Question

Write equations for the half-reactions that occur in the electrolysis of molten potassium bromide.

Short answer

The cathodic half-reaction is \(K^+ + e^- \rightarrow K(s)\). The anodic half-reaction is \(2 Br^- - 2 e^- \rightarrow Br_2(g)\).

Step-by-step solution

Identify Cathodic Half-Reaction

During electrolysis, the reduction happens at the cathode. For molten potassium bromide (KBr), potassium ions (K+) gain electrons to form potassium atoms. The cathodic half-reaction equation can be written as: \(K^+ + e^- \rightarrow K(s)\)

Identify Anodic Half-Reaction

Oxidation occurs at the anode. In the case of potassium bromide, bromide ions (Br-) lose electrons to form bromine molecules. The anodic half-reaction can be expressed as: \(2 Br^- - 2 e^- \rightarrow Br_2(g)\)

Key concepts

Cathodic Half-Reaction

In electrolysis, the cathodic half-reaction plays a pivotal role as it is the site of reduction, where positive ions gain electrons and transform into a neutral state. When dealing with molten potassium bromide (KBr), the electrolyte provides potassium ions (\text{K}^+), which migrate to the cathode. Here, they undergo a critical transformation. Each potassium ion accepts an electron (\text{e}^-) to neutralize its positive charge, leading to the formation of solid potassium (\text{K(s)}).

The equation representing this process can be written succinctly as: \[ K^+ + e^- \rightarrow K(s) \] This equation clearly shows the direct consumption of electrons, which is indicative of reduction. It is essential to understand that in the broader context of electrolysis, the cathodic half-reaction is where the electrical energy provided to the system is converted into chemical energy, manifesting as a change in oxidation state of the element involved, potassium in this case.

Anodic Half-Reaction

Conversely, at the anode, the story is about oxidation in the anodic half-reaction. Here, we see bromide ions (\text{Br}^-), which are negatively charged, flowing toward the anode. The ions surrender their extra electrons to the anode and in doing so, they are oxidized. There's a distinct change as the bromide ions come together in pairs to form bromine molecules (\text{Br}_2), a diatomic gas.

The chemical transformation at the anode can be articulated with the following equation: \[ 2 Br^- - 2 e^- \rightarrow Br_2(g) \] It is essential to highlight that this equation symbolizes the release of two electrons by each pair of bromide ions, hence constituting an oxidation process. The electrons given up by bromide ions are the same electrons that are later used at the cathode for the reduction of potassium ions. In essence, the process at the anode complements the cathodic half-reaction, maintaining the electrical neutrality of the system.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often referred to as redox reactions, are the heart of electrolytic processes like the electrolysis of molten potassium bromide. Redox reactions involve the transfer of electrons between chemical species. Oxidation describes the loss of electrons, while reduction refers to the gain of electrons. It is critical to emphasize that these two half-reactions always occur simultaneously; a species cannot undergo oxidation without another undergoing reduction.

In the context of our exercise, the cathodic and anodic half-reactions represent the two halves of the redox reaction taking place during electrolysis: \[ \text{Cathode (Reduction):} K^+ + e^- \rightarrow K(s) \] \[ \text{Anode (Oxidation):} 2 Br^- - 2 e^- \rightarrow Br_2(g) \] Thus, the potassium ions are reduced and the bromide ions are oxidized during electrolysis. The interdependency of these processes demonstrates the core principle of redox reactions - they are inseparable and establish a balance within a chemical reaction. The collective understanding of these concepts provides students with a cohesive picture of how electrolysis operates on a molecular level.