# How is iodine oxidized to iodate ion by nitric acid?

May 28, 2017

With conc. nitric acid...........

$5 N {O}_{3}^{-} + \frac{1}{2} {I}_{2} \left(s\right) + 4 {H}^{+} \rightarrow 5 N {O}_{2} + I {O}_{3}^{-} + 2 {H}_{2} O \left(l\right)$

#### Explanation:

Elemental iodine is oxidized up to iodate, $I {O}_{3}^{-}$......

$\frac{1}{2} {I}_{2} \left(s\right) + 3 {H}_{2} O \left(l\right) \rightarrow I {O}_{3}^{-} + 6 {H}^{+} + 5 {e}^{-}$ $\left(i\right)$

And nitric acid is reduced to $N {O}_{2}$:

$\stackrel{V}{N} {O}_{3}^{-} \rightarrow \stackrel{I V}{N} {O}_{2}$, i.e.

$N {O}_{3}^{-} + 2 {H}^{+} + {e}^{-} \rightarrow N {O}_{2} + {H}_{2} O \left(l\right)$ $\left(i i\right)$

And in the usual way we cross-multiply the individual redox equations, so that electrons do not appear in the final reaction.

$\left(i\right) + 5 \times \left(i i\right) =$

$5 N {O}_{3}^{-} + \frac{1}{2} {I}_{2} \left(s\right) + 4 {H}^{+} \rightarrow 5 N {O}_{2} + I {O}_{3}^{-} + 2 {H}_{2} O \left(l\right)$

Which looks balanced with respect to mass and charge to me. Is it? All care taken but no responsibility admitted!