Year 11 Chemistry Module 3 Module Quiz

Module 3 Quiz — Reactive Chemistry

A comprehensive assessment covering all 12 lessons of Module 3: chemical change, reaction types, precipitation, combustion, acid-base, redox, galvanic cells, collision theory, and rate factors.

⏱ ~35 min 15 MC · 2 Extended Response All 12 Lessons 35 marks total
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Module 3 at a Glance

L01
Physical & Chemical Change
L02
Synthesis & Decomposition
L03
Precipitation & Solubility
L04
Combustion Reactions
L05
Acid-Base Reactions
L06
Indigenous Detoxification
L07
Metal Activity Series
L08
Redox & Oxidation States
L09
Galvanic Cells
L10
Inert Electrodes & Cathodic Protection
L11
Collision Theory & Rate
L12
Factors Affecting Rate

Section I — Multiple Choice (15 marks)

Allow about 20 minutes. Select the best answer for each question. 1 mark each.

L01 — Evidence of Chemical Change

1. A student heats copper metal in air and observes the surface turning black. This is best described as a chemical change because:

A The copper melts and changes state
B The colour changes, which always indicates a chemical change
C A new substance (copper(II) oxide) with different properties is formed, and the process is not easily reversed
D Energy is absorbed from the surroundings as the copper heats up
L02 — Balancing Equations

2. What are the correct coefficients (in order) for the balanced equation: —Al + —O₂ → —Al₂O₃?

A 1, 1, 1
B 2, 1, 1
C 4, 3, 2
D 2, 3, 2
L03 — Solubility Rules

3. Which of the following ionic compounds is insoluble in water according to the solubility rules?

A Potassium sulfate (K₂SO₄)
B Sodium carbonate (Na₂CO₃)
C Calcium carbonate (CaCO₃)
D Ammonium chloride (NH₄Cl)
L04 — Combustion

4. A hydrocarbon burns in a limited supply of oxygen, producing a yellow sooty flame. Which of the following best describes what has occurred?

A Complete combustion: CO₂ and H₂O are the sole products
B Complete combustion: the yellow colour is caused by excited sodium ions
C Incomplete combustion: only CO₂ and water vapour are produced but at lower temperatures
D Incomplete combustion: carbon monoxide and/or carbon soot are produced due to insufficient oxygen
L05 — Acid-Base Reactions

5. Which of the following represents the balanced equation for the reaction between sulfuric acid and potassium hydroxide?

A H₂SO₄ + KOH → KSO₄ + H₂O
B H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
C 2H₂SO₄ + KOH → K(SO₄)₂ + H₂O
D H₂SO₄ + KOH → K₂SO₄ + H₂
L06 — Reaction Classification

6. When cycad seeds are washed in running water, the water-soluble neurotoxin BMAA is removed. When Fe₂O₃ reacts with HCl to form FeCl₃ and H₂O, this is a separate reaction. Which of the following correctly classifies BOTH processes?

A Both are chemical changes involving new bond formation
B Both are physical changes: no new substances are formed
C Washing cycad seeds is a physical change (leaching); Fe₂O₃ + HCl is a chemical change (acid-base reaction)
D Washing cycad seeds is a chemical change (decomposition); Fe₂O₃ + HCl is a physical change
L07 — Activity Series Prediction

7. A student places a strip of aluminium metal in copper(II) sulfate solution. Based on the NESA activity series (K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au), which prediction is correct?

A Aluminium displaces copper; a red-brown solid forms and the blue solution fades: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu
B Copper displaces aluminium; aluminium dissolves and a blue solution intensifies
C No reaction; aluminium and copper have similar reactivity
D No reaction; aluminium is a less reactive metal than copper
L08 — Redox & Oxidation States

8. In the reaction Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), which of the following statements about oxidation and reduction is correct?

A Zinc is reduced (gains electrons) and acts as the oxidising agent
B Zinc is oxidised (oxidation state 0 → +2) and H⁺ is reduced (oxidation state +1 → 0)
C Chlorine is oxidised and zinc is the reducing agent
D H₂ is oxidised upon formation and zinc acts as the catalyst
L08–L09 — Half-Equations & Cell Notation

9. Which of the following correctly shows the two half-equations and the overall equation for the reaction in a zinc-copper galvanic cell?

A Anode: Cu → Cu²⁻ + 2e⁻; Cathode: Zn²⁻ + 2e⁻ → Zn; Overall: Cu + Zn²⁻ → Cu²⁻ + Zn
B Anode: Zn → Zn²⁻ + 2e⁻; Cathode: Cu²⁻ + 2e⁻ → Cu; Overall: Zn + Cu²⁻ → Zn²⁻ + Cu
C Anode: Zn²⁻ + 2e⁻ → Zn; Cathode: Cu → Cu²⁻ + 2e⁻; Overall: Cu + Zn²⁻ → Zn + Cu²⁻
D Anode: Zn + 2e⁻ → Zn²⁻; Cathode: Cu²⁻ → Cu + 2e⁻; Overall: Zn²⁻ + Cu → Zn + Cu²⁻
L09–L10 — E°cell and Spontaneity

10. Using standard reduction potentials: E°(Fe²⁻/Fe) = −0.44 V and E°(Ag⁺/Ag) = +0.80 V. A galvanic cell is constructed with an iron anode and a silver cathode. Which values are correct?

A E°cell = −0.44 − 0.80 = −1.24 V; non-spontaneous
B E°cell = 0.80 + 0.44 = +1.24 V by adding both values
C E°cell = E°cathode − E°anode = +0.80 − (−0.44) = +1.24 V; spontaneous
D E°cell = −0.44 + 0.80 = +0.36 V; spontaneous but iron is the cathode
L10 — Cathodic Protection

11. A zinc block is attached to the hull of a steel ship to prevent corrosion. Which statement correctly explains how this works?

A Zinc is less reactive than iron and forms a protective oxide layer over the steel
B Zinc has a lower reduction potential than iron, so zinc is preferentially oxidised; the steel hull acts as the cathode and is protected from corrosion
C Zinc ions from the block dissolve into the seawater and neutralise the corrosive salt ions
D Zinc transfers its electrons directly to the rust on the steel, chemically reducing it back to iron
L11 — Collision Theory

12. At a given temperature, only a small fraction of collisions between reactant molecules result in a reaction. Which of the following best explains why?

A Most molecules are moving too slowly to collide with each other
B Most molecules are too far apart to be influenced by each other’s electric fields
C Most collisions involve molecules with the correct orientation but insufficient energy
D Most collisions lack either sufficient energy (kinetic energy < Eₐ) or the correct orientation — both conditions must be satisfied simultaneously
L12 — Surface Area vs Concentration

13. A student investigates the reaction of limestone (CaCO₃) with HCl. She compares large limestone lumps with powdered limestone at the same concentration of HCl. Which of the following correctly identifies the effect of grinding on collision frequency and on the proportion of collisions that are effective?

A Collision frequency: unchanged; proportion effective: increases (higher energy particles exposed)
B Collision frequency: increases (more surface area); proportion effective: unchanged (same Eₐ, same temperature)
C Collision frequency: decreases (particles are smaller, harder to collide with); proportion effective: unchanged
D Both collision frequency and proportion effective increase when the surface area is increased
L12 — Catalyst Effect on Diagrams

14. On an energy distribution diagram (Maxwell-Boltzmann), adding a catalyst at constant temperature is shown by:

A Shifting the entire curve to the right, increasing the area beyond Eₐ
B Raising the curve peak, indicating more energetic particles
C Moving the Eₐ line to the right, reducing the proportion of reactive particles
D Moving the Eₐ line to the left (lower Eₐ); the curve is unchanged and more particles now exceed the new lower threshold
L07–L12 — Module Synthesis

15. A galvanic cell operates at 25°C using the reaction Mg(s) + Fe²⁻(aq) → Mg²⁻(aq) + Fe(s). E°(Mg²⁻/Mg) = −2.37 V; E°(Fe²⁻/Fe) = −0.44 V. A student then adds a catalyst to the cell electrolyte. Which correctly predicts the effect on E°cell and the rate of reaction?

A E°cell increases and rate increases; the catalyst raises both the cell voltage and the reaction rate
B E°cell decreases and rate increases; the catalyst lowers cell voltage by reducing activation energy
C E°cell is unchanged (+1.93 V) and rate increases; catalysts lower Eₐ (kinetics) but do not change ΔH or E°cell (thermodynamics)
D E°cell is unchanged and rate is unchanged; catalysts only affect equilibrium constants, not reaction rates

Section II — Extended Response (20 marks)

Allow about 15 minutes. Write extended answers in full sentences. Show all working for calculations.

L08–L10 — Galvanic Cells & Redox

Question 16 (10 marks)

A galvanic cell is constructed using the following half-reactions. Standard reduction potentials are provided.

Half-reactionE° (V)
Mg²⁻(aq) + 2e⁻ → Mg(s)−2.37
Fe²⁻(aq) + 2e⁻ → Fe(s)−0.44
Cu²⁻(aq) + 2e⁻ → Cu(s)+0.34
Ag⁺(aq) + e⁻ → Ag(s)+0.80

(a) For a galvanic cell constructed from a magnesium electrode and a copper electrode:

  • (i) Identify the anode and cathode. (1 mark)
  • (ii) Write the half-equation occurring at each electrode and the balanced overall cell equation. (2 marks)
  • (iii) Calculate E°cell and state whether the cell reaction is spontaneous. (1 mark)
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(b) Explain why a salt bridge is required in this galvanic cell and describe the direction of ion flow through it. (2 marks)

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(c) A new cell is constructed using a platinum electrode in a solution containing both Fe²⁻(aq) and Fe³⁻(aq) as one half-cell, and a silver electrode in AgNO₃(aq) as the other.

  • (i) Explain why platinum is used instead of an active metal at the Fe²⁻/Fe³⁻ half-cell. (1 mark)
  • (ii) Write the half-equation occurring at the platinum electrode. (1 mark)
  • (iii) Calculate E°cell for this cell. E°(Fe³⁻/Fe²⁻) = +0.77 V; E°(Ag⁺/Ag) = +0.80 V. State whether the reaction is spontaneous. (2 marks)
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L11–L12 — Collision Theory & Reaction Rate

Question 17 (10 marks)

A student investigates the reaction: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

She measures the volume of CO₂ produced over time under four different conditions, changing one variable at a time from the control (marble chips, 1.0 mol/L HCl, 25°C, no catalyst).

(a) The student records the following data for the control condition:

Time (s)0306090120150180
Volume CO₂ (mL)0183038434648
  • (i) Calculate the average rate of CO₂ production between 0 and 60 seconds. Include units. (1 mark)
  • (ii) Explain why the rate of reaction decreases between 0 and 180 seconds, using collision theory. (2 marks)
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(b) The student repeats the experiment at 45°C (all other variables unchanged).

  • (i) Describe, with reference to a Maxwell-Boltzmann energy distribution diagram, how the increased temperature changes the proportion of particles that can react. (3 marks)
  • (ii) Predict two observable differences between the 45°C experiment and the control. (1 mark)
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(c) In a separate experiment, a heterogeneous catalyst is added at 25°C.

  • (i) Explain how the catalyst increases the rate of reaction. (2 marks)
  • (ii) Explain whether adding the catalyst will change the total amount of CO₂ produced (assuming excess HCl). (1 mark)
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  • Q1 — C: A chemical change produces a new substance with different properties. Copper(II) oxide (black solid) has different properties from copper (orange metal) and is not easily converted back — this is the key test for a chemical change. Colour change alone (option B) is not sufficient evidence; melting and heating are physical effects.
  • Q2 — C: 4Al + 3O₂ → 2Al₂O₃. Check: Al left = 4; right = 4 ✓. O left = 6; right = 6 ✓.
  • Q3 — C: Carbonates are generally insoluble EXCEPT for Group 1 (Na, K, etc.) and ammonium carbonates. Potassium sulfate and ammonium chloride are soluble; sodium carbonate is soluble (Group 1 carbonate). Calcium carbonate is insoluble — this is why shells, limestone, and marble are solid.
  • Q4 — D: Yellow sooty flame is characteristic of incomplete combustion — insufficient oxygen causes carbon to be incompletely oxidised, producing CO and/or carbon soot (C) rather than complete oxidation to CO₂.
  • Q5 — B: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. Check: H 2+2=4; right 4 ✓. K 2=2 ✓. S 1=1 ✓. O 4+2=6; right 4+2=6 ✓.
  • Q6 — C: Washing seeds in water is a physical process — BMAA simply dissolves and diffuses out; no new substances are formed. Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O is an acid-base (chemical) reaction that produces new substances.
  • Q7 — A: Aluminium is above copper in the NESA activity series → Al is more reactive → Al displaces Cu from CuSO₄. 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu (check: Al 2=2; Cu 3=3; S 3=3; O 12=12 ✓).
  • Q8 — B: Zn goes from oxidation state 0 to +2 (loses electrons) → oxidised. H⁺ goes from +1 to 0 (gains electrons in H₂) → reduced. Zn is the reducing agent; H⁺ is the oxidising agent. Chloride does not change oxidation state.
  • Q9 — B: Oxidation (loss of electrons) occurs at the anode; reduction (gain of electrons) occurs at the cathode. In Zn/Cu cell: Zn is more reactive (higher in activity series) → anode. Cu²⁻ ions are reduced → cathode. Electrons flow from Zn to Cu in external circuit.
  • Q10 — C: E°cell = E°cathode − E°anode = +0.80 − (−0.44) = +1.24 V > 0 → spontaneous. Never add E° values directly (option B is wrong). Never multiply the anode’s sign (option A).
  • Q11 — B: E°(Zn²⁻/Zn) = −0.76 V < E°(Fe²⁻/Fe) = −0.44 V: zinc has the lower reduction potential and is therefore more readily oxidised. Zinc acts as the sacrificial anode — it is oxidised preferentially, protecting the steel hull (cathode) from oxidation/corrosion.
  • Q12 — D: For an effective collision, two conditions must BOTH be met: (1) kinetic energy ≥ Eₐ AND (2) correct orientation. At any given temperature, many collisions fail on one or both counts. Simply colliding is not enough.
  • Q13 — B: Grinding increases the exposed surface area of limestone → more CaCO₃ surface particles are accessible to H⁺ ions → collision frequency at the reaction interface increases. The activation energy is unchanged (same reaction, same temperature) → the proportion of those collisions that are effective (proportion exceeding Eₐ) is unchanged.
  • Q14 — D: A catalyst lowers Eₐ (moves the threshold line left) without changing the distribution of particle energies (temperature is constant, so the curve is unchanged). More of the existing distribution now lies to the right of the new (lower) threshold → larger fraction of particles can react.
  • Q15 — C: E°cell = −0.44 − (−2.37) = +1.93 V. This is a thermodynamic quantity determined by E° values (which depend on ΔH, ΔS) — catalysts do not change thermodynamics. A catalyst lowers the kinetic barrier (Eₐ) so the cell reaction proceeds faster, but E°cell and ΔH are unchanged.

Q16(a)(i): Anode = Mg (lower reduction potential, E° = −2.37 V; Mg is oxidised). Cathode = Cu (higher reduction potential, E° = +0.34 V; Cu²⁻ is reduced).

Q16(a)(ii): Anode: Mg(s) → Mg²⁻(aq) + 2e⁻. Cathode: Cu²⁻(aq) + 2e⁻ → Cu(s). Overall: Mg(s) + Cu²⁻(aq) → Mg²⁻(aq) + Cu(s).

Q16(a)(iii): E°cell = E°cathode − E°anode = +0.34 − (−2.37) = +2.71 V. Positive E°cell → reaction is spontaneous under standard conditions.

Q16(b): The salt bridge maintains electrical neutrality in both half-cell solutions. As the cell operates, Mg²⁻ ions build up in the anode compartment (making it positive) and Cu²⁻ ions are removed from the cathode compartment (making it negative). Without a salt bridge, this charge imbalance would stop the flow of electrons through the external circuit. The salt bridge allows ions to migrate: anions (e.g. Cl⁻ from KCl salt bridge) migrate toward the anode compartment; cations (e.g. K⁺) migrate toward the cathode compartment, maintaining charge balance.

Q16(c)(i): Platinum (an inert electrode) is used because the Fe²⁻/Fe³⁻ half-reaction involves only ions in solution — there is no solid metal electrode reactant or product. An active metal electrode would dissolve into the electrolyte and interfere with the cell reaction. Platinum conducts electrons without reacting.

Q16(c)(ii): At the platinum electrode, Fe²⁻ is oxidised: Fe²⁻(aq) → Fe³⁻(aq) + e⁻. This is the anode half-equation (oxidation occurs at the anode; anode E° = +0.77 V for the Fe³⁻/Fe²⁻ couple).

Q16(c)(iii): Anode = Pt|Fe²⁻/Fe³⁻ (E° = +0.77 V). Cathode = Ag⁺/Ag (E° = +0.80 V). E°cell = +0.80 − (+0.77) = +0.03 V. E°cell > 0 → spontaneous, but only marginally. The cell operates with iron(II) being oxidised at the platinum anode and silver ions being reduced at the silver cathode.

Q17(a)(i): Average rate = ΔV / Δt = (30 − 0) mL / (60 − 0) s = 30/60 = 0.50 mL/s.

Q17(a)(ii): The rate decreases because HCl is consumed as the reaction proceeds. As [HCl] decreases, there are fewer H⁺ ions per unit volume in solution. This means H⁺ ions collide with the CaCO₃ surface less frequently — collision frequency decreases. The activation energy is unchanged at constant temperature, so the proportion of collisions that are effective (energy ≥ Eₐ) is also unchanged. Therefore the number of effective collisions per second decreases → reaction rate decreases, approaching zero as one reactant is fully consumed.

Q17(b)(i): At higher temperature (45°C vs 25°C), the Maxwell-Boltzmann energy distribution changes as follows: (1) the curve shifts to the right — particles have higher average kinetic energy; (2) the peak of the curve decreases in height and becomes broader — the same number of particles is now spread over a wider range of energies; (3) the total area under the curve remains equal — the number of particles is unchanged. The activation energy line (Eₐ) does not move — it is fixed by the reaction mechanism. However, the area under the curve to the right of Eₐ is now significantly larger at 45°C. This means a much greater proportion of particles have kinetic energy ≥ Eₐ and can undergo effective collisions. More effective collisions per second → reaction rate increases significantly.

Q17(b)(ii): At 45°C compared to 25°C: (1) the initial rate is steeper / CO₂ is produced faster initially; (2) the total time to complete the reaction is shorter (reaction reaches plateau earlier). Note: the total amount of CO₂ produced is the same in both experiments (same mass of CaCO₃ and excess HCl).

Q17(c)(i): The heterogeneous catalyst provides an alternative reaction pathway with a lower activation energy. On a Maxwell-Boltzmann energy distribution diagram at 25°C, the curve is unchanged but a new Eₐ(cat) line is drawn to the left of the original Eₐ. A greater proportion of particles now have kinetic energy ≥ Eₐ(cat) → more effective collisions per second → rate increases. The catalyst is not consumed (it is regenerated at the end of each catalytic cycle).

Q17(c)(ii): The total amount of CO₂ produced will be unchanged. The total amount of CO₂ produced depends only on the amount of CaCO₃ (limiting reagent with excess HCl) — once all CaCO₃ is consumed, the reaction stops. The catalyst only affects the rate (how quickly the reaction proceeds), not the thermodynamic outcome (what products are formed or how much). ΔH, the overall equation, and therefore the stoichiometric yield are all unchanged by the catalyst.

Track Your Score

Section I — Multiple Choice (Q1–15) / 15
Section II — Q16 Galvanic Cells / 10
Section II — Q17 Reaction Rate / 10
Total — / 35
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