In the electrolysis of brine using neutral electrode, which ion is discharged at the anode?

Analysis of the Question

The question asks for the ion discharged at the anode during the electrolysis of brine (concentrated sodium chloride solution) using inert (neutral) electrodes.

Concept Explanation

In the electrolysis of brine, we have the following ions present in the solution:

• From NaCl: $Na^+$ and $Cl^-$
• From $H_2O$: $H^+$ and $OH^-$

The Anode (Positive Electrode): The anions ($Cl^-$ and $OH^-$) are attracted to the anode. According to the electrochemical series, the discharge of ions depends on their concentration and position in the series.

• In a concentrated solution of sodium chloride (brine), the concentration of $Cl^-$ ions is significantly higher than that of $OH^-$ ions.
• Because of this high concentration, the $Cl^-$ ions are preferentially discharged at the anode instead of the $OH^-$ ions.

The Reaction at the Anode: $2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$

Step-by-Step Solution

1. Identify the ions at the anode: The anions present are $Cl^-$ and $OH^-$.
2. Apply the rule of selective discharge: Although $OH^-$ is lower in the electrochemical series and would normally be discharged in dilute solutions, the high concentration of $Cl^-$ in brine forces the chloride ions to be discharged.
3. Determine the product: The $Cl^-$ ions lose electrons (oxidation) to form chlorine gas ($Cl_2$).
4. Conclusion: The ion that is discharged at the anode is the chloride ion ($Cl^-$).

Exam Tip

Remember the mnemonic "Concentration Conquers" for the electrolysis of brine. Even though $OH^-$ is theoretically easier to discharge, the sheer number of $Cl^-$ ions in concentrated brine makes them the winners at the anode!