Note: This is the first thing I ever wrote on this subject and it does contain some inaccurate information.
MJ

# WATER, PART 1

I can't believe the amount of negative response that I got by suggesting that we talk about using H2O as a fuel source. I thought that I would just start at the beginning and explain everything step by step but I couldn't even get past the first paragraph or two without a major argument so lets do this a different way. I'm going to explain the whole thing as I see it in as simple a manner as I can.
Argument #1 was that it takes just as much energy to electrolyze a given quantity of water as is released when you burn the products of that reaction. Ok, fair enough, let's explore that statement first.
The following problem shows how much energy is released
in burning one mole of H2 (in watts):  62,000 BTU per pound of H2 burned
in  air = 137.65663 BTU to the gram of H2 so then:
450.396 Grams per pound
(137.65663 BTU per gram of H2)*(1.008 grams per mole of H2)= 138.75788
BTU per mole of H2 and therefore:
(138.75788 BTU per mole of H2)*(778.26 ftlbs per BTU)= 107989.7 ftlbs
per mole of H2 so:
107989.7 ftlbs per mole of H2 = 3.2724151 hp per mole of H2 so then:
33,000 ftlbs per hp
(3.2724151 hp per mole of H2)*(746 watts per hp)= 2441.2216 watts per
mole of H2 and:
2441.2216 watts per mole of H2 = 2.4412216 kilowatts
1000 watts per kilowatt
Let's remember that shall we? Because now we are going to see how much energy is needed to electrolyze an equal quantity of water. Well actually we need to double the above figure to account for the fact that we are using 2H2 to react with O2 and form two molecules of H2O so it is really 4.8824432 kw. Following is the problem that shows how much energy is required to electrolyze one mole of water:
(68.32 kcal to electrolyze one mole of water)*(4.18*1000 [i.e.: ten
to the third power])= 285,850.88 joules to the mole of water so:
285,850.88 joules to the mole of water = 4764.1813 watts to electrolyze
one mole in one minute
1 min * 60 seconds
or 4.7641 kilowatts to do the job.
Let's review our findings, shall we? We found out above that it takes  4.7641 kilowatts of energy to electrolyze our mole of water right? Go ahead,  check my math....... Right then, we also found that when burned our same  mole of H2 will yield 4.8824432 kilowatts of energy didn't we? Sure we  did and what does that mean? Well let's see if you round up just a bit
it means that it takes 5 kilowatts of energy to electrolyze a mole of water  into H2 and O2 gas and 5 kilowatts of energy is given off when we burn  those two gasses to form one mole of water in the form of superheated steam.
So there we have AGAIN proved the old axiom that it takes just as much energy  to electrolyze water as is given off when you burn  the products of that reaction and OPEC can breathe a sigh of relief.  But wait...actually, if we don't round the numbers off,  there IS a difference between how much energy is required to separate water  into it's component parts and how much is given off  isn't there?  I grant you it isn't much of a difference but it could mean a great  deal in any attempts to use H2O as a viable source of pollution free fuel.
The best possible scenario in the electrolysis cell would  be for the electrodes to be made of platinum as it is an inert metal and  will not participate as a reactant as, for example, copper could.  The preferred electrolyte will be Na2SO4 (Sulfuric Acid) as the only products of this  reaction are H2 at the cathode and O2 at the anode. The water must be circulated  regularly to prevent the buildup of 2OH- molecules at the cathode and 4H+4e- at the anode which will ultimately interfere with the optimum operation of the cell. When the water is circulated these ions rejoin to form H2O molecules and the only products of the electrolysis reaction are the H2 and O2 gas.
I use copper electrodes and NaCl as reactants in this  paper for simplicity in duplication. The product at the cathode is still  H2 gas but at the anode it would be chlorine gas or the copper could cause metallic sodium to be the product there. It is probably actually safer to do it this way as an amateur because you don't have the gaseous O2 floating  around to pose a potential danger to you.
Stay tuned for the next installment.