Quote: |
You really need an oxidizer when making iodine from potassium iodide.
KI + H2SO4 -> KHSO4 + HI
4HI + O2 -> 2H2O + I2 |
Really ? Does it happen that way ? I'd say since HI is way stronger than H2SO4 sulphuric acid shouldn't usually be able to
protonate an Iodide. I first thought it would work like that, too that the Acid produces HI and Peroxide would oxidize it but when I look at the
Redox-Potentials:
2I- -> I2 + 2e- => +054
H2O2 + 2 H3O+ + 2e- -> 4 H2O => + 1,78
Or the one you were referring to
O2 + 4 H3O+ + 4e- -> 6 H2O => 1,23
So both of these Systems would be strong enough to oxidize Iodide to Elemental Iodine. Ehm...if I remember correctely Redox Systems with Acids or
Bases can be calculated with the Nernst Equation, there was one that deals with pH, too. So for these Systems the conc. of Acid is important, too.
Stochiometrically from the equations above it needs quite some acids to produce these quite high Redox Potentials. And that might be the reason why
Acids are needed.
I might be totally mistaking here but for me it just sounds better to produce Iodine via a Redox Reaction than protonating a stronger Acid with a
weaker one.
Does anyone see the mistake I'm making here, it's been a while since I had to calculate out stuff like this (my high school finals to be honest xD ).
[Edited on 29-9-2015 by fluorescence] |