I tried to balance the following redox reaction

$$\ce{S^0 + OH- -> S^2- + S2O3^2-}$$

by ion electron method, but I am unable to identify the oxidation and reduction half reactions. Please provide the correct method.


closed as off-topic by airhuff, Jan, paracetamol, Nilay Ghosh, Tyberius Nov 15 '17 at 15:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.


Genarally, sulfur doesn't disproportionate like that: in fact, it typically behaves very similarly to chlorine (see Balancing the redox disproportionation of chlorine by half-reactions), which upon heating with alkali disproportionates to chloride $\ce{Cl-}$ and chlorate $\ce{ClO3-}$:

$$\ce{3$\overset{0}{\ce{Cl2}}$ + 6 OH- ->[$\Delta$] 5 $\overset{\mathrm{-I}}{\ce{Cl}}${}^- + $\overset{\mathrm{+V}}{\ce{Cl}}$O3- + 3 H2O}$$

Without heating instead of chlorate $\ce{ClO3-}$ hypochlorite $\ce{ClO-}$ is formed, so this is an irrelevant process in this case. Back to sulfur: when reacting with alkali, sulphide $\ce{S^2-}$ and sulfite $\ce{SO3^2-}$ are formed as the primary products according to the scheme:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+IV}}{\ce{S}}$O3^2-} \tag{1}$$

Half-reactions and net equation for (1):

\begin{align} \ce{2 S + 4 e- &-> 2 S^2-} \tag{red1}\\ \ce{S + 6OH- &-> SO3^2- + 4e- + 3H2O} \tag{ox1}\\ \hline \ce{3S + 6 OH- &-> 2 S^2- + SO3^2- + 3 H2O} \tag{redox1} \end{align}

Note that (red1) has been multiplied by 2 to even the number the electrons in both (red1) and (ox1). The net equation (redox1) is obtained as a sum of both half-reactions.

Now, lets assume that the reaction you've written takes place, where where thiosulfate is formed instead of sulfite. In fact, it might as well happen when boiling solutions of sulfites with sulfur for a long time:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+II}}{\ce{S}}${}_2O3^2-} \tag{2}$$

Half-reactions and net equation for (2):

\begin{align} \ce{2 S + 4 e- &-> 2 S^2-} \tag{red2}\\ \ce{2S + 6OH- &-> S2O3^2- + 4e- + 3H2O} \tag{ox2}\\ \hline \ce{4S + 6 OH- &-> 2 S^2- + S2O3^2- + 3 H2O} \tag{redox2} \end{align}

Reaction (redox2) would be the net reaction for the process you asked about, even though I'd say it's more likely a hypothetical one. Keep in mind that for as long as the reaction takes place in aqueous solution and the redox process involves acid-base transformations (like formation of oxoanions in this case), you need to take water into account.

  • 2
    $\begingroup$ As far as I know (and Wikipedia confirms), thiosulphate is formed by reaction sulphite and sulphur under alkaline conditions:$$\ce{S + SO3^2- ->[OH-] S2O3^2-}$$So the second net reaction is not as unlikely as you make it sound. $\endgroup$ – Jan Nov 15 '17 at 7:45
  • $\begingroup$ @Jan Yeah, as a brutto reaction it's totally fine, I just wanted to underline the fact that it's a secondary process. $\endgroup$ – andselisk Nov 15 '17 at 7:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.