# Question #3ef9d

Jul 3, 2017

You have got TWO separate oxidation reactions, and I PRESUME that your starting material is $F e S$ NOT $F e {S}_{2}$.......

#### Explanation:

$\text{Ferrous sulfide}$ is oxidized to $\text{ferric oxide}$........

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

$\text{Oxygen gas}$ is reduced to $\text{oxide dianion}$

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

And $\text{sulfide anion}$ is oxidized to $S {O}_{2}$.........

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

And so (purely to eliminate the electrons on each side) let's take........

$\left(i\right) + \left(i i i\right) + 2 \times \left(i i\right)$.............

$2 F e S \left(s\right) + {H}_{2} O \left(l\right) + 2 {O}_{2} \left(g\right) \rightarrow F {e}_{2} {O}_{3} + {S}^{2 -} + 2 {H}^{+} + S {O}_{2} \left(g\right)$

And finally..........

$2 F e S \left(s\right) + {H}_{2} O \left(l\right) + 2 {O}_{2} \left(g\right) \rightarrow F {e}_{2} {O}_{3} \left(s\right) + {H}_{2} S \left(g\right) + S {O}_{2} \left(g\right)$

Sulfide and $F e \left(+ I I\right)$ are oxidized to give $S {O}_{2}$ and $F e \left(I I I +\right)$, oxygen is reduced.............

Do you follow my reasoning........? It's balanced, but whether it corresponds to an actual chemical process is beyond me.......