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

1 Answer
Apr 23, 2017

Answer:

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)#