Question

The first electrolytic process to produce sodium metal used molten NaOH as the electrolyte. Write probable half-equations and an overall equation for this electrolysis.

Short answer

The half-reactions are: Oxidation (anode): 2OH- → O2 + H2O + 4e-; Reduction (cathode): Na+ + e- → Na. The overall reaction is: 2NaOH → 2Na + O2 + H2O.

Step-by-step solution

Identify the half reactions

In electrolysis, chemical compounds are broken down into their component elements using electricity. In this case, we have NaOH (sodium hydroxide) being broken down into Na (sodium) and OH (hydroxide ions). So, we can write the half reactions as follows: \n\n- Oxidation (anode): 2OH- → O2 + H2O + 4e- \n- Reduction (cathode): Na+ + e- → Na

Write the overall reaction

We can combine these half-reactions to write the overall electrolysis reaction. We balance the number of electrons in the half-equations and combine them to give the overall equation: 2NaOH → 2Na + O2 + H2O

Key concepts

Sodium Metal Production

Sodium metal production through electrolysis involves the separation of sodium from its various compounds. In early methods, molten sodium hydroxide (NaOH) was used. During the process, the ionic compounds are melted, enabling the ions to move freely. This movement sets the stage for electrolysis to take place. Applying an electric current across the melted substance drives the separation of ions. When producing sodium metal specifically, sodium ions (Na+) gain electrons at the cathode and are converted into sodium metal. The final product is essential in many industrial processes. Sodium's reactivity and importance in forming compounds make it vital for the chemical industry. For example, sodium is used in the production of various chemicals, glass, and metals. Its reactive nature also demands careful handling, especially when it is in its pure metal state. Understanding the electrolytic processes involved in its production is crucial for industries reliant on sodium.

Half-Reactions

In electrolytic processes, understanding half-reactions is key as these show how the overall reaction is broken down into two parts. Each represents a different aspect of the reaction- oxidation and reduction. At the anode, the oxidation half-reaction occurs, where hydroxide ions (OH⁻) are oxidized to form oxygen gas and water, releasing electrons. The half-reaction at anode can be described as:

• Oxidation (anode): \(2OH^- \rightarrow O_2 + H_2O + 4e^-\)

At the cathode, a reduction half-reaction takes place. Sodium ions (Na⁺) accept electrons to form sodium metal. This reaction can be written as:

• Reduction (cathode): \(Na^+ + e^- \rightarrow Na\)

The individual half-reactions highlight how electrons are transferred during electrolysis. Electrons lost in the oxidation reaction are gained in the reduction reaction. This interplay ensures the conservation of charge and the completion of the overall electrolysis process.

Electrolysis

Electrolysis is a fundamental process used to decompose chemical compounds into their elements using electrical energy. In the context of sodium metal production, it involves applying an electrical current to molten sodium hydroxide, resulting in the movement of ions toward the electrodes. Upon reaching the electrodes, these ions undergo redox (reduction-oxidation) reactions. The entire electrolytic cell is tailored to ensure efficient reactions. For sodium production, the setup ensures that sodium ions gain electrons at the cathode, forming pure sodium metal. Similarly, hydroxide ions lose electrons at the anode, resulting in the release of oxygen gas. Understanding electrolysis is critical for various industrial applications:

• It allows for the production of elements that are not easily separated by chemical means alone.
• It is a cost-effective method of obtaining pure metals.
• It plays a significant role in batteries and electroplating processes.

This knowledge helps in optimizing processes across different sectors where electrolysis is integral.