Re: copper etching

From: n2mp (mcteguy_at_nospam.club-internet.fr)
Date: 12/31/04 

Date: Fri, 31 Dec 2004 20:05:19 +0100

Hello everyone,

<farooq_w@hotmail.com> a écrit dans le message de news:
1104407235.365465.13570@z14g2000cwz.googlegroups.com...
> Could you kind intrepret this observation?
> Though hydorgen peroxide is a reducing as well as oxidizing agent; eg
> it would reduce KMnO4 to Mn(II), but under no circumtances it would
> oxidize Mn(II) back to permanganate i.e it is oxidizing agent towards
> certain regeants or reducing agent towards some other; but it can not
> be both for a certain compound in the SAME pH range.

If H2O2 can act either as an oxidant or as a reductor, it's because it is
involved in two different redox couple : O2/H2O2 and H2O2/H2O. In the
former, it is the reduced form while, in the latter, it is the oxidized
form. In the acidic solutions, H2O2 is unstable and disproportionate into
H2O and O2.
Moreover, there is no reason for H2O2 to oxidize MnII (let's say Mn2+) to
MnO4-, even if it
reduces the latter for the former.Let's imagine, in given conditions (Eeq
refers to equilbrium potential, according to Nernst equations) :
O2 + 2H+ + 2e- <----------> H2O2 Eq1
MnO4- + 5e- + 8H+ <------------> Mn2+ + 4 H2O Eq2
H2O2 + 2H+ + 2e- <------------> 2 H2O Eq3

If H2O2 reduce MnO4-, this imply that Eq2 > Eq1 and we have the following
reaction :
5 H2O2 + 2 MnO4- + 6H+ --------> 5 O2 + 2Mn2+ + 8 H2O

Now, to oxydize Mn2+ back to MnO4- with H2O2, this would imply :
5 H2O2 + 2 Mn2+ ----------> 2 MnO4- + 6 H+ + 2 H2O,
which in turn implies that Eq3 > Eq2. If not, the reaction is impossible.

Another point, if the later was possible, then : Eq3 > Eq2 > Eq1. Then,
since according to the oxydoreduction rule, the oxydant of the redox couple
having the highest Eq reacts spontaneously with the reductor of the redox
couple having the lowest Eq, then H2O2 will preferably disproportionate into
O2 and H2O instead for oxydizing Mn2+ or reducing MnO4-. This make sense...
Otherwise, you would be able to create energy from nothing, just by the
miracle of mixing MnO4- with H2O2. Good try !...

Now let's consider that Eq2 > Eq3, then H2O2 will never oxydize Mn2+ back to
MnO4-. We have seen that Eq2 > Eq 1. Now, we have two possibilities : Eq2 >
Eq 1 > Eq3 or Eq2 > Eq3 > Eq1
In the first case, MnO4- would spontaneously oxidize water to produce H2O2.
As far as I know, this never happen.
In the second case, MnO4- oxydize H2O2 until all the additives are
consummed. Then, if it remains some MnO4-, it would oxydize H2O to H2O2, and
in turn H2O2 until no more MnO4- could exists. This seems also impossible.

Last chance to clear-up the mess : overlap the Pourbaix diagrams for both
MnO4-/Mn2+ and O2/H2O/H2O2 systems. To happen, a reaction must involved two
species that exists simultaneously in your operating conditions and the main
reaction will involve the oxident of the couple which has the highest Eq in
these corresponding operating conditions with the reductor of the couple
which has the lowest Eq.

> The surprising part is that H2O2 chemistry is that "immediately"
> reduces gaseous chlorine to chloride in acidic medium, this is well
> established in the literature.

So we would have :
2 Cl- --------> Cl2 + 2e- Eq4
This means that : H2O2 + 2Cl- + 2H+ ------------> Cl2 + 2 H2O and that Eq3 >
Eq4.
However, as far as I know, the H2O2 is not stable in acidic medium and
disproportionate into O2 and H2O2. But this has to be checked in Pourbaix
diagrams.

>Now how can hydrogen peroxide oxidize
> chloride back to chlorine (in the same pH range i.e acidic medium) as
> Wilco has shown
> that concentrated HCl with 20% H2O2 evolves gaseous chlorine after
> sometime, first the solution turns light green (most likely due to
> saturated Cl2(aq)) and then chlorine gas is evolved.
>Though the
> mechanism of chlorine evolution seems to be multitstep one...perhaps
> intermediated formation of chlorine species in higher oxidation states
> and subsequent reduction by HCl resulting in gaseous chlorine
> production. Since this reaction takes place (as Wilco Oelen) has noted
> when HCl is in excess!

Are you sure that Cl- are oxydized by the remaining H2O2 ? The "after
sometime" teases me. Is the experiment performed in open atmosphere ? Surely
due to Cl2 toxicity. What about O2 from air ?

Here again to clear up the mess, you need to overlap the Pourbaix diagrams
of both Cl and O2/H2O2/H2O systems.

Best regards. 

-- 
Philippe MOÇOTÉGUY
Vous pouvez me retrouver sur mon site internet à l'adresse suivante :
http://membres.lycos.fr/artzamendi/index.html
--------------------------------------------------------------------------------------------------------------------
You can also visit my webpage at the following address :
http://membres.lycos.fr/artzamendi/generale.html

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