Why doesnt deuterium oxide (heavy water) ionize suffiently to allow electrolysis?

Pure heavy water (D2O) was first obtained by electrolysis in 1933 by Gilbert N. Lewis, the mentor of Harold Urey, discoverer of deuterium (Urey had been concentrating deuterium by distilling liquid hydrogen).

The electrolysis method works as a depletion procedure (elimination of protium in gaseous form): Protium is electrolyzed at a lower potential (crudely estimated below) than deuterium. So, when almost all the water has been electrolyzed, what's left behind in liquid form is essentially pure heavy water (the completion of the procedure is indicated by a jump of hundreds of millivolts in the electric potential between the electrodes).

It all boils down to estimating the difference in ionization potentials that are due to mass differences for the nuclei involved, respectively, in an HO bond and a DO bond. I'll start with one back-of-envelope calculation that's fairly easy to do for energy of ionization of isolated atoms (although that's not directly applicable to what we're after):

As is the case for energy formulas in classical mechanics, the quantum energy levels of a classical electron around a nucleus are, strictly speaking, not proportional to the mass m of the electron itself but to the so-called "reduced mass" mM/(m+M) where M is the mass of the nucleus. This is due to the fact that the electron does not revolve around the center of the nucleus but around the center of mass of the system. If M is doubled (as is roughly the case when deuterium replaces protium) the energy levels are multiplied by

[(M+m)/mM] . [2mM/(m+2M)] = 2(M+m)/(m+2M)

which is rouglhly 1+m/2M = 1.0002723

More precisely it's about 1.0002724 (if the neutron-proton difference is taken into account)

Thus, if the energy of ionization is 13.598 eV for protium,

it's about 13.602 eV for deuterium.

In electrolysis that would translate directly into a potential difference of about 4 mV.

A covalent H-O or D-O bond is certainly very different from the above electron-proton bound state (lone hydrogen atom) but we may still use the similarity for a rough estimate of what difference in energy is due to the fact that a deuterium atom moves closer to the center of mass of the molecule than does a protium atom. If we consider that the masses of protium and deuterium are 1 and 2 while the "rest of the molecule" is 17, the relevant ratio is:

(2+17) / (1+17) = 1.0555...

A 6% difference in binding energy translates into hundreds of millivolts which would effectively prevent the ionization of deuterium until most of the protium is gone...

I recommend the video in the last link below.