Question

Which species is oxidized in the reaction below?

`I^(-)(aq) + ClO^(-) (aq) -> IO^(-) (aq) + Cl^(-) (aq)`

Answer 1

Iodine is oxidized to the conjugate base of hypiodous acid, `HOI`.

Explanation:

The iodide ion, `I^-` clearly has a formal oxidation number of `-I`. On the other hand, iodine in hypoiodous acid, clearly has a formal oxidation number of `+I`. If this is not clear to you ask for clarification; iodine is certainly LESS electronegative than oxygen, and oxygen is presumed to get the electron when we distribute the bonding electron.

So `"oxidation:"`

`I^(-) + H_2O rarr IO^(-) +2H^(+) +2e^-`

And `"reduction:"`

`ClO^(-)+ 2H^(+)+ 2e^(-) rarr Cl^(-) +H_2O`

And `"overall:"`

`ClO^(-)+ I^(-) rarr Cl^(-) +IO^(-)`

Which species is reduced in the above reaction?