Displacement as Redox

$\mathrm{Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)}$

Copper metal forms while zinc enters the solution as zinc ions. This is called displacement because the more reactive zinc replaces the less reactive copper.

The reactivity comparison tells us whether displacement can happen. It does not yet explain the change at particle level. For that, we must follow the electrons.

Looking beneath the full equation

In water, copper(II) sulfate and zinc sulfate contain separate, mobile ions. Writing these ions explicitly gives:

$\mathrm{Zn(s) + Cu^{2+}(aq) + SO_4^{2-}(aq) \rightarrow Zn^{2+}(aq) + SO_4^{2-}(aq) + Cu(s)}$

The sulfate ions appear unchanged on both sides. They are spectator ions, so they can be removed to leave the ionic equation:

$\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$

This equation exposes the essential change: zinc atoms become zinc ions, while copper(II) ions become copper atoms.

Following the electrons

A neutral zinc atom becomes a $\mathrm{Zn^{2+}}$ ion by losing two negatively charged electrons:

$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$

Loss of electrons is oxidation, so zinc is oxidised.

A $\mathrm{Cu^{2+}}$ ion becomes a neutral copper atom by gaining two electrons:

$\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$

Gain of electrons is reduction, so copper(II) ions are reduced. The two equations are called half-equations because each shows one part of the overall electron transfer.

The electrons lost by zinc are exactly the electrons gained by the copper(II) ions. Electrons are transferred rather than created or destroyed, so oxidation and reduction must occur together. That is why the displacement reaction is a redox reaction.

Connecting reactivity with redox

A more reactive metal has a greater tendency to form positive ions. In a metal–salt displacement reaction, its atoms lose electrons and are oxidised. Ions of the less reactive metal accept those electrons and are reduced to atoms:

• more reactive metal atoms $\rightarrow$ positive ions: oxidation;
• less reactive metal ions $\rightarrow$ metal atoms: reduction.

For example, zinc displaces lead from lead(II) nitrate because zinc is more reactive than lead. The reacting-particle equation is:

$\mathrm{Zn(s) + Pb^{2+}(aq) \rightarrow Zn^{2+}(aq) + Pb(s)}$

Zinc loses two electrons, while lead(II) ions gain those two electrons. Nitrate ions remain unchanged and therefore do not appear in the ionic equation.

The reverse pairing does not react: copper cannot displace lead from lead(II) nitrate because copper is less reactive than lead. Without a displacement, there is no corresponding electron transfer between those substances.

A reliable explanation method

To explain any metal–salt displacement as redox:

1. Identify the added metal and the metal ion in solution.
2. Decide whether the added metal is more reactive. If it is not, there is no displacement.
3. Remove unchanged spectator ions to obtain the ionic equation.
4. Write a half-equation for the added metal losing electrons.
5. Write a half-equation for the displaced metal ion gaining electrons.
6. State explicitly that loss is oxidation, gain is reduction, and both occur together—therefore the reaction is redox.

Always balance charge as well as atoms. In the zinc example, two electrons are needed because both $\mathrm{Zn^{2+}}$ and $\mathrm{Cu^{2+}}$ carry a charge of magnitude two.