Electrolysis of Water - Equations

Equations

In pure water at the negatively charged cathode, a reduction reaction takes place, with electrons (e−) from the cathode being given to hydrogen cations to form hydrogen gas (the half reaction balanced with acid):

Reduction at cathode: 2 H+(aq) + 2e− → H₂(g)

At the positively charged anode, an oxidation reaction occurs, generating oxygen gas and giving electrons to the anode to complete the circuit:

Anode (oxidation): 2 H₂O(l) → O₂(g) + 4 H+(aq) + 4e−

The same half reactions can also be balanced with base as listed below. Not all half reactions must be balanced with acid or base. Many do, like the oxidation or reduction of water listed here. To add half reactions they must both be balanced with either acid or base.

Cathode (reduction): 2 H₂O(l) + 2e− → H₂(g) + 2 OH-(aq)

Anode (oxidation): 4 OH- (aq) → O₂(g) + 2 H₂O(l) + 4 e−

Combining either half reaction pair yields the same overall decomposition of water into oxygen and hydrogen:

Overall reaction: 2 H₂O(l) → 2 H₂(g) + O₂(g)

The number of hydrogen molecules produced is thus twice the number of oxygen molecules. Assuming equal temperature and pressure for both gases, the produced hydrogen gas has therefore twice the volume of the produced oxygen gas. The number of electrons pushed through the water is twice the number of generated hydrogen molecules and four times the number of generated oxygen molecules.