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 that are observed in a beaker after an hour:

1. So, actually $\mathrm{Fe}$ usually tend to displace the $\mathrm{Cu}$ from ${\mathrm{CuSO}}_4$ solution in the beaker A, and this happens due to the reason that $\mathrm{Fe}$ is more reactive than $\mathrm{Cu}$.
2. And further as a result , in the beaker A a reddish-brown layer of Copper will get deposited on the iron nail and so the blue-colored solution will convert into the yellowish green due to the formation of ${\mathrm{FeSO}}_4$.
3. Reaction for this would be: $\underset{\mathrm{Iron}}{\mathrm{Fe}\left(\mathrm{s}\right)}+\underset{\underset{(\mathrm{blue}\mathrm{color})}{\mathrm{Copper}\mathrm{sulfate}}}{{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)}\to \underset{\underset{(\mathrm{Blue}-\mathrm{green})}{\mathrm{Ferrous}\mathrm{sulfate}}}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$.
4. Basically, in beaker no reaction will take place as $\mathrm{Cu}$ is less reactive and can not displace $\mathrm{Fe}$ from the ${\mathrm{FeSO}}_4$ solution.
5. And so , the given solution will remain in the yellowish-green color as it was before, as no reaction took place.
6. And for this chemical reaction would be: $\underset{\left(\mathrm{Copper}\right)}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\underset{(\mathrm{Blue}-\mathrm{green})}{\mathrm{Ferrous}\mathrm{sulfate}}}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}\to \mathrm{no}\mathrm{reaction}$

Therefore, the above mentioned are the changes which will be observed in the two beakers after an hour.