A more reactive metal displaces a less reactive metal from its salt solution, for example Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s); the ionic equation is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
A more reactive metal displaces a less reactive metal from a solution of its salt. The reactive metal goes into solution as ions while the less reactive metal is deposited as the free metal. Because electrons are transferred from one metal to the ion of the other, displacement is a redox reaction, and it is one of the clearest illustrations of the reactivity series in the Redox chapter of 4541.
The balanced equation
The general pattern is: more reactive metal + salt of less reactive metal → salt of more reactive metal + less reactive metal. With correct formulae and state symbols:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
Mg(s) + ZnSO4(aq) → MgSO4(aq) + Zn(s)
The ionic equation removes the spectator sulfate ion and shows the real change, the metal atom gives its electrons to the metal ion:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
The two half-equations show the electron transfer clearly. Zinc is oxidised (it loses electrons) and copper(II) ions are reduced (they gain electrons):
Zn(s) → Zn2+(aq) + 2e−
Cu2+(aq) + 2e− → Cu(s)
Conditions required
No heating or catalyst is needed. A strip or piece of the more reactive metal is simply placed in a solution of a salt of the less reactive metal at room temperature. The metal chosen must be higher in the reactivity series than the metal in the salt, otherwise no reaction occurs, for example, copper placed in zinc sulfate solution does nothing.
What you observe
With zinc in blue copper(II) sulfate solution, the blue colour fades towards colourless as copper(II) ions are used up, a brown or reddish solid (copper) coats the zinc, the zinc strip becomes thinner as it dissolves, and the mixture warms up because the reaction is exothermic. The pattern is the same for any pair: the salt colour changes as one metal ion replaces another, and a deposit of the displaced metal forms on the added metal.
Where it appears in the SPM exam
In 4541/1 you predict whether a displacement will occur from the reactivity series or identify the products. In 4541/2 you write the balanced equation, the ionic equation and the two half-equations, identify the oxidising and reducing agents, and state the change in oxidation number. Displacement also appears in electrochemistry, where the more reactive metal is the negative terminal of a simple cell, and in the practical paper as an experiment to arrange metals in order of reactivity.
How we teach it
Our teachers make sure you can label the redox roles precisely: the more reactive metal is the reducing agent and is oxidised, while the less reactive metal ion is the oxidising agent and is reduced. Students most often lose marks by writing a reaction that cannot happen because the added metal is less reactive, by forgetting the spectator ion in the ionic equation, or by mislabelling oxidation and reduction. Anchoring every answer to the reactivity series and “oxidation is loss of electrons” keeps the whole topic consistent.
Quick summary
A more reactive metal displaces a less reactive metal from its salt solution in a redox reaction. Learn the balanced equation, the ionic equation with the spectator ion removed, the two half-equations, the colour and deposit observations, and the reactivity-series rule, and this dependable redox topic is fully covered.
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