I2 + NO2+ + H2O -> 2HI + NO3+

redox potentials:

I2 + 2e -> 2I-     0.5355 V

NO3- + H2O + 2e -> NO2- + 2OH-     0.01 V

Many thanks, Florin Omota.

But I think those potentials involve the nitrite-nitrate pair.

Since NO2+ must be an oxidant, I would expect iodine to become positively charged, but I haven't found a reference for that... maybe iodate formation in the presence of moisture?

Best regards

Indeed, you mentioned NO2+ that is oxidant. Note reaction of HNO3+H2SO4+I2 forming HIO3. This reaction is mentioned in Wikipedia:  https://en.wikipedia.org/wiki/Iodine#Redox_reactions

"Iodine is oxidized to iodate by nitric acid as well as by chlorates:[18]I2 + 10 HNO3 → 2 HIO3 + 10 NO2 + 4 H2O"

I suppose you will get NO2+ from HNO3 in H2SO4 that can oxidize I2 by the above reaction. Note that standard potentials of HNO3 / NOx(g) and HIO3 / I2(s) does not allow the above reaction and in diluted solution the reverse reaction will take place:

HIO3 +NOx  -->  I2 + HNO3

Standard redox potentials:

IO3− + 5 H+ + 4 e−  -->  HIO(aq) + 2 H2O                  +1.13

2 IO3− + 12 H+ + 10 e−  -->  I2(s) + 6 H2O                 +1.20

NO3−(aq) + 2 H+ +  e−  -->  NO2(g) +  H2O               +0.80

NO3−(aq) + 4 H+ + 3 e−  -->  NO(g) + 2 H2O             +0.958

shows IO3- is stronger oxidant than HNO3, at standard conditions.

I suppose NO2+ is not stable in aqueous solutions and perhaps that's why redox potential is not available:    NO2+ + H2O  -->  HNO3 + H+