas collected at the cathode is twice the volume of gas collected at the anode. Explain this observation using appropriate half-equations. [3]
(c) State the role of sodium sulfate in this electrolysis. [1]
10. An object is to be electroplated with silver. The object to be plated is one electrode; a silver rod is the other electrode.
(a) State whether the object to be plated should be connected to the positive or negative terminal of the battery. Explain your answer. [2]
Section C: Data Interpretation and Application (10 marks)
Answer all questions in this section. Questions 11-15.
11. The table below shows the products formed during the electrolysis of three different aqueous solutions using inert electrodes.
| Solution | Product at Anode | Product at Cathode |
|---|---|---|
| A | Oxygen | Hydrogen |
| B | Chlorine | Hydrogen |
| C | Bromine | Metal X |
(a) Identify a possible solute for Solution A. Explain your answer. [2]
(b) Solution B is a concentrated solution of a chloride salt. Explain why chlorine, rather than oxygen, is produced at the anode. [2]
(c) Metal X is less reactive than hydrogen. Name a metal that could be Metal X. [1]
12. The following reactions represent two different types of cells.
Cell 1: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) Cell 2: Hydrogen fuel cell: 2H₂(g) + O₂(g) → 2H₂O(l)
(a) State one advantage of using a hydrogen fuel cell over a simple chemical cell. [1]
(b) In a hydrogen fuel cell, state where oxidation occurs and write the half-equation for this reaction. [2]
13. A student purifies a sample of impure copper by electrolysis. The impure copper is used as the anode, and a thin strip of pure copper is used as the cathode. The electrolyte is aqueous copper(II) sulfate.
(a) Explain what happens to the impure copper anode during electrolysis. [2]
(b) State what happens to the impurities that are less reactive than copper. [1]
14. Calculate the mass of pure copper deposited at the cathode when a current of 2.0 A is passed through the electrolyte for 30 minutes. [Faraday constant = 96,500 C/mol; Relative atomic mass of Cu = 63.5] [3]
15. A student adds a piece of magnesium ribbon to aqueous copper(II) sulfate. A reaction occurs.
(a) State two observations the student would make. [2]
(b) Write the ionic equation for the reaction, including state symbols. [2]
Section D: Extended Response and Applications (10 marks)
Answer all questions in this section. Questions 16-20.
16. Explain why iron nails are often coated with zinc (galvanised) rather than tin to prevent rusting. Refer to the reactivity series and the concept of sacrificial protection. [3]
17. A student investigates the electrolysis of dilute aqueous copper(II) chloride using inert electrodes.
(a) State the product formed at the anode and explain why it is formed. [2]
(b) Write the half-equation for the reaction occurring at the cathode. [1]
18. Describe how the reactivity series can be used to predict the products of electrolysis of an aqueous solution. Use an example to illustrate your answer. [3]
19. A simple cell is constructed using magnesium and copper electrodes in dilute hydrochloric acid.
(a) Identify the positive electrode and explain your choice. [2]
(b) Write the overall ionic equation for the reaction that occurs in the cell. [1]
20. Rusting of iron is an electrochemical process. Explain the role of water and oxygen in the rusting of iron, and suggest one method, other than galvanising, to prevent rusting. [3]
END OF QUIZ
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Answers
Secondary 3 Chemistry Quiz - Redox Electrochemistry — Answer Key
Total Marks: 40
Section A: Short Answer and Structured Response (10 marks)
1. Define oxidation in terms of electron transfer. [1]
- Answer: Oxidation is the loss of electrons (from an atom, ion, or molecule).
- Marking note: Accept "removal of electrons" or "giving away electrons." Must reference electron transfer.
2. State the oxidation state of sulfur in SO₂. [1]
- Answer: +4
- Marking note: Accept "+4" or "4+". Do not accept "4" without charge indication.
3. Identify the oxidising agent in the reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) [1]
- Answer: Cu²⁺ (or copper(II) ions / CuSO₄)
- Marking note: Accept "copper(II) sulfate" or "CuSO₄". The oxidising agent is the species that is reduced (gains electrons).
4. Aqueous potassium iodide added to iron(III) chloride solution.
(a) Ionic equation for the reaction: [2]
- Answer: 2I⁻(aq) + 2Fe³⁺(aq) → I₂(aq) + 2Fe²⁺(aq)
- Marking note: 1 mark for correct species, 1 mark for correct balancing and state symbols. Accept I₂ as "brown solution."
(b) Explain which species is reduced in terms of electron transfer: [1]
- Answer: Fe³⁺ is reduced because it gains an electron (to form Fe²⁺).
- Marking note: Must identify Fe³⁺ and mention electron gain.
5. Chemical test to distinguish between an oxidising agent and a reducing agent. [3]
- Answer:
- Reagent: Add aqueous potassium iodide (KI) to both solutions.
- Observation with oxidising agent: Solution turns brown/yellow-brown (due to formation of I₂).
- Observation with reducing agent: No colour change (or remains colourless).
- Alternative answer:
- Reagent: Add acidified potassium manganate(VII) (KMnO₄).
- Observation with reducing agent: Purple solution turns colourless.
- Observation with oxidising agent: No colour change (remains purple).
- Marking note: 1 mark for reagent, 1 mark for observation with oxidising agent, 1 mark for observation with reducing agent. Accept other valid tests (e.g., acidified potassium dichromate(VI)).
Section B: Electrolysis Principles (10 marks)
6. Electrolysis of molten lead(II) bromide using inert electrodes.
(a) Product at the anode: [1]
- Answer: Bromine (Br₂)
- Marking note: Accept "bromine gas" or "Br₂(l/g)". Do not accept "bromide."
(b) Half-equation at the cathode: [2]
- Answer: Pb²⁺(l) + 2e⁻ → Pb(l)
- Marking note: 1 mark for correct species (Pb²⁺ + 2e⁻ → Pb), 1 mark for correct state symbols and balancing.
(c) Why must lead(II) bromide be molten for electrolysis: [2]
- Answer: In the solid state, the ions (Pb²⁺ and Br⁻) are held in fixed positions in the ionic lattice and cannot move freely. When molten, the ions become mobile and can move to the electrodes, allowing conduction of electricity and electrolysis to occur.
- Marking note: 1 mark for stating ions are not mobile in solid state, 1 mark for explaining ions become mobile when molten.
7. Electrolysis of aqueous copper(II) sulfate using copper electrodes.
(a) Observation at the anode: [1]
- Answer: The copper anode dissolves / becomes smaller / decreases in mass.
- Marking note: Accept "anode dissolves" or "copper goes into solution."
(b) Why concentration of copper(II) sulfate remains constant: [2]
- Answer: At the anode, Cu(s) is oxidised to Cu²⁺(aq), adding Cu²⁺ ions to the solution. At the cathode, Cu²⁺(aq) is reduced to Cu(s), removing Cu²⁺ ions from the solution. The rate of dissolution at the anode equals the rate of deposition at the cathode, so the concentration of Cu²⁺ remains unchanged.
- Marking note: 1 mark for describing what happens at each electrode, 1 mark for linking rates to constant concentration.
8. Simple cell with zinc and copper in dilute sulfuric acid.
(a) Metal acting as negative electrode: [1]
- Answer: Zinc (Zn)
- Marking note: Zinc is more reactive; it loses electrons more readily.
(b) Half-equation at the positive electrode: [1]
- Answer: 2H⁺(aq) + 2e⁻ → H₂(g)
- Marking note: Accept "hydrogen ions gain electrons to form hydrogen gas." Must include electron gain.
9. Electrolysis of aqueous sodium sulfate.
(a) Gas collected at the cathode: [1]
- Answer: Hydrogen (H₂)
- Marking note: Accept "hydrogen gas."
(b) Explanation of gas volume ratio (cathode:anode = 2:1): [3]
- Answer:
- At the cathode: 4H⁺(aq) + 4e⁻ → 2H₂(g) [or 2H⁺ + 2e⁻ → H₂]
- At the anode: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ [or 4OH⁻ → O₂ + 2H₂O + 4e⁻]
- From the half-equations, 4 electrons produce 2 molecules of H₂ at the cathode and 1 molecule of O₂ at the anode. Therefore, the volume of H₂ produced is twice the volume of O₂ produced (since equal volumes of gases contain the same number of molecules at the same temperature and pressure).
- Marking note: 1 mark for each correct half-equation, 1 mark for linking electron transfer to gas volume ratio.
(c) Role of sodium sulfate in this electrolysis: [1]
- Answer: Sodium sulfate acts as an electrolyte to increase the conductivity of the solution / to allow current to flow.
- Marking note: Accept "provides ions to conduct electricity" or "increases conductivity."
10. Electroplating with silver.
(a) Connection of object to be plated and explanation: [2]
- Answer: The object to be plated should be connected to the negative terminal (cathode). This is because reduction occurs at the cathode, where Ag⁺ ions gain electrons to form a layer of solid silver (Ag) on the object: Ag⁺(aq) + e⁻ → Ag(s).
- Marking note: 1 mark for correct terminal (negative/cathode), 1 mark for explanation referencing reduction and silver deposition.
Section C: Data Interpretation and Application (10 marks)
11. Electrolysis products table.
(a) Possible solute for Solution A and explanation: [2]
- Answer: Sodium sulfate (Na₂SO₄) or sulfuric acid (H₂SO₄) or any sulfate of a reactive metal.
- Explanation: At the anode, oxygen is produced because sulfate ions (SO₄²⁻) are not discharged; instead, OH⁻ ions from water are discharged. At the cathode, hydrogen is produced because the metal ion (e.g., Na⁺) is more reactive than hydrogen and is not discharged; instead, H⁺ ions from water are discharged.
- Marking note: 1 mark for correct solute, 1 mark for explanation referencing selective discharge.
(b) Why chlorine is produced at the anode in Solution B: [2]
- Answer: In a concentrated chloride solution, the concentration of chloride ions (Cl⁻) is high. According to selective discharge, chloride ions are discharged in preference to hydroxide ions (OH⁻) when the solution is concentrated, producing chlorine gas.
- Marking note: 1 mark for mentioning high concentration of Cl⁻, 1 mark for linking to selective discharge.
(c) Metal X less reactive than hydrogen: [1]
- Answer: Copper (Cu) or silver (Ag) or gold (Au) or any metal below hydrogen in the reactivity series.
- Marking note: Accept any valid metal less reactive than hydrogen.
12. Cell comparisons.
(a) Advantage of hydrogen fuel cell over simple chemical cell: [1]
- Answer: Hydrogen fuel cells produce only water as a by-product (no pollution) / reactants can be continuously supplied (do not run out) / more energy-efficient.
- Marking note: Accept any one valid advantage.
(b) In a hydrogen fuel cell, where oxidation occurs and half-equation: [2]
- Answer:
- Oxidation occurs at the anode (negative electrode).
- Half-equation: H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻ [or H₂ → 2H⁺ + 2e⁻ in acidic electrolyte]
- Marking note: 1 mark for stating anode, 1 mark for correct half-equation.
13. Purification of copper by electrolysis.
(a) What happens to the impure copper anode: [2]
- Answer: The impure copper anode undergoes oxidation. Copper atoms lose electrons to form Cu²⁺ ions, which enter the electrolyte solution: Cu(s) → Cu²⁺(aq) + 2e⁻. The anode gradually dissolves / decreases in size.
- Marking note: 1 mark for stating oxidation occurs, 1 mark for correct half-equation or description of dissolution.
(b) What happens to impurities less reactive than copper: [1]
- Answer: Impurities that are less reactive than copper (e.g., silver, gold) do not oxidise. They fall off the anode and collect as "anode sludge" or "anode mud" below the anode.
- Marking note: Accept "fall to the bottom as sludge/mud" or "do not dissolve."
14. Calculation of mass of copper deposited. [3]
- Answer:
- Charge (Q) = Current (I) × Time (t) = 2.0 A × (30 × 60) s = 3600 C
- Number of moles of electrons = Q / F = 3600 C / 96,500 C/mol ≈ 0.0373 mol
- Half-equation: Cu²⁺ + 2e⁻ → Cu, so 2 moles of electrons deposit 1 mole of Cu.
- Moles of Cu deposited = 0.0373 mol / 2 = 0.01865 mol
- Mass of Cu = moles × Ar = 0.01865 mol × 63.5 g/mol ≈ 1.18 g (or 1.2 g)
- Marking note: 1 mark for correct charge calculation, 1 mark for correct mole ratio, 1 mark for correct final answer with unit. Accept 1.18 g or 1.2 g.
15. Magnesium and copper(II) sulfate reaction.
(a) Two observations: [2]
- Answer:
- The blue colour of the solution fades / becomes colourless.
- A brown/pink solid (copper) deposits on the magnesium / at the bottom of the beaker.
- The magnesium ribbon dissolves / decreases in size.
- Marking note: 1 mark for each correct observation, up to 2 marks.
(b) Ionic equation with state symbols: [2]
- Answer: Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
- Marking note: 1 mark for correct species, 1 mark for correct state symbols and balancing.
Section D: Extended Response and Applications (10 marks)
16. Galvanising vs. tin coating for rust prevention. [3]
- Answer: Zinc is more reactive than iron, while tin is less reactive. If the coating is scratched, zinc acts as a sacrificial anode and corrodes preferentially, protecting the exposed iron (sacrificial protection). Tin does not provide sacrificial protection; if the coating is scratched, the exposed iron (more reactive) corrodes faster than the tin.
- Marking note: 1 mark for relative reactivity, 1 mark for explaining sacrificial protection with zinc, 1 mark for explaining why tin fails to protect when scratched.
17. Electrolysis of dilute aqueous copper(II) chloride with inert electrodes.
(a) Product at anode and explanation: [2]
- Answer: Oxygen gas (O₂) is formed. In a dilute solution, hydroxide ions (OH⁻) are discharged in preference to chloride ions (Cl⁻) according to selective discharge, producing oxygen.
- Marking note: 1 mark for oxygen, 1 mark for explanation referencing selective discharge of OH⁻ over Cl⁻ in dilute solution.
(b) Half-equation at cathode: [1]
- Answer: Cu²⁺(aq) + 2e⁻ → Cu(s)
- Marking note: Accept correct half-equation with state symbols.
18. Using reactivity series to predict electrolysis products. [3]
- Answer: The reactivity series helps determine which ions are discharged at the electrodes. At the cathode, the ion of the less reactive element is discharged. For example, in aqueous copper(II) sulfate, Cu²⁺ ions are discharged in preference to H⁺ because copper is less reactive than hydrogen. At the anode, the ion of the less reactive non-metal is discharged, but concentration also matters (e.g., in concentrated NaCl, Cl⁻ is discharged over OH⁻).
- Marking note: 1 mark for explaining cathode prediction, 1 mark for explaining anode prediction, 1 mark for a relevant example.
19. Simple cell with magnesium and copper in dilute hydrochloric acid.
(a) Positive electrode and explanation: [2]
- Answer: Copper is the positive electrode. Magnesium is more reactive than copper, so magnesium loses electrons more readily and acts as the negative electrode. Copper is therefore the positive electrode.
- Marking note: 1 mark for identifying copper, 1 mark for explanation based on reactivity.
(b) Overall ionic equation: [1]
- Answer: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
- Marking note: Accept correct equation with state symbols.
20. Role of water and oxygen in rusting and prevention method. [3]
- Answer: Iron rusts in the presence of both water and oxygen. Water acts as an electrolyte, allowing ions to flow, while oxygen acts as the oxidising agent, accepting electrons from iron. Iron is oxidised to Fe²⁺, and oxygen is reduced to OH⁻. A method to prevent rusting (other than galvanising) is painting/oiling/greasing, which creates a barrier preventing water and oxygen from reaching the iron surface.
- Marking note: 1 mark for role of water, 1 mark for role of oxygen, 1 mark for a valid prevention method.
END OF ANSWER KEY