The Importance of VLR Cells in Electrolysis
                                                           By Mike Johnston
                                                             Copyright 2001

 In my last paper I came to the conclusion that electrolysis cells, as commonly designed and used, are actually a form of battery (power source) in their own right. The evidence to support this idea is readily available from any chemistry or electrochemistry textbook. Like primary cells, electrolysis cells depend on an oxidation/reduction reaction to create an electrical charge and like a secondary cell they store up the electricity produced as potential chemical energy, in this case as the H2 gas which is released at the cathode.

 They differ from primary cells though because, in the traditional electrolysis cell, the electrodes are both made of the same material and so create no potential difference between them and also unlike primary cells the metal of the anode in an electrolysis cell is not oxidized. Instead the negative ions themselves are oxidized (OH- + OH-  ---> 2H+ + O2(g) + 2e-) (one of many possible reactions based on the electrolyte used) and produce O2 gas instead of an oxide of the metal of the anode as the product. Other than that the same chemical reactions which create an electric current in a primary cell also exist in an electrolysis cell. In both cases electrons are released to the anode as a product of oxidation and released from the cathode as an effect of reduction.

 What is a "VLR" Cell then? That is my abbreviation of the term; "Very Low Resistance" and refers to the internal resistance of the cell. Obviously, in order to allow the maximum amount, or better yet, a specific, preplanned amount of gas to be produced per cell, in a series of cells, the ability to control the internal resistance  of each cell is is one of the most important factors that must be taken into consideration in order to achieve maximum efficiency in the use of the charge delivered, across the system. The methods to create such a cell may be devised by identifying the specific resistance producing components within the cell and the ways in which that resistance could be deliberately modified. This would then be the next step to take and that has been so in my own research and the results of that research to date are the focus of this paper.

 As an example, if you have a single cell which draws 6amps of current from a power source, it would produce a quantity of gas which is "equal" to that predicted by Faraday's formula of mass equivalence ( m= atomic number * a * t). So then, if you were to
construct two cells who's combined resistances were equal to that of the aforementioned single cell then, since both cells would be carrying the same 6 amps at the same voltage as the single cell carrying 6 amps at that voltage (according to Ohm's law) you would effectively double the amount of gas produced by that 6 amp charge.

 This concept can be applied to any number of cells with the same result. The more cells that you have at a given current, the more gas you produce at that current. I don't think  there is a whole lot of room for argument with that statement but I believe that it does merit further explanation. That is because voltage is the other factor which must be taken into consideration here and that is the main stumbling block which people seem to encounter when trying to create a really efficient electrolysis system.

 The reason for that difficulty is twofold, on the one hand there must exist a certain "Potential Difference" (voltage) between the electrodes in a cell for electrolysis to take place (whether provided by the electrodes of the cell or provided by an outside power source) and on the other hand, because (in a standard electrolysis cell) there is always going to be a voltage drop across the cell due to internal resistance.

 Some people seem to believe that these two figures (necessary potential difference and voltage drop) are one and the same. This is not the case though. It is true that you can design and construct a cell which will "use" nearly all of the input energy (in that nearly all of the input voltage is dropped across the cell) and I have seen examples where such a design was used to show high efficiency in converting electrical energy into the "stored" energy of hydrogen gas.

 What is wrong with that picture though? At first glance it seems true but if you look closer you realize that it is an example of the lowest efficiency that you could achieve and still have the cell functioning at all. Voltage being defined as the "push" behind electrons as they move through a circuit and, since ( due to the laws of chemical equilibrium) as many electrons must leave the cell at the anode as enter at the cathode, then by using up all your "push" in one cell you are simply wasting much of it. Especially when, by a better design, you could use half that push in one cell and the rest in another, etc and by so doing exponentially increase the total amount of gas produced.

 You see, if you look at hydrogen as an energy carrier, you must admit that the energy which it stores can only logically be equated with the amperage of the cell used to produce it, NOT the voltage! This is because amperage is defined as a specific number of electrons (6.28 x 10 18 ),  moving  past a specific point, per second and if that point is the surface of your electrode in an electrolysis cell then the number of those available electrons which are picked up by waiting H+ ions is the percentage of your input charge which is stored as H2 gas. Faraday's mass equivalence formula (quoted above) makes no provision for voltage, only mass, amperage and time. The formula is equally valid whether you are using it on an electrolysis cell or a primary cell. In both cases the minimum "push" must be there but, other than that, it is is not really relative to the conversion.

 Further, if you consider the fact that; whether you supply said amperage to a cell at 12 volts or 110 volts while you are producing the hydrogen, when you later recover the energy stored in that hydrogen, either by combustion or by combination in a fuel cell, the energy that you get back from an equal quantity of gas produced in both cells is in no way greater in either of them due to the voltage used to produce them. You can't get a fuel cell to produce a current at 110 volts just because you used 110 volts to produce the gas from water in the first place! The voltage of the cell is determined solely by the design of the cell and is usually limited to 2 volts or less. If you recombine the H2 and O2 by combustion the same holds true. You don't get a bigger bang for your buck just because you produced the gas at a higher voltage. The amount of heat energy released at combustion is equal to the amperage used during production. You can't really convert it to watts because you have to invoke voltage to do that and it will always appear that you used more to produce it than what you got out of it that way.

 So then it should be obvious that you don't gain any advantage by dropping your entire supplied voltage over a single cell isn't it? In fact just the opposite is true. The lower the amount of voltage that you lose per cell, the more of that voltage that is available to the next cell in a series and the more gas that can be produced with the same current.

 While I was figuring out the above information I was working exclusively with cells that had electrodes made of the same metal (stainless steel). So for most of this paper I will be focusing on those, although I myself have since switched the focus of my research to a different type of setup the information presented here will be valid for either.

    First off, in primary cells the formula for calculating the output voltage of the cell is V=E-Ir where V is the output voltage of the cell, either alone or it's own in combination with energy input from other power sources and E is the total input voltage whether that produced by the cell under no load or the incoming voltage from another power source or the combination of the two, and I is the current carried/produced in the cell and r is the internal resistance of the cell. This formula applies to either primary cells or electrolysis cells. I found this out by experiment. I couldn't get the readings from my cells to match up with what was predicted by Ohm's formula for calculating the voltage drop across resistors ( v = IR) but they did match perfectly with the V = E - Ir formula.

 So the V = IR formula cannot be used effectively in electrolysis cells any more than it can in primary cells. Why? Simply because neither primary cells nor electrolysis cells contain only resistance, BOTH ARE POWER SOURCES. It is true that electrolysis cells are different from any other power source because a) they require energy to be input from another, separate power source and b) because they cannot produce any more (add to the) energy (as increased voltage) than what is supplied to them from the outside power source.

 They still do qualify as power sources though. In their favor, consider the process by which electrons move within the cell. Electrons enter at the cathode and are either picked up by waiting ions or expend their energy as heat into the system. They aren't ferried across the cell by ions. That was and perhaps is one "accepted" explanation but it really doesn't hold water (hehe) in an electrolysis cell as it does in a secondary cell. So that's it then, your entire incoming charge is "used up" at the cathode. NONE of it leaves as current (the electrons go in but they don't come out).

 But electrons DO come out, just as many as go in, in fact (minus the number who gave up their energy as resistance heat at the cathode). So they have to be coming from somewhere, right? Sure, they are being released by the negative ions at the anode and chemical equilibrium is maintained. The fuel that is consumed in the operation of this power source is the water within the cell. That brings up the question of how voltage can be transmitted across the cell then if the same electrons don't carry it, doesn't it? In a normal resistor it seems acceptable that the electrons go in under a certain "push" and some of that push is used up on the way through the resistor but in this case the same electrons don't leave and yet the ones that do have their own push. Some process must be imparting that push to them.

 I am speaking in general terms here for illustrative purposes, the same exact electrons as enter a regular resistor aren't believed to be the ones that leave it there either as the charge is thought to be passed from free electron to free electron but in an electrolytic solution that cannot be the case because there are no free electrons to pass the push along. Chemical equilibrium explains how the same number of electrons are released to the anode as are taken up at the cathode but doesn't really explain how the voltage (push) is transmitted as well (most of it anyway).

 I would like to advance one possible answer to that question based on my own observations. While I was experimenting with a series of four electrolysis cells with stainless electrodes and a solution of NaOH, KOH and water as the electrolyte under a current of 12 volts, 15 amps (it was supposed to be 6 amps but my resistance was too low) I decided to include a light bulb in the setup, in various configurations, to see the results (if any). Mainly because the cell resistance was low enough that, even with four in series, the system drew so much power that it kept tripping the internal circuit breaker on my power source every few seconds (even at the 6 volt setting) and that made extended observation of the workings of the system difficult at best. At the 12 volt setting I still had over 10 volts leaving the series of cells and that was just too much for the power source to handle. I figured that a light bulb would provide enough resistance to allow the system to operate for longer periods. I used a 12 volt resistor bulb. It was the kind you see on the side of a truck trailer as marker lights, the amber ones.

 I first connected the light in series with the cells, before the cells. I observed that the light glowed brightly and gas production in the cells decreased. When I measured the power consumption I noted that half the available voltage was dropped across the light and the remainder (most of it) over the four cells. There was still 3 volts leaving the last cell though.
 Next I wired the light in series with the cells AFTER the cells. The voltage being delivered to the light was around 10.4 volts and the amperage was the same as it had been when the light was connected before the cells.The light glowed less brightly than it had before because it was operating at 10 volts now which could be expected but the really interesting thing was that all obvious gas production in the cells ceased. This was sort of puzzling at first. If the light glowed then there must be current being delivered to it but, if there was no gas being produced in the cells (how current is passed in the cells), then how was this possible?

 The only obvious answer to present itself was that the cells were acting as capacitors. It made sense because electrolytic capacitors do exist and, since the cells were of lower resistance than the light, the cells must have been triggered into acting as capacitors in order that enough charge could be delivered to the light to keep the circuit operational. This facet of cell behavior has become more important as my research has gone on and will be addressed more fully later.

 The parts of a cell which appear to contribute to it's internal resistance and can most easily be modified are a) the size, shape and composition of the electrodes and b) the electrolyte solution. The resistance for the electrodes can be figured from a commonly available set of tables of conductivity and resistance and it isn't that hard to do. Here is an example of the range through which the resistance of a copper electrode could be modified. Not that copper is a good or bad electrode material but since the tables are based on copper I use it for purposes of the example only.

   The table says that the resistance of 1000' of 1 cm thick copper wire is .070 ohms. If you divide that length down to where you have a cube of copper 1cm on a side you end up with a resistance of .0292969 ohms for that cube. If you then expand the surface area of that cube to further reduce the resistance the results look like this, remembering that for two electrodes of the same size the value would need to be doubled;
Surface area (sq/in)                                Resistance in Ohms
1/2"                                                                .05
1"                                                                   .025
2"                                                                   .0125
4"                                                                   .00625
8"                                                                   .003125
16"                                                                 .0015625
32"                                                                 .0007812
64"                                                                 .0003906
128"                                                               .0001953
256"                                                               .0000976
512"                                                               .0000488
1024"                                                             .0000244
2048"                                                             .0000122
4096"                                                             .0000061

 This table is only for illustrative purposes and may not be accurate because it doesn't make any provision for any resistance source other than the metal of the electrodes, etc. Just for fun though, what if the above numbers did represent total cell resistance? What kind of gas output would this provide and over how many cells? Electrodes could be viewed as either two single plates or multiple plates in each cell. I will take the lowest resistance and use that and assume a 12 v charge is delivered and a 6a current desired through the series.
 So then, in a cell with an internal resistance of .0000061 ohms, the cell would draw;               I = V/R so I = 12/.0000061 = 1,967,213.1 amps. Ok, so that's a bit much. Let's divide that by 6 amps per cell to be reasonable and we find that we could have 327,868.85 cells in our series. If each of those cells had that resistance then at 12 volts each cell would "pass" a 6 amp current. How much H2 would those cells produce? 20.655737 gm/sec or 1239.3441 gm/min or 74360.646 gm/hr or 1784655.5 gm/day. Not bad for a 12v 6a charge considering that the amount of H2 in a gallon of water is 475.2 grams.

 The next factor to consider will be the electrolyte. There are tables which will tell you the resistance of each electrolyte and they seem to be accurate but they refer to the resistance of the electrolyte at a specific concentration in a specific quantity of water(usually 1m/liter). This can be misleading. That is because if you only consider it from that perspective then you are doomed to a life of inescapable resistance. But how can the tables of resistance be accurate and yet have it be possible to modify the resistance of an electrolytic solution?

 Glad you asked. To answer that question we have to look again at how an electrical charge is transmitted through an electrolytic solution. In a solid metal conductor the level of conductivity is determined by the number of free electrons in each different type of metal. So it is possible to make a piece of iron (for example) be equally conductive as a piece of copper simply by increasing the size and/or adjusting the surface area and the thickness of the piece of iron until the number of free electrons present in it are equal to the number of free electrons present in your piece of copper.

 Unlike a metallic conductor there are not thought to be any free electrons in an electrolytic solution. Instead there are ions which, in the case of a secondary cell, serve as electron "carriers" and in a primary or electrolysis cell serve to join with electrons at the cathode and, in turn, deposit them at the anode without carrying them across the cell. The ions themselves do the traveling in the case of electrolytic cells, as more water is broken apart. But even that travel of the ions is not necessary for the electrolysis process to occur.

 This fact was first demonstrated by Faraday in experiments where he basicly connected two otherwise separate containers of electrolyte with a salt bridge (a piece of paper or string soaked with solution) or a tube containing another chemical. When an electrode wasplaced in each container and a current developed electrolysis occured. After a sufficient period of time had passed the container with the cathode showed a buildup of negative ions and the one with the annode a buildup of positive ions. That was because the ions couldn't migrate across the salt bridge. and yet electrolysis continued until the buildup of oppositely charged ions in each container became too great.

 In a solution of electrolyte and water there are a certain number of positive and negative ions present. This number is determined by the amount of electrolyte present, the concentration of the electrolyte and the disassociation level of the electrolyte at that concentration as well as the temperature of the solution ( with conductivity increasing as temperature rises through a certain range). A little more complex, with a few more variables than metals but simple enough once you understand it and equally subject to deliberate manipulation.

 In order to "transfer" an electric current between the cathode and the anode of a cell the first requirement is that there must be an ion present at the surface of the cathode to pick up every electron that is available. The number of electrons available would be determined by the amperage that is being used. Each amp is defined as 6.28 x 10 18 electrons per second. So then it could be assumed that, to reduce resistance in the solution, there should be at least an equal number of ions present to react with those electrons. Not that there must be that many available but that if there were then the reaction could proceed at an equal pace in both the metal of the electrode and in the electrolytic solution as well with the least resistance possible.

 To insure that a sufficient number of ions is available you must consider each requirement (as outlined above) and how to modify them. First the concentration of electrolyte and disassociation of same.  As I said, the standard concentration for an electrolyte used in figuring the values given in tables of electrolytic resistance is 1m/liter or one mole of electrolyte per liter of water. At this concentration HCL and most salts are considered to be 100% disassociated, as well as Na and K (into Na+ OH- and K+ OH-) so it is (more or less) simple to figure out the actual number of ions which are present at this concentration and the one I will use in this explanation (electrolyte will be assumed to be 100% disassociated).

 The next step in finding the number of ions present in our one liter of solution is to find the number of atoms present. To do this I will rely on the Avagadro number. This idea says that there are 6.0235 x 10 23 atoms in one gram/atom of a substance. A gram/atom is defined as the weight (in grams) of one mole of the substance. So one gram/atom of sodium would weigh 22.991gm and one gram atom of potassium would weigh 39.10 grams. In a compound the weights of each individual component must be added together and so one gram atom of NaOH would weigh 39.999gm because 22.991gm (one mole Na) + 16gm (one mole O) + 1.008gm (one mole H) equals 39.999gm  of NaOH.

 Now that we know how many electrons we are moving per second and we can figure out how many atoms/ions are present in a given quantity of electrollytic solution at a specific concentration it should be easy to make sure that there is indeed at least one ion present for every electron that we want to move. For my experiments I simplified it by accepting that, if I used the ratio of one gm/atom of electrolyte per liter of water for every amp of electricity I wished to move, then the conductivity of a 100% disassociated electrolyte should match that of the electrodes.

 This of course means that, to pass a 20a current (for example), through such a cell with the lowest resistance possible you would need 20 gm/atoms of electrolyte in 20 liters of water in each cell. Obviously size might become a problem when using this formula in practical applications but for experimental purposes size is not nearly as important as function.