When molten lead(II) bromide (PbBr\(_2\)) is electrolyzed, it dissociates into lead ions (Pb\(^{2+}\)) and bromide ions (Br\(^-\)). The electrolysis involves two half-reactions, one at the cathode (reduction) and one at the anode (oxidation).
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At the Cathode (Reduction): Lead ions (Pb\(^{2+}\)) gain electrons to form solid lead (Pb). The half-equation for the reaction at the cathode is: \[ \text{Pb}^{2+} (l) + 2e^- \rightarrow \text{Pb} (s) \]
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At the Anode (Oxidation): Bromide ions (Br\(^-\)) lose electrons to form bromine (Br\(_2\)). Since two bromide ions are required to form one molecule of bromine gas, the half-equation for the reaction at the anode is: \[ 2 \text{Br}^- (l) \rightarrow \text{Br}_2 (g) + 2e^- \]
In summary:
- Cathode half-equation: \(\text{Pb}^{2+} (l) + 2e^- \rightarrow \text{Pb} (s)\)
- Anode half-equation: \(2 \text{Br}^- (l) \rightarrow \text{Br}_2 (g) + 2e^-\)
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