IGCSE Chemistry 0620: Electrolysis of Copper(II) Sulfate with Copper and Inert Electrodes

A metal recycling company purifies impure copper recovered from scrap electrical cable by electrolysis. A large block of impure copper and a thin strip of pure copper are placed in a solution of copper(II) sulfate and connected to a direct current supply.

(a) State which electrode, the impure copper block or the pure copper strip, is connected as the anode. Describe what happens to the mass of each electrode as electrolysis proceeds. [2]

(b) Write ionic half-equations, including state symbols, for the reaction at each electrode.

(i) the anode [1]

(ii) the cathode [1]

(c) Describe, in terms of the movement of electrons in the external circuit and the movement of ions in the electrolyte, how charge is transferred around the complete circuit during this electrolysis. [2]

(d) The company then tests a fresh sample of the same copper(II) sulfate solution, but using two inert graphite electrodes instead of copper electrodes.

(i) State the products formed at each electrode in this case. [2]

(ii) Describe one visible change to the colour of the solution as electrolysis proceeds. [1]

Part (a): Identifying the anode and mass changes

The impure copper block is connected as the anode (positive electrode), and the pure copper strip is the cathode (negative electrode).

As current flows, copper atoms at the anode lose electrons and dissolve into solution as $\text{Cu}^{2+}$ ions, so the anode loses mass. At the same time, $\text{Cu}^{2+}$ ions from the solution gain electrons at the cathode and are deposited as solid copper, so the cathode gains mass. This is exactly how impure copper is refined into pure copper.

Part (b): Ionic half-equations

(i) At the anode (oxidation. Copper atoms lose electrons): $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$

(ii) At the cathode (reduction. Copper ions gain electrons): $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Part (c): Charge transfer around the circuit

In the external circuit (the wires and power supply), charge is carried by the flow of electrons, moving from the anode, through the circuit, to the cathode.

Inside the electrolyte, there are no free electrons; instead, charge is carried by the movement of ions. Positive $\text{Cu}^{2+}$ ions move through the solution towards the cathode, while negative sulfate ions move towards the anode. Together, electron flow in the wires and ion movement in the solution form one complete circuit.

Part (d): Using inert graphite electrodes instead

(i) With inert electrodes, the anode itself cannot dissolve. Copper is still deposited at the cathode (as before), but at the anode, hydroxide ions/water are discharged instead, releasing oxygen gas, rather than copper dissolving.

(ii) Because the graphite anode no longer replaces the $\text{Cu}^{2+}$ ions removed at the cathode, the concentration of $\text{Cu}^{2+}$ ions in solution steadily falls. As a result, the blue colour of the solution fades as electrolysis continues.

Final answers

(a) Impure copper block = anode (loses mass); pure copper strip = cathode (gains mass).
(b)(i) $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$
(b)(ii) $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$
(c) Electrons flow through the external wires; Cu2+ ions and sulfate ions carry charge through the electrolyte.
(d)(i) Copper still forms at the cathode; oxygen forms at the anode.
(d)(ii) The blue colour fades.