- #1
HelloCthulhu
- 151
- 3
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!
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!