AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2021: Question 7

17 marks · Standard Demand difficulty · Short Answer

Explain differences between electrolysis and chemical cells, complete half equations and electrolysis products, evaluate an experiment measuring mass of copper deposited including graph analysis, and calculate the number of copper atoms formed.

Practise this question

Question

Question 7 covers chemical reactions and electricity. Part 07.1 asks to explain the difference between electrolysis and a chemical cell. Part 07.2 asks to complete and balance the half equation for bromide ions forming bromine. Part 07.3 asks to complete a table of products at the positive and negative electrodes for aqueous copper nitrate and potassium iodide. Figure 4 shows an electrolysis setup with a beaker containing copper nitrate solution, two electrodes, an ammeter, and a DC power supply. An 8-step experimental procedure is given to measure mass change over time and current. Part 07.4 asks how to find total copper mass if some copper falls off into the beaker. Figure 5 shows a graph of total mass of copper produced versus time in minutes for three currents (0.3 A, 0.6 A, 0.9 A), each showing a straight line through the origin. Parts 07.5 and 07.6 ask how the graph supports direct proportionality to time and current. Part 07.7 asks why the blue colour fades. Part 07.8 asks to determine the number of copper atoms produced after 20 minutes at 0.6 A to 3 significant figures, given Ar(Cu)=63.5 and the Avogadro constant.
Question text

07 This question is about chemical reactions and electricity.

07.1 Electrolysis and chemical cells both involve chemical reactions and electricity.

Explain the difference between the processes in electrolysis and in a chemical cell.

[2 marks]

07.2 A teacher demonstrates the electrolysis of molten lead bromide.

Bromine is produced at the positive electrode.

Complete the half equation for the production of bromine.

You should balance the half equation.

[2 marks]

–

Br → +

07.3 Two aqueous salt solutions are electrolysed using inert electrodes.

Complete Table 4 to show the product at each electrode.

[3 marks]

Table 4

Product at Product at

Salt solution

positive electrode negative electrode

Copper nitrate copper

Potassium iodide 20

Some students investigated the electrolysis of copper nitrate solution using

inert electrodes.

Figure 4 shows the apparatus.

Figure 4

The students investigated how the mass of copper produced at the negative electrode

varied with:

• time

• current.

This is the method used.

1. Weigh the negative electrode.

2. Set up the apparatus shown in Figure 4.

3. Adjust the power supply until the ammeter shows a current of 0.3 A

4. Switch off the power supply after 5 minutes.

5. Rinse the negative electrode with water and allow to dry.

6. Reweigh the negative electrode.

7. Repeat steps 1 to 6 for different times.

8. Repeat steps 1 to 7 at different currents.

07.4 Some of the copper produced did not stick to the negative electrode but fell to the

bottom of the beaker.

Suggest how the students could find the total mass of copper produced.

[4 marks]

The students plotted their results on a graph.

Figure 5 shows the graph.

Figure 5

A student correctly concluded that the total mass of copper produced is directly

proportional both to the time and to the current.

07.5 How do the results in Figure 5 support the conclusion that the total mass of copper

produced is directly proportional to the time?

[1 mark]

07.6 How do the results in Figure 5 support the conclusion that the total mass of copper

produced is directly proportional to the current?

Use data from Figure 5 in your answer.

[1 mark]

07.7 Copper nitrate solution is blue.

Suggest why the blue colour of the copper nitrate solution fades during the

electrolysis.

*22* [1 mark]

07.8 Determine the number of atoms of copper produced when copper nitrate solution is

electrolysed for 20 minutes at a current of 0.6 A

Give your answer to 3 significant figures.

Use Figure 5.

Relative atomic mass (Ar): Cu = 63.5

The Avogadro constant = 6.02 × 1023 per mole

[3 marks]

Number of atoms (3 significant figures) =

Mark scheme

Show the mark scheme Mark scheme for Question 7 provides marking guidance. 07.1 awards 1 mark for electrolysis using electricity to produce a chemical reaction and 1 mark for cells using a chemical reaction to produce electricity. 07.2 awards 2 marks for 2 Br⁻ → Br₂ + 2 e⁻. 07.3 awards marks for oxygen at positive electrode for copper nitrate, and iodine at positive, hydrogen at negative for potassium iodide. 07.4 awards 4 marks for: filter the mixture, wash and dry the copper residue, weigh the collected copper, and add to the increase in mass of the electrode. 07.5 awards 1 mark for straight line through the origin. 07.6 awards 1 mark for showing mass doubles when current doubles at a given time using data. 07.7 awards 1 mark for copper ions being discharged/removed from solution. 07.8 awards 3 marks: finding mass from graph (0.24 g) to calculate moles (0.00378), multiplying by Avogadro's constant (6.02 × 10²³), giving 2.28 × 10²¹ to 3 significant figures. Total: 17 marks.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 allow voltage for electricity AO1

allow potential difference for 4.4.3.1

electricity 4.5.2.1

allow (electrical) current for

electricity

electrolysis uses electricity to allow electrolysis uses electricity 1

produce a chemical reaction to decompose a compound /

electrolyte

(but) cells use a chemical 1

reaction to produce electricity

AO /

Spec. Ref.

07.2 2 Br– → Br + 2 e– allow multiples 2 AO2

4.1.1.1

allow 1 mark for Br and e– 4.4.3.1

4.4.3.2

4.4.3.5

AO / Spec.

Ref.

07.3 Product at AO2

Product at

Salt solution negative 4.4.1.2

positive electrode

electrode 4.4.3.4

(copper nitrate) oxygen (1) (copper) 1 RPA3

(potassium iodide) iodine (1) – hydrogen (1)HEMISTRY – 2 –

Question 7 continued

AO /

Spec. Ref.

07.4 filter the mixture 1 AO3

4.1.1.2

wash and dry the copper / 1

4.4.3.4 19

residue

RPA3

weigh the copper collected 1

add to the increase in mass of 1

the electrode

AO /

Spec. Ref.

07.5 (for given current) straight line allow (for given current) when 1 AO3

through the origin time doubles, mass doubles 4.4.3.4

AO /

Spec. Ref.

07.6 (for given time) when current 1 AO3

doubles, mass doubles with 4.4.3.4

supporting data

AO /

Spec. Ref.

07.7 copper ions are discharged allow the solution becomes less 1 AO3

(from the solution) concentrated 4.4.3.1

allow copper ions are removed

(from the solution)

allow copper ions are used up

(from the solution)

– HEMISTRY – –

Question 7 continued

20 AO /

Spec. Ref.

07.8 0.24 1 AO2

(number of moles = 63.5 =)

4.3.2.1

3.78 × 10–3 or 0.00378

(number of atoms =) allow correct use of an 1

0.00378 × 6.02 × 1023 incorrectly calculated number of

moles

= 2.28 × 1021 allow a correct evaluation to 3 1

significant figures of an incorrect

expression which involves only

a mass from the graph, the Ar of

copper and the Avogadro

constant

Total 17

How to answer it

Electrolysis, Chemical Cells & Quantitative Analysis

📋 WHAT THIS QUESTION TESTS

This question assesses fundamental and quantitative concepts across AQA Electrolysis (Specification 4.4.3) and Chemical Calculations (Specification 4.3.2):

  • Comparing Electrolysis & Cells: Contrasting energy transfers (using electrical energy vs generating electricity).
  • Electrode Reactions & Half Equations: Balancing charges and diatomic halogen molecules at the anode.
  • Aqueous Electrolysis Rules: Predicting products at inert electrodes using reactivity and halide rules.
  • Experimental Practical Skills: Improving recovery methods to account for lost mass during electrolysis.
  • Graphical Analysis: Defining direct proportionality from lines through the origin and numerical pairing.
  • Moles & Avogadro's Constant: Calculating the number of atoms using mass, Aᵣ, and 6.02 × 10²³ to 3 significant figures.

Part 07.1: Electrolysis vs Chemical Cells

Explain the difference between the processes in electrolysis and in a chemical cell [2 marks]

✅ Model Answer

  • Electrolysis: Uses electricity (electrical energy) to produce a chemical reaction (decomposing an electrolyte). [1 mark]
  • Chemical cells: Use a chemical reaction to produce electricity (voltage / electrical current). [1 mark]

🧠 Exam Technique

Always state the direction of energy change for both systems clearly. Use reciprocal phrasing: "Electrolysis uses electricity to cause a reaction; a cell uses a reaction to produce electricity."

Mark breakdown: 1 mark for function of electrolysis; 1 mark for function of a chemical cell. Allow 'voltage' or 'current' in place of 'electricity'.

Part 07.2: Half Equation for Bromine Production

Complete and balance the half equation for the production of bromine from molten lead bromide [2 marks]

✅ Model Answer

2Br⁻ → Br₂ + 2e⁻

or: 2Br⁻ - 2e⁻ → Br₂

💡 Key Knowledge

Bromide ions (Br⁻) are negative anions. They are attracted to the positive electrode (anode) where they lose electrons (oxidation) to form diatomic bromine molecules (Br₂).

❌ Common Errors

  • Writing Br instead of diatomic Br₂ .
  • Balancing with a single electron: 2Br⁻ → Br₂ + e⁻ (charge must balance!).
  • Putting electrons on the wrong side (e.g. 2Br⁻ + 2e⁻ → Br₂ ).
Mark breakdown: 1 mark for correct species ( Br₂ and e⁻ ); 1 mark for fully balancing ( 2Br⁻ and 2e⁻ ).

Part 07.3: Products of Aqueous Electrolysis

Complete Table 4 to show the product at each electrode [3 marks]

✅ Model Answer

  • Copper nitrate: Product at positive electrode = oxygen [1 mark]
  • Potassium iodide: Product at positive electrode = iodine [1 mark]
  • Potassium iodide: Product at negative electrode = hydrogen [1 mark]

💡 Rules for Aqueous Solutions

  • Negative electrode (Cathode): Hydrogen gas is produced unless the metal is less reactive than hydrogen (Cu, Ag, Au). Copper is less reactive, so copper forms. Potassium is more reactive, so hydrogen forms.
  • Positive electrode (Anode): Halide ions (Cl⁻, Br⁻, I⁻) form halogens. If no halide is present (e.g. nitrate NO₃⁻, sulfate SO₄²⁻), hydroxide ions (OH⁻) are discharged to form oxygen.
Mark breakdown: 1 mark per correct box in Table 4. Total = 3 marks.

Part 07.4: Method to Determine Total Mass of Copper

Suggest how the students could find the total mass of copper produced [4 marks]

✅ Model Answer (4-Step Method)

  1. Filter the mixture/solution from the beaker to collect the fallen copper pieces. [1 mark]
  2. Wash and dry the copper residue collected on the filter paper. [1 mark]
  3. Weigh the collected dry copper (e.g. using a balance). [1 mark]
  4. Add this mass to the increase in mass of the negative electrode. [1 mark]

🧠 Exam Technique

Questions asking to find a total mass when precipitate or solid has fallen off always follow standard filtration and gravimetric steps. Don't forget that wet copper contains water, so drying is essential before weighing!

Mark breakdown: 1 mark each for: filter mixture, wash and dry residue, weigh residue, add to electrode mass gain.

Parts 07.5 & 07.6: Evaluating Proportionality from Graph

✅ 07.5: Proportional to Time [1 mark]

Answer: The line of best fit is a straight line passing through the origin (0,0) (for a given current).

Key rule: Direct proportionality on a graph must have both conditions: 1) Straight line, and 2) Passes through (0,0).

✅ 07.6: Proportional to Current [1 mark]

Answer: At any fixed time, when the current doubles, the mass produced doubles.

Supporting data example (at 30 minutes):

  • At 0.3 A, mass = 0.18 g
  • At 0.6 A, mass = 0.36 g (0.18 × 2 = 0.36 g)
Mark breakdown: 07.5: 1 mark for straight line through origin. 07.6: 1 mark for showing doubling current doubles mass using correct coordinate pairs.

Part 07.7: Solution Colour Change

Suggest why the blue colour of the copper nitrate solution fades during electrolysis [1 mark]

✅ Model Answer

Copper ions (Cu²⁺) are discharged / removed from the solution (they gain electrons at the cathode to form copper atoms, so the concentration of copper ions decreases).

❌ Common Errors

  • Saying "copper is being used up" without specifying copper ions or the solution.
  • Claiming the water evaporates or dilutes the solution.
Mark breakdown: 1 mark for copper ions are discharged / removed / concentration decreases.

Part 07.8: Calculation of Number of Copper Atoms

Determine the number of atoms of copper produced (20 mins, 0.6 A) to 3 significant figures [3 marks]

📐 Step-by-Step Calculation

  1. Step 1: Read the mass from Figure 5
    Find 20 minutes on the horizontal axis and read up to the 0.6 A line.
    Mass of Cu = 0.24 g
  2. Step 2: Calculate moles of copper produced
    Moles = Mass ÷ Aᵣ = 0.24 ÷ 63.5 = 0.0037795... mol (or 3.78 × 10⁻³ mol) [1 mark]
  3. Step 3: Calculate the number of copper atoms
    Number of atoms = moles × Avogadro constant
    = 0.0037795... × (6.02 × 10²³) = 2.27527... × 10²¹ [1 mark]
  4. Step 4: Round to 3 significant figures
    2.28 × 10²¹ atoms [1 mark]

❌ Common Calculation Traps

  • Reading the wrong curve: Reading the 0.3 A or 0.9 A line instead of 0.6 A.
  • Premature rounding: Rounding to 0.0038 mol too early can lead to rounding discrepancies. Keep calculator values until the final step.
  • Significant figures penalty: Forgetting the instruction to quote the answer to 3 significant figures loses the final mark.

🧠 Exam Technique: Error Carried Forward (ECF)

If you misread the graph (e.g. reading 0.25 g instead of 0.24 g), you can still receive the remaining 2 marks if your mole calculation and multiplication by Avogadro's constant are executed correctly to 3 sig figs.

Mark breakdown: 1 mark for calculating moles (0.00378 mol); 1 mark for multiplying moles by 6.02 × 10²³; 1 mark for final answer of 2.28 × 10²¹ (3 sig figs).

Topics

Chemistry · Required Practicals · C3: Quantitative Chemistry · C4: Chemical Changes · C5: Energy Changes · Required Practicals

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.