I was electrolysing water using a copper anode and zinc cathode. The water was from the tap and where I live is very hard. I was using about 20V between the two electrodes. As I applied the voltage, a white precipitate formed around the anode. What is strange is that I can't replicate this using any other type of metal as the anode. Any ideas on what this might be? enter image description here Or, I suppose a better question is: are there any ideas on how I can find out what this was?

  • $\begingroup$ What other anode materials have you tried? $\endgroup$
    – hBy2Py
    Dec 16 '15 at 15:51
  • $\begingroup$ I've tried iron, aluminum, and graphite. $\endgroup$ Dec 16 '15 at 15:59
  • $\begingroup$ The aluminum probably just anodized and stopped the current flow -- I'm guessing there was no bubbling there? As for iron, was it actually iron, or was it a steel of some kind? $\endgroup$
    – hBy2Py
    Dec 16 '15 at 16:14
  • $\begingroup$ Yes, it was steel $\endgroup$ Dec 17 '15 at 1:27
  • $\begingroup$ Aha --- stainless steel? $\endgroup$
    – hBy2Py
    Dec 17 '15 at 1:28

The slight bluish tinge to that white precipitate makes it look to me more like copper(II) hydroxide than anything to do with the hardness minerals, especially since you haven't observed it with other anode metals thus far.

One way to test this would be to make a whole bunch of the precipitate, by running the current for a long time, and then taking a small sample of the liquid with precipitate in it and adding an equal volume of distilled vinegar (5% acetic acid). If it is indeed $\ce{Cu(OH)2}$, I think it should dissolve to form a clear blue solution of copper(II) acetate.

The only problem might be if 5% acetic acid isn't able drop the pH far enough, in which case you'd have to track down a more concentrated acid. One choice might be $\ce{HCl}$, which is available at, e.g., aquarium supply stores under its historical name of muriatic acid. You'd then get a solution of copper(II) chloride, the color of which will vary depending on the $\ce{Cl-}$ concentration. Another option, per MaxW's comment, would be pickling vinegar (10% acetic acid) -- in fact, a quick search just revealed that Amazon has several vinegar options available in higher acidities.

  • $\begingroup$ How would I determine if it was produced? I'm a bit new to electrochemistry but does it have something to do with the overvoltage I used? $\endgroup$ Dec 16 '15 at 16:00
  • $\begingroup$ @MichaelStachowsky As re overvoltage, not really. Anodic electrolysis of copper in neutral aqueous media is likely to make $\ce{Cu(OH)2}$ regardless of the applied potential, as long as it's large enough to get electrons flowing. $\endgroup$
    – hBy2Py
    Dec 16 '15 at 16:11
  • $\begingroup$ Since the water was noted as hard, I'd guess some sort of copper (II) carbonate would be in the ppt as well. // "Picking vinegar" is 10% acetic acid, as opposed to "regular vinegar" which is 5%. $\endgroup$
    – MaxW
    Dec 16 '15 at 21:48
  • 1
    $\begingroup$ @MaxW Agreed, $\ce{CuCO3}$ is certainly possibility, depending on the type of hardness in OP's water. Either way, acidification should solubilize it, I figure? $\endgroup$
    – hBy2Py
    Dec 16 '15 at 21:51

Metallic copper at the anode works as an electrode producing $\ce{Cu(OH)2}$, the slight blue pricipitate you observed in the course of reaction. The $\ce{Cu(OH)2}$ gradually releases water creating a black precipitate of, CuO. Please read our work in which we described the homemade reactions to synthesized copper(II) aspirinate and acetate:


  • $\begingroup$ This may be a matter of language but a sacrificial anode is not what copper is in this case. $\endgroup$
    – A.K.
    May 16 '16 at 22:57

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