Prove that the reaction below is a displacement reaction: Pb_(s) + CuSO4(aq) → PbSO4(aq) + Cu_(s)

This is a displacement reaction because lead ($\mathrm{Pb}$) displaces copper ($\mathrm{Cu}$) from copper sulphate solution. The more reactive metal (lead) displaces the less reactive metal (copper).

The reaction is:

$$\mathrm{Pb}_{(s)} + \mathrm{CuSO}_{4(aq)} \longrightarrow \mathrm{PbSO}_{4(aq)} + \mathrm{Cu}_{(s)}$$