Quick answer: Electrolysis uses direct current to split a molten or dissolved ionic compound. This Chemistry 9-10 lab on electrolysis gives the aim, the equipment, the safety points, a step-by-step method, sample results and 7 analysis questions; the answers are on a separate page and the lab sheet is a free printable PDF.
When a current flows through copper(II) sulfate solution, copper moves from one plate to the other. How much copper is moved, and does the amount depend on the current? You measure the mass changes at five currents, then use the numbers to find how many electrons are needed for each copper atom and how much charge one mole of electrons carries.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4
Aim
To find how the mass of copper deposited on the cathode depends on the current when copper(II) sulfate solution is electrolysed with copper electrodes for a fixed time.
Equipment
- Low-voltage variable power supply (0 to 6 V d.c.)
- Ammeter (0 to 1 A) and connecting leads
- Two copper plates, about 2 cm × 6 cm
- 250 cm³ beaker with 150 cm³ of 0.5 mol/dm³ copper(II) sulfate solution
- Balance reading to 0.01 g, stopwatch, fine sandpaper, water and paper towel
Safety
Wear safety goggles. Copper(II) sulfate solution is harmful if swallowed and irritates the eyes, so wash hands afterwards. Keep the supply at 6 V or less and switch off before touching the plates.
Method
- Clean both plates with sandpaper, rinse with water, dry and weigh each. Record the masses.
- Set up the circuit as shown, with the ammeter in series, and mark which plate is joined to the negative terminal.
- Switch on and adjust the supply until the ammeter reads 0.20 A. Start the stopwatch.
- Keep the current steady for 1800 s (30 minutes), adjusting the supply if it drifts.
- Switch off, remove the plates, rinse gently, dry them and reweigh both.
- Use fresh clean plates and repeat for 0.40, 0.60, 0.80 and 1.0 A.
Results
Sample results for 1800 s at each current:
| Current / A | Time / s | Cathode gain / g | Anode loss / g |
|---|---|---|---|
| 0.20 | 1800 | 0.11 | 0.12 |
| 0.40 | 1800 | 0.23 | 0.24 |
| 0.60 | 1800 | 0.34 | 0.36 |
| 0.80 | 1800 | 0.45 | 0.47 |
| 1.00 | 1800 | 0.57 | 0.59 |
Analysis questions
Question 1 Plot the graph
Plot the cathode gain (y-axis) against current (x-axis) and draw the line of best fit.1. Describe the shape of the graph.
2. State the relationship between current and mass.
Question 2 Find the gradient
Use the point at 1.00 A to calculate the gradient in g per ampere for 1800 s. State what the gradient means.Question 3 Moles of electrons
Using the 1.00 A result (Ar: Cu = 64):1. Calculate the moles of copper deposited.
2. Write the cathode half equation.
3. Calculate the moles of electrons that flowed.
Question 4 Charge per mole of electrons
Charge in coulombs equals current × time.1. Calculate the charge for the 1.00 A run.
2. Calculate the charge carried by one mole of electrons.
3. Compare with the accepted value of 96 500 C/mol and find the percentage difference.
Question 5 Anode and cathode
The anode lost slightly more mass than the cathode gained at every current.1. Write the half equation at the anode.
2. Suggest two reasons for the small difference.
Question 6 Uncertainty
The balance reads to 0.01 g.1. Calculate the percentage uncertainty in the 0.11 g reading.
2. Calculate the percentage uncertainty in the 0.57 g reading.
3. Explain which result is more reliable.
Question 7 Conclusion and improvements
State the conclusion of the lab and suggest two improvements to make the results more accurate.
