How is iodine oxidized to iodate ion by nitric acid?

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How is iodine oxidized to iodate ion by nitric acid?

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Answer 1 · 2017-05-28T10:55:16

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

$5 N O_{3}^{-} + \frac{1}{2} I_{2} (s) + 4 H^{+} \to 5 N O_{2} + I O_{3}^{-} + 2 H_{2} O (l)$ $5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)$

Explanation:

Elemental iodine is oxidized up to iodate, $I O_{3}^{-}$ $IO_3^(-)$ ......

$\frac{1}{2} I_{2} (s) + 3 H_{2} O (l) \to I O_{3}^{-} + 6 H^{+} + 5 e^{-}$ $1/2I_2(s) + 3H_2O(l) rarr IO_3^(-) + 6H^(+) + 5e^(-)$ $(i)$ $(i)$

And nitric acid is reduced to $N O_{2}$ $NO_2$ :

$V N O_{3}^{-} \to I V N O_{2}$ $stackrel(V)NO_3^(-) rarr stackrel(IV)NO_2$ , i.e.

$N O_{3}^{-} + 2 H^{+} + e^{-} \to N O_{2} + H_{2} O (l)$ $NO_3^(-) +2H^(+) +e^(-) rarrNO_2 +H_2O(l)$ $(i i)$ $(ii)$

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

$(i) + 5 \times (i i) =$ $(i) + 5xx(ii)=$

$5 N O_{3}^{-} + \frac{1}{2} I_{2} (s) + 4 H^{+} \to 5 N O_{2} + I O_{3}^{-} + 2 H_{2} O (l)$ $5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)$

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

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How is iodine oxidized to iodate ion by nitric acid?

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