Question

Given two test tubes, A and B, A is CuSO${}_4$ solution (blue) and B is FeSO${}_4$ solution (green). When few iron nails are dipped in both the solutions, it is observed that the colour of solution A changes to green but the colour of solution B remains the same. This is because:

(More than one option can be correct)

• A. Fe displaces Cu from CuSO${}_4$ solution (test tube A)
• B. Fe does not displace Fe from FeSO${}_4$ solution (test tube B)
• C. Fe is more reactive than Cu
• D. Cu is more reactive than Fe.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Fe displaces Cu from CuSO${}_4$ solution (test tube A)
B Fe does not displace Fe from FeSO${}_4$ solution (test tube B)
C Fe is more reactive than Cu

The blue coloured solution in test tube A is ${\text{CuSO}}_4$ while the green coloured solution in test tube B is ${\text{FeSO}}_4$. When an iron nail is dipped in test tube A, $\text{Fe}$ displaces $\text{Cu}$ from ${\text{CuSO}}_4$ solution as $\text{Fe}$ is more reactive than $\text{Cu}$. This causes the solution to turn green. The reaction occurs as follows:
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

However, when an iron nail is dipped in test tube B containing ${\text{FeSO}}_4$ solution, the colour of the solution remains the same. This is because, $\text{Fe}$ of the nail cannot displace $\text{Fe}$ of ${\text{FeSO}}_4$ solution.