Question

Keep two carbon rods immersed in a copper sulphate solution. Then pass electricity through the solution.

i. At which electrode does colour change occur anode or cathode?

ii. Is there any change in the blue colour of the copper sulphate solution?

iii. Write down chemical equations for the changes occurring here.

Answer 1

(i) The colour change occurs at the cathode. The electrolyte ${\mathrm{CuSO}}_4$ dissociates into ${\mathrm{Cu}}^{2+}\mathrm{and}{\mathrm{SO}}_4^{2-}.$ So at the cathode${\mathrm{Cu}}^{2+}$, converts into Cu which is deposited on the cathode and the colour will change at the cathode.
${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2e^{–}\to \mathrm{Cu}\left(s\right)$

(ii) Yes, the intensity of the blue colour of ${\mathrm{CuSO}}_4$ will decrease as the copper ions are converted to the copper deposit on the cathode.

(iii) Reaction at the cathode: ${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2e^{–}\to \mathrm{Cu}\left(s\right)$

Reaction at the anode: The negative sulphate ions (${\mathrm{SO}}_4^{2-}$) and the hydroxide ions (${\mathrm{OH}}^{–}$) are attracted to the positive electrode. But the sulphate ion is too stable and nothing happens. So either hydroxide ions or water molecules are oxidised to form oxygen.
$4{\mathrm{OH}}^{–}\left(\mathrm{aq}\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)+4e^{–}\mathrm{Or}2H_{2}O\to O_2\left(g\right)+4H+\left(\mathrm{aq}\right)+2e^{–}$