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by stamps 2343 days ago
A lot of RO systems come with a sediment filter, carbon filter, and UV as part of the package.

Seems about as close as you can get to completely clean water.

1 comments

Electrolyze water in one vessel. Combust the gases in another vessel, and condense the water vapor.

Hilariously inefficient use of energy, but I think the only contaminants you are likely to encounter would be... nitrous oxide (N2O)?

I'd be less worried about the contaminants in the water, and more worried about the contaminants in the air.

As I understand it electrolysis of water converts chloride salts (table salt) to chlorine gas (poisonous), fluoride salts (rarer) to fluoride gas (really poisonous), etc.

Electrolysis only produces chlorine to the extent that chloride ion is present in the water. If the charge-carrying ions are metallic positive and hydroxide negative, the chlorine produced will be negligible.

So to promote electrolysis, add KOH or NaOH (lye) instead of NaCl (table salt). Or don't add anything, and let your charge carriers be H3O+ and HCO3- from dissolved CO2.

The chlorine will come off first, and if you have a reducible anode, it will attack that first. If you have a graphite anode, you will get Cl2, which will oxidize pure hydrogen just fine. You'll get HCl in the combustion, which isn't dangerous when diluted (it's the same acid that's produced in your stomach).

Fluorine gas is not produced from aqueous fluoride salts because the oxygen is always oxidized first.

The half-reactions:

  2 H2O + 2 e- --> H2 + 2 [OH-]       (-0.83 V)
  2 H2O        --> O2 + 4 [H+] + 4 e- (-1.23 V)
  2 [Cl-]      --> Cl2 + 2 e-         (-1.36 V)
  2 [F-]       --> F2 + 2 e-          (-2.87 V)
Even if you do get a F2, it immediately reacts with the electrolysis water to produce oxygen gas:

  2 F2 (g) + 2 H2O (l) --> 4 HF (aq) + O2 (g)
In short, this is not much of a problem.

By the redox half-reactions, it looks like oxygen should be produced before chlorine, but there is also something called "overpotential" at the electrode. On a graphite electrode, the activation overpotentials for H2, O2, and Cl2 are -0.62V, +0.95V, and +0.12V, respectively. There is also a "bubble overpotential" of the tiny bubbles that form on the electrode. That pushes the preference towards chlorine gas. Tiny oxygen bubbles will form, but the energy required to grow them large enough to detach from the electrode and rise to the surface exceeds that required to form and grow a chlorine bubble.