How is iodine oxidized to iodate ion by nitric acid?

1 Answer
May 28, 2017

Answer:

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

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

Explanation:

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!