Background:
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Note, as to the underlying mechanics of Cupric oxide formation, I speculate that the NaOCl/HOCl first acts electrochemically on Cu to form Cu2O. The
latter cuprous forms a soluble complex with say chloride followed by a Fenton-type redox creating cupric. My take:
Cu(l) + HOCl --> Cu(II) + ClOH•-
Followed by either of the following reactions, depending on pH or presence of bicarbonate:
ClOH•- --> Cl- + OH• k = 6.1×10^9
ClOH•- + H+ --> Cl• + H2O k = 2.1×10^10
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[Edited on 14-6-2018 by AJKOER] |