How would dichromate ion, $C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ , oxidize sulfide ion, $S^{2 -}$ $S^(2-)$ , to give elemental sulfur? The reduction product is $C r^{2 +}$ $Cr^(2+)$ .

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Answer 1 · 2017-04-23T16:59:20

Are you sure that the dichromate is reduced to $C r (+ I I)$ $Cr(+II)$ .........?

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

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ $Cr_2O_7^(2-) +14H^+ + 6e^(-) rarr 2Cr^(3+) +7H_2O$

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

$S^{2 -} \to S (s) + 2 e^{-}$ $S^(2-) rarr S(s) + 2e^(-)$

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

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 3 S^{2 -} \to 2 C r^{3 +} + 3 S + 7 H_{2} O$ $Cr_2O_7^(2-) + 14H^(+) + 3S^(2-) rarr 2Cr^(3+) + 3S + 7H_2O$

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

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

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 8 e^{-} \to 2 C r^{2 +} + 7 H_{2} O (l)$ $Cr_2O_7^(2-) +14H^+ + 8e^(-) rarr 2Cr^(2+) +7H_2O(l)$

How would dichromate ion, Cr2O2−7Cr_2O_7^(2-), oxidize sulfide ion, S2−S^(2-), to give elemental sulfur? The reduction product is Cr2+Cr^(2+).