# How does "thiosulfate ion", S_2O_3^(2-) give rise to sulfur and sulfur dioxide upon disproportionation?

Jun 4, 2017

This is a disproportionation reaction........

${S}_{2} {O}_{3}^{2 -} + 2 {H}^{+} \rightarrow S + S {O}_{2} + {H}_{2} O$

#### Explanation:

Thiosulfate, ${S}_{2} {O}_{3}^{2 -}$ has formal $\stackrel{- I I}{S}$ and $\stackrel{+ V I}{S}$ oxidation states, i.e. an average oxidation state of $\frac{- I I + V I}{2} = + I I$.

It is reduced to $\stackrel{0}{S}$, i.e. zerovalent sulfur, and oxidized to ${\stackrel{+ I V}{\text{SO}}}_{2}$.

And thus reduction............:

${S}_{2} {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-} \rightarrow 2 S + 3 {H}_{2} O$ $\left(i\right)$

And oxidation............:

${S}_{2} {O}_{3}^{2 -} + {H}_{2} O \rightarrow 2 S {O}_{2} + 2 {H}^{+} + 4 {e}^{-}$ $\left(i i\right)$

For both $\left(i\right)$ and $\left(i i\right)$, charge and mass are balanced as is absolutely required. Are they balanced? Don't trust my arithmetic.

We add $\left(i\right) + \left(i i\right)$ to remove the electrons...........

$2 {S}_{2} {O}_{3}^{2 -} + 4 {H}^{+} \rightarrow 2 S + 2 {H}_{2} O + 2 S {O}_{2}$

And of course we could halve this, as per Nam D.'s suggestion..

${\underbrace{{S}_{2} {O}_{3}^{2 -} + 2 {H}^{+}}}_{\text{sulfurous acid} \cdot {H}_{2} S {O}_{3}} \rightarrow S + {H}_{2} O + S {O}_{2}$

Again, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, $2 \times S$, $3 \times O$, and $2 \times H$.........so balanced with respect to mass and charge.

What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of $S {O}_{2}$.

Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give ${H}_{2} {S}_{2} {O}_{3}$. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. $S \left(- I I\right)$. And so we got $S \left(V I +\right)$ and $S \left(- I I\right)$...the average oxidation state is still $S \left(+ I I\right)$. And we compare this with $S {O}_{4}^{2 -}$, i.e. $S \left(V I +\right) + 4 \times O \left(- I I\right)$...