Displacement Reactions Gcse Chemistry Coursework

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The Reactivity Series

MetalDisplacement Reactions.

A metal will displace (take the place of) a lessreactive metal
in a metalsalt solution.
This is similar to the non-metal displacement reactions
seen on page 23 of  the periodic table.

For example,
iron    +    copper(II) sulfate      ironsulfate  + copper.
Fe(s)      +      CuSO4(aq)            FeSO4(aq)  +   Cu(s)

Copper(II) sulfate is blue, ironsulfate is colourless.
During the reaction the blue solution loses its colour
and the iron metal is seen to turn pink-brown
as the displacedcopper becomes deposited on it.

If a less reactivemetal is added to a metalsalt solution
there will be noreaction - nothing will happen!
For example, iron is less reactive than magnesium.
iron + magnesiumsulfate  noreaction.

In these displacement reactions
the metals are competing for the non-metal anion.
In the above examples the non-metal anion is sulfate - SO42-.
Reactions using chlorides or nitrates can also be used.
The order of the metals in the reactivity series
can be worked out by using these type of reactions.

For example,
tin would be seen to displacelead from leadchloride
but would notreact with iron(II) chloride.

tin   +   leadchloride        tinchloride  +  lead.
Sn(s)   +   PbCl2(aq)            SnCl2(aq)   +   Pb(s)

tin  +  iron(II) chloride        noreaction.
Sn(s)    +     FeCl2(aq)                           
Therefore tin must be above lead but belowiron
in the reactivity series.

Copper would be seen to displacesilver from silvernitrate
but would notreact with leadnitrate.

copper  +  silvernitrate         coppernitrate  +  silver.
Cu(s)   +   2AgNO3(aq)          Cu(NO3)2(aq)   +   2Ag(s)

copper  +  leadnitrate          noreaction.
Cu(s)    +    Pb(NO3)2(aq)                       
Thereforecoppermust be above silver but belowlead
in the reactivity series.

These reactions can be written as ionicequations.

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The reactivity of the halogens – the Group 7 elements - decreases as you move down the group. This can be shown by looking at displacement reactions.


When chlorine (as a gas or dissolved in water) is added to sodium bromide solution, the chlorine takes the place of the bromine. Because chlorine is more reactive than bromine, it displaces bromine from sodium bromide.

The solution turns brown. This brown colour is the displaced bromine. The chlorine has gone to form sodium chloride.

In this equation, the Cl and Br have swapped places:

chlorine + sodium bromide → sodium chloride + bromine

Cl2(aq) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq)

This type of reaction happens with all the halogens. A more reactive halogen displaces a less reactive halogen from a solution of one of its salts.

Reactivity series

If you test different combinations of the halogens and their salts, you can work out a reactivity series for Group 7:

  • the most reactive halogen displaces all of the other halogens from solutions of their salts, and is itself displaced by none of the others
  • the least reactive halogen displaces none of the others, and is itself displaced by all of the others

It doesn’t matter whether you use sodium salts or potassium salts – it works the same for both types.

The slideshow shows what happens when chlorine, bromine and iodine are added to various halogen salts:

Adding chlorine, bromine and iodine to halogen salts

Chlorine water is added to three solutions

The result of adding chlorine to the three solutions

Bromine water is added to three solutions

The result of adding bromine to the three solutions

Iodine water is added to three solutions

The result of adding iodine to the three solutions

Redox reactions involve both oxidation (loss of electrons) and reduction (gain of electrons). You could remember it as: OIL RIG – Oxidation Is Loss of electrons, Reduction Is Gain of electrons.

Halogen displacement reactions are redox reactions because the halogens gain electrons and the halide ions lose electrons.

When we consider one of the displacement reactions, we can see which element is being oxidised and which is being reduced.

bromine + potassium iodide → iodine + potassium bromide

Br2 + 2KI → I2 + 2KBr

As an ionic equation (ignoring the ‘spectator’ potassium ions):

Br2 + 2I- → I2 + 2Br-

We can see that the bromine has gained electrons, so it has been reduced. The iodide ions have lost electrons, so they have been oxidised.

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