Which one of the following statements is correct for electrolysis of brine solution?

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Given: Electrolysis of brine solution.

Find: Which statement is correct for the products formed at the electrodes.

In electrolysis of brine, chloride ions are oxidized at the anode and water is reduced at the cathode.

\text{NaCl(aq.)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)}

At anode:

2\text{Cl}^-\text{(aq.)} \rightarrow \text{Cl}_2\text{(g)} + 2e^-

A minor competing reaction can produce oxygen:

2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4e^-

At cathode:

2\text{H}_2\text{O(l)} + 2e^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}

Then:

2\text{Na}^+ + 2\text{OH}^- \rightarrow 2\text{NaOH}

Therefore, $OH^-$ is formed at the cathode. The solution states this conclusion, although it also incorrectly labels the option as A. Since the statement " $OH^-$ is formed at cathode" corresponds to option D, the correct option is D.

Given: Brine is aqueous sodium chloride solution.

Find: Which species is formed at the cathode.

At the anode, oxidation occurs, so chloride ions lose electrons to form chlorine gas.

At the cathode, reduction occurs, so water gains electrons and forms hydrogen gas and hydroxide ions.

Hence, chlorine is not formed at the cathode, oxygen is not the cathode product here, and hydrogen is not formed at the anode. The correct statement is that $OH^-$ is formed at the cathode.

Therefore, the correct option is D.

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