Question

10. Two beakers A and B contain Iron (II) sulphate solution. In the benker A is placed a
small piece of copper and in the beaker B is placed a small piece of zinc. It is found
that a grey deposit forms on the zinc but not on the copper. What can be concluded
from these observations? (3).​

Answer 1

Required Answer:

Short Answer:

→ In beaker B in which zinc was placed, due to displacement reaction grey deposit forms on the zinc when compound of zinc in solution displaced iron . Whereas, in beaker A in which copper was placed, displacement reaction doesn't occur as copper can't displace iron because it is less reactive than iron.

Detailed Answer [Explanation] :

As per the given information :

• Both A and B beakers contain Iron sulphate. $\sf { (Fe{SO}_{4}) }$
• Small piece of copper is placed in the Beaker A.
• Small piece of zinc is placed in the Beaker B.
• In beaker B , green deposits forms on the zinc but not on the copper.

_______________________________

• In beaker A : Iron Sulphate + Copper

• In beaker B : Iron Sulphate + Zinc

________________________________

Reaction :

Here, displacement reaction will occur in which a more reactive metal displaces a less reactive metal from its compound in solution.

★ $\boxed {\sf{ Metal \: A + Salt \: Solution \: of \: B \longrightarrow Salt \: Solution \: of \: A + Metal \: B }}$

In Beaker A :

$\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} \longrightarrow \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} }$

As copper is less reactive than iron. So, it can't displace iron from its compound in solution. Reactants and product will be same.

In Beaker B:

$\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Zn}_{Zinc} \longrightarrow \underbrace{ZnSO_4}_{Zinc \: Sulphate} + \underbrace{Fe}_{Iron} }$

Here, zinc is more reactive metal than iron. So, when zinc (Metal A) reacts with Iron sulphate (Salt solution of B) , zinc displaced iron from its compound in solution and forms zinc sulphate (Salt solution of A) and Iron (Metal B).There will be a grey deposit forms on the zinc.

______________________________________

Thus, it can be concluded from these observations that

• Zinc is more reactive than copper, that's why it displaced iron and a grey deposit forms on the zinc or it became zinc sulphate.

• Whereas, in the case of copper , copper couldn't displace iron as it is less reactive than the iron. That's why a grey deposit doesn't form on the copper.

Answer 2

Answer:

beaker B in which zinc was placed, due to displacement reaction grey deposit forms on the zinc when compound of zinc in solution displaced iron . Whereas, in beaker A in which copper was placed, displacement reaction doesn't occur as copper can't displace iron because it is less reactive than iron.

Detailed Answer [Explanation] :

As per the given information :

Both A and B beakers contain Iron sulphate. \sf { (Fe{SO}_{4}) }(FeSO

4

)

Small piece of copper is placed in the Beaker A.

Small piece of zinc is placed in the Beaker B.

In beaker B , green deposits forms on the zinc but not on the copper.

_______________________________

• In beaker A : Iron Sulphate + Copper

• In beaker B : Iron Sulphate + Zinc

________________________________

Reaction :

Here, displacement reaction will occur in which a more reactive metal displaces a less reactive metal from its compound in solution.

★ \boxed {\sf{ Metal \: A + Salt \: Solution \: of \: B \longrightarrow Salt \: Solution \: of \: A + Metal \: B }}

MetalA+SaltSolutionofB⟶SaltSolutionofA+MetalB

In Beaker A :

\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} \longrightarrow \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} }

IronSulphate

FeSO

4

+

Copper

Cu

⟶

IronSulphate

FeSO

4

+

Copper

Cu

As copper is less reactive than iron. So, it can't displace iron from its compound in solution. Reactants and product will be same.

In Beaker B:

\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Zn}_{Zinc} \longrightarrow \underbrace{ZnSO_4}_{Zinc \: Sulphate} + \underbrace{Fe}_{Iron} }

IronSulphate

FeSO

4

+

Zinc

Zn

⟶

ZincSulphate

ZnSO

4

+

Iron

Fe

Here, zinc is more reactive metal than iron. So, when zinc (Metal A) reacts with Iron sulphate (Salt solution of B) , zinc displaced iron from its compound in solution and forms zinc sulphate (Salt solution of A) and Iron (Metal B).There will be a grey deposit forms on the zinc.

______________________________________

Thus, it can be concluded from these observations that

• Zinc is more reactive than copper, that's why it displaced iron and a grey deposit forms on the zinc or it became zinc sulphate.

• Whereas, in the case of copper , copper couldn't displace iron as it is less reactive than the iron. That's why a grey deposit doesn't form on the copper.