# How would dichromate ion, Cr_2O_7^(2-), oxidize sulfide ion, S^(2-), to give elemental sulfur? The reduction product is Cr^(2+).

Apr 23, 2017

Are you sure that the dichromate is reduced to $C r \left(+ I I\right)$.........?

#### Explanation:

Normally, we would expect dichromate to give $C {r}^{3 +}$ as its reduction product:

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$

Sulfide anion is oxidized to elemental sulfur.......:

${S}^{2 -} \rightarrow S \left(s\right) + 2 {e}^{-}$

Overall................

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 3 {S}^{2 -} \rightarrow 2 C {r}^{3 +} + 3 S + 7 {H}_{2} O$

PS I am not saying you are wrong. I would check your sources or lab manual.

We could formulate reduction to $\text{chromous ion}$ as follows:

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 8 {e}^{-} \rightarrow 2 C {r}^{2 +} + 7 {H}_{2} O \left(l\right)$