On the reduction:
Low Acid Concentration:
5 KI + KIO3 + 6 HCl --> 6 KCl + 3 I2 + 3 H2O
But in high acid concentration:
2 KI + KIO3 + 6 HCl --> 3 KCl + 3 ICl + 3 H2O
and ICl + H2O --> HOI + HCl
But immediately:
5 HOI --> 2 H2O + 2 I2 + HIO3
Source, per my notes, is Mellor, around page 118.
I suspect your source of KIO3 actually already has some KI. My speculation is based on:
3 I2 + 3 H2O <-----> 5 HI + HIO3
and upon adding KOH, produces KI and KIO3. Potassium iodate is reportedly prepared by the anodic oxidation of a KI solution. My attempt at this:
KI + H2O <---> KOH + HI
4 HI + O2 --> 2 H2O + 2 I2
and again we have Iodine and water plus KOH.
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Working with solid KIO3, one could attempt a thermal decomposition (between 655 to 736 K):
KIO3 --> KI + 3/2 O2
Also, a reported secondary reaction:
4 KIO3 --> 2 K2O + 2 I2 + 5 O2
so be prepared to condensed escaping Iodine.
Source: "High temperature properties and thermal decomposition of inorganic salts ..." by Kurt H. Stern, page 248.
Link:
http://books.google.com/books?id=2BpMo7HpXzIC&pg=PA248&a...
Then, treat the Potassium iodide with Chlorine:
2 KI + Cl2 --> KCl + I2
or, one could theoretically use the KI to seed the first procedure that requires some KI.
Let me know what happens.
[Edited on 13-1-2012 by AJKOER] |
