When we run a current through water (pure or not), $\ce{O_2}$ forms at the anode while $\ce{H_2}$ forms at the cathode. If I understand correctly, as water is already slightly ionized, $\ce{H}$'s snap together (and so do $\ce{O}$'s) during the electron exchange. Is it possible to apply enough heat or enough pressure to water to prevent $\ce{H}$'s and $\ce{O}$'s from snapping together and so we could thus form bubbles of ionized hydrogen and ionized oxygen?


The electrolysis will not form any ionized oxygen or ionized hydrogen, but instead proceeds through radical species. Some experiments have attempted to show the presence of $H^{\bullet}$ and $OH^{\bullet}$ radicals at the anode / cathode respectively (see P.H. Kasai, D. McLeod J. Am. Chem. Soc (1978)). Provided that the surface concentration is sufficient, these radical species will very quickly recombine to lead to $H_{2}$ and $O_{2}$. The surface of the electrode is very likely to influence the process, as the radicals species are likely to interact with it and not be released in high concentration in the solution.

Other factors are also influential, and the radical species might also react with other constituents of the solution (see for instance T. Mokudai, K. Nakamura, T. Kanno, Y. Niwano, PLOS One (2012) on the formation of hydroxyl radical vs formation of hypochlorous acid through electrolysis).

Heating the solution is likely to increase radical concentration, but the temperature should be extremely high to prevent radical recombination within the gas. As for pressure, my guess would be that a reduced pressure is likely to improve radical isolation by displacement of the solvation equilibrium.

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