Question

1) A student filled four beakers with solutions. Metals were added to each solution. In which
beaker would a displacement reaction take place. Write equation involved.
Beaker
Solution
Metal
Added
1
Fe
2
CU
ZnSO.
FeSO
CuSO
CuSO
3
Ag
Zn
4​

Answer 1

Answer:

Fe + CuSO4 ------> FeSO4 + Cu (ppt)

Displacement rxn occur

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Answer 2

Answer:

In a second beaker displacement reaction would take place between Cu and FeSO4.

Reaction:

Cu + FeSO4 → CuSO4 + Fe

Concept:

• Oxidation: Increase in the Oxidation number is termed as oxidation reaction.
• Reduction: Decrease in the oxidation number is termed as reduction reaction.
• Electric Potential energy :    A energy needed to move a charge against the electric field.
• The main condition for displacement reaction to occur is that the metal which is in the electrolyte solution form should have electric potential energy value compare to the  free metal.
• We can take the electric potential energy values from electrochemical series.
• So when a free metal mixed with the metal having higher electric potential energy value , i.e., more reactive metal atom push it's electrons to the less reactive metal atom and get reduced.

Explanation:

• Here, in this case, Fe is having higher electric potential energy than the Cu an placed above the Cu in the electrochemical series.

• Fe acts as a oxidizing agent and Cu displaces the Fe atom from FeSO4 and convert it to CuSO4 and itself get reduced to Fe.

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