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O Level Chemistry Redox Electrochemistry Quiz
Free O Level Chemistry Redox Electrochemistry quiz, HY3 Exam version, with questions, answers, and O Level-style practice for Singapore students.
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O-Level Chemistry Quiz - Redox Electrochemistry: Answer Key
Total Marks: 40
Topic: Redox Electrochemistry
Section A: Multiple Choice
1. C [1]
Teaching note: Reduction is gain of electrons. Cu²⁺ + 2e⁻ → Cu shows Cu²⁺ gaining 2 electrons to become Cu(s). A and D are oxidation (loss of e⁻); B is oxidation of Fe²⁺ to Fe³⁺.
2. C [1]
Teaching note: Potassium is highest in the reactivity series among the options; most reactive metals are strongest reducing agents (easily lose e⁻).
3. B [1]
Teaching note: At cathode (negative), Pb²⁺ ions gain electrons: Pb²⁺ + 2e⁻ → Pb. Bromide is oxidised at anode.
4. D [1]
Teaching note: K = +1, O = -2 × 4 = -8. For neutral compound: +1 + Mn + (-8) = 0 → Mn = +7.
5. A [1]
Teaching note: Fuel cell is an electrochemical cell that converts chemical energy of reactants directly to electrical energy.
Section B: Structured Short Answers
6. [1] Oxidation is loss of electrons (OIL).
Marking: 1 mark for "loss of electrons" or "electrons are removed".
7. [1] 4OH⁻ → O₂ + 2H₂O + 4e⁻
Marking: Correct formula and e⁻ on right. State symbols optional at this level but encourage (aq).
8. [2]
(a) [1] A redox reaction is one where both reduction and oxidation occur simultaneously.
(b) [1] Carbon monoxide (CO).
Note: CO reduces Fe₂O₃ by accepting oxygen; itself oxidised to CO₂.
9. [2]
(a) [1] Reddish-brown copper deposits on nail / blue solution fades.
(b) [1] Fe + Cu²⁺ → Fe²⁺ + Cu
Note: Fe is more reactive, displaces Cu²⁺.
10. [2]
Product: Chlorine gas (Cl₂). [1]
Explanation: At anode, Cl⁻ is preferentially discharged over OH⁻ in concentrated NaCl due to higher concentration of Cl⁻; 2Cl⁻ → Cl₂ + 2e⁻. [1]
11. [2]
(a) [1] -2 (H is +1, so 2(+1) + S = 0 → S = -2)
(b) [1] +4 (2(+1) + S + 2(-2) = 0 → S = +4)
12. [2]
Zinc is more reactive than iron, so it acts as sacrificial protector; [1] even if coating is scratched, Zn oxidises first, preventing Fe rusting. [1]
13. [3]
Oxidation: 2Br⁻ → Br₂ + 2e⁻ [1]
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ [1]
Overall: Cl₂ + 2Br⁻ → Br₂ + 2Cl⁻ [1]
14. [3]
(a) [2] Cathode: Aluminium (Al); Anode: Oxygen (O₂).
(b) [1] Cryolite lowers melting point of Al₂O₃, saving energy.
15. [2]
- Molten: products are Na and Cl₂; aqueous: products are H₂, Cl₂, NaOH. [1]
- In molten, Na⁺ reduced; in aqueous, H⁺ (from water) reduced instead at cathode. [1]
Section C: Data and Calculation
16. [3]
Step 1: Cu²⁺ + 2e⁻ → Cu, so 2 mol e⁻ deposit 1 mol Cu. [1]
Step 2: mol Cu = 0.025 / 2 = 0.0125 mol [1]
Step 3: mass = 0.0125 × 63.5 = 0.79375 ≈ 0.794 g [1]
Answer: 0.794 g
17. [3]
(a) [2] Highest voltage between Zn²⁺/Zn (-0.76) and Cu²⁺/Cu (+0.34): Cell = Zn | Zn²⁺ || Cu²⁺ | Cu, Voltage = 0.34 - (-0.76) = 1.10 V.
(b) [1] Zn + Cu²⁺ → Zn²⁺ + Cu
18. [2]
Step 1: From graph, at 10 min volume = 60 cm³. [1]
Step 2: Rate = 60 / 10 = 6.0 cm³/min. [1]
Answer: 6.0 cm³/min
19. [2]
(a) [1] 2H₂ + O₂ → 2H₂O
(b) [1] No CO₂ / pollutant emissions (only water produced).
20. [4]
(a) [1] Mg + Pb²⁺ → Mg²⁺ + Pb
(b) [3]
Step 1: mol Mg = 0.120 / 24.3 = 0.004938 mol [1]
Step 2: 1 mol Mg displaces 1 mol Pb, so mol Pb = 0.004938 mol [1]
Step 3: mass Pb = 0.004938 × 207.2 = 1.023 g ≈ 1.02 g [1]
Answer: 1.02 g