Sodium Polysulfide production from NaOH and Sulfur Does the sulfur also oxidize the $\ce{OH-}$ radicals after the $\ce{Na2S4}$ chains form? Something like this:
$$\ce{9S + 4NaOH -&gt; 2Na2S4 + SO3 + H2O}$$
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: $\ce{S -&gt; SO3}$
Reduction: $\ce{S -&gt; S4^2-}$


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