AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2019: Question 5

💡 Key knowledge

• Electrolysis is used when a metal is too reactive to be extracted by carbon.
• Molten ionic compounds contain ions only, so they can carry charge.
• At the negative electrode, cations gain electrons (reduction).
• At the positive electrode, anions lose electrons (oxidation).

🧠 Exam technique

• Use the mark scheme wording where possible.
• For graphs, always show working: read values, calculate change, then divide.
• Give units for gradient: usually g h⁻¹ , mg min⁻¹ , etc.

❌ Common errors

• Saying electrolysis is used because the metal is unreactive — this is the opposite.
• Mixing up molten and solution.
• For half-equations, forgetting electron numbers or charges.
• Calling a line of best fit a dot-to-dot line.

✅ Correct answer

Because the metal is more reactive than carbon / too reactive to be extracted by reduction with carbon.

💡 Key knowledge

Reactive metals such as aluminium need electricity to force the metal ions to gain electrons and form the metal.

🧠 Exam technique

Even though the mark scheme shown for 05.1 is for a different phrasing, the key idea being tested is always the need for electrolysis for reactive metals.

❌ Common errors

Do not say “because it is less reactive” or “because it is a compound”. The exam wants the reason electrolysis is needed.

✅ Correct answer

1. Aluminium oxide (or aluminium compound in the melt)
2. Cryolite

💡 Key knowledge

In the Hall-Héroult process, aluminium oxide is dissolved in molten cryolite. Cryolite lowers the melting point and helps the electrolyte conduct electricity.

🧠 Exam technique

If asked for the molten mixture, think about the industrial extraction process: the compound being electrolysed plus the substance added to make it easier.

❌ Common errors

Do not write “aluminium and oxygen” — those are products, not the substances in the molten mixture.

✅ Correct answers

Negative electrode: Cu²⁺ + 2e⁻ → Cu

Positive electrode: 2Cl⁻ → Cl₂ + 2e⁻

💡 Key knowledge

• Positive ions go to the negative electrode.
• Negative ions go to the positive electrode.
• Electrons are gained at the cathode and lost at the anode.

🧠 Exam technique

Check that atoms and charge balance on both sides. The number of electrons must match the ion charge.

❌ Common errors

• Writing Cu²⁺ → Cu²⁺ or leaving out electrons.
• For chlorine, forgetting it is diatomic: Cl₂ , not Cl .
• Putting electrons on the wrong side of the equation.

✅ Correct answer points

• The current may have been different from the expected/current setting.
• The process may not have run for exactly the expected time.
• The electrolyte concentration may have changed.
• The electrode may not have been fully clean/dry.

Any two sensible reasons gain credit.

💡 Key knowledge

Mass deposited depends on the amount of charge passed through the solution: larger current or longer time gives more copper.

🧠 Exam technique

For this style of question, give practical reasons linked to the experiment, not general chemistry ideas.

❌ Common errors

Do not say “because of energy” or “because the copper was pure”. The mark scheme rewards explanations about the experiment conditions.

📐 Calculations: step-by-step

1. Read two clear points from the line, for example 30 min → 6 mg.
2. Find the rate per minute: 6 ÷ 30 = 0.2 mg per minute.
3. Convert 24 hours to minutes: 24 × 60 = 1440 minutes.
4. Multiply: 1440 × 0.2 = 288 mg.

Mass = 288 mg

✅ Correct answer

288 mg

Method marks are for reading the graph accurately and scaling up correctly.

🧠 Exam technique

• Use the graph scale carefully.
• Show the rate first, then scale to 24 hours.
• Keep units consistent: minutes and mg, then convert to the final answer.

❌ Common errors

• Reading off 24 as if it were 24 minutes instead of 24 hours.
• Using the gradient directly but forgetting to multiply by 1440.
• Mixing up mg and g.

✅ Correct answer

The graph shows the mass at 0 hours is 5 g.

🧠 Exam technique

For a start value, read the y-intercept where the line crosses the vertical axis.

❌ Common errors

Do not read the final value at 10 hours. The question asks for the start.

📐 Calculations: step-by-step

1. Choose two points on the line. A good pair is (0 h, 5 g) and (10 h, 7 g).
2. Work out the change in mass: 7 − 5 = 2 g.
3. Work out the change in time: 10 − 0 = 10 h.
4. Gradient = 2 ÷ 10 = 0.2 g h⁻¹.

Gradient = 0.2
Unit = g h⁻¹

✅ Correct answer

Gradient = 0.2

Unit = g h⁻¹

💡 Key knowledge

Gradient means rise ÷ run. Here it tells you how much the electrode mass increases each hour.

❌ Common calculation traps

• Using the wrong scale on the graph.
• Forgetting to subtract to find the change.
• Writing the wrong unit, such as g only, or g/min .
• Giving a negative gradient when the line rises.

🧠 Top-level responses

• Use precise scientific wording: electron transfer, ions, cations, anions.
• Explain graph answers with clear working, not just a final number.
• For rate questions, mention that the reaction slows because there are fewer particles and therefore less frequent collisions.

💡 Mark scheme-linked note

In the rate explanation, the mark scheme rewards: fewer reactant particles as the reaction progresses, so collisions become less frequent, and the reaction eventually stops when a reactant is used up.

❌ Final reminder

Do not explain the rate change using energy or temperature unless the question asks for it. Here, the reason is about particle concentration and the limiting reactant.