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 questionQuestion
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
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
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."
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 ).
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.
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)
- Filter the mixture/solution from the beaker to collect the fallen copper pieces. [1 mark]
- Wash and dry the copper residue collected on the filter paper. [1 mark]
- Weigh the collected dry copper (e.g. using a balance). [1 mark]
- 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!
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).
✅ 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)
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.
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
- 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 - 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] - Step 3: Calculate the number of copper atoms
Number of atoms = moles × Avogadro constant
= 0.0037795... × (6.02 × 10²³) = 2.27527... × 10²¹ [1 mark] - 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.
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.