2
$\begingroup$

I am doing electrolysis of water to collect the hydrogen and I'm running into a problem: on the positive side I'm using graphite from a pencil (I cannot use any other material because it corrodes too fast) and on the negative side I was using copper. Copper was working really fine, the water was clear and no problem. But I don't have copper with a big surface area so I decided using aluminum (from coca cans) and my solution started to get dirty because of the aluminum and the aluminum is slowly disappearing.

Why? Any other eletrode that I used for the negative pole was not corroded (only on the positive pole everything corrodes). Why is aluminum making my water dirty? Do you think if I use iron this problem will not happen?

Improve this question
$\endgroup$
3
  • $\begingroup$ Aluminum is pretty active. Once the passivation layer is breached, it would react with water quite readily. $\endgroup$
    – Ivan Neretin
    Oct 23 '16 at 15:58
  • $\begingroup$ Please don’t use shorthands like $-$ or $+$ for negative and positive poles/cathode and anode, respectively. It took me quite a while to understand that this question is actually valid. Also, please don’t resort to colloquialisms such as cause and try not to forget any apostrophes. Thanks. $\endgroup$
    – Jan
    Oct 23 '16 at 20:47
  • $\begingroup$ You'll probably need to add more information in order to get any useful answers. What, if anything, did you add to your water? What potential are you using? Could you take a picture of the aluminum cathode and the water after a test where you've observed the aluminum to dissolve? $\endgroup$
    – hBy2Py
    Jan 2 '17 at 16:13
1
$\begingroup$

If you are using salt to make the solution more conductive the process will cause the creation of Sodium Hydroxide, which will react with aluminium.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.