The more electropositive element should displace metal from it's salts' solution. This can be seen in reaction of copper salts with iron. But, if you try using for example potassium metal as the more electropositive metal in a displacement reaction you don't get a result since potassium will react with water before it has a chance to displace much of the other metal.

  1. What makes dissolution in water favour displacement by the more electropositive metal, even if calculated gibbs free energy for the reaction would be positive? Is it the higher strength of hydrogen bonding of the more electropositive metal, or is it the lack of lattice energy of the salt that otherwise needs to be overcome?

  2. This will depend on the answer to the first question, but is it possible to actually displace, say, Calcium with Potassium metal from any of it's salts provided that this salt is soluble and the solvent is inert towards Calcium and Potassium metal? If you had to do this in reality, what would you use, perhaps a phase transfer catalyst?

I imagine that if it's mainly the lattice energy making the difference one should be able to just use a different solvent, as it would be only important to dissociate the salt

  • $\begingroup$ Some of this was already answered when you asked earlier chemistry.stackexchange.com/questions/118031/… $\endgroup$
    – Mithoron
    Nov 9, 2019 at 23:46
  • $\begingroup$ I realize (and mentioned it in the post) that water makes all the difference, now im asking how does it work? I can write the same equation for the reaction dry and in aqueous solution, and obtain the same gibbs free energy change, which would make it seem like it wouldn't happen even in solution. Yet, something makes the reaction favourable in water (or would make it favourable if not the mentioned reaction of potassium with the solvent). Is it because of stronger hydrogen bonding of the K+ ion? Or is the lattice energy the culprit and it's only important that the salt is dissociated? $\endgroup$
    – Francis L.
    Nov 10, 2019 at 0:05
  • $\begingroup$ I feel I explained what im asking better in the comment, so im going to edit the post to match. $\endgroup$
    – Francis L.
    Nov 10, 2019 at 0:16


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