International General Certificate of Secondary Education (IGCSE) Chemistry Practice Exam

What is the ionic half-equation for the cathode reaction during the electrolysis of lead(II) bromide?

1. 2Br⁻ -> Br₂ + 2e⁻

2. Pb²⁺ + 2e⁻ -> Pb

3. Br₂ + 2e⁻ -> 2Br⁻

4. Pb + 2e⁻ -> Pb²⁺

The correct answer is: Pb²⁺ + 2e⁻ -> Pb

During the electrolysis of lead(II) bromide (PbBr₂), the cathode reaction involves the reduction of lead ions (Pb²⁺) to form lead metal (Pb). At the cathode, cations migrate towards the negatively charged electrode where they gain electrons, a process known as reduction. In this specific case, the half-equation for the cathode reaction can be represented as Pb²⁺ + 2e⁻ → Pb. This reaction shows that each lead ion (Pb²⁺) gains two electrons (2e⁻) to be converted into solid lead (Pb). This is a crucial reduction reaction in the process of electrolysis and leads to the deposition of lead at the cathode. The other options represent different processes that do not accurately depict the cathodic reaction occurring during the electrolysis of lead(II) bromide. For instance, the first option involves bromide ions being oxidized to bromine gas, which occurs at the anode, not the cathode.