Upper secondary · 15 min

Electrolysis of copper(II) chloride

Pass a current through $\ce{CuCl2}$ solution: copper grows on the cathode and chlorine bubbles off the anode.

Goal

Identify the products at each electrode and link the gas volume to the current and time via Faraday’s law.

Apparatus and reagents

Virtual electrolysis cell filled with blue-green $\ce{CuCl2}$ solution, two inert electrodes and a current slider.

Procedure

Switch on the current at $I = 1{,}5$ A and watch where the bubbles appear.
Compare the two electrodes: chlorine gas collects at the anode (oxidation) while a copper layer darkens the cathode (reduction).
Raise the current and note that the bubbling rate increases proportionally.

What to observe

Cathode ( $-$ ): $\ce{Cu^{2+} + 2e^- -> Cu}$ — the electrode gains a reddish copper coating.
Anode ( $+$ ): $\ce{2Cl^- -> Cl2 + 2e^-}$ — bubbles of pale green chlorine.

Explanation

In solution, $\ce{Cu^{2+}}$ is reduced before water because its reduction potential is higher, so copper plates the cathode; $\ce{Cl^-}$ is oxidised to $\ce{Cl2}$ at the anode. The overall reaction is $\ce{CuCl2 ->[electrolysis] Cu + Cl2}$ . By Faraday’s first law, the deposited mass is $m = MIt/(nF)$ : doubling $I$ or $t$ doubles the copper produced.

Chemists behind it

Michael Faraday