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Electrolysis of Water

Virtual lab · IGCSE 0620 · GCSE Chemistry

2H2O → 2H2 + O2H2 : O2 = 2 : 1
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Hydrogen · cathode
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Oxygen · anode
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Volume ratio
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Current
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Charge passed
0C
Gas volume indicator collect some gas to compare
H₂
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O₂
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Cathode · negative · reduction
2H⁺ + 2e⁻ → H₂
H⁺ ions are attracted to the negative electrode, gain electrons and pair up as hydrogen molecules.
Anode · positive · oxidation
4OH⁻ → O₂ + 2H₂O + 4e⁻
OH⁻ ions from the water lose electrons. SO₄²⁻ also travels here but is not discharged.
Overall
2H₂O → 2H₂ + O₂
Four electrons make two H₂ but only one O₂ - so hydrogen appears at twice the rate.

What the experiment actually shows

Six points examiners look for.

Pure water is a poor conductor Water self-ionises only very slightly, so there are far too few ions to carry a useful current. The ammeter reads almost zero.
The acid supplies the ions Dilute sulfuric acid ionises fully: H₂SO₄ → 2H⁺ + SO₄²⁻. Those extra ions let charge move through the solution.
The acid is not split SO₄²⁻ ions migrate to the anode but are never discharged - OH⁻ ions are discharged in preference. The acid stays in solution and slowly becomes more concentrated as water is removed.
Ions carry charge inside, electrons outside Charge crosses the solution as moving ions. Electrons only travel through the wires and electrodes - never through the electrolyte.
Why the ratio is 2 : 1 Making one O₂ needs 4 electrons; making one H₂ needs only 2. The same charge therefore produces twice as many moles of hydrogen - and equal moles of gas occupy equal volumes.
Testing the gases Hydrogen gives a squeaky pop with a lit splint. Oxygen relights a glowing splint.
A note on the anode equation Some specifications write the anode step as 2H₂O → O₂ + 4H⁺ + 4e⁻ instead of 4OH⁻ → O₂ + 2H₂O + 4e⁻. Both describe the same overall change - water being oxidised - so either is acceptable unless your board asks for one.
Quick check
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What this experiment shows

Passing a direct current through acidified water splits it into its elements: hydrogen at the cathode and oxygen at the anode, in a volume ratio of exactly two to one. That ratio is the formula of water made visible. A little acid is added because pure water barely conducts - the acid supplies the free ions the current needs, but is not itself used up.

2H2O → 2H2 + O2

What you need

  • A Hofmann voltameter, or two inverted test tubes over electrodes
  • Distilled water with a little dilute sulfuric acid
  • Inert electrodes - platinum or graphite
  • A d.c. supply, an ammeter and a stopclock

How it is done

  1. Fill the apparatus with acidified water and make sure both tubes are full to the top.
  2. Connect the d.c. supply and note which electrode is which.
  3. Let the current run and watch bubbles form at both electrodes.
  4. Read the volume in each tube at the same moment.
  5. Test each gas: a lit splint at the cathode gas, a glowing splint at the anode gas.

What you should see

  • Twice as much gas collects at the cathode as at the anode.
  • The cathode gas burns with a squeaky pop - hydrogen.
  • The anode gas relights a glowing splint - oxygen.
  • The volumes follow Faraday: the charge passed fixes how much gas is made.

Where marks are lost

  • Getting the electrodes the wrong way round. Reduction happens at the cathode - hydrogen there.
  • Saying the acid is used up. It is not; it only makes the water conduct.
  • Quoting a 1:2 ratio the wrong way round. Hydrogen is the larger volume.
  • Using copper electrodes. Copper is not inert and takes part in the reaction.

Questions students ask

Why add acid to the water?
Pure water contains very few ions, so it hardly conducts. A little acid supplies ions to carry the current without changing the products.
Why twice as much hydrogen?
Each water molecule contains two hydrogen atoms for every oxygen atom, and equal volumes of gas contain equal numbers of molecules, so the volume ratio matches the formula.
How are the gases identified?
Hydrogen burns with a squeaky pop from a lit splint. Oxygen relights a glowing splint.

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Written for Cambridge IGCSE Chemistry 0620 and CBSE Class 9-12 by Ajay Shekhawat, founder of MrChemCoach. Run the simulator above, then check yourself against the questions.