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Secondary 3 Chemistry Redox Electrochemistry Quiz

Free Sec 3 Chemistry Redox Electrochemistry quiz, HY3 Exam version, with questions, answers, and O Level-style practice for Singapore students.

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Secondary 3 Chemistry From Real Exams Generated by Tencent HY3 Free Updated 2026-08-17

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Secondary 3 Chemistry Quiz - Redox Electrochemistry: Answer Key

Total Marks: 40
Teaching notes are provided for each question.


Section A

1. [1] Redox (or reduction-oxidation)
Teaching: A redox reaction is one where reduction and oxidation happen simultaneously. "Redox" is the combined term.

2. [1] Zn (zinc atom)
Teaching: Zn loses electrons to become Zn2+\text{Zn}^{2+} (oxidation). The species oxidised is the one that loses electrons.

3. [1] +7
Teaching: K is +1, O is -2 (×4 = -8). For neutral KMnO4\text{KMnO}_4: +1 + x - 8 = 0 → x = +7.

4. [1] Galvanic cell (or electrochemical cell / voltaic cell)
Teaching: A galvanic cell uses spontaneous redox to produce electricity.

5. [1] from the anode (negative) to the cathode (positive)
Teaching: Electrons flow from the electrode where oxidation occurs (anode, negative) to where reduction occurs (cathode, positive).


Section B

6. [3]
(a) [1] Oxidation is loss of electrons.
(b) [1] Reduction is gain of electrons.
(c) [1] OIL RIG (Oxidation Is Loss, Reduction Is Gain).
Teaching: Electron transfer defines redox at Sec 3 level.

7. [3]
(a) [1] Cl2\text{Cl}_2 (it gains electrons to become Cl\text{Cl}^-).
(b) [2] 2Fe2++Cl22Fe3++2Cl2\text{Fe}^{2+} + \text{Cl}_2 \rightarrow 2\text{Fe}^{3+} + 2\text{Cl}^-
Teaching: Multiply Fe half-equation by 2 to balance electrons; add with Cl half-equation.

8. [4]
(a) [1] Reddish-brown copper metal deposits on Mg strip.
(b) [2] Mg+Cu2+Mg2++Cu\text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu}
(c) [1] Cu2+\text{Cu}^{2+} (it gains electrons, gets reduced).
Teaching: Mg is more reactive, displaces Cu²⁺.

9. [3]
(a) [2] Hydrogen gas at 1 atm, H+\text{H}^+ at 1 mol dm31\text{ mol dm}^{-3}, 25C25^\circ\text{C}, Pt electrode. (any two)
(b) [1] It is assigned a potential of 0.00 V as standard reference.
Teaching: SHE defines the zero point for measuring other potentials.

10. [3]
(a) [1] Zinc
(b) [1] Zn loses electrons more readily than Ag.
(c) [1] ZnZn2++2e\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-
Teaching: Better reducing agent = easier to oxidise = more negative EE^\circ.

11. [4]
(a) [2] Anode = Zn electrode (left), Cathode = Cu electrode (right).
(b) [1] From Zn to Cu (left to right).
(c) [1] ZnZn2++2e\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-
Teaching: Diagram must show Zn oxidised; electrons leave Zn through wire to Cu.

12. [3]
(a) [1] Species that accepts electrons / causes oxidation of another.
(b) [1] Acidified KMnO4\text{KMnO}_4 (or Cl2\text{Cl}_2, etc.)
(c) [1] It gains electrons from the other species, oxidising it.
Teaching: Oxidising agent itself is reduced.

13. [4]
(a) [1] Pb2++2ePb\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}
(b) [1] 2BrBr2+2e2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-
(c) [2] Bromine; test with starch-iodide paper turns blue-black, or vapour is red-brown.
Teaching: Molten salt electrolysis: cations to cathode, anions to anode.

14. [3]
(a) [1] Purple to colourless.
(b) [1] Mn2+\text{Mn}^{2+}
(c) [1] +7 to +2.
Teaching: In acid, manganate(VII) reduced from Mn(VII) to Mn(II).

15. [3]
(a) [1] I\text{I}^- (iodide ion)
(b) [1] -1 to 0.
(c) [1] Br2\text{Br}_2 gains electrons to become Br\text{Br}^-, oxidising I\text{I}^-.
Teaching: Reducing agent is oxidised; oxidising agent is reduced.


Section C

16. [4]
(a) [2] Ecell=EcathodeEanode=(0.44)(2.37)=+1.93 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.44) - (-2.37) = +1.93\text{ V}
(b) [2] Mg+Fe2+Mg2++Fe\text{Mg} + \text{Fe}^{2+} \rightarrow \text{Mg}^{2+} + \text{Fe}
Teaching: More negative electrode is anode (Mg). Cell potential positive for spontaneous.

17. [4]
(a) [2] n=c×V=0.020×(25.0/1000)=5.00×104 moln = c \times V = 0.020 \times (25.0/1000) = 5.00 \times 10^{-4}\text{ mol}
(b) [2] From equation 1:5 ratio → 5×5.00×104=2.50×103 mol Fe2+5 \times 5.00\times10^{-4} = 2.50\times10^{-3}\text{ mol Fe}^{2+}
Teaching: Convert cm³ to dm³ by ÷1000 before using n=cVn=cV.

18. [4]
(a) [1] Li (most negative EE^\circ = strongest reducer)
(b) [2] LiLi+F2,FPt\text{Li}|\text{Li}^+ || \text{F}_2,\text{F}^-|\text{Pt} (or LiLi+F2F\text{Li}|\text{Li}^+ || \text{F}_2|\text{F}^- with Pt)
(c) [1] 2.87(3.04)=5.91 V2.87 - (-3.04) = 5.91\text{ V}
Teaching: Max voltage from most negative and most positive couples.

19. [4]
(a) [2] Cu at cathode: Cu2++2eCu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}
(b) [2] O2\text{O}_2 at anode: 4OHO2+2H2O+4e\text{4OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^- (water oxidised, not sulphate).
Teaching: With inert electrodes in aqueous CuSO₄, Cu²⁺ reduced preferentially; OH⁻ oxidised at anode.

20. [4]
(a) [2] nZn=1.00/65.4=0.0153 moln_{\text{Zn}} = 1.00 / 65.4 = 0.0153\text{ mol}
(b) [2] 1:1 ratio → mCu=0.0153×63.5=0.971 gm_{\text{Cu}} = 0.0153 \times 63.5 = 0.971\text{ g}
Teaching: Mole ratio from equation is 1:1; use m=n×Mrm = n \times M_r.