rikkitikkitavi - 6-8-2003 at 10:28
I was letting my mind play (tricks) this sunny evening at my balcony...
If you have a diaphragm cell and fill it with say 5 M CaN solution and start a current through it the following will happen
anode 2H2O =>4 H+ + O2 + 4e-
cathode 4e- + 2H2O =>4 OH- + H2
nitrate will also be reduced at the cathode to NO2-, NH3 as a parallell reaction, depending on H2 overvoltage potential on the cathode.
since we have a 5M CaN , Ca(OH)2 will preciptate in the catholyte,thus removing OH- from solution. If we then let a small flow go from the cathode to
the anode, sufficient large, it will prevent the H+ formed at the anode to reach the catholyte and react with the solid Ca(OH)2 .
Simultaneously we transport NO3- away from the cathode, into the anolyte where we now have Ca2+ , NO3- and H+ 
With a proper physical layout you can probably let the Ca(OH)2 sedimentate at the bottom of the cathode compartment.
You probably cant expect ALL CaN to be converted, this because we need to increase the flow to prevent the H+ migrating. Exactly how much CAN be
calculated by ion migrating velocity and voltage, but simple trial and error also work. I did a rough estimate and 20-30 % conversion of a 5 M CaN is
possible.
The anolyte now can be recovered, evaporated, distilled and any H2O and HNO3 recovered. The remaining is CaN solution wich can be processed again.
Comments? Execpt from being a messy way 
to make HNO3 ?
/rickard
Yes, comments, comments...
Theoretic - 7-8-2003 at 06:00
Yes! It won't work! 
Apart from clogging the electrode and stopping the current from flowing (Edit: that's why electrolytic manufacturing of Ca(ClO3)2 doesn't
work), Ca(OH)2 is also a powerful base. If you try to dissolve it in water, 0.2g/l dissolve and create a pH of 11.5! Not exactly acid!
Ca(OH)2 reacts with acids, just like KOH or NaOH do.
You just can't make HNO3 that way, sorry.
[Edited on 7-8-2003 by Theoretic]
rikkitikkitavi - 8-8-2003 at 10:26
Theoretic, solubility of Ca(OH)2 in 5M CaN solution is very low...so you wont dissolve 0,2 g/l.
but as for the clogging, yes that is probably something that makes this process difficult. I found info about that after this post. But it also
depends on the conditions in the cathode compartment , oversaturation, metastable zone, crystall size and formation. With a low supersaturation
perhaps most of the Ca(OH)2 can be preciptated in the bulk instead of on the electrodes.
I have to think a bit more about this.
/rickard
Theoretic - 11-8-2003 at 08:09
Well, Ca(OH)2 will still act as an insoluble base - compare with CaCO3.
It would still create a HIGH pH on its surface, which would nicely react with acid that diffuses to it.
And anyway, 0.2 g/l is still classified as very low solubility.
rikkitikkitavi - 12-8-2003 at 04:06
thanks for the advice about the clogging,which I think is going to be the main problem.theoretic, but otherwise it is obvious that you havent
understood what I wrote.
Anyway, it is meaningless discussing this further, some experimenting needs to be done. But like you say, it is probably futile without cation-anion
membranes.
/rickard
markx - 13-8-2003 at 00:11
Well, this must be the most complicated way of producing HNO3 and it is very questionable if it works at all.
I personally think it doesn't but experimentation will probably bring some answers.