Question

The following reactions occur at the anode during the electrolysis of aqueous sodium chloride solution: \[ Cl^-_{(aq)} \rightarrow \frac{1}{2}Cl_2(g)+e^- \qquad E^\circ_{\text{cell}}=1.36\,V \] \[ 2H_2O(l) \rightarrow O_2(g)+4H^+(aq)+4e^- \qquad E^\circ_{\text{cell}}=1.23\,V \] Which reaction is feasible at the anode and why?

Hint:

In electrolysis of aqueous NaCl (brine): \[ \boxed{\text{Cathode: } H_2 \text{ gas is produced}} \] \[ \boxed{\text{Anode: } Cl_2 \text{ gas is produced}} \] Although water can give oxygen, chloride ions are preferentially oxidised due to: \[ \boxed{\text{High } Cl^- \text{ concentration + oxygen overvoltage}} \]

Answer 1

Step 1: Competing reactions at the anode.
Two oxidation reactions compete: $Cl^-_{(aq)} \rightarrow \frac{1}{2}Cl_2(g) + e^-$ ($E^\circ = 1.36\,V$) and $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$ ($E^\circ = 1.23\,V$).
Step 2: Thermodynamic prediction vs reality.
Lower $E^\circ$ at the anode means easier oxidation, so water oxidation ($1.23\,V$) appears thermodynamically favoured over $Cl^-$ oxidation ($1.36\,V$).
Step 3: Concentration and overvoltage factors.
In concentrated NaCl solution $[Cl^-]$ is very high. Moreover, $O_2$ evolution requires a substantial overvoltage, making water oxidation kinetically disfavoured. These two effects together override the thermodynamic prediction.
Step 4: Feasible reaction.
Chloride ions are preferentially oxidised at the anode. \[ \boxed{Cl^-_{(aq)} \rightarrow \frac{1}{2}Cl_2(g) + e^-} \]