# How does iodine oxidize sulfide ion to sulfate ion?

May 16, 2016

${S}^{2 -} + 4 {I}_{2} + 4 {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 8 {I}^{-} + 8 {H}^{+}$

#### Explanation:

Sulfide ion is oxidized to sulfate $\left(S \left(- I I\right) \rightarrow S \left(V I\right)\right)$:

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

And elemental iodine is reduced to iodide:

${I}_{2} + 2 {e}^{-} \rightarrow 2 {I}^{-}$ $\left(i i\right)$

To balance, we remove the electrons: $\left(i\right) + 4 \times \left(i i\right) :$

${S}^{2 -} + 4 {I}_{2} + 4 {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 8 {I}^{-} + 8 {H}^{+}$

This is balanced with respect to mass and charge. Whether it is valid chemically is another matter.