3.1.4 - Laboratory Activity: Displacement of metals from its salt solution

Laboratory Activity 3.1.4: Displacement of metals from its salt solution

Aim: To understand the displacement of metals from their salt solution.
Problem statement: How does displacement reactions relate to redox reaction?
Hypothesis:

	»	Metallic ions will formed if a metal acts as a reducing agent.
	»	Metal will formed if a metallic ion acts as an oxidising agent.

Variable:

	»	Fixed variable : Concentration of the solution
	»	Manipulated variable : Metals and salt solutions
	»	Responding variable : Deposition and change of colour

Material: » Zinc strips » Copper strips » Copper (II) sulphate » Silver nitrate solution

Apparatus: » Test tubes » Sand paper

Procedure:

	►	(A) Zinc and copper (II) sulphate 1. Use a sand paper to cleaned the zinc strip. 2. The clean zinc strip is dipped into a copper (II) sulphate solution. 3. Observe the changes to the copper (II) sulphate solution and the zinc strip.
	►	(B) Copper and silver nitrate solution. 1. Use a sand paper to cleaned the copper rod. 2. The clean copper rod is dipped into a silver nitrate solution. 3. Observe the changes to the silver nitrate solution and the copper rod.

Observation:

	►	(A) Zinc and copper (II) sulphate ○ The copper(II) sulphate solution becomes paler until the become colourless. ○ The zinc strip became thinner and smaller. ○ Orange colour deposits were found on the zinc surface ○ Oxidation and Reduction - Zinc Rod
	►	(B) Copper and silver nitrate solution. ○ The solution changes colour from colourless to blue. ○ The copper rod become thinner and smaller. ○ Grey deposits were found.

Discussion:

	►	(A) Zinc and copper (II) sulphate ○ Half ionic equation: ■ Zn(s) → Zn2+(aq) + 2e− (oxidation → reducing agent: zinc) ■ Cu2+(aq) + 2e− → Cu(s) (reduction → oxidising agent: copper (II) ions, Cu2+) ○ Overall equation: ■ Zn(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) ○ Zinc strips became thinner and smaller as it dissolved in the copper (II) sulphate solution. Zinc is oxidised to zinc ions. Copper ions was reduced to copper forming an orange deposits.
	►	○ Half ionic equation: ■ Cu(s) → Cu2+(aq) + 2e− (oxidation → reducing agent: copper) ■ Ag+(aq) + e− → Ag(s) (reduction → oxidising agent: silver ions ions, Ag+) ○ Overall equation: ■ Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s) ○ Copper rod becomes thinner and smaller as it dissolved in the silver nitrate solution. The copper was oxidised to copper ions. Silver ions was reduced to silver, forming a grey deposit.

Conclusion:

	►	In a displacement reaction, the more electropositive metal will displace the less electropositive metal.
	►	The ion of the less electropositive metal will be reduced, acting as the oxidising agent.
	►	The more electropositive metal will be oxidised, acting as a reducing agent.

⇲ For exercise(objective and subjective), download for free on Android OS.