# How is "thiosulfate anion", S_2O_3^(2-), oxidized by "permanganate anion", MnO_4^(-)?

Jan 27, 2016

$5 {S}_{2} {O}_{3}^{2 -} + 8 M n {O}_{4}^{-} + 14 {H}^{+} \rightarrow 8 M {n}^{2 +} + 10 S {O}_{4}^{2 -} + 7 {H}_{2} O \left(l\right)$

#### Explanation:

Thiosulfate is oxidized to sulfate $S \left(I I\right) \rightarrow S \left(V I\right)$:

${S}_{2} {O}_{3}^{2 -} + 5 {H}_{2} O \rightarrow 2 S {O}_{4}^{2 -} + 10 {H}^{+} + 8 {e}^{-}$ $\left(i\right)$

Permanganate ($M n \left(V I I\right)$) is reduced to $M {n}^{2 +}$:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O \left(l\right)$ $\left(i i\right)$

If we take $5 \times \left(i\right) + 8 \times \left(i i\right)$ (we do this cross-multiplication so that electrons, ${e}^{-}$, do not appear in the reaction), we get (finally!):

$8 M n {O}_{4}^{-} + 5 {S}_{2} {O}_{3}^{2 -} + 14 {H}^{+} \rightarrow 8 M {n}^{2 +} + 10 S {O}_{4}^{2 -} + 7 {H}_{2} O \left(l\right)$

This equation (I think) is balanced with respect to mass and charge (I did a double take when I realized there were 47 oxygens on each side!). The reaction is also self-indicating: deep purple permanganate ion is reduced to (almost) colourless $M {n}^{2 +}$. I leave it to you to give the complete reaction (i.e. where $M n {O}_{4}^{-}$ appears as $K M n {O}_{4}$ and thiosulfate appears as $N {a}_{2} {S}_{2} {O}_{3}$).

Note that I designated the AVERAGE oxidation state of sulfur in thiosulfate as $+ I I$. I could have treated the terminal sulfur as the analogue of oxygen in sulfate, i.e. a $- I I$ oxidation state, and the central sulfur as the analogue of the central sulfur in sulfate, i.e. $+ V I$. The average oxidation state of $S$ of course is the same, $+ I I$ in each case.