Question
In a fume cupboard molten lead(ll) bromide is electrolyzed using graphite.Write half equations (states are required) for the reactions at the cathode and anode?
Answers
GPT-4o mini
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).
1. **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:
\[
\text{Pb}^{2+} (l) + 2e^- \rightarrow \text{Pb} (s)
\]
2. **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: \(2 \text{Br}^- (l) \rightarrow \text{Br}_2 (g) + 2e^-\)
1. **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:
\[
\text{Pb}^{2+} (l) + 2e^- \rightarrow \text{Pb} (s)
\]
2. **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: \(2 \text{Br}^- (l) \rightarrow \text{Br}_2 (g) + 2e^-\)