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.
