Frank,

Spent a little more time reviewing the H2O2/UV reaction.

I should have said that under the illumination of UV light H2O2 is reduced
or split into 2 OH radicals.

This is what is found in practice. Excess H2O2 will be oxidised by the OH*
radicals already present in the solution to water and oxygen gas. If this
reaction goes to completion, there will be either OH* radicals or H2O2 left
in solution unless the amount of UV irradiation is carefully controlled.

H2O2 + hv(UV) ---> 0H* + OH*
(2)OH* + H2O2 ---> (2)H2O + O2

Under alkaline conditions the H2O2 decomposes directly to H2O + O2, the OH*
radical production occurs most readily at neutral pH.

When H2O2 is added to water a small amount of OH* radicals are formed
directly.

I confess I don't know what concentration of OH radical molecules constitute
an oxidative danger if ingested. Small concentrations of H2O2 are to be
preferred.

By the way, the above reaction will proceed quite readily in sunlight.

Ivan.

> -----Original Message-----
> From: Frank Key [mailto:[email protected]]
> Sent: Sunday, 5 August 2001 04:00
> To: *Silver-List*
> Subject: Re: CS>What Happens to CS Particle Size & DistributionWhenH2O2
> is Added
>
>
> Ivan wrote:
>
> > I think you will find that UV light will not reduce H2O2 at
> all, but rather
> > oxidise it further to the highly reactive OH- radical.
>
> What we have found is that practice confirms theory:
>
> H2O2 + UV -> H2O + O2
>
> Ultraviolet light reduces H2O2 to water (H2O) and free oxygen O2.
>
> To properly radiate the solution with UV requires that the
> solution be pumped
> through quartz tubing that is in close proximity to the the UV lamp.
>
> frank key


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