Does the sulfur also oxidize the $\ce{OH-}$ radicals after the $\ce{Na2S4}$ chains form? Something like this: $$\ce{9S + 4NaOH -> 2Na2S4 + SO3 + H2O}$$

  • $\begingroup$ If so then I can expect a mixed solution of Sodium Sulfate and Sodium Polysulfide(s) after boiling of the Sodium Hydroxide/Sulfur mixture. $\endgroup$ Commented Jul 26, 2015 at 16:45

3 Answers 3


If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a disproportionation. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.

Oxidation numbers:

Element    Reactant     Product
S           0            -1/2 in polysulfide
S           0            +6 in sulfur trioxide
Na         +1            +1
O          -2            -2 in sulfur trioxide and water
H          +1            +1

The half reactions (unbalanced) are:

  • Oxidation: $\ce{S -> SO3}$
  • Reduction: $\ce{S -> S4^2-}$

You might get some sodium hydrogen sulfate forming through the followin; $$\ce{NaOH + SO3 <=> NaHSO4}$$

  • $\begingroup$ Now to the real challenge, extracting the sulfides from the rest of the products/leftover reactants. The leftovers should be minimal as a sulfur excess existed after solution cool down to ambient. The liquid portions were drained off and the remnants disposed of. Currently I am involved with reducing 1M H2SO4 in hopes of using its desiccant behavior to carbonize sugar impregnated cotton cloth for use in the electrodes. It would seem that the MathJax does not work in the comments. $\endgroup$ Commented Jul 27, 2015 at 1:15
  • $\begingroup$ MathJax still works in comments $\ce{H2SO4}$ is produced by $\ce{H2SO4}$ just like normal. $\endgroup$
    – Ben Norris
    Commented Jul 27, 2015 at 1:31
  • $\begingroup$ I left out the $ signs $\endgroup$ Commented Jul 27, 2015 at 1:33

Note that $\ce{SO3}$ reacts strongly and rapidly with $\ce{H2O}$ to give $\ce{H2SO4}$. I'd expect this to consume two more equivalents of $\ce{OH-}$ (giving $\ce{SO4^2- + 2H2O)}$, and that those reactions would be more rapid than the disproportionation reactions.

Dr. Nathan E. Stott's proposed reactions which involve sulfur oxidizing hydroxide to give oxygen aren't favoured energetically, under normal conditions. Oxygen's a much stronger oxidant than sulfur.

  • $\begingroup$ This seems to be rather a comment than an answer. $\endgroup$
    – Poutnik
    Commented Nov 15, 2021 at 8:28
  • $\begingroup$ In short, the answer to the original question is "no, the proposed reaction wouldn't come out like that". $\endgroup$ Commented Nov 27, 2023 at 22:32

Here are a list of equations that could potentially occur for this reaction depending on the amounts of sulfur and sodium hydroxide used as reactants:

4 NaOH + 2 S8 $\rightarrow$ 2 Na2S8 + 2 H2O + O2

4 NaOH + S8 $\rightarrow$ Na2S3 + Na2S4 + 2 H2O + SO2

4 NaOH + S8 $\rightarrow$ Na2S2 + Na2S5 + 2 H2O + SO2

4 NaOH + S8 $\rightarrow$ Na2S3 + Na2S5 + 2 H2O + O2

4 NaOH + S8 $\rightarrow$ Na2S2 + Na2S6 + 2 H2O + O2

4 NaOH + S8 $\rightarrow$ 2 Na2S4 + 2 H2O + O2

6 NaOH + S8 $\rightarrow$ 2 Na2S3 + Na2S2O3 + 3 H2O

12 NaOH + S8 $\rightarrow$ 4 Na2S + 2 Na2S2O3 + 6 H2O

  • 1
    $\begingroup$ Displacing oxygen? Really? $\endgroup$ Commented Nov 16, 2021 at 15:54

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