I have some idea about the question. I think it has to do with oxidation/reduction potentials but I am not able to find any resource for the exact values or how to make sense of them in this context. Any reference would be appreciated.
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3$\begingroup$ The Chemistry SE site policy recommends to use just plain text for the question titles, for indexing/searching reasons. $\endgroup$ – Poutnik Nov 8 '20 at 21:22
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1$\begingroup$ Instead of writing e.g. H_3PO_3 in MathJax, write e.g. \ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}. \ce{} invokes mhchem MathJax extension. It additionally uses expected upright font. See $$\ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}$$ $\endgroup$ – Poutnik Nov 8 '20 at 21:24
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Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions.
Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as:
$\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$
that implies the further conversion of $\ce{S}$ to $\ce{SO2}$ to $\ce{SO3}$, which in alkaline conditions is represented by sulfate formation.
Note, the cited source notes only in acidic conditions, to quote:
Neutral-Acid pH: $\ce{3 H2S + 2 KMnO4 → 3S + 2 MnO2 }$
Wikipedia on KMnO4 supports this pH effect, to quote:
The Mn-containing products from redox reactions depend on the pH. Acidic solutions of permanganate are reduced to the faintly pink manganese(II) ion (Mn2+) and water. In neutral solution, permanganate is only reduced by three electrons to give manganese dioxide (MnO2), wherein manganese is in a +4 oxidation state.
Interestingly, the corresponding reaction replacing KMnO4 with either H2O2 or HOCl, can results in $\ce{SO2}$, ..., in acidic conditions.
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1$\begingroup$ Thank you Ajkoer for your answer. It is very informative and an interesting read. Unfortunately (or fortunately) it challenges my knowledge. You have said the following things that seem confusing: 1. Although $MnO_4$ is a weaker oxidizing agent in alkaline conditions, it is still able to turn $S$ to $SO_2$ and $SO_3$. 2. Adding peroxide or hypochloric acid works the other way (by making the solution even more oxidizing) to obtain $SO_2$ or $SO_3$. I checked out the references and indeed your argument is in agreement with them. I think something more interesting is happening. $\endgroup$ – Rakesh Arya Nov 8 '20 at 21:09
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1$\begingroup$ I wonder if it is the enthalpy of salt formation/solvation that pushes the S to Sulphate form in alkaline conditions. $\endgroup$ – Rakesh Arya Nov 8 '20 at 21:10
