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I was trying to remove the electroplating from a aluminum rod using inverse electrolysis and everything went wrong. I'm definitely not a chemistry expert, so I'd like to know what happened here.

I used hydrochloric acid (30%) as the electrolyte, the aluminum rod as the cathode and a common copper wire as the anode. I ran it through a 12V 1A power supply, in a small glass jar.

When I turned the power on a lot of bubbles started to appear at the surface of the rod, as expected. After aproximately 30 seconds the bubbling was too strong, so I decided to turn off the power supply, however, the bubbling didn't stop. As I know that electrolysis of hydrochloric acid releases chlorine, I decided to put the jar to outside, with both electrodes still inside it. After aproximately five minutes, to my surprise, the jar was bubbling extremely strong, spilling over and the electrolyte had turned black, and quite frankly it was kind of scary.

I removed the electrodes and the reaction ceased, but I still have no clue of what happened.

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    $\begingroup$ You used 30% hydrochloric acid as the entirety of the electrolytic bath, or did you just add a few drops of the concentrated acid to an amount of water? If it's the former, then it's pretty clear what happened. $\endgroup$
    – Nicolau Saker Neto
    Dec 22, 2013 at 0:54

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As I recall, $\ce{Cu}$ is inert to $\ce{HCl}$, however $\ce{Al}$ will react with it to produce $\ce{H2}$ and $\ce{AlCl3}$, resulting in a turbid, dark gray solution that bubbles vigorously as the $\ce{H2}$ gas is evolved. The reaction may not have been instantaneous, especially if the $\ce{Al}$ had a passivation layer of oxides on it.

Once initiated, that reaction would've continued even after removal of the external current, and is probably the source of the black precipitate you describe.

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  • $\begingroup$ Aluminium has absolutely no protection against hydrochloric acid. Its only limited by hydrogen formation. Its slower at lower temperature- the electrolysis heated the water and increased the potential to speed the reaction. $\endgroup$
    – user2617804
    Dec 22, 2013 at 13:33

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