Electrolysis of Water | Department of Chemistry

Summary

Sodium sulfate solution is electrolyzed to make hydrogen and oxygen gas. A ignited wood stick shown to glow more brightly in the presence of concentrated oxygen gas (collected in a test tube). Hydrogen gas collected in a test tube is ignited using a bunsen burner.

Materials

Electrolysis Apparatus
D.C. Power Supply
Dilute (0.5M or so) Na₂SO₄ solution (500 mL minimum to fill apparatus). Phenolphthalein may be added to show formation of hydroxide ion at the cathode.
Splint
Matches
Test tubes

Procedure

Open stopcocks, making sure that the sodium sulfate solution fills each compartment to the tip. Close stopcocks. Attach apparatus to D.C. power supply.
Do a check to make sure the demo works.
As soon as class starts, turn on the power supply. Operate at about 10-12V.
Hydrogen may be collect at the cathode in a small test tube and ignited. Oxygen may be collected at the anode. Add a glowing splint to show that oxygen supports combustion.
Rinse apparatus immediately after use to avoid corrosion of the electrode.

Hint: Too much voltage will blow the Variac fuse, best to keep demo under 15V.

Sodium sulfate solution will eventually mold.

Discussion

The overall reaction:

$2 H_{2} O_{(l)} ⟶ 2 H_{2}_{(g)} + O_{2}_{(g)}$

Reduction at the cathode:

$2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2 OH^{-}$

Oxidation at the anode:

$4 OH^{-} ⟶ 2 H_{2} O + O_{2} + 4 e^{-}$

By multiplying the equation for the reaction at the cathode by two and then combining it to the equation for the reaction at the anode you get:

$4 H_{2} O + 4 e^{-} + 4 OH^{-} ⟶ 2 H_{2} O + O_{2} + 4 e^{-} + 2 H_{2} + 4 OH^{-}$

Simplifying the equation by cancelling out the electrons, hydroxide ions, and by reducing the water by two units on both sides gives you the overall equation above.