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...........

$5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)$

Elemental iodine is oxidized up to iodate, $IO_3^(-)$ ......

$1/2I_2(s) + 3H_2O(l) rarr IO_3^(-) + 6H^(+) + 5e^(-)$ $(i)$

And nitric acid is reduced to $NO_2$ :

$stackrel(V)NO_3^(-) rarr stackrel(IV)NO_2$ , i.e.

$NO_3^(-) +2H^(+) +e^(-) rarrNO_2 +H_2O(l)$ $(ii)$

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

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

$5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)$

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