What percentage of deuterium might be removed from water after a typical electrolysis procedure? From a post on this site I read that, "In the first stage, $\ce{NaOH}$ solution (initially $0.5\rm{M}$) is subject to electrolysis, until only $\frac{1}{32}$ of the original volume remains. This increased the original concentration of deuterium by a factor of $12$."

This seems to be, for the electrolyzed hydrogen, an approximate $38\%$ decrease from the original concentration. If we assume the original concentration was $150\rm{ppm}$ then after the first pass, we might expect to have $93\rm{ppm}$ in the deuterium reduced hydrogen. If my estimate is valid, may I expect that each successive pass would reduce the deuterium content by a similar percentage?

  • $\begingroup$ Looks like so. Also, the procedure is likely optimized for getting deuterium; if your goal is opposite, you might want to stop at an earlier point. $\endgroup$ Mar 1, 2016 at 6:06

1 Answer 1


You are quoting from my answer here.

Looking at the source reference Inexpensive Equipment for the Preparation and Concentration of Pure D2O or of Deuterium-Rich Water. Description of Construction and Operation of the Former Ohio State University Heavy Water Plant The Ohio Journal of Science. v41 n5 (September, 1941), 357-365:

With NaOH electrolyte we obtained, with Armco iron, an electrolytic separation factor[footnote 8] of a little more than 8, whose magnitude was practically independent of current density and of the deuterium concentration in the electrolyte. This is the true separation factor applicable to the dry gas. The practical separation factor is somewhat less than this (about 6 or 7) due to water vapor carried out with the electrolytic gases.

So, like Ivan Neretin is commenting, it is not optimal to electrolyze most of the first stage solution if the goal is obtaining deuterium depleted hydrogen. The initial hydrogen coming from natural abundance water will be deuterium depleted by a factor of "about 6 or 7". Since natural abundance water is essentially free, it would be best to only electrolyze a small fraction of a large amount of water. Then you will deplete deuterium down to about 25 ppm in one stage.

After the first stage, if you attempt a second stage, burning the water to make deuterium depleted water, you will then have to add electrolyte, which, if it is NaOH, would introduce more deuterium. So if you used a hydrogen-free electrolyte it would be better. By the second stage, the starting material is valuable, and you would want to electrolyze a significant fraction of it, but not 31/32.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.