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Interaction between cationic ions

 

  I want to ask you about cationic metal ions. For example, suppose that FeCl2 and MgCl2 are mixed in distilled water. Fe2+ is more likely to be oxidized to Fe3+. Does Mg2+ prevent Fe2+ from being oxidized? Could you tell me whether cationic ions affect each other’s reactions or not?
  

U (Jan 2003)

 
 

Answer:
  First of all, the following basic conditions are required in discussing the question above:

  • All ions are diluted.

  • When Fe2+ is treated, use slightly concentrated acid, especially acid which do not tend to form complexes with metal ions, i.e., sulfuric acid and perchloric acid. 

  The influence of pH and the co-existence of anionic ions is also not negligible.  In addition, it is important whether the experiment is done under air or not.  Magnesium ion and zinc ion do not precipitate under basic and neutral conditions, but Fe3+ precipitate easily.  Thus, Fe2+ is more easily oxidized by oxygen in air under such conditions.
 
  Based on the conditions above, let’s discuss the interaction between cationic metal ions.  I assume that this questioner would wonder whether there is electrostatic interaction between these ions since they have more than one charge.  However, these ions are actually surrounded by water molecules, and also by anionic ions.  As a result, the electric field surrounding the cationic ion is cancelled close to the cationic ion. Therefore, as long as the concentration is not too high, these ions exert only a slight influence on one another. This is also the case for Fe2+.
 
  Let me give you one more example related to this question. Fe2+ is oxidized at the anode (positive electrode) and becomes Fe3+. Why is Fe2+ able to come close to the anode?  The reason is that potential drops only in the extremely narrow region (called “electric double layer”) near the electrode. Therefore, Fe2+ can come close to the electrode and pass an electron to it, without the influence of potential gradient (electric field).
 
 
Acknowledgement
  We would like to thank an anonymous professor for his careful teaching and kind support for this answer.
 
 

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