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i performed electrolisis on a sodium chloride solusion for half an hour and was left with a rusty coulored sludge at the bottomwhat is it?
I have definitely seen Cu2O from electrolysis of salt water using Cu electrodes.Electrolytically deposited copper is, in my observation, generally black (like most finely powdered metals).
Maybe rust? You didn't write what the electrodes are made of. If it's iron or steel...
About Cu2O, there are several ways to test it. One of them consists in taking it (after washing several times with water) and adding an excess of concentrated HCl (in case heating gently): it goes in solution as dichloro copper complex.
Unfortunately, lightarrow, Cu metal will react exactly the same way with warm concentrated hydrochloric acid. A stream of hydrogen gas will slowly be evolved, but this will not be easy to notice.I know that a lot of textbooks, especially the elementary ones, will say that copper does not react with concentrated hydrochloric acid, but try it and you will see.
It would be rather difficult to check if conc HCl dissolved copper because you would need to remove dissolved oxygen first.Anyway, another way of checking would be to add ammonia which will disolve half the Cu2O and turn the other half into Copper powder.It might be interesting to add a saturated solution of salt and see if it's converted to CuCl.However I don't agree with the initial supposed reason why it might be copper.The anode mud in electrolytic refining cells isn't copper.
Further to the reaction between copper and concentrated hydrochloric acid:I found the following web reference to a peer reviewed article by Swedish nuclear scientists (chemical engineers or chemists):http://www.stralsakerhetsmyndigheten.se/Global/Publikationer/SKI_import/010803/04318226039/98-19.pdfIn it I could find the following delta G°f values:CuCl2– = –246.0 kJ/molCl– = -131.2 kJ/moltherefore in the reaction Cu + H+ + 2Cl– --> CuCl2– + 1/2 H2delta G°r is +16.4 kJ/molKeq is therefore: exp{–16.4/(8.314 * 0.298)} = 0.00133Concentrated HCl as supplied is above 10M in H+ and Cl–Keq = [CuCl2–]/([H+].[Cl–]2)In 10 M HCl solution [CuCl2–]eq = 0.00133 x 1000 = 1.33 MClearly this reaction can proceed to a significant extent!(My apologies to forum readers who are not trained chemists, but there was a point here that needed to be addressed).