Question

Reaction of zinc with copper sulphate solution is:
A. $Zn\quad +\quad 2CuSO_{ 4 }\quad \rightarrow \quad ZnSO_{ 4 }\quad +\quad 2Cu$
B. $2Zn\quad +\quad CuSO_{ 4 }\quad \rightarrow \quad 2ZnSO_{ 4 }\quad +\quad Cu$
C. $Zn\quad +\quad CuSO_{ 4 }\quad \rightarrow \quad ZnSO_{ 4 }\quad +\quad Cu$
D. None of these

Answer 1

Hint: In the reaction of zinc with copper sulphate, zinc will replace copper from copper sulphate solution. Now you can choose the correct answer only if you know how to balance this reaction.

Complete answer:
On adding zinc to copper sulphate solution, zinc displaces copper from copper sulphate and forms zinc sulphate solution. It will undergo a redox reaction, called the displacement reaction of metal because zinc is more reactive than copper according to the reactivity series.
But firstly it depends on what is the oxidation state of the copper since copper is a transition metal and has varying oxidation states. There is copper (I) and copper (II) and even copper (III). Also, the copper sulphate has to be in aqueous state, dissolved in water in order for the reaction to occur.
So, we are talking about copper(II) sulphate, the full chemical equation would be as follows,
$Zn(s)\quad +\quad CuSO_{ 4 }(aq)\quad \rightarrow \quad ZnSO_{ 4 }(aq)\quad +\quad Cu(s)$
The Copper form will be the reddish brown deposit and the copper (II) sulphate solution will become less blue (paler) $CuSO_{ 4 }$ solution has a blue colour while ZnSO_{ 4 } solution is colourless.
The Ionic Equation is -
$Zn(s)\quad +\quad Cu^{ 2+ }(aq)\quad \rightarrow \quad Zn^{ 2+ }(aq)\quad +\quad Cu(s)$

Therefore, the correct answer for this question is option C.

Note: Any metal, which has low standard reduction potential values,can displace other metal having relatively large standard reduction potential value from its salt.
Standard reduction potential of zinc is lower than that of copper. Hence zinc can displace copper from copper sulphate. Thus zinc sulphate is obtained and copper gets deposited, hence the blue colour of $CuSO_{ 4 }$ disappears.