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# 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.

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Solution

## 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: $\mathrm{Fe}\left(\mathrm{s}\right)-2{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\left(\mathrm{oxidation},\mathrm{acts}\mathrm{as}\mathrm{reducing}\mathrm{agent}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)\left(\mathrm{reduction},\mathrm{acts}\mathrm{as}\mathrm{oxidising}\mathrm{agent}\right)$  Suggest Corrections  2      Similar questions  Explore more