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Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a greater standard electrode potential than a hydrogen ion will be reduced in its stead, and no hydrogen gas will be produced.

From what I understand, that bit states that certain electrolytes will stop the production of hydrogen gas and only oxygen gas and whatever gas from the electrolyte will be produced.

So I have two questions:

  1. What are some electrolytes that will stop the production of hydrogen gas in water electrolysis?
  2. What kind of water will be left behind if distilled water and an aforementioned electrolyte runs through electrolysis until all of the oxygen has been exhausted from the solution?

Please correct me if I'm misunderstanding anything.

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The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:

$$\ce{2 H+_{(aq)} + 2e^{–} ->H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$

So if a species has a has a reduction potential $>0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -> H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.

If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:

$$\ce{4OH^{−}_{(aq)} -> O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$

Similarly, if a species has a has a oxidation potential $>-.401\,\mathrm{V}$, then it will be be oxidized before the $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.

Wikipedia has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!

To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.

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