url: http://escribe.com/health/thesilverlist/m62787.html CS>electrons in making concentrated CS From: Reid Harvey Date: Thu, 25 Sep 2003 20:40:20
> Ole Bob, Marshall, Ode, Trem, Everybody, > I'm gearing up to big production of concentrated CS for use in > saturating our ceramic purifiers, but I'm feeling a little > ignorant as to exactly what's going on in the process. the link to > the generator is: > http://www.purifier.com.np/CS.html > As I've described it this is 27 volts with a 2.4 liter Erlenmeyer > flask in a double boiler, kept just under the boiling point. > Polarity reversal is every one minute or every two minutes, and > one gets an additional 50ppm for every hour of operation. What I > want to try to come to grips with is where the electrons go as > there's more and more ionic CS. Can someone kindly enlighten me? [...] > Can anyone tell me where the electrons go when I make this CS? > Then I'll feel better about recommending the generator to others. > The ultimate goal is to get the purifiers to the poor who are > vulnerable to the ravages of diarrheal illnesses. Hi Ried, There are a few simple experiments you can perform that may help understand what you are making. Start with plain cs made at room temperature using low current density at the electrodes. By low current density, I suggest 300 uA/sq.in. or less, although Trem's Silvergen or Ken's Silverpuppy would also work fine. You can tell how much silver is liberated at the anode by using Faraday's electrolysis equation. It is very simple, but the conversions between different units is highly error-prone. A DOS program called Mercury will eliminate this problem and solve the equations for you. It is available many places on the web. Here's two: http://www.mirror.ac.uk/collections/hensa-micros/collections/aeres/edsw/d-smath/mrcry209.zip http://archives.math.utk.edu/software/msdos/calculus/mrcry209/.html Now all you need are the unit conversions and some data. Here is a list of unit conversions suitable for making cs: Cou = I * sec ; total number of Coulombs esec = I / 1.60217733e-19; electrons per second gm = k * I * sec ; Faraday's equation isin = esec / sqin ; ions per sq. in. per sec isnm = isin / 6.45e14 ; ions per square nanometer per sec k = 107.868 / 96485 ; Coulombs required per gram of silver lt = 3.785 * gal ; convert gallons to litres lt = ml / 1000 ; convert millilitres to litres mg = gm * 1000 ; convert grams to milligrams ml = 29.57 * oz ; convert ounce to milliliters phr = ppm / hrs ; ppm per hour ppm = mg / lt ; 1 ppm is 1 milligram per litre sec = hrs * 3600 + mnt * 60 ; convert hours to seconds uAin = 1e6 * I / sqin ; current density in uA per sq in And here's some data I use for my Godzilla cs generator. You can change the parameters to suit your system. I = 1.544e-3 ; current ml = 2000 ; volume of dw mnt = 0 ; minutes ppm = 20 ; target ppm sqin = 11.5 ; wetted area Mercury's solution is to run a constant current of 1.544 mA for 6.44 hours. Some of the ions will combine to form silver oxide, so the actual ppm will be a bit lower. Let's say it is 96% ionic and 4% oxide. OK, now you have some high-ionic cs. Put about 2 inches in a small glass and heat it to about 140 degrees F until all the water evaporates. You should see the sides and bottom covered with black silver oxide. How did it get there? Let's say you had 2 ounces of 20 ppm cs. You can put this into Mercury and find the volume was 59 ml, and it took 1.057 Coulombs to liberate 20 ppm. Now, one Coulomb is one Ampere for one second. An electron has a charge of 1.6e-19 Coulomb, so one Coulomb is 6.25e18 electrons. It took this many electrons to liberate the silver, and it will take the same number of electrons to convert the ions into oxide. Where do they come from? Let's say it takes 6 hrs to evaporate the cs. This means the average current is 1 / (6 * 3600) = 4.6296296E-5, or 46 uA Glass is a very good insulator, and is used on high voltage transmission lines. So this current could not come through the glass. Air is a better insulator, even when it is humid. So the current could not come through the air. Where else do we have electrons? Let's look at the basic cs equations. At the anode, a silver atom gives up an electron to become an ion. The equation is Ag(s) - e --> Ag(+) At the cathode, hydrogen ions accept electrons to form hydrogen gas: 2H(+) + 2e --> H2(g) This means a molecule of water has to dissociate. The applied field is sufficient to ionize the water molecule: H2O --> H(+) + OH(-) So for every silver ion, one hydroxyl ion is produced. When the cs evaporates, the silver ions combine with the hydroxyl ions to form silver oxide. There are at least two paths. Path #1: One silver ion combines with one hydroxyl ion to form silver hydroxide: Ag(+) + OH(-) --> AgOH (silver hydroxide) The silver hydroxide dissociates to form silver oxide particles: 2AgOH --> Ag2O + H2O (silver oxide) Path #2: Two silver ions combine with two hydroxyl ions to form silver oxide: 2Ag(+) + 2OH(-) --> Ag2O + H2O So this explains where the electrons come from when the water evaporates. They are supplied by the hydroxyl ion. Now we can address your question about your process. When you make cs by electrolysis, a very thin layer of ions forms at each electrode. This is called the Nernst diffusion layer, and it may be 10 nanometers to 10 micrometers thick. The concentration of ions is highest next to the electrode, and falls off with distance. The concentration depends on the current density at the electrode. As the ions are generated, normal thermal diffusion causes them to leave the vicinity of the electrode, and eventually, the entire volume of dw is filled with ions. Some of them wander near the opposite electrode from where they were formed, and may happen to meet their opposite species and combine to form silver oxide as described in the above equations. The silver oxide particles are too small to affect visible light, but if the concentration is high enough, they can combine and form larger particles. This turns the cs a pale shade of yellow that gets deeper as more large particles become available. In your case, you are running at 200 degrees, just under the boiling point of water. This increases the thermal velocity of the ions, which greatly increases the probability that they will combine to form silver oxide. The high concentration of particles is what turns your solution red. So I think you are making concentrated silver oxide, not ionic silver. You can verify this by adding a small amount of H2O2. If the solution fizzes and makes lots of bubbles, the H2O2 is acting as a catalyst and converting the oxide back to ions. This reaction is very fast: 2Ag2O + H2O2 --> 4Ag(+) + O2(g) + H2O2 If the reaction runs very slowly and takes a very long time, you may have silver particles. The reaction is: 2Ag + 2H2O2 --> 2Ag(+) + O2(g) + 2H2O But it's difficult to see how pure silver particles could be generated, unless the silver oxide somehow dissociates due to the high temperature. it would be interesting if you performed this experiment and let us know your results. Best Regards, Mike Monett -- The silver-list is a moderated forum for discussion of colloidal silver. Instructions for unsubscribing may be found at: http://silverlist.org To post, address your message to: [email protected] Silver-list archive: http://escribe.com/health/thesilverlist/index.html List maintainer: Mike Devour <[email protected]>
