Why $\ce{I2}$ is formed when $\ce{HI}$ and $\ce{HNO3}$ are reacted?

I know that $\ce{HI}$ is more acidic than $\ce{HNO3}$ so nitric acid will accept protons from $\ce{HI}$, so $\ce{I-}$ (iodide ion ) should be formed and nitric acid on accepting proton would form $\ce{H2NO3+}$ but that is not formed why?

iodination of alkane

  • $\begingroup$ chemistry.stackexchange.com/questions/600/… $\endgroup$ Commented Aug 25, 2016 at 12:54
  • 4
    $\begingroup$ This is not about acidic strength and protons at all. This is a redox reaction. $\endgroup$ Commented Aug 25, 2016 at 13:00
  • $\begingroup$ @IvanNeretin in an weak acid strong acid reaction products are formed such that weak acid gets proton and strong acid looses proton but in redox reaction what is rule of thumb if reduction potentials are not known? $\endgroup$
    – JM97
    Commented Aug 25, 2016 at 13:04
  • 2
    $\begingroup$ Nobody remembers the exact potentials, but you are expected to know that $\ce{HNO3}$ is a pretty strong oxidant, while $\ce I$ is an element with relatively low electronegativity (for a non-metal, that is). That should suffice. $\endgroup$ Commented Aug 25, 2016 at 13:09
  • $\begingroup$ @IvanNeretin I2 formed will react with HNO3 to form hio3 and as seen in image in question this Hio3 will again produce I2 so where does overall reaction leads us to? $\endgroup$
    – JM97
    Commented Aug 25, 2016 at 13:27

1 Answer 1


Nitrate is a strong oxidant which oxidizes the iodide to iodine.

$$\ce{NO_3^- + 2I^- + 2H^+\rightarrow NO_2^- + I_2 \uparrow + H2O}$$

Note that the oxidation number of the nitrogen atom in $\ce{NO3-}$ is $+V$ and in $\ce{NO2-}$ it's $+III$, so over all, we have a reduction equation of:

$$\ce{NO_3^- + 2e^- + 2H^+\rightarrow NO_2^- + H2O}$$

On the oxidation side we want to form $\ce{I_2}$ out of $\ce{I^-}$, so the oxidation equation is:

$$\ce{2I^- \rightarrow I_2 + 2e^-}$$

  • $\begingroup$ Actually, I don't believe it will stop at I2. Also, pay attention to the formatting and use \ce{}. $\endgroup$ Commented Aug 25, 2016 at 13:12
  • $\begingroup$ In parts it will react further, but test it and you'll see, that a violet vapour ($\ce{I2}$) comes out of the beaker, so you definitely form iodine. $\endgroup$
    – Sam
    Commented Aug 25, 2016 at 13:16
  • $\begingroup$ No, iodine is oxidized only by concentrated (fuming) nitric acid. $\endgroup$
    – vapid
    Commented Aug 25, 2016 at 13:16
  • $\begingroup$ Well, I have made a different experience during my internship. When adding an iodide solution to an unknown, nitrate-containing substance violet vapors came out of my beaker, thus iodide was definitely oxidized to iodine. And that unknown solution was definitely not fuming $\ce{HNO3}$. $\endgroup$
    – Sam
    Commented Aug 25, 2016 at 13:21
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    $\begingroup$ Agreed. Diluted $\ce{HNO3}$ will give $\ce{I2}$, concentrated acid will push further. $\endgroup$ Commented Aug 25, 2016 at 13:32

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