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. What changes will they observe in the two beakers after an hour?

AcademicChemistryNCERTClass 8

In beaker A, a reddish-brown layer of copper will deposit on the iron nail and blue coloured solution of copper sulphate will become yellowish green due to the formation of iron sulphate. This is because iron is more reactive than copper and it displaces copper from its salt solution (copper sulphate solution).

In beaker B, no change is observed because copper being less reactive does not displace iron (more reactive) from ferrous sulphate solution.

Updated on 10-Oct-2022 13:24:25