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Author: Subject: Electrolysis of sodium bisulfate
vampirexevipex
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[*] posted on 15-5-2012 at 14:54
Electrolysis of sodium bisulfate


Hello im currently working on electrolysis of aqueous sodium bisulfate. Before i started electrolysing, the copper anode vigorously reacted with the compound (maybe because of the oxides around it?) leaving the anode a greyish purple and the solution a blue colour (perhaps copper sulfate?). After 2 hours of electrolysis i noticed a black blob on the cathone(copper hydroxide? But... Black??). My question is: is it possible to create copper sulfate this way? What is the blob glued to the cathone? (also it is conductive)
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bob800
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[*] posted on 15-5-2012 at 15:20


Quote: Originally posted by vampirexevipex  
What is the blob glued to the cathone? (also it is conductive)


I could be wrong, but I believe that "blob" is actually copper metal coated in some sort of basic copper compound (hydroxide, oxide?). During electrolysis, your copper anode is losing electrons and thus forming soluble copper(II) ions:

Cu ---> Cu++ + 2e

As these copper ions travel across your solution, they are eventually reduced at the cathode:

Cu++ + 2e ---> Cu

I don't know about the exact products of this electrolysis, but it's likely to be a mixture of several compounds.
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vampirexevipex
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[*] posted on 15-5-2012 at 15:42


Thanks, but whats the blue electrolyzed solution? Can it be copper sulfate and sodium hydroxide?
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[*] posted on 15-5-2012 at 23:01


At the anode you get the following reaction:

Cu - 2e --> Cu(2+)

At the cathode the primary reaction in the acidic solution is:

2H(+) + 2e --> H2


So, in solution you get copper(II) ions, hence the blue color. Your solution slowly changes from NaHSO4 to a mix of Na2SO4 and CuSO4 (each pair of H(+) ions is replaced by a single Cu(2+) ion).

If electrolysis is allowed to continue for a long time, then the concentration of H(+) will get lower and the concentration of Cu(2+) increases. In that case the reactions at the cathode change and you get a mix of reactions:

Cu(2+) + 2e --> Cu

2H2O + 2e --> 2OH(-) + H2, and the OH(-) further reacts with Cu(2+) to form Cu(OH)2, which in turn partly decomposes to CuO.

At the cathode you will collect a mix of compounds, Cu(OH)2, CuO and Cu. This mix consists of very fine particles and forms a mud, which is very dark, almost black. CuO is black.




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AndersHoveland
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[*] posted on 16-5-2012 at 16:49


I do not have the details, but I think I remember something about the production of sodium peroxymonosulfate and/or hydrogen peroxide from the electrolysis of sodium bisulfate.

[Edited on 17-5-2012 by AndersHoveland]
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