Molecule  Energy (eV)
 --------  ---------
 H         13.6
 H2        15.4
 OH        13.0
 H2O       12.6
 HO2       11.4
 H2O2      10.5


Fig. 1 - Molecular Ionization Potentials in Gas


The above table shows that water requires about the same but less potential to ionize than OH. Given an H2O or OH molecule is about 1 angstrom in length, a field gradient of over 1.3x10^11 V/m is sufficient to ionize OH to O + p + e-. Thus in a two step process at the anode surface we have:

   H2O   --->  OH + p + e-  --->  O + p + p + 2e-

If this can happen in a single step, but producing a paired electron 2e- boson, then this reaction is ideal for creating the dual electron catalysis ingredients.

If we assume the potential drop at the cathode is about 2 volts, and our cell operating voltage is 200 volts DC, the anode voltage drop can be spread over about 15 atomic layers. The anode interphase can be about 15 molecules thick and yet support the required free proton release at a diode layer breakdown voltage of 200 V, which is readily achievable. The formation of hydrogen peroxide should be suppressed at over 1.3x10^11 V/m.

Horace Heffner

Reply via email to