Question

If an electric current is passed through aqueous solutions of sodium chloride, sodium bromide, and sodium iodide, the elemental halogens are produced at one electrode in each case, with hydrogen gas being evolved at the other electrode. If the liquid is then evaporated from the mixture, a residue of sodium hydroxide remains. Write balanced chemical equations for these electrolysis reactions.

Short answer

For the electrolysis of aqueous solutions of sodium chloride (NaCl), sodium bromide (NaBr), and sodium iodide (NaI), the following balanced chemical equations are formed: Sodium chloride (NaCl) solution: \[2H_{2}O(l) + 2Cl^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + Cl_{2}(g)\] Sodium bromide (NaBr) solution: \[2H_{2}O(l) + 2Br^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + Br_{2}(l)\] Sodium iodide (NaI) solution: \[2H_{2}O(l) + 2I^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + I_{2}(s)\] When the liquid evaporates from the mixture, sodium hydroxide (NaOH) residue is formed with the following balanced equation: \[Na^⁺(aq) + OH^⁻(aq) \rightarrow NaOH(s)\]

Step-by-step solution

Half-reactions at the anode (negative electrode) for each solution

When an electric current is passed through aqueous solutions of sodium chloride (NaCl), sodium bromide (NaBr), and sodium iodide (NaI), the negatively charged halide ions (Cl⁻, Br⁻, or I⁻) lose electrons at the anode, turning into halogen molecules (Cl₂, Br₂, or I₂). For each halide ion, the half-reaction at the anode is: \[2Cl^⁻ \rightarrow Cl_{2}(g) + 2e^⁻\] \[2Br^⁻ \rightarrow Br_{2}(l) + 2e^⁻\] \[2I^⁻ \rightarrow I_{2}(s) + 2e^⁻\]

Half-reaction at the cathode (positive electrode) for each solution

At the cathode, water is reduced to form hydrogen gas (H₂) and hydroxide ions (OH⁻). The half-reaction for this process is: \[2H_{2}O(l) + 2e^⁻ \rightarrow 2OH^⁻(aq) + H_{2}(g)\] Notice that this half-reaction is the same for all three solutions.

Combining the half-reactions to form the overall balanced equations for each solution

Now, let's combine these half-reactions to form the overall balanced chemical equations for each solution. For sodium chloride (NaCl) solution: \[2H_{2}O(l) + 2Cl^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + Cl_{2}(g)\] For sodium bromide (NaBr) solution: \[2H_{2}O(l) + 2Br^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + Br_{2}(l)\] For sodium iodide (NaI) solution: \[2H_{2}O(l) + 2I^⁻(aq) \rightarrow 2OH^⁻(aq) + H_{2}(g) + I_{2}(s)\]

Formation of sodium hydroxide residue

After the electrolysis reactions, when the liquid is evaporated from the mixture, the sodium cations (Na⁺) and hydroxide anions (OH⁻) that were present in the solution combine to form sodium hydroxide (NaOH) residue. The balanced equation for this is: \[Na^⁺(aq) + OH^⁻(aq) \rightarrow NaOH(s)\]

Key concepts

Half-reaction

In electrolysis, a key aspect is understanding the concept of half-reactions. Half-reactions describe the processes occurring at each electrode during an electrolysis process. One reaction happens at the anode, and another distinct reaction occurs at the cathode.
For sodium chloride, sodium bromide, and sodium iodide, the negatively charged halide ions ( - \(^-\) ) move towards the anode, where they lose electrons (a process known as oxidation). This results in the transformation of the halide ions into their corresponding halogen molecules:

• Chloride ions become chlorine gas: \(2Cl^- \rightarrow Cl_2(g) + 2e^-\)
• Bromide ions become bromine liquid: \(2Br^- \rightarrow Br_2(l) + 2e^-\)
• Iodide ions become iodine solid: \(2I^- \rightarrow I_2(s) + 2e^-\)

This loss of electrons is crucial in completing the circuit in an electrolysis setup. These half-reactions ensure the production of elemental halogens and are vital in various industrial processes where these elements are needed.

Anode and Cathode

In any electrolysis system, the roles played by the anode and cathode are essential. The anode is the electrode where oxidation occurs, and the cathode is where reduction happens.
In electrolysing sodium halides, the anode is the site where halogen molecules are formed as halide ions lose electrons, as detailed in the half-reactions. This makes the anode crucial for extracting elements directly from their ions.
On the other hand, at the cathode, the water molecules undergo reduction. Each water molecule gains electrons, forming hydrogen gas and hydroxide ions. This process can be summarized in a familiar half-reaction:

• At the cathode: \(2H_2O(l) + 2e^- \rightarrow 2OH^-(aq) + H_2(g)\)

Thus, while the anode is oxidizing the halide ions, the cathode is where new substances are formed, such as hydrogen gas, which has various applications, including as a fuel source.

Sodium hydroxide formation

After the electrolysis process of sodium halide solutions is complete, an interesting development occurs. As the liquid evaporates, sodium hydroxide is left as a residue. This is a result of the reaction between sodium cations ( - \(^+\) ) and hydroxide ions ( - \(^-\) ) that were earlier produced at the cathode.

• Combination reaction: \(Na^+(aq) + OH^-(aq) \rightarrow NaOH(s)\)

This formation of solid sodium hydroxide is not just a chemical curiosity; it's extremely useful industrially. Sodium hydroxide, or lye, is pivotal in processes such as soap making, cleaning agents, and even in the food industry for processing certain types of foods. Thus, the entirety of the electrolysis process shows how seemingly simple ionic reactions can lead to the creation of an incredibly wide array of useful compounds.