Chemistry · Simulation and virtual lab Free
Electrolysis: what happens at each electrode?
Pass a current through different electrolytes, collect the products at each electrode and watch the ions move in the close-up.
- You will learn to
- Distinguish electrolytes from non-electrolytes.
- Identify the products at the cathode and the anode and write half-equations.
- Show that the amount of product is proportional to the charge (Faraday's first law).
- You will measure
- Current (A), time (s), charge Q = It (C), gas volumes (cm³) and mass of copper (mg).
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: DC supply, switch, ammeter, beaker, carbon and copper electrodes, test tubes, electrolytes, balance, stopwatch.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To identify the products of electrolysis at the cathode and the anode, and to show that the amount of product is proportional to the charge passed (Faraday's first law).
What to do
- Choose Sugar solution and close the switch. Read the ammeter. Then try Distilled water. Which are electrolytes?
- Choose Dilute sulfuric acid, 6 V, clock × 60. Close the switch and let it run. Which gas collects at each electrode? Compare the volumes.
- Choose Copper(II) sulfate, carbon electrodes. Watch the cathode and the colour of the solution.
- Choose Copper(II) sulfate, copper electrodes. Run for 10 minutes (600 s). Record the change in mass of the cathode and of the anode.
- Empty the tubes and repeat with a different voltage (current) or time. Record at least 4 results for the same electrolyte.
- Look at the graph of product against charge Q = It and use the gradient to estimate the Faraday constant.
Results
Questions
What is an electrolyte?
A compound that conducts electricity when molten or dissolved in water, and is decomposed by the current, because it contains ions that are free to move.
Why does sugar solution not conduct electricity?
Sugar dissolves as neutral molecules, not ions, so there are no charged particles free to carry the current.
Write the half-equations at the electrodes when dilute sulfuric acid is electrolysed with carbon electrodes.
Cathode: 2H⁺(aq) + 2e⁻ → H₂(g). Anode: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻ (or 2H₂O → O₂ + 4H⁺ + 4e⁻).
Why is the volume of hydrogen twice the volume of oxygen?
Each H₂ molecule needs 2 electrons but each O₂ molecule needs 4 electrons. The same charge flows through both electrodes, so twice as many moles of H₂ are formed.
Why does the blue colour fade with carbon electrodes but not with copper electrodes?
With carbon electrodes Cu²⁺ ions are removed (deposited as copper) and not replaced. With copper electrodes the anode dissolves (Cu → Cu²⁺ + 2e⁻) and replaces the Cu²⁺ ions as fast as they are deposited.
Give one use of electrolysis in industry.
Purifying copper, electroplating (e.g. chromium, silver, gold), extracting aluminium or sodium, making chlorine and sodium hydroxide from brine.
For teachers: expected results and how the model works
Model: I = (V − Ed) ÷ R with decomposition voltages 1.7 V (water/acid), 1.4 V (CuSO₄, carbon), 0.05 V (CuSO₄, copper), 2.2 V (brine) and cell resistances 3–6 Ω. Products follow Faraday's law exactly: 0.124 cm³ of H₂ per coulomb (0.062 cm³ of O₂) and 0.329 mg of Cu per coulomb. Example: dilute H₂SO₄ at 6 V gives 1.08 A; after 300 s, Q = 323 C, H₂ = 40.1 cm³, O₂ = 20.1 cm³. Sugar solution gives zero current; distilled water about 1 mA. The test tubes hold 50 cm³: the switch opens automatically when one is full.
Exam practice Exam set
National-exam style questions on this simulation (17 marks). Try each one on paper first, then open the marking scheme.
A current of 0.50 A is passed through copper(II) sulfate solution for 30 minutes. (a) Calculate the charge passed. (b) Calculate the mass of copper deposited. (Cu = 63.5, F = 96 500 C/mol) (5 marks)
Marking scheme: (a) Q = It = 0.50 × 30 × 60 (1) = 900 C (1). (b) Moles of electrons = 900 ÷ 96 500 = 9.33 × 10⁻³ mol (1); Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 4.66 × 10⁻³ mol (1); mass = 4.66 × 10⁻³ × 63.5 = 0.296 g (1).
Concentrated sodium chloride solution is electrolysed with carbon electrodes. (a) Name the product at each electrode. (b) Explain why sodium is not formed at the cathode. (c) Give a test for the gas at the anode. (5 marks)
Marking scheme: (a) Cathode: hydrogen (1); anode: chlorine (1). (b) H⁺ ions (from water) are discharged in preference to Na⁺ because hydrogen is lower in the reactivity series / H⁺ is more easily reduced (2). (c) Chlorine bleaches damp blue litmus paper (it first turns red) (1).
Explain how impure copper is purified by electrolysis. Include the half-equation at each electrode. (4 marks)
Marking scheme: Impure copper is the anode, a thin pure copper sheet is the cathode, copper(II) sulfate is the electrolyte (1). Anode: Cu → Cu²⁺ + 2e⁻ (1). Cathode: Cu²⁺ + 2e⁻ → Cu (1). Impurities fall below the anode as anode sludge (1).
In an experiment the volume of hydrogen collected was plotted against the charge passed. The gradient was 0.125 cm³/C. Calculate the Faraday constant. (1 mol of gas occupies 24 000 cm³) (3 marks)
Marking scheme: 1 C gives 0.125 cm³ = 0.125 ÷ 24 000 = 5.21 × 10⁻⁶ mol H₂ (1); this needs 2 × 5.21 × 10⁻⁶ = 1.04 × 10⁻⁵ mol of electrons (1); F = 1 ÷ 1.04 × 10⁻⁵ ≈ 96 000 C/mol (1).
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Electrolysis: what happens at each electrode?
Aim: To identify the products of electrolysis at the cathode and the anode, and to show that the amount of product is proportional to the charge passed (Faraday's first law).
Apparatus (real laboratory): DC supply, switch, ammeter, beaker, carbon and copper electrodes, test tubes, electrolytes, balance, stopwatch.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Sugar solution and close the switch. Read the ammeter. Then try Distilled water. Which are electrolytes?
- Choose Dilute sulfuric acid, 6 V, clock × 60. Close the switch and let it run. Which gas collects at each electrode? Compare the volumes.
- Choose Copper(II) sulfate, carbon electrodes. Watch the cathode and the colour of the solution.
- Choose Copper(II) sulfate, copper electrodes. Run for 10 minutes (600 s). Record the change in mass of the cathode and of the anode.
- Empty the tubes and repeat with a different voltage (current) or time. Record at least 4 results for the same electrolyte.
- Look at the graph of product against charge Q = It and use the gradient to estimate the Faraday constant.
Results
| # | Electrolyte | I (A) | t (s) | Q = It (C) | Cathode | Anode |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
Graph
Plot the amount at the cathode (volume of hydrogen in cm³, or mass of copper in mg) against the charge Q. Draw the best straight line through the origin.
Questions
- What is an electrolyte? [2]
- Why does sugar solution not conduct electricity? [2]
- Write the half-equations at the electrodes when dilute sulfuric acid is electrolysed with carbon electrodes. [2]
- Why is the volume of hydrogen twice the volume of oxygen? [2]
- Why does the blue colour fade with carbon electrodes but not with copper electrodes? [2]
- Give one use of electrolysis in industry. [2]
- A current of 0.50 A is passed through copper(II) sulfate solution for 30 minutes. (a) Calculate the charge passed. (b) Calculate the mass of copper deposited. (Cu = 63.5, F = 96 500 C/mol) [5]
- Concentrated sodium chloride solution is electrolysed with carbon electrodes. (a) Name the product at each electrode. (b) Explain why sodium is not formed at the cathode. (c) Give a test for the gas at the anode. [5]
- Explain how impure copper is purified by electrolysis. Include the half-equation at each electrode. [4]
- In an experiment the volume of hydrogen collected was plotted against the charge passed. The gradient was 0.125 cm³/C. Calculate the Faraday constant. (1 mol of gas occupies 24 000 cm³) [3]
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Electrolysis: what happens at each electrode?
Expected results
Model: I = (V − Ed) ÷ R with decomposition voltages 1.7 V (water/acid), 1.4 V (CuSO₄, carbon), 0.05 V (CuSO₄, copper), 2.2 V (brine) and cell resistances 3–6 Ω. Products follow Faraday's law exactly: 0.124 cm³ of H₂ per coulomb (0.062 cm³ of O₂) and 0.329 mg of Cu per coulomb. Example: dilute H₂SO₄ at 6 V gives 1.08 A; after 300 s, Q = 323 C, H₂ = 40.1 cm³, O₂ = 20.1 cm³. Sugar solution gives zero current; distilled water about 1 mA. The test tubes hold 50 cm³: the switch opens automatically when one is full.
Answers
- What is an electrolyte? [2]
A compound that conducts electricity when molten or dissolved in water, and is decomposed by the current, because it contains ions that are free to move.
- Why does sugar solution not conduct electricity? [2]
Sugar dissolves as neutral molecules, not ions, so there are no charged particles free to carry the current.
- Write the half-equations at the electrodes when dilute sulfuric acid is electrolysed with carbon electrodes. [2]
Cathode: 2H⁺(aq) + 2e⁻ → H₂(g). Anode: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻ (or 2H₂O → O₂ + 4H⁺ + 4e⁻).
- Why is the volume of hydrogen twice the volume of oxygen? [2]
Each H₂ molecule needs 2 electrons but each O₂ molecule needs 4 electrons. The same charge flows through both electrodes, so twice as many moles of H₂ are formed.
- Why does the blue colour fade with carbon electrodes but not with copper electrodes? [2]
With carbon electrodes Cu²⁺ ions are removed (deposited as copper) and not replaced. With copper electrodes the anode dissolves (Cu → Cu²⁺ + 2e⁻) and replaces the Cu²⁺ ions as fast as they are deposited.
- Give one use of electrolysis in industry. [2]
Purifying copper, electroplating (e.g. chromium, silver, gold), extracting aluminium or sodium, making chlorine and sodium hydroxide from brine.
- A current of 0.50 A is passed through copper(II) sulfate solution for 30 minutes. (a) Calculate the charge passed. (b) Calculate the mass of copper deposited. (Cu = 63.5, F = 96 500 C/mol) [5]
(a) Q = It = 0.50 × 30 × 60 (1) = 900 C (1). (b) Moles of electrons = 900 ÷ 96 500 = 9.33 × 10⁻³ mol (1); Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 4.66 × 10⁻³ mol (1); mass = 4.66 × 10⁻³ × 63.5 = 0.296 g (1).
- Concentrated sodium chloride solution is electrolysed with carbon electrodes. (a) Name the product at each electrode. (b) Explain why sodium is not formed at the cathode. (c) Give a test for the gas at the anode. [5]
(a) Cathode: hydrogen (1); anode: chlorine (1). (b) H⁺ ions (from water) are discharged in preference to Na⁺ because hydrogen is lower in the reactivity series / H⁺ is more easily reduced (2). (c) Chlorine bleaches damp blue litmus paper (it first turns red) (1).
- Explain how impure copper is purified by electrolysis. Include the half-equation at each electrode. [4]
Impure copper is the anode, a thin pure copper sheet is the cathode, copper(II) sulfate is the electrolyte (1). Anode: Cu → Cu²⁺ + 2e⁻ (1). Cathode: Cu²⁺ + 2e⁻ → Cu (1). Impurities fall below the anode as anode sludge (1).
- In an experiment the volume of hydrogen collected was plotted against the charge passed. The gradient was 0.125 cm³/C. Calculate the Faraday constant. (1 mol of gas occupies 24 000 cm³) [3]
1 C gives 0.125 cm³ = 0.125 ÷ 24 000 = 5.21 × 10⁻⁶ mol H₂ (1); this needs 2 × 5.21 × 10⁻⁶ = 1.04 × 10⁻⁵ mol of electrons (1); F = 1 ÷ 1.04 × 10⁻⁵ ≈ 96 000 C/mol (1).