Secondary 4 Combined Science Chemistry Redox Electrochemistry Quiz

Answer Key: Secondary 4 Combined Science Chemistry Quiz - Redox Electrochemistry

Total Marks: 40
Topic: Redox Electrochemistry
Note: This quiz is syllabus-first generated from LLM-inferred templates. Past-paper evidence for this topic was limited (52 references in Stage 2-1), so questions are designed to match syllabus expectations and familiar Singapore practice style, not claimed as exam-derived.

Section A (2 marks each)

1. C [2]
Oxidation is loss of electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain). Gain of electrons is reduction; loss of oxygen is reduction; decrease in oxidation state is reduction.

2. B [2]
$Cu^{2+}$ gains 2 electrons to become $Cu(s)$: $Cu^{2+} + 2e^- \rightarrow Cu$. Gain of electrons = reduction.

3. D [2]
In $KMnO_4$: K = +1, O = -2 (×4 = -8). Sum = 0, so Mn + 1 - 8 = 0 → Mn = +7.

4. B [2]
Zinc is more reactive (higher in series) so it is oxidised, releasing electrons; it is the negative terminal (anode).

5. C [2]
With inert electrodes and $Cu^{2+}$ present, $Cu^{2+}$ is discharged at cathode: $Cu^{2+} + 2e^- \rightarrow Cu(s)$.

Section B

6. [2]
A redox reaction is a reaction where both reduction and oxidation occur simultaneously (electron transfer).
Marking: 1 mark for simultaneous oxidation+reduction, 1 mark for electron transfer idea.

7. [2]
$Fe^{2+} \rightarrow Fe^{3+} + e^-$
Marking: correct species 1, electron on product side 1.

8. [3]
Let oxidation state of S = x.
H = +1 (×2 = +2), O = -2 (×4 = -8).
+2 + x - 8 = 0 → x = +6.
Answer: +6.
Marking: working 2, answer 1.

9. [3]
(a) Mg (anode, oxidised) [1]
(b) $Mg(s) + 2Ag^+(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)$ [2]
Mg is above Ag in reactivity series, so Mg loses electrons.

10. [2]
Anode: bromine ($Br_2$) [1]
Cathode: lead ($Pb$) [1]
Molten $PbBr_2$: $2Br^- \rightarrow Br_2 + 2e^-$; $Pb^{2+} + 2e^- \rightarrow Pb$.

11. [2]
A salt bridge completes the circuit by allowing ion flow / maintains electrical neutrality in half-cells.
Marking: complete circuit 1, ion/neutrality 1.

12. [3]
Yes, copper displaces silver ions [1] because Cu is above Ag in reactivity series [1], so Cu is oxidised and $Ag^+$ reduced [1].
Equation: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.

13. [3]
Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ (×5)
Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Balanced: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$
Marking: balancing Fe 1, balancing Mn 1, H+/H2O 1.

14. [2]
Gas: hydrogen ($H_2$) [1]
Test: lighted splint gives pop sound [1].
From diagram: cathode (-) produces H2 from water; pop test confirms.

15. [3]
(a) Silver (Ag) [1]
(b) $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$ [2]

Section C

16. [4]
(a) Anode: copper electrode dissolves: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ [2]
(b) Cathode: copper ions deposit: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ [2]

17. [3]
(a) $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = +1.10$ V [2]
(b) From Zn to Cu (Zn is anode, electrons flow out) [1]

18. [4]
At anode, possible ions: $Cl^-$ and $OH^-$ (from water). In concentrated KCl, $Cl^-$ concentration is high; $Cl^-$ is discharged preferentially over $OH^-$ because of lower discharge preference / higher concentration: $2Cl^- \rightarrow Cl_2 + 2e^-$. Oxygen is not formed because $Cl^-$ oxidises more readily under these conditions.
Marking: identify Cl- discharged 1, reason concentration 1, equation 1, contrast with OH-/O2 1.

19. [3]
Mass at 5 min = 0.25 g (from linear interpolation) [1]
Rate constant because graph is straight line / equal mass per equal time [2].

20. [6]
Voltaic cell: converts chemical energy to electrical energy (spontaneous redox), e.g., Zn-Cu cell, battery. [3]
Electrolytic cell: uses electrical energy to drive non-spontaneous reaction, e.g., electrolysis of water, electroplating. [3]
Marking: energy change each 2, example each 1.