By Mike Johnston
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
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.
So ok, now we see that we can indeed influence the level of internal resistance in an electrolysis cell, just as we can in a primary or secondary cell battery. The next step seemed to be to identify all of the sources of resistance within such a cell and to devise ways to modify each so that their effect on the operation of the cell could be minimized or increased as the requirements of a particular setup demanded.
Once I reached this point in the research I felt like I had arrived. My experiments were bearing out my initial theories. Unfortunately that feeling didn't last for long. I felt that I had made quite a bit of progress along the road to a better understanding of the electrolysis process, but it wasn't enough. Too many questions still remained hanging, without answers, and that annoys me.
With the realization that just as many electrons must exit the cell at the Anode as enter it at the Cathode (to maintain chemical equilibrium) there came the understanding that the only real obstacle that remained was to find a way to reduce or eliminate the voltage drop across the cell. Ok , fine. I thought. So therefore, by identifying ALL of the sources of such resistance within the cell and then eliminating them, or modifying them in order to reduce their negative impact, it should be possible to drastically improve the efficiency of the system. Once that was done you could pass a charge through a very long series of cells with little waste of energy and thereby exponentially increase the amount of energy delivered back to you as H2 gas.I know I mentioned this a few paragraphs ago but it's importance cannot be stressed enough. Understanding all of the sources of resistance within the cell and being able to modify them is the key to the whole process of producing H2 from water more effeciently.
In the first part of this paper I looked at electrode size and the number of ions available in the solution. These are two of the biggest contributors to cell resistance and, as I have demonstrated, both of these sources can be manipulated through a very wide set of paramaters.
Not yet satisfied I started to experiment with some new ideas. One thing that differentiates between electrolysis cells and wet cell batteries is that batteries use the potential difference created between two electrodes of different metals to provide the "push" needed to get a current flowing between them. Electrolysis cells, on the other hand, use electrodes of the same, chemically inert, metal and consequently no potential difference is thought to be developed between them.
Much of what determines the "potential" of a given type of metal is the number of free electrons which are thought to exist within the metal. Those metals which have more free electrons have more potential to give some off or to more easily move a charge through them and so are good conductors. Metals with fewer free electrons, conversely, have a lower tendency to give any off, better ability to accept more of them and are worse conductors. So with two identical pieces of metal one wouldn't tend to expect any difference in potential between them.
Is that "written in stone" though, I wondered? It would be SO useful to be able to create some type of potential difference (voltage) in the electrolysis cell itself, to further offset the losses created by the various sources of internal resistance. It would do this by adding to the voltage supplied by your power source much as each battery in a series adds it's owm voltage to that of the others which precede it in the series.
Footnotes: There are several other factors which contribute to or otherwise influence the internal resistance of a cell which I am not going to address here. Such as; distance between the electrodes, mixed electrolytes which don't share a common ion, temperature of the solution, polarization of the electrodes,etc.
Then it hit me. If "potential difference" is by and large determined by the number of free electrons in each different metal then perhaps a potential difference could be created between two pieces of the same metal. How? By making sure that one of the pieces had substantially more free electrons than the other. The only way I could figure how to easily do that was by adjusting the size ratio between the two electrodes.
For example; if you start with a piece of metal 1" square and 1/8" thick and, for the purposes of the example, we know that such a piece of that metal would have a billion free electrons. Then we could assume that a piece of the same metal which was twice as large (surface area, maintaining the 1/8" thickness) would have double the amount of free electrons (2 billion), right? And therefore the larger piece should have a greater potential to be able to give up electrons than the smaller piece, as well as having greater conductivity (due to size).
Armed with that theory I decided to devise a couple of simple experiments to test it's validity. The experiments:
1) I had actually done this experiment previously but the data I obtained from it is useful to mention at this point. What it involves is using two electrodes made of aluminum foil in a solution of NaOH/KOH and water. Most people who paid some small bit of attention in High School chemistry know that the above solution will react vigorously with aluminum to release H2. I wanted to see how using two aluminum electrodes, connected to each other by a wire, would affect the process.
For the experiment I used a clean, wide mouth Mason jar (1-qt) and filled it 2/3 full of warm tap water (distilled would be better). Then I added enough NaOH/KOH solution to bring the level of the liquid to within 1 1/2" of the top of the jar.
Next I took two pieces of aluminum foil, 6" square, and folded them over several times until I ended up with two electrodes about 1" wide and 6" long. I did this to give them a bit of rigidity so that I could keep them in position in the cell while the reaction was taking place.
Next I slid the aluminum electrodes down into the jar, on opposite sides. The longer sides being vertical, and secured them in place with alligator clips. Then I connected my multimeter to the electrodes and turned it on. I observed a potential difference between these two identical electrodes of .57 volts. I verified this potential difference by reversing the multimeter leads and the result was a negative reading on the dial. It was still climbing when I hooked a wire between the electrodes to form a complete circuit between them.
After several minutes it appeared that the the electrode which was serving as the cathode (as indicated by current direction) was decomposing at a much slower rate than the electrode which was serving as cathode. To further test this I hooked my power source to the cell and accelerated the process. The cathode piece was indeed decomposing at a much slower rate than the anode.
I allowed the reaction to proceed until the piece which was serving as the anode had completely dissolved, to the point where the remaining scrap of electrode was no longer in contact with the surface of the solution. At this point the circuit between the electrodes was broken. I observed then that the electrode which had been serving as the cathode then reverted to the anode function and began to decompose rapidly by oxidation as had the other electrode. Within a short amount of time this former cathode had also totally decomposed.
My conclusion from this experiment was that, under the right conditions, two pieces of the same metal, identical for the intents and purposes of this experiment, could indeed develop a potential difference between them, of a sufficient magnitude to cause electrolysis to take place spontaneously. The main drawback in this case being that; in order for the reaction to proceed spontaneously, one or ultimately both of the electrodes had to be sacrificed. Since this particular reaction would seem to offer neither the cost efficiency or ease of maintenance that I have decided are two of the main requirements for a successful, commercial hydrogen source, I did not pursue this reaction any further with these particular reactants other than to repeat it with the addition of a 12v fluctuating DC power source. I noted that gas output increases but so did the rate of decomposition. I noted that the phenomenon of Cathodic Protection, as observed in this setup was again present and might prove useful at some future point in my research.
Experiment #2: For this experiment I waanted to test the validity of my theory that two pieces of the same metal, unreactive to the electrolyte in a solution, could still produce a potential difference between them if one of the electrodes was substantially larger than the other and therefore had more free electrons than the smaller one.
From a piece of non-ferrous stainless I cut one piece 1/2" x 4" x 1/8" and another
6" x 4" x 1/8". I then bent the larger piece into a cyllinder, 4" high and placed it inside an 8oz container. It pretty much lined the inside perimeter of the container. I then filled the container with an electrolyte/water solution with 25% NaOH/KOH as the electrolyte and 75% water.
I then fastened the smaller electrode to a piece of wooden dowel rod and using the rod as a cross member. I then dangled the smaller electrode inyo the center of the cell.