3
$\begingroup$

In our discussion group, my friend mentioned that metal cations can be separated from other cations in solution using cryptands. After addition of the cryptands to the solution of the cations, chelation of the cations takes place and since the complexes formed are more soluble in organic solvent, addition of an organic solvent to this resultant solution would allow these complexes to be migrate to the organic layer. The organic layer can then be decanted and the metal ions have now been separated successfully from the aqueous solution. This was my friend's proposal.

From hearing this, I quickly thought of the possibility of creating a negatively-charged solution. By applying the method of using cryptands which my friend has proposed to separating sodium ions from a solution of sodium chloride, wouldn't the remaining chloride ions in solution give the resultant solution an overall negative charge? This led me to think that there must be something wrong with his idea as this should not be able to occur, based on my instinct. However, I do not know exactly how to refute my friend, to say his proposal does not work. Thus, I would like to ask if anyone knows the reason why my friend's method would not work.

|improve this question
$\endgroup$
  • $\begingroup$ If cations move anions are drawn with them. If anions can't do that then cations stay close to them. $\endgroup$ – Mithoron Jan 15 '18 at 17:59
  • $\begingroup$ Never, ever. The initial charge separation immediately leads to a small electrical potential, against which no further separation is possible. (The largest possible potential in any electrochemical system is in the range of 5V) If you think about a quantitative separation, that voltage difference would easily go into the megavolts, even for a rather dilute system. $\endgroup$ – Karl Jan 15 '18 at 22:23
3
$\begingroup$

For some reason this question reminds me of Maxwell's demon paradox. You don't really separate anions and cations here; in reality the resulting metal ($\ce{M+}$) cryptand ($\ce{L}$) complex is going to be associated with anionic part ($\ce{X-}$) in organic solvent ($\ce{s}$) after it's being extracted from aqueous phase ($\ce{w}$). For the detailed process, refer to [1], for example.

There are two main processes to consider:

$$ \begin{align} \ce{L_\mathrm{s} + M^+_\mathrm{s} &<=> LM^+_\mathrm{s}}\label{rxn:1}\tag{1}\\ \ce{\overline{\ce{L}} + M^+_\mathrm{w} + X^-_\mathrm{w} &<=> \overline{\ce{LMX}}}\tag{2} \end{align}$$

You would be correct if only $\eqref{rxn:1}$ occurs, however, this is not the case.

Reference

  1. Fyles, T. M. Can. J. Chem. 1987, 65 (4), 884–891. DOI 10.1139/v87-149 (Open Access).
|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.