O Level Chemistry Redox Electrochemistry Quiz

O-Level Chemistry Quiz - Redox Electrochemistry

Answer Key and Marking Scheme

Total Marks: 50

Section A: Multiple Choice and Short Answer (10 marks)

1. C. Zn → Zn²⁺ + 2e⁻
Oxidation is the loss of electrons. Zinc atoms lose two electrons to form Zn²⁺ ions.
[1 mark]

2. A. Fe²⁺ is oxidised and Cl₂ is reduced
Fe²⁺ loses an electron (oxidation state increases from +2 to +3); Cl₂ gains electrons (oxidation state decreases from 0 to -1).
[1 mark]

3. Oxidation is the increase in oxidation state of an element.
Accept: Oxidation is the gain of oxygen / loss of hydrogen / loss of electrons, but "increase in oxidation state" is specifically requested.
[1 mark]

4. +6
Oxygen has oxidation state -2. Total for four O atoms = -8. Overall charge is -2. Therefore: S + (-8) = -2, so S = +6.
[1 mark]

5. CuSO₄ (or Cu²⁺ ions)
The oxidising agent is the substance that is reduced. Cu²⁺ ions gain electrons to form Cu, so CuSO₄/Cu²⁺ is the oxidising agent.
[1 mark]

6.
Colour change: Colourless solution turns brown (or yellow-brown / reddish-brown).
Substance responsible: Iodine (I₂).
Iodide ions (I⁻) are oxidised to iodine (I₂), which is brown in aqueous solution.
[2 marks — 1 for colour, 1 for substance]

7. The unknown solution does not contain a reducing agent (or the solution cannot reduce KMnO₄).
Acidified potassium manganate(VII) is an oxidising agent. It is decolourised (turns from purple to colourless) by reducing agents. If the purple colour remains, no reducing agent is present.
[1 mark]

8. Electrolysis is the decomposition of an ionic compound (electrolyte) by the passage of an electric current, causing chemical changes at the electrodes.
Accept: Electrolysis is the process by which electrical energy is used to bring about a chemical change in a molten or aqueous ionic compound.
[1 mark]

9. Chlorine gas (Cl₂)
At the anode, chloride ions (Cl⁻) are oxidised: 2Cl⁻ → Cl₂ + 2e⁻.
[1 mark]

10. In solid sodium chloride, the Na⁺ and Cl⁻ ions are held in fixed positions in a giant ionic lattice and are not free to move. Therefore, solid NaCl cannot conduct electricity. When molten, the ions are free to move and can carry the electric current, allowing conduction.
Key points: ions fixed in solid (1 mark); ions free to move in molten state (1 mark).
[2 marks]

Section B: Structured Questions (18 marks)

11. (a) Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Correct formulae and state symbols required. Spectator ion SO₄²⁻ omitted.
[2 marks — 1 for correct species, 1 for state symbols]

(b) Magnesium (Mg) is oxidised. Magnesium atoms lose two electrons to form Mg²⁺ ions: Mg → Mg²⁺ + 2e⁻.
Must identify Mg and explain in terms of electron loss.
[2 marks — 1 for identification, 1 for explanation]

(c) The blue colour of copper(II) sulfate solution is due to the presence of Cu²⁺ ions. During the reaction, Cu²⁺ ions are reduced to copper metal (Cu), so the concentration of Cu²⁺ ions decreases. As the blue Cu²⁺ ions are removed from the solution, the blue colour fades.
Must link colour to Cu²⁺ ions and explain their removal.
[2 marks — 1 for linking colour to Cu²⁺, 1 for explaining decrease]

12. (a) The colourless solution turns brown (or yellow-brown / reddish-brown).
Bromide ions are oxidised to bromine, which is brown in aqueous solution.
[1 mark]

(b) Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)
Correct species and state symbols required.
[2 marks — 1 for correct species, 1 for balancing and state symbols]

(c) Chlorine is more reactive than bromine (chlorine is above bromine in Group 17). Chlorine atoms gain electrons more readily than bromine atoms. Therefore, chlorine oxidises bromide ions: each chlorine atom gains one electron (is reduced) to form Cl⁻, while each bromide ion loses one electron (is oxidised) to form Br₂. The reaction occurs because a more reactive halogen can displace a less reactive halogen from its salt solution.
Must reference relative reactivity (1 mark), explain electron transfer (1 mark), and link to displacement principle (1 mark).
[3 marks]

13. (a) The zinc electrode acts as the negative terminal. Zinc is more reactive than copper (zinc is higher in the reactivity series), so zinc atoms lose electrons more readily than copper atoms. Electrons flow from the zinc electrode through the external circuit to the copper electrode, making zinc the negative terminal.
Must identify zinc and explain using reactivity series.
[2 marks — 1 for identification, 1 for explanation]

(b) Zn(s) → Zn²⁺(aq) + 2e⁻
Accept: Zn → Zn²⁺ + 2e⁻ with state symbols.
[1 mark]

(c)
Advantage: Hydrogen fuel cells produce only water as a waste product (no greenhouse gases or pollutants) / more efficient energy conversion / continuous operation with fuel supply.
Disadvantage: Hydrogen is flammable/explosive (storage and transport safety concerns) / hydrogen production may require fossil fuels / expensive technology / lack of refuelling infrastructure.
Any one valid advantage and one valid disadvantage.
[2 marks — 1 for advantage, 1 for disadvantage]

14. (a) The iron spoon should be connected to the negative terminal (cathode). During electroplating, the object to be plated is made the cathode so that positively charged metal ions (Ag⁺) in the electrolyte are attracted to it and reduced to form a coating of silver metal on the spoon.
Must state cathode/negative terminal and explain attraction and reduction of Ag⁺.
[2 marks — 1 for electrode, 1 for explanation]

(b) Ag⁺(aq) + e⁻ → Ag(s)
Correct half-equation with state symbols.
[1 mark]

Section C: Data Analysis and Extended Response (22 marks)

15. (a) Mg > Zn > Fe > Cu > Ag
Most reactive to least reactive. Deduced from the number of reactions each metal undergoes: Mg reacts with 4 solutions, Zn with 3, Fe with 2, Cu with 1, Ag with 0.
[2 marks — all correct for 2 marks; one error for 1 mark]

(b) Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s)
Correct species, balancing, and state symbols.
[2 marks — 1 for correct species, 1 for state symbols]

(c) Copper is less reactive than zinc (copper is below zinc in the reactivity series). Copper atoms cannot lose electrons as readily as zinc atoms, so copper cannot displace zinc ions from zinc nitrate solution. For a displacement reaction to occur, the added metal must be more reactive than the metal in the compound.
Must reference relative reactivity and inability of Cu to displace Zn²⁺.
[2 marks — 1 for reactivity comparison, 1 for explanation of no displacement]

16. (a)
Product: Copper (Cu)
Half-equation: Cu²⁺(aq) + 2e⁻ → Cu(s)
Copper is less reactive than hydrogen, so Cu²⁺ ions are discharged at the cathode in preference to H⁺ ions.
[2 marks — 1 for product, 1 for half-equation]

(b)
Product: Oxygen gas (O₂)
Explanation: In aqueous solution, OH⁻ ions (from the dissociation of water) are discharged at the anode in preference to SO₄²⁻ ions because OH⁻ ions are more easily oxidised than SO₄²⁻ ions. The sulfate ions remain in solution.
Must name oxygen, reference OH⁻ vs SO₄²⁻ discharge, and explain selective discharge.
[3 marks — 1 for product, 2 for explanation]

(c)
At the cathode: A pink/brown deposit of copper metal forms on the electrode.
At the anode: Bubbles of a colourless gas (oxygen) are evolved.
Accept any reasonable observation descriptions.
[2 marks — 1 for each electrode]

(d) The blue colour of the electrolyte fades (becomes paler). The blue colour is due to Cu²⁺ ions. At the cathode, Cu²⁺ ions are discharged and removed from the solution as copper metal. At the anode, OH⁻ ions are discharged to form oxygen gas, but Cu²⁺ ions are not replaced. Therefore, the concentration of Cu²⁺ ions in the electrolyte decreases, causing the blue colour to fade.
Must link colour to Cu²⁺ concentration and explain decrease at cathode without replacement at anode.
[3 marks — 1 for observation, 1 for linking colour to Cu²⁺, 1 for explaining concentration decrease]

17. (a)
Anode: Impure copper
Cathode: Pure copper
Both required.
[1 mark]

(b) Cu(s) → Cu²⁺(aq) + 2e⁻
Oxidation of copper at the anode.
[1 mark]

(c) At the anode, copper atoms are oxidised to Cu²⁺ ions, which enter the electrolyte. At the cathode, Cu²⁺ ions from the electrolyte are reduced to copper atoms, which are deposited on the pure copper cathode. The rate at which Cu²⁺ ions enter the solution at the anode is approximately equal to the rate at which Cu²⁺ ions are removed from the solution at the cathode. Therefore, the concentration of Cu²⁺ ions remains approximately constant.
Must describe both electrode processes and explain the balance.
[2 marks — 1 for describing both processes, 1 for explaining the balance]

(d) The impurities fall to the bottom of the cell as anode sludge (or anode mud).
Accept: Impurities collect below the anode.
[1 mark]

18. (a) 2H₂(g) + O₂(g) → 2H₂O(l)
Correct equation with state symbols.
[1 mark]

(b) O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
Accept: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (if using alkaline electrolyte). Either version is acceptable.
[1 mark]

(c)

1. Hydrogen fuel cells produce only water as the product (no CO₂, SO₂, or NOₓ emissions), so they do not contribute to global warming or acid rain.
2. Hydrogen fuel cells are more efficient at converting chemical energy to electrical energy compared to combustion engines, reducing overall energy consumption.
   Accept any two valid environmental or efficiency reasons. Other acceptable answers: renewable if hydrogen is produced from water using renewable energy; no particulate emissions; quieter operation.
   [2 marks — 1 for each valid reason]

19.
Test 1: Add aqueous potassium iodide (KI) to the unknown solution.

• If the solution contains an oxidising agent: The colourless solution turns brown because iodide ions (I⁻) are oxidised to iodine (I₂).
• If the solution contains a reducing agent: No colour change is observed (or the solution remains colourless).

Test 2: Add acidified potassium manganate(VII) (KMnO₄) to the unknown solution.

• If the solution contains a reducing agent: The purple potassium manganate(VII) solution is decolourised (turns from purple to colourless) because MnO₄⁻ ions are reduced to Mn²⁺ ions.
• If the solution contains an oxidising agent: No colour change is observed (the purple colour remains).

Must describe both tests with reagents and expected observations for both types of agents. Award marks for: Test 1 reagent and observations (2 marks); Test 2 reagent and observations (2 marks).
[4 marks — 2 for each test with correct reagent and observations for both agent types]

20.
Molten sodium chloride:

• Cathode product: Sodium metal (Na). Half-equation: Na⁺ + e⁻ → Na(l)
• Anode product: Chlorine gas (Cl₂). Half-equation: 2Cl⁻ → Cl₂(g) + 2e⁻
• Only Na⁺ and Cl⁻ ions are present, so these are the only ions discharged.

Concentrated aqueous sodium chloride:

• Cathode product: Hydrogen gas (H₂). Half-equation: 2H⁺(aq) + 2e⁻ → H₂(g)
• Anode product: Chlorine gas (Cl₂). Half-equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
• At the cathode, H⁺ ions (from water) are discharged in preference to Na⁺ ions because H⁺ ions are more easily reduced than Na⁺ ions (sodium is more reactive than hydrogen).
• At the anode, Cl⁻ ions are discharged in preference to OH⁻ ions because the solution is concentrated (high concentration of Cl⁻ ions favours their discharge).

Differences at the anode:

• In both cases, chlorine gas is produced at the anode because chloride ions are present and are discharged. However, in dilute aqueous sodium chloride, OH⁻ ions would be discharged at the anode instead, producing oxygen gas. The concentration of chloride ions affects which ion is discharged.

Key comparison points:

• Cathode products differ: sodium (molten) vs. hydrogen (aqueous) due to the presence of H⁺ ions from water in the aqueous solution.
• Anode products are the same (chlorine) when the solution is concentrated, but would differ if the solution were dilute.
• The presence of water in aqueous electrolysis introduces H⁺ and OH⁻ ions, which compete with Na⁺ and Cl⁻ for discharge.

Marking scheme:

• Products at both electrodes for molten NaCl (2 marks)
• Products at both electrodes for aqueous NaCl (2 marks)
• Explanation of difference at cathode (1 mark)
• Explanation of difference/competition at anode (1 mark)
• Relevant half-equations (1 mark for at least two correct half-equations)
• Overall coherence and use of correct terminology (1 mark)

[6 marks]

END OF ANSWER KEY