In a displacement reaction I understand that the more reactive metal essentially takes the place of the less reactive metal. But I can't seem to understand how this more reactive metal can break the bonds of this compound. Why would this more reactive metal want to take the place of the less reactive metal in the compound, if the compounds is already very stable?
2 Answers
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This can be easily understood if one consider the electrochemical series. I can explain this concept with a simple and famous illustration. Consider the following reaction: $$ \ce{Zn + Cu^2+ -> Zn^2+ + Cu}\quad E^\circ=+1.1\ \mathrm V$$ Where, $E^\circ$ is the standard cell potential for a cell reaction. From thermodynamics we have a relation between $\Delta G^\circ$ and $E^\circ$ which is given by, $$\Delta G^\circ=nFE^\circ$$ where, $n$ is the number of electrons transferred and $F$ is known as Faraday and it is the magnitude of charge on one mole of electrons. So from this relation we will get $\Delta G^\circ\lt0$ for the above reaction. And this is the criterion for spontaneity or feasibility of the reaction. So this reaction would happen in forward direction. This illustration explains why a metal would displace another metal in such kind of reactions.
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User124053's explanation does not correspond to the reality. He or she says that
"the more reactive metal essentially takes the place of the less reactive metal. But I can't seem to understand how this more reactive metal can break the bonds of this compound"
No ! No metal "takes the place" of another one. No bonds are broken : the reaction takes place between ions and a metal in aqueous solution, as explained by Infinite. The only effect is a charge transfer between $\ce{Zn}$ and $\ce{Cu^{2+}}$.
