0
$\begingroup$

I'm working on a chemistry project and that features a reaction where $\ce{Br-}$ participates in a redox reaction as a reductor. Eventually, $\ce{HBrO2}$ competes with $\ce{Br-}$ to react with the same chemical as $\ce{Br-}$ reacts with.

Is is therefore true that $\ce{HBrO2}$ is also a reductor? If yes, what is the redox potential of $\ce{HBrO2}$?

$\endgroup$
  • 1
    $\begingroup$ Well, it obviously can be further oxidised - for example to BrO3 and BrO4 anions - with different potentials. $\endgroup$ – Mithoron Jun 10 at 16:43
2
$\begingroup$

Bromous acid ($\ce{HBrO2}$) is an inorganic compound, which is an unstable compound. However, the salts of its conjugate base, bromites (e.g., $\ce{NaBrO2.3H2O}$) have been isolated. Bromous acid is considered to be an oxidizer. However, I can't find its reduction potential, probably because of its unstable nature (Ref.1). Based on the studies of Faria, et al. (Ref.1), $\ce{HBrO2}$ decomposes according to following mechanism:

$$\ce{HBrO2 <=>[$K_\mathrm{a}$] H+ + BrO2-} \tag{E1}$$ $$\ce{HBrO2 + BrO2- ->[$k_1$] HOBr + BrO3-} \tag{R1}$$ $$\ce{HBrO2 + HBrO2 ->[$k_2$] HOBr + BrO3- + H+} \tag{R2}$$

It was found that $k_1 = \pu{39.1 ± 2.6 M-1 s-1}$ and $k_2 = \pu{800 ± 100 M-1 s-1}$ for reactions $(\mathrm{R1})$ and $(\mathrm{R2})$, respectively. Hence, authors have calculated and given equilibrium quotient for fast equilibrium $(\mathrm{E1})$ as $K_\mathrm{a} = \pu{(3.7 ± 0.9) \times 10^{-4} M}$ at ionic strength $\pu{0.06 M}$ and $\pu{25.0 ± 0.1 ^\circ C}$.

I think OP is working on Belousov-Zhabotinsky reaction (oscillator), which would involves both $\ce{HBrO2}$ and $\ce{Br-}$. For example, usual reaction conditions are for reaction volume of $\pu{28 mL}$ at $\pu{39.6 ^\circ C}$ are: $\ce{[NaBrO3]} = \pu{1.8 \times 10^{-3} M}$; $\ce{[CH2(CO2H)2]} = \pu{5.6 \times 10^{-3} M}$; $\ce{[CeSO4]} = \pu{5.8 \times 10^{-4} M}$; and $\ce{[H2SO4]} = \pu{1.5 M}$ (Ref.2; yet this reference did not mention adding $\ce{[NaBr]}$ or $\ce{[KBr]}$ solution to the mixture). Note that either $\ce{Ce(IV)}$, or $\ce{Ru(II)}$, or $\ce{Fe(II)}$ complex can be employed as the catalyst. The Ref.3 proposed three mechanisms for the reaction:

Process A (consumption of bromide ion): $$\ce{Br- + BrO3- + 2H+ -> HBrO2 + HOBr} \tag{A1}$$ $$\ce{Br- + HBrO2 + H+ <=> 2HOBr} \tag{A2}$$ $$\ce{Br- + HOBr + H+ -> Br2 + H2O} \tag{A3}$$

Process B (Oxidation of catalyst(R), autocatalytic reaction): $$\ce{2HBrO2 + 2H+ -> HOBr + BrO3- + H+} \tag{B1}$$ $$\ce{HBrO2 + BrO3- + H+ <=> 2BrO2 + H2O} \tag{B2}$$ $$\ce{BrO2 + cat.(R) + H+ <=> HBrO2 + cat.(O)} \tag{B3}$$

Process C (Reduction of catalyst(O), production of bromomalanoic acid): $$\ce{BrO2 + cat.(O) + H2O -> BrO3- + cat.(R) + 2H+} \tag{C1}$$ $$\ce{Br2 + CH2(CO2H)2 -> BrCH(CO2H)2 + Br- + H+} \tag{C2}$$ $$\ce{6cat.(O) + CH2(CO2H)2 + 2H2O -> 6cat.(R) + HCO2H + 2CO2 + 6H+} \tag{C3}$$ $$\ce{4cat.(O) + BrCH(CO2H)2 + 2H2O -> 4cat.(R) + HCO2H + Br- + 2CO2 + 5H+} \tag{C4}$$

As a consequence of this chain of reactions, it is evident by the equation $(\mathrm{A2})$ that $\ce{Br-}$ is oxidized to $\ce{HOBr}$ by $\ce{HBrO2}$ (which has formed as an intermediate):

$$\ce{Br- + H2O <=> HOBr + H+ + 2e-} \tag{A2'}$$ $$\ce{HBrO2 + 2H+ + 2e- <=> HOBr + H2O} \tag{A2"}$$

The sum of the equations $(\mathrm{A2'})$ and $(\mathrm{A2''})$ gives: $$\ce{Br- + HBrO2 + H+ -> 2HOBr} \tag{A2}$$

Since this reaction is spontaneous for oscillation to be continued, it is safe to say $\ce{HBrO2}$ is a strong oxidizer compared to $\ce{Br-}$ since $\ce{Br-}$ acts as a reducer in this particular reaction.

However, it is also evident by the equation $(\mathrm{B2})$ that $\ce{HBrO2}$ has reduced to $\ce{BrO3-}$ to $\ce{HBrO2}$ (all formed as intermediates except $\ce{BrO3-}$):

$$\ce{BrO3- + 2H+ + e- <=> BrO2 + H2O} \tag{B2'}$$ $$\ce{HBrO2 <=> BrO2 + H+ + e- } \tag{B2"}$$

The sum of the equations $(\mathrm{B2'})$ and $(\mathrm{B2''})$ gives: $$\ce{HBrO2 + BrO3- + H+ -> 2BrO2 + H2O} \tag{B2}$$

Since this reaction is also spontaneous for oscillation to be continued, it is safe to say $\ce{HBrO2}$ can be a reducer compared to $\ce{BrO3-}$ since it acts as a reducer in this particular reaction.

Also note that according to the equation $(\mathrm{R2})$, $\ce{HBrO2}$ self propagates. Therefore, whenever $\ce{HBrO2}$ forms, it could be acting as oxidizer or reducer, based on the conditions.


References:

  1. Robert de Barros Faria, Irving R. Epstein, Kenneth Kustin, “Kinetics of Disproportionation and pKa of Bromous Acid,” J. Phys. Chem. 1994, 98(4), 1363–1367 (https://doi.org/10.1021/j100055a051).
  2. Y. Pomeau, J. C. Roux, A. Rossi, S. Bachelart, C. Vidal, “Intermittent behaviour in the Belousov-Zhabotinsky reaction,” J. Physique Lett. 1981, 42(13), 271 – 273 (https://doi.org/10.1051/jphyslet:019810042013027100).
  3. Yu Chang, Nannan Zhang, Yuxin Yang, Jun Du, Xing Fan, Changyuan Tao, “Time-periodic oscillation reaction in an organic-solvent dominated electrolyte,” Physical Chemistry Chemical Physics 2017, 19(40), 27643-27650 (https://doi.org/10.1039/C7CP05414A).
| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you Mathew for your very extensive answer. You indeed guessed correct that it was about the BZ (or you looked at my other questions ;-) ). My question rose when I was reading this paper: scholar.rose-hulman.edu/rhumj/vol3/iss1/1, it says that ‘the concentration of bromide eventually falls below some critical level [Br−]cr. It is at this point that Process B begins to dominate Process A: the hypobromous acid (HBrO2) begins to compete with the bromide to reduce the bromate.’ (citation). So I thought, if Br- is a reductor, is HBrO2 that too? $\endgroup$ – ralphsmit Jun 12 at 10:20
  • $\begingroup$ @ralphsmit: So my guess is right! :-) By the way, the rose-hulman reference is very interesting. Yet, there are some errors I found such as not balancing chemical equations. And also, keep in mind that hypobromous acid is $\ce{HOBr}$, not $\ce{HBrO2}$. $\endgroup$ – Mathew Mahindaratne Jun 12 at 17:04
  • 1
    $\begingroup$ Thanks for checking it out. I already knew about the wrong name for HBrO2, someone else pointed it out in an other question. It does, however, say sth about the quality of it. The paper was my starting point for it. It is for secondary school (last year of pre-university-education). In the Netherlands it is common for sec. school students to do some limited kind of research as a preparation for university. As you can probably imagine, I didn't have any experience with ODEs (we're doing it about the Oregonator), so it was a good starting point. I've based most of it on more trustworthy sources. $\endgroup$ – ralphsmit Jun 13 at 10:51
2
$\begingroup$

If you need a good answer, you should not put an information embargo on the context of your question.


This is to be taken rather as an expanded comment than a full answer:

$\ce{HBrO2}$ oxidizes $\ce{Br-}$ to $\ce{Br2}$.

$$\ce{HBrO2 + 3 Br- + 3 H+ -> 2 Br2 + 2 H2O}$$

To tell if it may compete with $\ce{Br-}$ to react with the "chemical", we would have to know that chemical.

Regarding standard redox potentials, an exhaustive list is on Wikipedia, but $\ce{Br^{III}}$ species are not there.

But be aware $\ce{HBrO2}$ is unstable, strong oxidizer, while $\ce{Br-}$ is a weak reducer. I personally think the info may have been misinterpreted, having meaning $\ce{HBrO2}$ may compete with the chemical to react with $\ce{Br-}$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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