I have read that the reaction between $\ce{I2}$ and conc. $\ce{HNO3}$ gives $\ce{HIO3}$ and $\ce{NO2},$ but there is no mention of the other halogens' products. Do they all give halous acids and nitrogen dioxide? Specifically with chlorine, how does the reaction proceed?

  • $\begingroup$ @Waylander That website looks fishy to me. Is this journal even trustworthy? BTW, iMedPub is in the list of predatory publishers. $\endgroup$
    – andselisk
    May 26, 2020 at 7:43
  • $\begingroup$ @andselisk I know nothing about this journal other than Google found me a reference to the oxidation of bromine by nitric acid. In all honesty the world does not need yet another method for the oxidation of benzylic acohols, but it appeared to demonstrate the reaction the OP asked about. There are other references to Br2/Br- oxidation by nitric acid $\endgroup$
    – Waylander
    May 26, 2020 at 7:58
  • $\begingroup$ @ThenardRinmann From what Google is showing me, it appears that nitric acid will oxidise chloride to chlorine but no further. $\endgroup$
    – Waylander
    May 26, 2020 at 8:00
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    $\begingroup$ Previous question in chem.SE: (fluorine + nitric acid) and (iodine + nitric acid). $\endgroup$ May 26, 2020 at 8:51
  • $\begingroup$ I may want to have a look at the redox potential Wikipedia data page. $\ce{HNO3}$ does not have much chnace to oxidize chlorine. $\endgroup$
    – Poutnik
    May 26, 2020 at 8:52

2 Answers 2


This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:

$$ \begin{align} \ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\ \ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\ \ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3 \end{align} $$

Now, let's look at reduction potential of halides:

$$ \begin{align} \ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\ \ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\ \ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\ \ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\ \end{align} $$ Accordingly, we can list oxidizing power at standard conditions in decreasing order:

$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$

Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:

$$ \begin{align} \ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\ \ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O} \end{align} $$

Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).

In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

Bromine as brominating reagent

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.


  1. J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
  2. Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
  3. Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).

$\ce{HNO_3}$ does not react with $\ce{Cl_2}$, as $\ce{Cl_2}$ is produced in the famous "aqua regia", which is a mixture of nitric and chlorhydric acid, having reacted according to :$$\ce{HNO_3 + 3 HCl -> NOCl + Cl_2 + 2 H_2O}$$ This equation is given in Cotton and Wilkinson, Chapter 5-13. In Chapter 12-8, it is stated that NOCl is never pure. It is decomposed to $\ce{Cl_2 + NO}$ to the extent of $0.5$% at room temperature.

NOBr is produced in a mixture $\ce{NO + Br_2}$ and it is an equilibrium, where ~ $7$ % of NOBr is decomposed at room temperature and $1$ atm.

Ref.: F. Albert Cotton, Geoffrey Wilkinson, Advanced Inorganic Chemistry, A Comprehensive Text, 3rd Ed., 1972, Interscience Pub., John Wiley & Sons, Inc. New York, London.

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    $\begingroup$ I'm not sure why you decided to delete your previous answer and re-post it again with minor changes instead of editing it, but this is still not exactly the answer to the question IMO. I don't understand the logic behind "$\ce{HNO3}$ does not react with $\ce{Cl2},$ as $\ce{Cl2}$ is produced in the famous 'aqua regia'" statement, and how the paragraphs after the reaction are relevant to the question. $\endgroup$
    – andselisk
    May 26, 2020 at 11:30
  • $\begingroup$ @ Andselisk. The question was : How does the reaction $\ce{HNO_3 + Cl_2}$ proceed ? My reasoning is that those two substances do not react at all. If $\ce{Cl_2}$ would react with the solvant containing nitric acid, it would not be produced in the aqua regia. It would immediately be transformed into something else, and would not exist in aqua regia. This reasoning is a little bit similar to the stupid equations that poor students sometimes establish "showing" that an acid reacting with something produces a hydroxide. Sorry for the silly comparison ! $\endgroup$
    – Maurice
    May 26, 2020 at 11:41
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    $\begingroup$ Took me a while to obtain the edition you are citing, but I haven't found the reaction you provided in chapter 5-13, only the statement on p. 179: "Aqua regia (~3 vol. of cone. $\ce{HCl}$ + 1 vol. of conc. $\ce{HNO3})$ contains free chlorine and $\ce{ClNO},$ and it attacks gold and platinum metals, its action being more effective than that of $\ce{HNO3}$ mainly because of the complexing function of chloride ion." $\endgroup$
    – andselisk
    May 26, 2020 at 12:19
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    $\begingroup$ Keep in mind that the reaction at the room temperature is in fact reversible, and doesn't involve instant formation of chlorine gas, which IMO sort of defeats your reasoning: $$\ce{HNO3(conc) + 3 HCl(conc) <=> NOCl + 2 Cl^0 + 2 H2O}$$ The irreversible reaction occurs when the mixture is heated: $$\ce{HNO3(conc) + 3 HCl(conc) ->[\pu{100–150 °C}] NO(g) + 1.5 Cl2(g) + 2 H2O}$$ I'm not saying your answer is wrong, but to me, the explanation is not convincing and it also only covers the reaction with chlorine, ignoring other halogens (see the question's title). $\endgroup$
    – andselisk
    May 26, 2020 at 12:21
  • $\begingroup$ The main thing is that Chlorine $\ce{Cl_2}$ cannot be oxidized by nitric acid to some higher oxidation number. The production of halous acid and nitrogen oxides, as suggested by Thenard Rinmann, simply does not happen with nitric acid. $\endgroup$
    – Maurice
    May 26, 2020 at 14:03

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