A student prepared an aqueous solution of $CuSO_4$ in beaker X and an aqueous solution of $FeSO_4$ in beaker Y. He then dropped some iron pieces in beaker X and some zinc pieces in beaker Y. After about 10 hours he observed that the solutions in X and Y respectively appear:

In Beaker X, iron ($Fe$) is added to a blue copper sulphate ($CuSO_4$) solution. Since iron is more reactive than copper (as per the reactivity series), it displaces copper from its solution to form ferrous sulphate ($FeSO_4$), which is greenish in colour:

$$Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s)$$

Thus, the blue solution in beaker X changes to greenish.

In Beaker Y, zinc ($Zn$) is added to a pale green ferrous sulphate ($FeSO_4$) solution. Since zinc is more reactive than iron, it displaces iron from its solution to form zinc sulphate ($ZnSO_4$), which is a colourless solution:

$$Zn(s) + FeSO_4(aq) \rightarrow ZnSO_4(aq) + Fe(s)$$

Thus, the pale green solution in beaker Y changes to colourless.

Therefore, after 10 hours, the solutions in X and Y appear greenish and colourless respectively.

Distractor Analysis:

• Option A is incorrect because it suggests no reaction or incorrect products (blue and green).
• Option B is incorrect because it reverses the observations (colourless for X and pale green for Y).
• Option C is incorrect because it suggests X becomes colourless and Y becomes light blue, which is chemically incorrect.