The standard electrode potential for the reduction of oxygen to water is: \[ E^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.23 \, \text{V} \] While for the reduction of chloride ions to chlorine gas: \[ E^\circ_{\text{Cl}^-/\text{Cl}_2} = +1.36 \, \text{V} \]
Although the standard electrode potential for chloride ions is slightly more positive than that for oxygen, this does not automatically mean that oxygen will be reduced to water at the anode in the electrolysis of aqueous NaCl.
The reason why chlorine gas (\( \text{Cl}_2 \)) is liberated at the anode during electrolysis is due to the **concentration of ions** in the solution. Chloride ions (\( \text{Cl}^- \)) are present in **much higher concentrations** compared to oxygen in the solution.
The **concentration effect** plays a significant role in determining which ion is more readily oxidized. Even though oxygen has a more positive reduction potential, the abundance of chloride ions ensures that they are more readily oxidized to form chlorine gas at the anode.
During the electrolysis of aqueous NaCl, the **anode** is the site of oxidation, where electrons are released. At the anode, chloride ions (\( \text{Cl}^- \)) are oxidized to chlorine gas (\( \text{Cl}_2 \)): \[ 2\text{Cl}^- (aq) \rightarrow \text{Cl}_2 (g) + 2e^- \] At the **cathode**, water is reduced to form hydroxide ions (\( \text{OH}^- \)) and hydrogen gas (\( \text{H}_2 \)): \[ 2\text{H}_2\text{O} (l) + 2e^- \rightarrow \text{H}_2 (g) + 2\text{OH}^- \]
Despite the standard electrode potential being more positive for oxygen reduction than for chloride ion reduction, chlorine is preferentially liberated at the anode during the electrolysis of aqueous NaCl. This is because the concentration of chloride ions in the solution is much higher, making them more readily available for oxidation.
Conductivity of CH\(_3\)COOH decreases on dilution.
(a) State the following:
(i) Kohlrausch law of independent migration of ions