Question

A strip of copper is placed in four different colorless solutions. They are: $KN{O_3},AgN{O_3},Zn{(N{O_3})_2},Ca{(N{O_3})_2}$
Which of the following solutions will finally turn blue?
A) $Ca{(N{O_3})_2}$
B) $KN{O_3}$
C) $AgN{O_3}$
D) $Zn{(N{O_3})_2}$

Answer 1

Hint: To turn the solution blue we need to see that copper will react with which of the following solutions and we see that by determining the position of copper in the activity series. The reactivity series of metals also known as the activity series, refers to the arrangement of metals in the decreasing order of their reactivity.

Complete step-by-step answer:
So, the reaction will take place between copper and $AgN{O_3}$ solution as we know that copper is more reactive than silver and will easily displace silver in the silver nitrate solution and will eventually form copper nitrate solution and the solution will therefore turn blue in color. The blue color develops due to the presence of copper ions.

The reaction proceeds as:
$Cu + 2AgN{O_3} \to 2Ag + Cu{(N{O_3})_2}$

This reaction is known as the single displacement reaction. A single displacement reaction is a chemical reaction in which one element is replaced by another element. Copper is unable to displace potassium, calcium and zinc from their nitrate solution as they are present at higher positions in the reactivity series as compared to copper.

Therefore, the correct answer is option C.

Note: This reaction can also be explained on the basis of electrochemical series. The metal with more standard potential displaces metal with less standard potential. Here the electrode potential of copper is $ + 0.337\nu $ and that of silver is $ + 0.222\nu $ therefore copper will displace silver from silver nitrate solution.