How does iodine oxidize sulfide ion to sulfate ion?

1 Answer
May 16, 2016

Answer:

#S^(2-) + 4I_2 + 4H_2O rarr SO_4^(2-) + 8I^(-) + 8H^(+)#

Explanation:

Sulfide ion is oxidized to sulfate #(S(-II) rarrS(VI))#:

#S^(2-) + 4H_2O rarr SO_4^(2-) + 8H^(+) + 8e^-# #(i)#

And elemental iodine is reduced to iodide:

#I_2 +2e^(-) rarr 2I^-# #(ii)#

To balance, we remove the electrons: #(i) +4xx(ii):#

#S^(2-) + 4I_2 + 4H_2O rarr SO_4^(2-) + 8I^(-) + 8H^(+)#

This is balanced with respect to mass and charge. Whether it is valid chemically is another matter.