Topic 4 Exam Practice: Chemical Changes — AQA GCSE Chemistry

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📋 Structured Questions

These questions test key concepts from Topic 4. Attempt each question on paper, then click "Show Mark Scheme" to check your answer.

(a) Explain what happens to the pH of an acid as the acid is diluted with water. [2]

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pH increases [1]
(because) the concentration of hydrogen ions decreases [1]

Examiner tip: Many students think diluting an acid decreases pH — it's the opposite! As H⁺ concentration drops, pH rises (towards 7). Remember: pH is logarithmic, so a 10× dilution raises pH by about 1.

Nickel is extracted from nickel oxide by reduction with carbon: NiO + C → Ni + CO

(a) Explain why carbon can be used to extract nickel from nickel oxide. [2]

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Carbon is more reactive (than nickel) [1]
(so) carbon will displace nickel (from nickel oxide) OR carbon will remove oxygen (from nickel oxide) [1]

Examiner tip: Don't just say "carbon reacts with oxygen" — you must specify that it removes oxygen from the nickel oxide. Avoid answering in terms of electron transfer here as students often make errors with that approach.

Fe₂O₃ + 3C → 2Fe + 3CO

(a) Which substance in the equation is reduced? Give one reason for your answer. Answer in terms of oxygen. [2]

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Fe₂O₃ / iron oxide [1]
(Fe₂O₃) loses oxygen [1]

Examiner tip: Read the command words! The question says "in terms of oxygen", so mentioning electron transfer will be ignored. A common mistake is identifying "Fe" (the element) instead of "Fe₂O₃" (the compound) as the substance being reduced.

(a) Aqueous sodium chloride solution is electrolysed to produce an alkaline solution. Explain how the alkaline solution is produced. You should refer to the processes at the electrodes. [3]

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Sodium ions and hydroxide ions are left (in solution) [1]
(because) hydrogen ions are discharged / reduced at the negative electrode to form hydrogen [1]
(and because) chloride ions are discharged / oxidised at the positive electrode to form chlorine [1]

Examiner tip: Don't confuse molten vs aqueous electrolysis rules. In aqueous NaCl, water provides H⁺ and OH⁻ ions. Because sodium is more reactive than hydrogen, H₂ is produced at the cathode — leaving Na⁺ and OH⁻ behind to form NaOH (alkaline).

Links to: Topic 2 — Ionic Bonding (ions must be free to move for electrolysis to work)

A student wants to make pure, dry crystals of copper(II) sulfate from solid copper(II) oxide and dilute sulfuric acid.

(a) Describe the method the student should use. [4]

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Add copper(II) oxide to the sulfuric acid and stir until the copper(II) oxide is in excess / until solid remains [1]
Filter the mixture to remove the excess copper(II) oxide [1]
Heat the filtrate / copper sulfate solution to evaporate some of the water / until crystallisation point is reached [1]
Leave the solution to cool and crystallise, then pat the crystals dry with filter paper [1]

Examiner tip: Never write "evaporate to dryness" or "boil the acid" — you must specify heating the solution to crystallisation point or evaporating some of the water. Ensure you explicitly mention filtering to remove the excess solid, not just "impurities."

Aqueous sodium chloride solution (brine) is electrolysed using inert electrodes.

(a) Explain the chemical processes occurring during this electrolysis. In your answer, you should name the products formed at each electrode, explain how they are formed using half-equations, and state what remains in the solution. [6]

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Level 3 (5–6 marks): Accurate chemical understanding demonstrated. Ion movement linked to correct electrodes, products correctly identified, oxidation and reduction described with balanced half-equations.

Level 2 (3–4 marks): Correct products at electrodes and remaining solution identified, with an attempt to link ion discharge, but may lack half-equations.

Level 1 (1–2 marks): Basic information given, such as identifying a single product or stating ions move to oppositely charged electrodes.

Indicative content:

Ions present: Na⁺, Cl⁻, H⁺, and OH⁻
Cathode (−): hydrogen gas produced (H is less reactive than Na); 2H⁺ + 2e⁻ → H₂ (reduction)
Anode (+): chlorine gas produced (halide ion present); 2Cl⁻ → Cl₂ + 2e⁻ (oxidation)
Remaining solution: Na⁺ and OH⁻ ions remain → sodium hydroxide (NaOH)

Examiner tip: Use OILRIG — Oxidation Is Loss, Reduction Is Gain of electrons. Ensure you assign oxidation to the anode and reduction to the cathode. Always state "gas" when naming electrode products, and mention why hydrogen is produced instead of sodium (reactivity).

A student makes magnesium chloride crystals from magnesium oxide and dilute hydrochloric acid.

(a) Give a reason for each of these three steps:

(i) Step 2: Warm the hydrochloric acid. [1]

(ii) Step 5: Add magnesium oxide until it is in excess. [1]

(iii) Step 6: Filter the mixture. [1]

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(Step 2) To speed up the reaction [1]
(Step 5) To make sure all the (hydrochloric) acid reacts [1]
(Step 6) To remove the excess magnesium oxide [1]

Examiner tip: Students memorise the method but often can't explain why each step is done. Adding excess base ensures all the acid is neutralised (no hazardous unreacted acid). "Filtering to remove impurities" is too vague — you must say "to remove excess unreacted solid".

A student wants to prepare a pure, dry sample of copper sulfate crystals from copper oxide and sulfuric acid.

(a) Describe the key steps in the method to produce pure, dry copper sulfate crystals. [4]

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Warm the sulfuric acid gently; add excess copper oxide and stir [1]
Filter the mixture to remove the excess (unreacted) copper oxide [1]
Pour the filtrate into an evaporating basin and heat gently until crystals start to form [1]
Leave to cool and crystallise; pat dry with filter paper [1]

Examiner tip: The reason for adding excess insoluble base is to ensure all the acid has reacted — this means no acid remains in the final product. If a student says "add copper oxide until no more dissolves," this is also accepted.

Links to: Topic 3 — Reacting Masses (calculating how much reactant is needed)

Copper sulfate crystals can be prepared by reacting copper oxide (an insoluble base) with dilute sulfuric acid.

(a) Describe the full method for preparing a pure, dry sample of copper sulfate crystals. Explain the purpose of each key step. [6]

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Level 3 (5–6 marks): Detailed, logically sequenced description of the full method with clear explanations of the purpose of adding excess base, filtering, and crystallisation.

Level 2 (3–4 marks): Describes most steps correctly but may not fully explain why each step is performed (e.g. states filtering but not why).

Level 1 (1–2 marks): Basic steps such as "mix acid and base" without procedural detail or explanation.

Indicative content:

Warm dilute sulfuric acid gently using a Bunsen burner
Add excess copper oxide (insoluble base) and stir — purpose: ensures all acid is used up so no unreacted acid remains
When excess solid remains at the bottom (no more dissolves), filter the mixture — purpose: removes the unreacted copper oxide
Pour the blue filtrate (copper sulfate solution) into an evaporating basin
Heat gently until half the water has evaporated (do not boil dry)
Leave to cool slowly to allow crystals to form; dry with filter paper

Examiner tip: Explaining the purpose of each step is what separates L2 from L3. Don't just describe what you do — explain why. For example: "excess base is added to ensure all the acid reacts."

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Topic 4: Chemical Changes — Exam Practice

Topic 4: Chemical Changes Exam Practice

📋 Structured Questions

Question 1: The pH Scale and Dilution

Question 2: Extraction of Metals

Question 3: Oxidation and Reduction

Question 4: Electrolysis of Aqueous Solutions

Question 5: Making a Pure Dry Salt

Question 6: Electrolysis of Brine ⭐ Extended Response

Question 7: Making Salts (Required Practical)

Question 8: Required Practical — Making a Soluble Salt 🔬

Question 9: Required Practical — Preparing Pure Dry Crystals ⭐🔬 Extended Response