The gas produced at the cathode during electrolysis of brine is

Analysis of the Question

The question asks for the product formed at the cathode during the electrolysis of brine. Brine is a concentrated aqueous solution of sodium chloride ($\text{NaCl}$).

Concept Explanation

During the electrolysis of brine (using inert electrodes like carbon or platinum), there are two competing ions at each electrode:

1. At the Cathode (Negative Electrode): • The ions present are $\text{Na}^+$ (from $\text{NaCl}$) and $\text{H}^+$ (from the dissociation of water, $\text{H}_2\text{O}$). • According to the electrochemical series, the ion that is lower in the series is discharged more easily. $\text{H}^+$ is lower than $\text{Na}^+$, so hydrogen ions are preferentially discharged. • The reaction is: $2\text{H}^+_{(aq)} + 2e^- \rightarrow \text{H}_{2(g)}$

2. At the Anode (Positive Electrode): • The ions present are $\text{Cl}^-$ and $\text{OH}^-$. • Due to the high concentration of chloride ions in brine, $\text{Cl}^-$ is preferentially discharged over $\text{OH}^-$. • The reaction is: $2\text{Cl}^-_{(aq)} \rightarrow \text{Cl}_{2(g)} + 2e^-$

Step-by-Step Solution

• Step 1: Identify the ions moving to the cathode. These are $\text{Na}^+$ and $\text{H}^+$.
• Step 2: Apply the rule of selective discharge. Hydrogen ions ($\text{H}^+$) have a lower discharge potential than sodium ions ($\text{Na}^+$).
• Step 3: Write the reduction half-equation at the cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$.
• Step 4: Conclude that the gas produced is hydrogen.

Conclusion

The gas produced at the cathode is hydrogen, while the gas produced at the anode is chlorine. Therefore, the correct option is D.

Exam Tip

Remember the mnemonic "PANIC": Positive Anode, Negative Is Cathode. Also, always remember that in the electrolysis of brine, the cathode produces Hydrogen and the anode produces Chlorine—this is a very common JAMB question!