Question

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

Answer 1

Are you sure that the dichromate is reduced to `Cr(+II)`.........?

Explanation:

Normally, we would expect dichromate to give `Cr^(3+)` as its reduction product:

`Cr_2O_7^(2-) +14H^+ + 6e^(-) rarr 2Cr^(3+) +7H_2O`

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

`S^(2-) rarr S(s) + 2e^(-)`

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

`Cr_2O_7^(2-) + 14H^(+) + 3S^(2-) rarr 2Cr^(3+) + 3S + 7H_2O`

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

We could formulate reduction to `"chromous ion"` as follows:

`Cr_2O_7^(2-) +14H^+ + 8e^(-) rarr 2Cr^(2+) +7H_2O(l)`