- #1
HelloCthulhu
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Greetings forum!
I'm experimenting with saltwater electrolysis and have a few questions. I'm using a salt solution I've prepared overnight. The solution has about 2 parts water, 6 parts salt, so I will essentially be electrolyzing brine at the beginning of the experiment (For safety, the experiment is done outside with a fume hood.). At this point, the solution is very dry and flaky, with a voltage of about 1V without current. I'll be using a current of about 1A with a voltage peaking around 5V. The overall reaction should be:
2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + H2(g) + Cl2(g)
However, I will be adding another prepared solution of HCl + H2O to my original solution. How will adding the protonated water to the brine during electrolysis change the reactions? Thank you for your assistance!![Smile :smile: :smile:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I'm experimenting with saltwater electrolysis and have a few questions. I'm using a salt solution I've prepared overnight. The solution has about 2 parts water, 6 parts salt, so I will essentially be electrolyzing brine at the beginning of the experiment (For safety, the experiment is done outside with a fume hood.). At this point, the solution is very dry and flaky, with a voltage of about 1V without current. I'll be using a current of about 1A with a voltage peaking around 5V. The overall reaction should be:
2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + H2(g) + Cl2(g)
However, I will be adding another prepared solution of HCl + H2O to my original solution. How will adding the protonated water to the brine during electrolysis change the reactions? Thank you for your assistance!