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I was electrolysing water with a 12v DC power supply at about 3 milliamps, using copper wire for both my anode and cathode. The solution was about 17 grams of table salt in 750 milliliters of purified water. After running the electrolysis cell for 1 minute, I left it overnight. The next day, I found a light blue powder at the bottom of the cell, which was still full of solution. Could this be copper hydroxide? If not, what could it be?

I should also note that the anode and cathode had become red, similar to what iron oxide looks like.

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    $\begingroup$ I just want to tell you that this experiment is pretty unsafe in general. Okay the currents are low so the amount of products formed are low, but you're forming hydrogen gas and possible chlorine gas. I would recommend you to not do this experiment for extended periods or higher currents. $\endgroup$
    – Noah
    Feb 14 at 22:34

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Yes. The blue precipitate is $\ce{Cu(OH)2}$. Let's explain how it is produced. At the cathode, the following reaction occurs :$$\ce{2 H2O + 2e^- -> H2 + 2 OH^-}$$ So the solution becomes basic, due to the new $\ce{OH^-}$ ions being produced by the reaction. At the anode, the following reaction occurs : $$\ce{Cu -> Cu^{2+} + 2 e-}$$ which produces new $\ce{Cu^{2+}}$ ions in solutions. The new $\ce{Cu^{2+}}$ and $\ce{OH^-}$ ions are attracting one another, and when they join, they react and produce an insoluble precipitate, made of $\ce{Cu(OH)2}$ according to $$\ce{Cu^{2+} + 2 OH^- -> Cu(OH)2}$$

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    $\begingroup$ To note, given the reaction conditions that are given (pH 7) it is unlikely that Cu2+ is formed free in solution and more likely that the reaction going on was direct to the hydroxide. The hydroxide could have formed dendrites or other mechanically unstable structures (or just simple delamination) and just have broken off from the electrode. $\endgroup$
    – Noah
    Feb 14 at 22:31

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