Question

A student was given two metal strips X and Y alongwith colourless silver nitrate solution to perform two separate displacement reactions. When the student placed metal strip X in silver nitrate solution for a considerable time, he observed that the solution turned blue and a coating of silver metal was formed on the strip. And when the student immersed metal strip Y in silver nitrate solution for an equal time, he observed that the solution turned light green with the formation of a coating of silver metal on the strip. The correct conclusion of the student about the identity of metals X and Y is :
(1) X is copper and Y is magnesium
(2) X is zinc and Y is copper
(3) X is iron and Y is copper
(4) X is copper and Y is iron

Answer 1

For X, when silver nitrate is treated with X it gives the blue colour solution which means that that solution may contain copper. Copper is more reactive than silver, it will displace silver from the silver nitrate and form blue colour compound copper nitrate.
$\mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)\to \mathrm{Cu}\left({\mathrm{NO}}_3{)}_2\right(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})$

For Y, when silver nitrate is treated with Y it gives the light green colour solution which means that that solution may contain iron. Iron is more reactive than silver, it will displace silver from the silver nitrate and form green colour compound iron nitrate.
$\mathrm{Fe}\left(\mathrm{s}\right)+{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)\to {\mathrm{FeNO}}_3\left(\mathrm{aq}\right)+\mathrm{Ag}\left(\mathrm{s}\right)$

Hence, the correct answer is option 4.