Question

Paheli prepared a blue coloured solution of copper sulphate in beaker A and placed an iron nail in it. Boojho prepared a yellowish green solution of ferrous sulphate in beaker B and placed a copper wire in it. Changes they observe in the two beakers after an hour will be :

• A. in beaker A, no change ; in beaker B, change to green
• B. in beaker A, no change ; in beaker B, no change
• C. in beaker A, changes to green ; in beaker B, no change
• D. in beaker A, changes to green ; in beaker B, change to blue

Answer 1

Answer: (C)

in beaker A, changes to green ; in beaker B, no change

Fe displaces the Cu from $CuS{O}_4$ solution in beaker A as Fe is more reactive than Cu.

Therefore the blue colour of the solution changes to green due to the formation of $FeS{O}_4$.

In beaker B, no reaction takes place as $Cu$ being less reactive cannot displace $Fe$ from $FeS{O}_4$ solution.

Hence the correct option is C.