What happens during a displacement reaction? [closed]








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When Zn metal is added to a solution of Cu(II)SO4, Zinc displaces Cu to form ZnSO4 as Zn is more reactive than Cu. What I don't understand is how.


1- What I'm assuming is: Zn metal collides with [SO4]2- ions and because it wants to be stable (by losing 2 electrons), it forces those 2 of its electrons onto [SO4]2- which then goes on to lose those 2 (of its previous) electrons as it also wants to be stable. Then, those 2 extra electrons collide with [Cu]2+ ions and are attracted to the ion because of the charge difference and then they form Cu metal.


2- But then, what stops Cu metal colliding with [SO4]2- ions (with Zn electrons) and making [Cu]2+ ions again.


3- Is it just the fact that the collisions are so frequent that Zn wins in the 'battle of displacement' because of its higher reactivity and forms some sort of equilibrium?


4- Surely, the electrons of Zn metal aren't different/stronger than the electrons of Cu metal to resist being moved out if the mechanism by which Zn metal reacts is by forcing its electrons to be accepted.


5- In the solution, the ions are separated due to the polarity of water molecules, so it's not like the Zn metal is next to the CuSO4 ionic compound, it would only be next to / 'collide' with the [SO4]2- ion. The only way for a displacement reaction I can think of then, is by the mechanism I'm describing in 1.


Please keep the answers relatively simple, I'm only in college. No quantum mechanics magic, ideally. I'm assuming the mechanism is completely wrong, but I can't find another way for it to happen (at least not in a non complex form).




Salts like $\ce{ZnSO4}$ or $\ce{CuSO4}$ do not exist in water solution as dissolved molecules. In fact, such molecules do not exist in the solid salts either (anhydrous salts or salt hydrates $\ce{ZnSO4 \cdot 7 H2O}$ or $\ce{CuSO4 \cdot 5 H2O}$). Those formulas just reflect the stoichiometric ratio of ions in a solid phase or solution. $\ce{ZnSO4}$ (or $\ce{CuSO4}$) exist in water solution as independent ions $\ce{Zn^2+(aq)}$ ( or $\ce{Cu^2+(aq)}$) and $\ce{SO4^2-(aq)}$. (aq) means hydrated state.


When zinc dissolves, there is ongoing reaction $\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$, with electrons remaining in the metallic zinc, charging it slightly negatively. These electrons, when the potential of the zinc electrode drops low enough, start reacting with copper ions, that more easily accept electrons than zinc ions: $\ce{Cu^2+(aq) + 2 e- -> Cu(s)}$, depositing metallic copper on the zinc surface.


The sulphate ions in the solution are just spectators/bystanders, "watching" indirect exchange of electrons between zinc atoms and hydrated copper ions.


 4

The reason for this is the fact that Zinc is a **more reactive** metal.


(Note that, as Poutnik mentions (and I forgot to), $\ce{CuSO4}$ and $\ce{ZnSO4}$ do not exist in the liquid state but in the dissolved form and in this form they are ionised, i.e. they break up into their respective ions $\ce{Cu}^{2+}$,$\ce{Zn}^{2+}$, and $\ce{SO4}^{2-}$ )


What does more reactive mean? It means that zinc requires less energy to remove the 2 electrons than copper does.


So the difference between the *energy released* when zinc forms an ionic bond with $\ce{SO4}^{2-}$ and the *energy required* to remove electrons from zinc is larger than in the case of copper, i.e. **more energy is released** in the formation of $\ce{ZnSO4}$ than in the formation of $\ce{CuSO4}$.


As a rule of thumb, lesser energy means more stability. Hence $\ce{ZnSO4}$ is more stable and the reaction favours the formation of $\ce{ZnSO4}$.


You can also consider the electrochemical series (as Maurice has mentioned) in which zinc has a greater value of standard reduction potential than that of copper (which basically means that zinc can get oxidised more easily than copper) and hence zinc reduces copper and itself gets oxidised.


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