In a message dated 6/20/01 6:35:00 AM EST, [email protected] writes:
<< Subj: RE: CS>Theorizing About the Species Produced at Electrodes
During LVDC CS Pro...
Date: 6/20/01 6:35:00 AM EST
From: [email protected] (Ivan Anderson)
Reply-to: [email protected] (*Silver-List*)
To: [email protected]
Well I'll send it to the list then.
The original heading was tech alert...so Marshalee, don't read this!
I've been sitting exams, which is why I have been scarce!
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.
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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++
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?
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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-
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Initial concentration 0 1.0E-9 0
Final concentration solid 1.0E-9 x
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[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
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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.
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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 (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
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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
Given enough time for OH- migration to reach the anode, isn't the reaction
above just as likely to occur as the well known Ag+ migration to the cathode
where it gets reduced back to silver metal? Note that this half cell reaction
consumes 1 mole of OH per electron RELEASED just as the cathodic reaction
CONSUMES 1 electron per mole of OH- produced. Since most LVDC CS methods I'm
aware of have a final pH around neutral, aren't these half cell reactions
sufficient to explain CS generation as well as the final pH?
Roger
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