Question

A solution of ${\mathrm{CuSO}}_4$ was kept in an iron pot. After a few days, the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.

Answer 1

• From the reactivity series we know that iron is more reactive than copper.
• As a result when Copper sulphate is kept in the iron pot, the iron displaces the copper in the Copper sulphate solution and forms Iron sulphate.
• Also a portion of the iron is dissolved during this process which results in holes formed in the pot.
• The complete reaction is:

$\underset{\mathrm{Iron}}{\mathrm{Fe}\left(\mathrm{s}\right)}\underset{\mathrm{Copper}\mathrm{sulphate}}{+{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Ferrous}\mathrm{sulphate}}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$

• This reaction is called displacement reaction as the Iron being more reactive than Copper displaces it and forms Ferrous sulphate.
• There is another type of reaction occurring which is a redox reaction(oxidation and reduction happening together).
• Iron loses 2 electrons and thus undergoes oxidation, and Copper gains 2 electrons and undergoes reduction:

$$\begin{gathered} \mathrm{Fe}\left(\mathrm{s}\right)-2{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)(\mathrm{oxidation},\mathrm{acts}\mathrm{as}\mathrm{reducing}\mathrm{agent}) \\ {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)(\mathrm{reduction},\mathrm{acts}\mathrm{as}\mathrm{oxidising}\mathrm{agent}) \end{gathered}$$