2
$\begingroup$

Sometimes universal indicator solution, or whatever acid-base indicators, is added into an electrolytic cell to indicate the change in pH at electrodes over time.

I wonder, if universal indicator solution will get electrolyzed in the set-up, leading to errors in the experiment.

My lecturer commented that universal indicator solutions are generally stable and do not involve in reactions. Moreover, we can observe that the products we get match what we have expected, so universal indicator solution has not been involved in the electrolysis.

But then, I am still confused. For example, we don't know if the substances in the indicator are electrolyzed to produce the same products expected.

Can somebody provide me with some ideas?

$\endgroup$
1
  • $\begingroup$ I guess the relative acidity of the solution is the same if you add the indicator before or after electrolysis. It means that the indicator is not electrolyzed. Have you tried these two approaches ? Is there any significative difference ? $\endgroup$
    – Maurice
    Commented Nov 24, 2023 at 10:57

2 Answers 2

2
$\begingroup$

Products being obtained and indicators being changed are two independent processes. There is no universal answer if the indicator can be reduced or oxidized, as it depends on the indicator and the potentials on either of electrodes.

Some indicators may be vulnerable to Kolbe electrolysis

$\ce{2 R-COO- ->[- 2 CO2][- 2e-]-> R-R}$

and generally to oxidation at anode, as there may be present reactive oxoradicals or atomic chlorine.

$\endgroup$
1
$\begingroup$

It really depends on the stability of your indicator under the electrochemical conditions you apply in your cell, like Poutnik has mentioned. And actually it is a good remark that you mention the formation of the to be expected products from electrolyzing your indicator. If you do the charge balance at a half-reaction, you will see you need to consume or produce some ions. In watery solutions this is typically hydroxide or protons, so a pH indicator is typically not really telling you what you're forming.

Besides the stability of the indicator, the main issue I would have throwing an indicator in my solution is the possibility of it to adsorb on the electrodes. So even if the indicator is completely stable, you will get deviation in your current-potential behavior.

$\endgroup$

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