Question

Paheli prepared a blue-coloured solution of Copper sulfate 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?

Answer 1

The changes observed in the two beakers after an hour:

• $\mathrm{Fe}$usually tend to displace the $\mathrm{Cu}$ from the ${\mathrm{CuSO}}_4$​ solution in beaker A, which happens due to the reason that $\mathrm{Fe}$ is more reactive than$\mathrm{Cu}$.
• As a result of this, in beaker A a reddish-brown layer of Copper will get deposited on the iron nail and the blue-colored solution will convert into yellowish green due to the formation of${\mathrm{FeSO}}_4$.
• For this, the reaction can be given as :

$\underset{\left(\mathrm{Iron}\right)}{\mathrm{Fe}\left(\mathrm{s}\right)}+\underset{\left(\mathrm{blue}\right)}{\underset{\left(\mathrm{Copper}\mathrm{sulfate}\right)}{{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)}}\to \underset{\left(\mathrm{reddish}-\mathrm{brown}\right)}{\underset{\left(\mathrm{Ferrous}\mathrm{sulfate}\right)}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}}+\underset{\left(\mathrm{Copper}\right)}{\mathrm{Cu}\left(\mathrm{s}\right)}$

• In beaker B, no reaction will take place as $\mathrm{Cu}$ being less reactive cannot displace $\mathrm{Fe}$ from the ${\mathrm{FeSO}}_4$solution.
• Hence, the given solution will remain yellowish-green as it was before, in color due to the occurrence of no reaction.
• For which the chemical reaction can be given as :

$\underset{\left(\mathrm{Copper}\right)}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\left(\mathrm{yellowish}-\mathrm{green}\right)}{\underset{\left(\mathrm{Ferrous}\mathrm{sulphate}\right)}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}}\to \mathrm{no}\mathrm{reaction}$

Hence, above mentioned are changes that will observe in the two beakers after an hour.