# Why is chloride oxidised instead of water in copper chloride electrolysis?

For the Electrolysis of Copper Chloride:

Cathode: $\ce{Cu^{2+} + 2e- <=> Cu}$
Anode: $\ce{2Cl <=> Cl2 + 2e- }$

I am confused about the reaction taking place at the anode. Wouldn't $\ce{H2O}$ rather undergo oxidation since it has an increased reducing ability? eg. It is above $\ce{2Cl-}$ on the Table of Standard Reduction Potentials.

Am I missing something? Or do all halogens discharge more readily than water?

The standard redox potential of chlorine is $E^\circ = +1.358\ \mathrm{V}$; the actual potential depends on the concentration of $\ce{Cl-}$:

$$\ce{Cl2 + 2e- <=> 2 Cl- }\quad\quad E^\circ = +1.358\ \mathrm{V}$$ \begin{aligned} E&=E^\circ+\frac{RT}{2F}\cdot\ln\frac{1}{\left[\ce{Cl-}\right]^2}\\ &=1.358\ \mathrm{V}-0.05916\ \mathrm{V}\times\log{\left[\ce{Cl-}\right]} \end{aligned}

The standard redox potential of oxygen is $E^\circ = +1.229\ \mathrm{V}$; the actual potential depends on $\mathrm{pH}$:

$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad\quad E^\circ = +1.229\ \mathrm{V}$$ \begin{aligned} E&=E^\circ+\frac{RT}{4F}\cdot\ln\left[\ce{H+}\right]^4\\ &= 1.229\ \mathrm{V}-0.05916\ \mathrm{V}\times\mathrm{pH} \end{aligned}

The redox potentials $(E_{\ce{O2}} < E_{\ce{Cl2}})$ suggest that oxidation of $\ce{H2O}$ to $\ce{O2}$ at the anode should be preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$.

However, depending on the material and shape of the anode, the overpotential of oxygen at the anode can be very large. Therefore, the oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ at the anode is feasible.

Nevertheless, at high $\mathrm{pH}$ and at a low concentration of $\ce{Cl-}$, the oxidation of $\ce{H2O}$ to $\ce{O2}$ is still preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$. Hence, if also $\ce{H2}$ and $\ce{OH-}$ are produced during the electrolysis of aqueous solutions (this is less relevant for the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ which is considered in the question, but for example important for the chloralkali process) $$\ce{2H2O + 2e- -> H2 + 2OH-}$$ the electrolysis of chloride cannot be made complete in a single step since $\mathrm{pH}$ is increased and the concentration of $\ce{Cl-}$ is decreased during the electrolysis: $$\ce{2 H2O + 2 Cl- -> H2 + 2 OH- + Cl2}$$

It depends upon the concentration of chloride. If a dilute NaCl solution is electrolysed, $\ce{O2}$ is produced at first, along with $\ce{H2}$. As the volume of the solution decreases $\ce{[NaCl]}$ increases, and eventually $\ce{Cl2}$ is formed. You can use the Nernst equation to find out the chloride concentration at which $\ce{Cl2}$ is formed over $\ce{O2}$.

Yes water has a higher reduction potential. But what is happening at your anode isn't reduction, it's oxidation. So a lower reduction potential for $\ce{Cl-}$means that it's more readily oxidised than water.

For additional information, see my other answer on this recent question on electrolysis and $\ce{Cl2}$ production.

• Really? I thought water is more likely to undergo oxidation than Cl−. My chem teacher cleared me up and said that the halogens will rather oxidise than water (Cl- is an exception in the table due to its position on the Standard Reduction Potentials Table) – Luke Taylor Aug 22 '14 at 17:07
• Water does oxidise some of the time due to its abundance in the cell (it's completely surrounding the electrode, right?) however Cl- preferentially oxidises over water due to having a lower reduction potential (therefore a tendency to oxidise more than water). It's a thermodynamic argument: Cl- oxidation is preferred on thermodynamic grounds, however due to rapid kinetics water will also oxidise. – user7232 Aug 22 '14 at 17:27
• As for "My chem teacher cleared me up and said that the halogens will rather oxidise than water" you need to be careful what you're stating, by halogen do you mean anion of halogen, or diatomic halogens etc.? If you're referring to Cl-, there are no exceptions here. Cl- prefers to oxidise over water. – user7232 Aug 22 '14 at 17:28