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guy
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[*] posted on 30-1-2006 at 17:54
Oxidation of lead question

I know that Pb++ and hypochlorite can form lead dioxide, but looking at the redution potentials, the E is negative suggesting unspontaneous reaction. Is there an explanation for this?
Pb2+ + 2H2O <-----> PbO2 + 4H+ + 2e- E= -1.455V
ClO- + H2O + 2e-<------> Cl- + 2OH- E= 0.89V
----------------------------------------------------
Pb2+ + ClO- + H2O <-----> PbO2 + 2H+ Cl- E= -0.555V

Could it be that the reaction between H+ + OH- --> H2O be the driving force? If so how much will that increase the E?



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BromicAcid
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[*] posted on 30-1-2006 at 19:53


The oxidation potentials as you found them are only for specific conditions and therefore not entirely applicable to the situation you describe, the oxidation usually taking place under basic conditions and as you described not only is it the hydroxide ions doing their part, but just that most everything is different.

[Edited on 1/31/2006 by BromicAcid]



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12AX7
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[*] posted on 30-1-2006 at 21:02


Yeah...

Those potentials probably are for acidic or neutral conditions. I know personally that PbO2 can be formed from chlorine, and that the exact opposite happens too! :)

I once cleaned a flask stained with PbO2 using hydrochloric acid and it gave off a bit of pungent fizzing: chlorine gas.

Tim



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[*] posted on 30-1-2006 at 23:24


So, both of the equations need to be in basic solution?? The redcution potential is even more as stated on this site http://www.webelements.com/webelements/elements/text/Pb/redn... But that's for PbO2 + 2e --> PbO, and I see that its less favorable because its a solid -> solid. Is is much more favorable if the reaction is Pb++ ----> PbO2 + 2e?



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