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A2 · Practice questions · Electrolysis and the Faraday constant

Charge in, moles out.

Six original Cambridge-style questions on electrolysis calculations and products.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Q = It, then divide by F, then use the half-equation.

01
Calculation
[3 marks]

A current of 1.50 A is passed through aqueous copper(II) sulfate for 1.00 hour. Calculate the mass of copper deposited on the cathode.

  • Q = It = 1.50 × 3600 = 5400 C. ✓
  • n(e−) = 5400 ÷ 96 500 = 0.05596 mol, so n(Cu) = 0.02798 mol. ✓
  • Mass = 0.02798 × 63.5 = 1.78 g. ✓
02
Calculation
[3 marks]

Concentrated aqueous sodium chloride is electrolysed with a current of 3.00 A for 20.0 minutes. Calculate the volume of chlorine produced at room conditions (24.0 dm3 mol−1).

  • Q = 3.00 × 1200 = 3600 C; n(e−) = 3600 ÷ 96 500 = 0.03731 mol. ✓
  • 2Cl− → Cl2 + 2e−, so n(Cl2) = 0.01865 mol. ✓
  • Volume = 0.01865 × 24.0 = 0.448 dm3. ✓
03
Calculation
[3 marks]

Aluminium is extracted by electrolysis of Al3+. How long, in seconds, must a current of 10.0 A flow to deposit 2.70 g of aluminium?

  • n(Al) = 2.70 ÷ 27.0 = 0.100 mol; Al3+ + 3e− → Al, so n(e−) = 0.300 mol. ✓
  • Q = 0.300 × 96 500 = 28 950 C. ✓
  • t = Q ÷ I = 28 950 ÷ 10.0 = 2.90 × 103 s (2895 s). ✓
04
Explain
[3 marks]

Explain, using E⊖ values, the products at each electrode when dilute aqueous sodium chloride is electrolysed with inert electrodes, and how they change when the solution is concentrated.

  • Cathode: E⊖(Na+/Na) = −2.71 V is far more negative than for water, so water is reduced: 2H2O + 2e− → H2 + 2OH−. ✓
  • Anode (dilute): E⊖(O2/H2O) = +1.23 V is less positive than E⊖(Cl2/Cl−) = +1.36 V, so water is oxidised to oxygen. ✓
  • Concentrated: the high [Cl−] makes Cl− oxidation easier, so chlorine is the main anode product; hydrogen still forms at the cathode. ✓
05
Analysis
[4 marks]

A student passes 0.500 A for 40.0 minutes through copper(II) sulfate solution using copper electrodes. The cathode gains 0.395 g. Using e = 1.60 × 10−19 C, calculate a value for the Avogadro constant and state two measurements that must be made accurately.

  • Q = 0.500 × 2400 = 1200 C; number of electrons = 1200 ÷ 1.60 × 10−19 = 7.50 × 1021. ✓
  • n(Cu) = 0.395 ÷ 63.5 = 0.006220 mol, so n(e−) = 0.01244 mol. ✓
  • L = 7.50 × 1021 ÷ 0.01244 = 6.03 × 1023 mol−1. ✓
  • Measure the mass of the cathode before and after (washed and dried), and keep the current constant while timing accurately. ✓
06
Recall
[2 marks]

Write the half-equations for the reactions at each electrode when aqueous copper(II) sulfate is electrolysed using inert electrodes.

  • Cathode: Cu2+ + 2e− → Cu. ✓
  • Anode: 2H2O → O2 + 4H+ + 4e− (or 4OH− → O2 + 2H2O + 4e−). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.