# Electrolysis of Water

## 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) NaSO 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

1. Open stopcocks, making sure that the sodium sulfate solution fills each compartment to the tip. Close stopcocks. Attach apparatus to D.C. power supply.
2. Do a check to make sure the demo works.
3. As soon as class starts, turn on the power supply. Operate at about 10-12V.
4. 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.
5. 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:

$$\ce{ 2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)} }$$

Reduction at the cathode:

$$\ce{2H2O + 2e- -> H2 + 2OH-}$$

Oxidation at the anode:

$$\ce{4OH- -> 2H2O + O2 + 4e-}$$

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:

$$\ce {4H2O + 4e- + 4OH- -> 2H2O + O2 + 4e- + 2H2 + 4OH-}$$

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.