When one mole of $\ce{H3PO2}$ is treated with excess $\ce{I2}$ in acidic solution, one mole of $\ce{I2}$ is reduced; on making the solution alkaline, a second mole of $\ce{I2}$ is consumed.

I am supposed to find the reaction corresponding to the description.

Here is the answer given to me by my teacher:

$$\ce{H2O + H3PO2 + I2 -> H3PO3 + 2H+ + 2I-}$$

I do not understand why this occurs in acidic media, as $\ce{H+}$ is formed in the products.

  • $\begingroup$ By this logic, reactions producing water should not happen in water solutions either. But they do happen. $\endgroup$
    – Poutnik
    Oct 6, 2023 at 7:19
  • $\begingroup$ Similarly, many reactions ongoing in alkalic media produce $\ce{OH-}$. $\endgroup$
    – Poutnik
    Oct 6, 2023 at 11:26

1 Answer 1


You answer your own question! Hypo phosphorous acid H3PO2 is readily oxidized to phosphorous acid, H3PO3 [really H2HPO3] even in acidic solution [there probably is a limit]. Phosphorous acid is less readily oxidized to phosphoric acid so the acid present or formed inhibits the reaction. Removing the acid generates the anion, the anion is more readily oxidized. You must write the Second equation[s a neutralization and oxidation] to see this.

This is a general trend to oxyanions such as MnO4-, CrO4= and Cr2O7=, NO3- even sulfate and phosphate. In acidic solutions they are [strong] oxidizing agents; In basic solutions they are much weaker and can be formed from their reduced counterparts by a sufficiently strong oxidizer such as O2 or peroxide or in the case of phosphite iodine, I2.

  • 2
    $\begingroup$ Using = as 2 negative charges is the invention I have not seen yet. $\endgroup$
    – Poutnik
    Oct 7, 2023 at 3:18
  • 1
    $\begingroup$ someone has to be innovative. $\endgroup$
    – jimchmst
    Oct 7, 2023 at 8:28
  • 2
    $\begingroup$ You can also replace the usual $\ce{PO4^3-}$ by Unicode U+2261 PO4≡ . :-P Who would use that boring mhchem anyway? $\endgroup$
    – Poutnik
    Oct 7, 2023 at 8:45

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