# Which species is oxidized in the reaction below?

## ${I}^{-} \left(a q\right) + C l {O}^{-} \left(a q\right) \to I {O}^{-} \left(a q\right) + C {l}^{-} \left(a q\right)$

Oct 31, 2016

Iodine is oxidized to the conjugate base of hypiodous acid, $H O I$.

#### 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 $\text{oxidation:}$

${I}^{-} + {H}_{2} O \rightarrow I {O}^{-} + 2 {H}^{+} + 2 {e}^{-}$

And $\text{reduction:}$

$C l {O}^{-} + 2 {H}^{+} + 2 {e}^{-} \rightarrow C {l}^{-} + {H}_{2} O$

And $\text{overall:}$

$C l {O}^{-} + {I}^{-} \rightarrow C {l}^{-} + I {O}^{-}$

Which species is reduced in the above reaction?