The electrolysis of water uses electricity to dissociate water into hydrogen gas and oxygen gas. In this article, we are concerned with the following cases:
1) Electrolysis of pure water
Water is a poor conductor of electricity, as it has a very small auto-ionisation constant (Kw = 10-14), i.e. water exists predominantly in its molecular form H2O instead of H+ and OH– ions. Therefore, the electrochemical reaction at the anode involves molecular H2O rather than OH– and the reaction at the cathode involves H2O instead of H+.
At the anode: 2H2O (l) → O2 (g) + 4H+ (aq) + 4e– Eo = -1.23 V
At the cathode: 2e– + 2H2O (l) → H2 (g) + 2OH– (aq) Eo = -0.83 V
Overall reaction: 2H2O (l) → O2 (g) + 2H2 (g) Eo = -2.06 V
In the bulk solution, H+ and OH– from the autodissociation of water migrate to the cathode and anode respectively, neutralising the OH– and H+ produced at the respective electrodes. However, due to the very small amounts of H+ and OH– from the autodissociation of water, H+ (from the anodic reaction) and OH– (from the cathodic reaction) accumulate at the anode and cathode respectively over time. This produces a counter potential that retards and eventually stops the electrolytic processes at both electrodes.
2) Electrolysis of water with a salt
As described above, the electrolysis of pure water cannot proceed after a while, rendering the process useless. This is why we add compounds to water to allow its electrolysis to proceed unhindered. If we add NaCl to water, the electrode half-cell equations remain the same. However, substantial amounts of Na+ and Cl– are able to migrate to the cathode and anode respectively to compensate the negatively charged OH– and positively charged H+, allowing the electrolysis to continue. The larger the amount of NaCl added, the longer the electrolysis of water proceeds. If NaCl becomes too concentrated, chlorine will be produced at the anode instead of O2, as Cl– will be oxidised preferentially.
3) Electrolysis of acidified water
If we add an acid, e.g. H2SO4, instead of NaCl to water, we have
At the anode: 2H2O (l) → O2 (g) + 4H+ (aq) + 4e– Eo = -1.23 V
At the cathode: 2e– + 2H+ (aq) → H2 (g) Eo = 0.00 V
Overall reaction: 2H2O (l) → O2 (g) + 2H2 (g) Eo = -1.23 V
4) Electrolysis of alkaline water
If we add a base, e.g. NaOH, to water, we have
At the anode: 4OH– (aq) → O2 (g) + 2H2O (l) + 4e– Eo = -0.40 V
At the cathode: 2e– + 2H2O (l) → H2 (g) + 2OH– (aq) Eo = -0.83 V
Overall reaction: 2H2O (l) → O2 (g) + 2H2 (g) Eo = -1.23 V
Note that the overall standard electrode potential for the electrolysis of acidified water is the same as that for the electrolysis of alkaline water. Since most water electrolysis experiments involve either the addition of an acid or a base to water, the typical overall standard electrode potential quoted is Eo = -1.23 V instead of Eo = -2.06 V.
