Lower secondary · 15 min

Hydrogen by electrolysis of water

Add a pinch of sodium hydroxide to water, switch on the current, and collect twice as much hydrogen as oxygen.

Goal

Observe the 2 : 1 volume ratio of $\ce{H2}$ to $\ce{O2}$ and understand why $\ce{Na+}$ is not reduced.

Apparatus and reagents

Virtual electrolysis cell with water made conducting by $\ce{NaOH}$ , two electrodes and a current slider.

Procedure

Switch on a moderate current and watch bubbles form on both electrodes.
Compare the two electrodes: the cathode produces hydrogen, the anode oxygen — hydrogen appears roughly twice as fast.
Raise the current: both gas rates grow together. Why does the 2 : 1 ratio survive?

What to observe

Cathode: $\ce{2H2O + 2e^- -> H2 + 2OH^-}$ ; anode: $\ce{2H2O -> O2 + 4H+ + 4e^-}$ .
The volume of $\ce{H2}$ is twice that of $\ce{O2}$ because producing one $\ce{O2}$ needs $4e^-$ while one $\ce{H2}$ needs only $2e^-$ .

Explanation

Pure water barely conducts, so a pinch of $\ce{NaOH}$ provides ions. Sodium is far too reactive to be plated out: $\ce{Na+}$ stays in solution because water is easier to reduce ( $E° = -2{,}71$ V for $\ce{Na+/Na}$ ). The same reluctance explains why alkali metals are only ever produced by electrolysis of molten salts, never from aqueous solution — Humphry Davy isolated Na and K this way in 1807. Overall: $\ce{2H2O -> 2H2 + O2}$ , with $V(\ce{H2}) = 2\,V(\ce{O2})$ .

Chemists behind it

Humphry Davy Michael Faraday