Deuterium removal from water
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.