Question

In the process of electrolysis copper II tetraoxosulphate IV solution (aq) using copper electrode, Explain throughly with appproprate equation what happens at both the anode and the cathode

Answer 1

Answer on Question #48134 - Chemistry – Other

In the process of electrolysis copper II tetraoxosulphate IV solution (aq) using copper electrode, Explain thoroughly with appropriate equation what happens at both the anode and the cathode.

Answer

$\mathrm{CuSO_4}$ ionises in aqueous solution as:

$$\mathrm{CuSO_4}(s) \xrightarrow{\text{water}} \mathrm{Cu^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq)$$

Cathode reaction: $\mathrm{Cu^{2+}}$ ions undergo reduction

$$\mathrm{Cu^{2+}}(aq) + 2\mathrm{e^-} \longrightarrow \mathrm{Cu}(s); \quad \mathrm{E^{\circ}} = 0.34\ \mathrm{V}$$

Anode reaction: Cu atoms from the copper electrode get oxidised because Cu lies below H in activity series

$$\begin{array}{c} \mathrm{Cu}(s) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{e^-}; \\ \mathrm{E^{\circ}}_{\text{oxi}} = -0.34\ \mathrm{V} \end{array}$$

Thus, copper dissolves from anode and get deposited at cathode. Mass of cathode increases and that of anode decreases. Blue colour of the solution does not fade because concentration of $\mathrm{Cu^{2+}}$ ions in solution remains unaltered. The pH of the solution also does not change as the electrolysis proceeds.

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