Question 2 – Electrolysis of dilute sulphuric acid
A student carried out an electrolysis of dilute sulphuric acid and collected the gases formed.
- Draw a labelled diagram to show the apparatus used. Show the gases collected in their correct ratio.
- Give the formulae of all the ions present in the electrolyte.
- Write half equations for the reactions at the anode and cathode. Write down an overall equation for the reaction.
- Describe the tests and results for any gases evolved.
Take note of the ratio of hydrogen to oxygen. Hydrogen produced is twice as much as the oxygen produced.
- H+, SO42- and OH–.
Anode reaction: 4OH–(aq) – 4e → 2H2O(l) + O2(g)
OH– is lower than SO42- in the order of preferential discharge so is discharged first.
Cathode reaction: 2H+(aq) + 2e → H2(g)
Overall reaction: 2H2O(l) → 2H2(g) + O2(g)
Hydrogen gas is produced at the cathode.
Test for hydrogen gas: Bring a burning splint to the mouth of the test-tube containing hydrogen gas.
Result The gas burns with a loud pop.
Oxygen gas is produced at the anode.
Test for oxygen: Dip a glowing splint in the test-tube containing the gas.
Result: Oxygen gas relights a glowing splint.