Question

Beakers A, B and C contain zinc sulphate, silver nitrate and iron (II) sulphate solutions respectively. Copper pieces are added to each beaker. Blue colour will appear in case of
A. beaker A
B. beaker B
C. beaker C
D. all the beakers.

Answer 1

Answer:

B. Beaker B

In the case of Beaker B, blue colour will appear.

Explanation:

Reactivity series is the basis of this question.

The following is the Reactivity series :

K > Na > Ca > Mg > Al > Zn > Fe >

Li > H > Cu > Hg > Ag > Au > Pt

From the above Reactivity series, we can clearly see that

Zinc (Zn) is more reactive than

Iron (Fe) , Iron (Fe) is more reactive than Copper (Cu) and Copper is more reactive than Silver (Ag).

So,

Zn > Fe > Cu > Ag

According to single displacement reaction, a more reactive metal displaces a less reactive metal from its compound or salt solution.

Also, we know that sulphate or nitrate solution formed by Copper is blue in colour.

So here we need to find among which solutions given, does copper replace the constituent metal.

Thus from the reference of the reactivity series we get,

FOR BEAKER A :

Here the solution is of Zinc

sulphate (ZnSO4.7H2O) .

Here constituent metal is zinc and zinc is more reactive than copper.

So here if copper is added to the zinc sulphate solution, displacement reaction will not take place because copper cannot displace zinc from its solution .

Thus no change will take place in the solution because no reaction will take place .

Since the initial colour of the zinc sulphate solution is colourless, so after adding copper the final colour will also remain colourless because no change has taken place.

Thus here the colour of solution will not change to blue.

FOR BEAKER B :

Here the solution is Silver

Nitrate ( AgNO3 ) .

Here constituent metal is Silver and Copper is more reactive than Silver.

So if Copper is added to the silver nitrate solution,then copper will displace silver from its solution. So the displacement reaction will take place.

Thus change in properties of the solution will take place as there happens a chemical reaction.

The initial colour of silver nitrate solution is colourless, but when copper is added,since copper displaces silver from its solution, the colour of the solution will turn blue due to the formation of Copper Nitrate { Cu(NO3)2} .

Cu(s) + AgNO3(aq) = Cu(NO3)2 (aq)

+ Ag (s)

Also colour of Copper Nitrate solution is blue in colour . Thus the final solution turns blue.

FOR BEAKER C :

Here the solution is Iron (II)

Sulphate ( FeSO4.7H2O ) .

Here the constituent metal is Iron and Iron is more reactive than Copper.

So if copper is added to the Iron Sulphate solution , it will not displace Iron from its solution and displacement reaction will not take place.

Thus no change in the solution will take place as no reaction will occur.

Since the initial colour of Iron (II) Sulphate solution is white in colour, so after adding Copper to the solution the colour of final solution will also remain white only because no reaction has take place .

Thus here the colour of solution will not change to blue.

From the trial and error method above, we get that colour of the solution after adding copper changes blue in colour only in Beaker B.

Hence the required answer is

B. Beaker B

Answer 2

Answer:

Hello Dear, this is your answer

Explanation:

Given that,

Beaker,

A=ZnSO

4

B=AgNO

3

C=FeSO

4

When copper pieces added in each beaker, only less reactive solution will be displaced by Cu.

According to reactivity series Cu is more reactive than silver and less reactive than Zn and Fe.

2Cu+

Solution

Beaker(B)

2AgNO

3

→

Blue colour

Cu(NO

3 )

2

+2Ag