In a message dated 6/21/01 7:30:12 AM EST, [email protected] writes: << Subj: RE: CS>Theorizing About the Species Produced at Electrodes During LVDC CS Pro... Date: 6/21/01 7:30:12 AM EST From: [email protected] (Ivan Anderson) Reply-to: [email protected] To: [email protected] > -----Original Message----- > From: [email protected] [mailto:[email protected]] > Sent: Thursday, 21 June 2001 05:08 > To: [email protected] > Subject: Re: CS>Theorizing About the Species Produced at Electrodes > During LVDC CS Pro... > > Another property of Lewis acids: > Some metal cations form Lewis acids (including silver), and the total > dissolved concentration of the metal depends upon the > concentration of the > complexing ion...say Cl-... as the concentration of Cl- increases the > concentration of Ag+ decreases (common ion effect) until a point > is reached > where Ag+ redissolves as variations of AgCl(aq) are formed eg > Ag++ Ag2Cl and > so on. > ------------------------------------------------------------------ > ------------ > > ----------------------------------------- > Ivan: I assume you are referring to several other equilibria > being setup, two > of which I've written below: > > [Ag+] + [Cl-] = AgCl(aq) > > [Ag++] + [Cl-] = AgCl2(s) > > Please tell me the likely oxidizing agent (in the blood, I guess) that > oxidized Ag+ to Ag++ There is no oxidising agent, this range of complexes is entirely due to the concentration of the ligand (the thing that is joined to the metal). xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: Well, then how did Ag+ (by far THE most stable ion of silver) get to Ag++, and what is present to allow it to remain in the +2 oxidation state? xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > In any case, wouldn't these additional equilibria tend to remove > even more > FREE ionic silver from the bloodstream since a finite amount of > silver ions > entered the body in the first place, and these extra equilibria > will consume > additional free silver ions to form the various compounds and molecules > required by these additional equilibria? I am not really saying anything about the body, just noting the the reason for an interesting phenomenom, namely the redissolving of silver due to increasing Cl- (in this case) concentration.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: OK. Let's look at a scenario that takes into account the formation of molecular AgCl(aq) after AgCl(s) has already formed. Add AgCl to water until it's saturated. Now add NaCl solution. Due to the common ion effect, the Ag+ conc. will continue to decrease until Cl- reaches a concentration sufficient to produce AgCl(aq) which represents molecular NOT crystalline AgCl. Now how can the Ag+ concentration increase when AgCl(aq) forms when an instant before there was no AgCl(aq) present? Are you saying that the activity coefficient of Ag+ decreases sharply when AgCl(aq) forms? If so do you have any actual Ag+ measurements to support this view? xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx [Ag+] + [Cl] = AgCl(s) [Ag] + [Cl] = AgCl(aq) Rather than consuming more free Ag ions, more is actually dissolved, each complex having its own equalibrium steals Cl- from the original solubility calculation. > ------------------------------------------------------------------ > ------------ > > ------------------------------------------- > > > Quantitative Chemical Analysis 5th Ed. - Harris > has this worked example: > > What is the maximum Cl- concentration at equilibrium in a > solution in which > [2Hg2+] is somehow FIXED at 1.0E-9 M? Our concentration table looks like > this: > > Hg2Cl2(s)<===>Hg(2)2+ + 2Cl- > --------------------------------------------- > Initial concentration 0 1.0E-9 0 > Final concentration solid 1.0E-9 x > --------------------------------------------- > [Hg(2)2+] is not x in this example, so there is no reason to set > [Cl-]=2x. > The problem is solved by plugging each concentration into the solubility > product: > [Hg(2)2+][Cl-]^2 = Ksp > (1.0E-9)(x)^2 = 1.2E-18 > x = [Cl-] = 3.5E-5 M > ------------------------------------------------------------------ > ------------ > > ------------------------------------------ > Ivan: Based on your stipulation, I guess I can't disagree even > though I'm at > a loss as to how the concentration of Hg(2)2+ can remain constant WHILE > Hg2Cl2(s) is formed since, according to your scenario, this salt > wasn't there > in the first place and it will take SOME finite amount of Hg(2)2+ > to form it. I copied this straight from the book. It demonstrates how a solubility equation is handled if there is NO solid at equilibrium, and it seems to fit the manner in which silver ions are introduced into water...directly, and not as the product of salt dissolution. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: As I said, I don't disagree. I just wonder how one could perform such an experiment in the laboratory. Perhaps using some buffering agent that maintains the concentration of Hg(2)2+ constant may work. Can you propose an actual experiment? xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > ------------------------------------------------------------------ > ------------ > > -------------------------------------------- > Here is my thinking; > For every Ag+ ion released into the water an OH- ion is created > (due to the > release of H2 gas) so the [OH-] (concentration) increases at the > same rate > as the [Ag+]. > > Ivan: So far so good. > > Now if [H+] was to remain the same as found in the water That should be [OH-] > (1.0E-7) then using > the template described above (except that x is not squared > because there is > only 1 Ag+ in AgOH) the equation looks like this: > > [OH-][Ag+] = Ksp > (1.0E-7)[Ag+] = 1.52E-8 > [Ag+] = 0.152M or 16,416 mg/L > ------------------------------------------------------------------ > ------------ > > ----------------------------------------- > Ivan: If the concentration of [OH-]continues to rise so that the > product of > [OH-] and [Ag+] reaches the Ksp of AgOH near the anode then why > couldn't the > following anodic reaction occur to reduce the buildup of [OH-]? > > 4[OH-] = O2(g) + 2H2O + 4e E = Eo (-0.40) + overvoltage No, because to have water oxidised at the anode would mean the reduction of water at the cathode, the net result being zero. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: Perhaps I'm missing something here, but if you put a little salt in DW and apply an EMF similar to that used to generate CS, wouldn't you get the following anodic and cathodic reactions? Anodic Reaction: [OH-] = 1/2H2O + 1/4O2(g) + 1e Eo = -0.40 Cathodic Reaction: [H+] + 1e = 1/2H2(g) Eo = +0.83 Net result: No net pH change as water is lost as hydrogen at the cathode, and oxygen at the anode. Now apply this same electrochemistry to the electrolysis of silver in DW. (1)Anodic Reaction (1) Ag(s) = [Ag+] + 1e Eo = -0.80 (2)Anodic Reaction (2) [OH-] = 1/2H2O + 1/4O2(g) + 1e Eo = -0.40 (3)Cathodic Reaction (1) [H+] + 1e = 1/2H2(g) Eo = +0.83 (4)Cathodic Reaction (2) [Ag+] +1e = Ag(s) Eo = +0.80 As I recall, the half cell reaction that occurs first, is the one with the lowest E (which equals Eo + the overvoltage to drive the electrochemical reaction to the right) To my way of thinking, the only reason Reaction(2) above MAY not go to the right is because the overvoltage is so high that it significantly exceeds the -0.80 (+ its overvoltage), the voltage associated with the oxidation of metallic silver to form silver ions. I don't see why it is important to consider what is going on at the other electrode. This type of analysis looks at the electrochemical potential at EACH HALF CELL in terms of "E" to decide which reaction(s) is most likely considering that the electrolysis of water will eventually -- if not immediately -- limit each half cell voltage . So please tell me in terms of "E" why any of the above reactions will not go to the right. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Roger -- The silver-list is a moderated forum for discussion of colloidal silver. To join or quit silver-list or silver-digest send an e-mail message to: [email protected] -or- [email protected] with the word subscribe or unsubscribe in the SUBJECT line. To post, address your message to: [email protected] Silver-list archive: http://escribe.com/health/thesilverlist/index.html List maintainer: Mike Devour <[email protected]>
