Re: CS>Silver-Colloids responds
From: Marshall Dudley
Date: Mon, 28 Mar 2005 09:12:37
http://escribe.com/health/thesilverlist/m79030.html
> How on earth do you reach that conclusion. I have already outlined
> how ionic silver most likely works when ingested, silver hydroxide
> -> silver chloride -> (maybe ammonial silver complexes) -> silver
> particles.
Marshall,
Welcome back. We missed you. Some questions for reference:
1. Has Frank responded to your analysis on 26 Jan 2005, showing the
solubility of AgCl in blood is 0.9ppm?
http://escribe.com/health/thesilverlist/m77225.html
2. Please show the balanced equations for converting Ag(+) to Ag as
you describe above.
3. I use sublingual absorption, which I find to be much more
effective. How does this affect your analysis?
4. In http://escribe.com/health/thesilverlist/m79033.html, you state
> Yep, done that myself, the black powder is finely divided silver
> which will be black. This can be easily confirmed by applying
> pressure to the powder, and since silver is very mallable, the
> powder will be forced together forming a larger particle which
> will regain it's silver appearance.
What are the balanced equations for converting Ag(+) to Ag in dw?
4. You state:
> Yep, done that myself. As I have reported before H2O2 will oxidize
> silver metal producing silver oxide which dissolves quite readily.
What are the balanced equations?
> No, what appears on the cathode is silver metal being precipitated
> out.
> When current is applied to pure water the Ag+ goes toward the
> cathode and a hydrogen is released at the cathode and the
> remaining OH- goes toward the anode. At that point you have silver
> hydroxide in the water.
> As electrolysis continues then some of the OH- makes it to the
> anode, and some of the Ag+ makes it to the cathode. At the cathode
> the Ag+ gains an electron and precipitates out as very finely
> grained particles which will be black. At the anode the OH- will
> combine with a second OH-, lose two electrons to the anode, and
> produce H2O and O. The O being very reactive will tend to react
> with the silver making a tan colored silver oxide at the anode.
Marshall, I think you will find it impossible to write balanced
equations describing the above. But try to surprise me:)
I'll leave the rest for when you have time to answer the above.
Mike Monett
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