How does iodine oxidize sulfide ion to sulfate ion?

Question

3703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-16T09:40:26

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

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.