In acidic conditions, I− tends to get oxidised by atmospheric oxygen to give iodine. This is evident in iodine based titration, where upon leaving the setup, blue colour of starch reappears as iodine is formed again.

$$\ce{4I- + O2 + 4H+ -> 2I2 + 2H2O}$$

This makes me believe that I− has less tendency to exist as the anion(at least at moderate temperature), since ΔH of reaction is of the order -150 kJ.

So my question is: How is it that HI is a strong acid if in acidic medium I− has tendency to get oxidised by atmospheric oxygen, reducing the pH in the process.


  • Chemistry Textbook of NCERT
  • J.D. Lee Concise Inorganic Chemistry, for JEE Main and Advanced, 2021
  • 5
    $\begingroup$ These two things are pretty much unrelated. Strong acid can be reducer as well as oxidant, eg. HNO3. $\endgroup$
    – Mithoron
    May 7, 2023 at 20:49
  • $\begingroup$ I was able to find $\ce{4I- + O2 + 4H+ -> 2I2 + 2H2O}$, but the source is not very authentic. Could you provide a reference for the reaction you have mentioned? $\endgroup$
    – ananta
    May 8, 2023 at 3:18
  • $\begingroup$ @Mithoron although I doubt it, but if the above reaction is correct, then wouldn't the acidic strength of $\ce{HI}$ reduce? $\endgroup$
    – ananta
    May 8, 2023 at 4:00
  • 2
    $\begingroup$ @ananta and OP. Again, some tendency to getting oxidised with air is hardly an issue - solution can get a bit less acidic if bottle is left opened, and so what? Nitric acid somewhat decomposes and is it an issue? HCl vaporises, H2SO4 sucks H2O from air... all this decreases acidity of conc. solutions. $\endgroup$
    – Mithoron
    May 8, 2023 at 15:33
  • 1
    $\begingroup$ The dissociation equilibrium and the oxidation equilibrium are mutually independent. Therefore, acidic strength and reductive strength are mutually independent either. $\endgroup$
    – Poutnik
    May 8, 2023 at 17:08

1 Answer 1



  1. Gas phase reaction (the fastest way to react two substances) between $\ce{HI}$ and $\ce{O2}$ is rather slow and often requires a catalyst.
  2. Low amounts of dissolved oxygen in the solution will reduce the rate of the reaction to negligible amounts.
  3. $\ce{HI(aq)}$ is commercially available and is expected to be stable with a long shelf life.

The details

Although, according to various sources, including Wikipedia, $\ce{HI}$ does react with $\ce{O2}$ to form $\ce{I2}$ and $\ce{H2O}$, I am not able to find sources that claim such a reaction in aqueous state; however, several studies have been conducted using catalysts. For example, the reference$^1$ suggests the following reaction mechanism for the reaction between $\ce{HI}$ and $\ce{O2}$:

heterogeneous reaction of hydrogen iodide and oxygen over a surface catalyst

where $S$ is a surface site. The same article does mention a concurrent slow reaction between $\ce{HI}$ and $\ce{O2}$:

homogenous reaction between gaseous hydrogen iodide and oxygen

Do note that the reaction has been suggested as being rather slow. Thus, in the absence of a catalyst, the reaction between $\ce{HI}$ and $\ce{O2}$ is not very fast. This study considers very large amounts of oxygen, while oxygen only feebly dissolves in water. The reference-suggested (important) transition state for the reaction between $\ce{HI}$ and $\ce{O2}$ is:

Transitions state for reaction of hydrogen iodide with oxygen showing that oxygen attaches with hydrogen

In aqueous state, such a transition state wouldn't be possible since the ions are dissociated and solvated.

Further, since Merck and other commercial chemical manufacturers provide the hydrogen iodide solution, I expect it to be stable with a long shelf life.


  1. Shum, L.G.S. and Benson, S.W. (1983), The oxidation of HI at low temperatures and the heat of formation of HO2. Int. J. Chem. Kinet., 15: 323-339. https://doi.org/10.1002/kin.550150403
  • 2
    $\begingroup$ Please do not use images (such as the first one) and typeset it in the style of the site, as it is (1) not searchable; and (2) not cited here in the body of your post. You have not addressed the OP's query, which is how is $\ce{HI}$ considered a strong acid given the conditions they gave. $\endgroup$ May 8, 2023 at 11:27
  • $\begingroup$ @ToddMinehardt I have referred the source of the images. These (cited) images are from a research article published in the International Journal of Chemical Kinetics. I have added appropriate image descriptions to make the image searchable. Thank you for the suggestion though, I will typeset it as soon as I find time. I am not sure if you have been following this question, but me and OP have been discussing this question for a few days now, and this answers OP's question, which is why $\ce{HI}$ does not degrade due to presence of dissolved/atmospheric oxygen, remaining a strong acid. $\endgroup$
    – ananta
    May 8, 2023 at 12:44
  • 2
    $\begingroup$ "Example" here is not the proper citation, it's a hyperlinked word. I also second the Todd's remark on posting text as images. And you are doing it again here: answering the question not really posed by OP, but rather addressing a different loosely-related issue, and pleasing the OP who is seemingly happy not to see the forest for the trees. $\endgroup$
    – andselisk
    May 8, 2023 at 13:28
  • $\begingroup$ @andselisk I couldn't find any studies answering the OP's question exactly (probably because this is a question about something that doesn't happen). In such cases, I refer to articles that are closely related and provide a good explanation. Again, the question is not why $\ce{HI}$ is a strong acid, but why it doesn't degrade into a weaker (more dilute) acid due to reaction with dissolved/atmospheric oxygen, which, I feel, I have answered to a satisfactory degree. I have also mentioned that I will typeset the images, and provide a proper reference, as soon as I find time. $\endgroup$
    – ananta
    May 8, 2023 at 13:45

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.