What can be used as an electrolyte in water to perform electrolysis other than salt?

Question

I am looking for an electrolyte to use in water to perform electrolysis. It can't be salt, as salt ($\ce{NaCl}$) would cause the chemical reaction of $$\ce{2H2O + 2NaCl -> Cl2 + H2 + 2NaOH}$$ or

$$\ce{H2O + NaCl -> HCl + NaOH}$$

Which both make products with poisonous chemicals (ex: $\ce{NaOH}$ = Sodium Hydroxide (aka lye) which is poisonous). I am looking for a chemical I can use to create the pure forms of hydrogen and oxygen from the electrolysis process. I understand how electrolysis works, and have tried it before, but am looking for something that could be used as a non-poisonous electrolyte. I read that baking soda could be used, would that produce what I am looking for?

Answer 1

You can actually just use salt. It is because that the chemical reaction on the top is for one condition which is that your halide solution ($\ce{NaCl}$) has a very high concentration.

If you halide solution has lower concentration,the hydrogen ion and hydroxide ion in the water will discharged and the two half equation will be

$\ce{4H^+ + 4e^- -> 2H_2}$

$\ce{4OH^- -> 2H_2O + O_2 + 4e^-}$

So, the overall equation is

$\ce{4H^+ + 4OH^- -> 2H_2 + 2H_2O + O_2}$

Thus, is okay to use salt.

Besides Salt

You can also use hydrochloric acid besides salt. Just remember that don't use a high concentration hydrochloric acid as the anode will just discharge the chloride ion instead of hydroxide ion.

Answer 2

I assure you that your worries are without basis. For the hydrolysis of water, the half reactions are: $$\ce{2H2O_{(l)} ->O2_{(g)} +4H+_{(aq)} +4e-}~~~~\varepsilon^0=-1.23~\mathrm{V}$$ $$\ce{4H+_{(aq)} +4e- ->2H2_{(g)}}~~~~\varepsilon^0=0~\mathrm{V}~\mathrm{(by~definitoin)}$$

The $\varepsilon^0$ of the reaction would be:

$$\ce{2H2O_{(aq)} ->2H2_{(g)} +O2_{(g)}}~~~~\varepsilon^0=-1.23~\mathrm{V}$$

With the proposed half reactions your are suggesting might take place:

$$\ce{2H2O_{(l)} +2e- ->H2_{(g)} +2OH-_{(aq)}}~~~~\varepsilon^0=-0.83~\mathrm{V}$$ $$\ce{2Cl-_{(aq)} ->Cl2_{(g)} + 2e-}~~~~\varepsilon^0=-1.36~\mathrm{V}$$

The $\varepsilon^0$ of the reaction would be:

$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} -> Cl2_{(g)} + H2_{(g)} + 2OH-_{(aq)}}~~~~\varepsilon^0=-2.19~\mathrm{V}$$

(Note that the $\ce{Na+}$ ions are present in the solution, though it is not listed in the reaction)

Because the hydrolysis of water requires less energy than the production of $\ce{Cl2}$, $\ce{H2}$, and $\ce{NaOH}$ from the ions present in the solution, the second reaction will not occur.

The reaction $\ce{H2O_{(l)} + Cl^{-}_{(l)} -> HCl_{(g)} + OH-_{(aq)}}$ would not occur. $\ce{HCl}$ could not evolve from the solution, as there would be no mechanism forcing the $\ce{H+}$ ions onto the electronically stable $\ce{Cl-}$ ions. Rather, it would be much more likely that the $\ce{H+}$ ions would re-react with the $\ce{OH-}$ ions to reform water, meaning no net reaction.

In short, go right ahead using $\ce{NaCl}$ in your hydrolysis reactions. It's actually probably one of the safest salts you could use!

Answer 3

Mainly anything that contains ions (i.e. electrolytes) is great to use in the electrolysis of water. Some examples of electrolytes you can use in electrolysis are baking soda (sodium bicarbonate), squirted lemon, hell, I bet you could use Gatorade since it contains electrolytes. Hope this answer helped, I've been researching the electrolysis of water for like 2 weeks now, and I know it can be a little frustrating to understand.