When $\ce{HI}$ reacts with $\ce{H2SO4}$, it can be represented by the following chemical equation:

$$\ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}$$

However, when $\ce{HBr}$ reacts with $\ce{H2SO4}$, the sulphate ion is reduced to sulphur dioxide instead:

$$\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}$$

Could anyone account for the difference here? Is there any more cases where the sulphate ion is reduced to $\ce{H2S}$ instead of $\ce{SO2}$?


2 Answers 2


The book intitled Qualitative Analysis emitted by F. P. Treadwell in $1924$ mentions in the chapter Iodine (p. 323) that the reaction depends on the relative amount of $\ce{H2SO4}$. In the presence of a great excess of $\ce{H2SO4}$, its reaction on $\ce{HI}$ or iodides produces $\ce{SO2}$ apart from $\ce{I2}$. In the presence of an excess of iodides, the reaction produces $\ce{H2S}$ as mentioned by Jelly Qwerty.

  • $\begingroup$ Is there any explanation to this? Or is it purely empirical $\endgroup$ May 15, 2023 at 16:26
  • 2
    $\begingroup$ If an explanation is to be found about this phenomena, it may be said that the first reaction between $\ce{H2SO4}$ and $\ce{HI}$ (or $\ce{KI}$) reduces sulfur from +$6$ to +$4$. The equation is $$\ce{H2SO4 + 2 HI -> SO2 + I2 + 2 H2O}$$ Then, in case of an excess of $\ce{HI}$, the Sulfur is further reduced from +$4$ to -$2$ according to the following equatoion : $$\ce{SO2 + 6 HI -> H2S + 3 I2 + 2 H2O}$$ $\endgroup$
    – Maurice
    May 15, 2023 at 18:31

As @Oscar said, hydrogen iodide (or iodide ion) is much stronger reducing agent than hydrogen bromide (or bromide ion). So the sulfuric acid reduce through a rather complicated pathway than just ending up with $\ce{SO2}$. The reduction pathway is somewhat like this:

$$\ce{H2SO4 -> SO2 -> S -> H2S}$$

Sulfurous acid is an intermediate product in the reaction which takes part in the transient reactions forming sulfur, hence driving the reduction reaction forward. $$\ce{SO2 (or H2SO3) + 2H2S -> 3S + 2H2O}$$

...and yes, if one of the reagent is a very strong reducing agent, you can end up with hydrogen sulfide instead of sulfur dioxide.

You can read the following articles:

  1. https://www.chemguide.co.uk/inorganic/group7/halideions.html
  2. Sulphuric Acid and Hydriodic Acid, Florence Bush, The Journal of Physical Chemistry 1929 33 (4), 613-620, DOI: 10.1021/j150298a010

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