Question

Before the development of the Downs cell, the Castner cell was used for the industrial production of Na metal. The Castner cell was based on the electrolysis of molten \(\mathrm{NaOH}\). (a) Write balanced cathode and anode half-reactions for this cell. (b) A major problem with this cell was that the water produced at one electrode diffused to the other and reacted with the \(\mathrm{Na}\). If all the water produced reacted with \(\mathrm{Na}\), what would be the maximum efficiency of the Castner cell expressed as moles of Na produced per mole of electrons flowing through the cell?

Short answer

The cathode reaction is \( \text{Na}^+ + e^- \rightarrow \text{Na} \), and the anode reaction is \[ 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^- \]. Efficiency is 0% since no net Na is produced.

Step-by-step solution

Determine the cathode half-reaction

At the cathode, \( \text{Na}^+ \) ions are reduced to sodium metal. The reduction half-reaction for sodium is \[ \text{Na}^+ + e^- \rightarrow \text{Na} \]

Determine the anode half-reaction

At the anode, hydroxide ions \( \text{OH}^- \) are oxidized to produce oxygen gas and water. The oxidation half-reaction is \[ 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^- \]

Write the overall cell reaction

Combine the cathode and anode half-reactions to give the overall balanced chemical reaction taking place in the Castner cell, \[ 4\text{Na}^+ + 4\text{OH}^- \rightarrow 4\text{Na} + 2\text{H}_2\text{O} + \text{O}_2 \]

Assess the maximum efficiency

If all the water produced reacts with the sodium, the side reaction would be \[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \] . Note that 2 moles of \( \text{Na} \) are consumed to regenerate 2 moles of \( \text{NaOH} \). So the efficiency would be zero as there would be no net production of \( \text{Na} \).

Calculate the moles of Na produced per mole of electrons

From the cathode reaction, each mole of \( \text{Na} \) produced uses 1 mole of electrons. If all water produced reacts with Na, it would involve equal moles of Na, making the net production zero. Hence, the efficiency is \( 0 \% \).

Key concepts

Electrolysis

Electrolysis is a process that uses electrical energy to drive a non-spontaneous chemical reaction. It involves passing an electric current through an electrolyte, causing the movement of ions to the respective electrodes where they undergo redox (reduction-oxidation) reactions. In the case of the Castner cell, molten \(\text{NaOH}\) serves as the electrolyte.

Cathode Half-Reaction

The cathode is the electrode where reduction occurs in the electrolysis process. In the Castner cell, sodium ions \(\text{Na}^+\) move towards the cathode because they are positively charged. At the cathode, they gain electrons (a reduction process) and form sodium metal: \[ \text{Na}^+ + e^- \rightarrow \text{Na} \]. This half-reaction shows the direct conversion of sodium ions into metallic sodium, which is the primary goal of the Castner cell.

Anode Half-Reaction

The anode is the electrode where oxidation occurs. In the Castner cell, hydroxide ions \(\text{OH}^-\) are attracted to the anode. These hydroxide ions lose electrons (an oxidation process) and produce water and oxygen gas: \[ 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^- \]. This reaction essentially describes how hydroxide ions are converted to produce water and oxygen, releasing electrons in the process.

Overall Cell Reaction

The overall cell reaction combines both the cathode and anode half-reactions to give a complete picture of the chemical processes occurring in the Castner cell. The balanced equation for the overall reaction is: \[ 4\text{Na}^+ + 4\text{OH}^- \rightarrow 4\text{Na} + 2\text{H}_2\text{O} + \text{O}_2 \]. This shows how sodium ions and hydroxide ions are transformed into metallic sodium, water, and oxygen gas.