[xshadow]

Have to balance this redox with ionic  half reaction  :

FeI₂+ HNO₃ ---> Fe(NO₃)₃ + HIO₃ + NO + H₂O

method a) FeI2 insoluble

OX: FeI₂ --> Fe³⁺ + 2 IO^- + 12 H⁺ + 13e^-    X 3
RID:  NO₃^- + 4 H⁺ + 3e^- --> NO + 2H₂O      X 13

I'll get:
3 FeI₂ + 13 NO₃^- + 16 H⁺ --->3 Fe ³⁺ + 6 IO₃^- + 13 NO + 8 H₂O



NOW if I considered FeI2 as a ""soluble species "" (even if it's less soluble)  I have to consider two oxditaion half reaction:

OX1 Fe ²⁺ ---> Fe³⁺ + 1e-

OX2   2I^- + 6H₂O--> 2 IO₃^- + 12 H⁺ + 12 e- 
   
OX TOT = OX1 + OX2


Or OX2 is it :

I^- + 3H₂O ---> IO₃^- +  6 H⁺ + 6e-

In other word when I consider the iodine half reaction I have to consider I- ---> IO3-  or 2I- ---> IO3-
Because I think only in the second way  it works


Thanks

[mjc123]

You have to get the correct stoichiometry between OX2 and OX1. If 2 I^- react for every Fe²⁺ that reacts, then OX1 and OX2 are as you first wrote them, so that OX = OX1 + OX2 works out correctly (i.e. 13 electrons are produced for every Fe²⁺ that reacts).

Also note as a general point, here and in your other post, that half reactions must balance, just like ordinary reactions. You cannot write e.g.

OX: FeI2 --> Fe3+ + 2 IO- + 12 H+ + 13e-    X 3

Where are the O and the H on the LHS?

[Borek]

There two separate oxidation processes here: Fe²⁺ oxidation, and I^- oxidation. That usually means you can't write a single redox reaction equation, unless there is some additional information - in this case, you know you have two I^- for each Fe²⁺, so the total reaction is a sum of one Fe²⁺ oxidation plus two I^- oxidation reactions.

[xshadow]

So I have to consider the dissolution stechiometry

Thanks