0
$\begingroup$

I am using the electrolysis of water to provide middle school students with a phenomena to explore chemical and physical change. The set up is producing much more hydrogen gas than oxygen gas and I am wondering if this is caused by the salt water solution I am using? I was looking to get a 2;1 ratio of hydrogen to oxygen.

$\endgroup$
1
  • 4
    $\begingroup$ Dump the salt and use sodium sulfate or magnesium sulfate. You do not want chlorine being produced from the electrolysis of saltwater. $\endgroup$
    – Ed V
    Aug 24, 2022 at 17:13

1 Answer 1

3
$\begingroup$

$\ce{NaCl}$ is not a good choice for showing the electrolysis process to high school students. The electrolysis of $\ce{NaCl}$ starts by producing $\ce{Cl2}$ (and a little $\ce{O2}$) at the anode, plus $\ce{NaOH + H2}$ at the cathode. But $\ce{Cl2}$ is a bit soluble in water, so its gaseous volume is not consistent with the stoichiometric proportions. Furthermore, after some time the dissolved $\ce{Cl2}$ reacts with the dissolved $\ce{NaOH}$ according to : $$\ce{Cl2 + 2 NaOH -> NaClO + NaCl + H2O}$$ As a consequence the solution contains now a new substance, sodium hypochlorite $\ce{NaClO}$ or its ions $\ce{Na^+ + ClO^-}$. The gaseous volume observed at the anode is much less than $1/2$ times the volume of $\ce{H2}$ produced at the cathode.

So if you want to show that the gaseous volumes produced at the cathode is twice what is produced at the anode, it is better to use a solution of sodium sulfate $\ce{Na2SO4}$, which produces exactly the gaseous volumes in a proportion $2 : 1$. The secondary reactions observed with $\ce{NaCl}$ do not occur with sodium sulfate.

$\endgroup$
1
  • $\begingroup$ As @Maurice pointed, $\ce{Na2SO4}$ is the best option. However, your results might be also affected by the electrodes material. This is discussed in a recent article from JCE : Kihyang Kim et al., Water Electrolysis Accompanied by Side Reactions, Journal of Chemical Education 2021 98 (7), 2381-2386. $\endgroup$
    – PAEP
    Sep 1, 2022 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.