Reactivity series of metals — predict which metal displaces which

Question

Using the reactivity series of metals, predict whether the following reactions will occur:

Iron nail placed in copper sulphate solution
Copper wire placed in iron sulphate solution
Zinc placed in magnesium sulphate solution

Explain your reasoning.

Solution — Step by Step

The reactivity series (also called the activity series) ranks metals from most to least reactive:

K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Au > Pt

A metal higher in the series can displace a metal lower in the series from its salt solution. Think of it as a competition — the more reactive metal “wins” and replaces the less reactive one.

From the series: Fe is above Cu (Fe is more reactive than Cu).

Therefore, iron will displace copper from copper sulphate solution:

\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}

Observable change: The blue colour of $\text{CuSO}_4$ solution fades (as $\text{Cu}^{2+}$ is consumed). A reddish-brown deposit of copper forms on the iron nail. The solution turns pale green (from $\text{FeSO}_4$ ).

Reaction occurs. ✓

From the series: Cu is below Fe (Cu is less reactive than Fe).

A less reactive metal cannot displace a more reactive metal. Copper cannot displace iron from iron sulphate:

\text{Cu} + \text{FeSO}_4 \rightarrow \text{No reaction}

No observable change — the copper wire remains unchanged, the iron sulphate solution stays the same colour.

Reaction does not occur. ✗

From the series: Mg is above Zn (Mg is more reactive than Zn).

Zinc is less reactive than magnesium, so zinc cannot displace magnesium from its salt solution:

\text{Zn} + \text{MgSO}_4 \rightarrow \text{No reaction}

Reaction does not occur. ✗

Why This Works

The reactivity series reflects how easily metals lose electrons (get oxidized). A more reactive metal has a stronger tendency to give up electrons and become a positive ion. When a more reactive metal enters a solution containing ions of a less reactive metal, it gives away its electrons to those ions — displacing them from solution and depositing them as the metal.

This is an oxidation-reduction (redox) reaction:

Iron gets oxidized: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$
Copper(II) ion gets reduced: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

The driving force is the difference in reduction potentials. Metals higher in the reactivity series have more negative standard electrode potentials — they are stronger reducing agents.

Alternative Method

You can also predict displacement using standard electrode potentials ( $E^\circ$ ):

Lower $E^\circ$ (more negative) = more reactive = stronger reducing agent
$E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}$ ; $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$

Since Zn has a lower (more negative) $E^\circ$ than Cu, Zn can reduce $\text{Cu}^{2+}$ — confirming Fe can displace Cu. This method is used in Class 12 electrochemistry but gives the same predictions as the reactivity series.

For CBSE Class 10 exams, you need to memorize the first 10-12 metals in the reactivity series in order. A classic mnemonic: “Please Stop Calling Me A Zany Iron Knight, Come Home, Sir Gold, Platinum” → K, Na, Ca, Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Hg, Ag, Au, Pt. The position of hydrogen is important — metals above H react with dilute acids; metals below H (Cu, Ag, Au) do not react with dilute $H_2SO_4$ or HCl.

Common Mistake

Students sometimes think that ANY metal can displace ANY other metal — that displacement reactions always occur. The direction matters: displacement only works downward in the reactivity series. Also, some students confuse the colour changes in these reactions. For the iron + copper sulphate reaction: the solution turns from blue to green (not colourless), because $\text{FeSO}_4$ is light green, not colourless. Getting the colour observation wrong in CBSE practical-based questions costs marks.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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