Molten lead(II) bromide can be electrolysed.
What is the correct half-equation for the reaction at the negative electrode (cathode)?
2Br−−2e−→Br22\text{Br}^- - 2\text{e}^- \rightarrow \text{Br}_22Br−−2e−→Br2
Pb2++2e−→Pb\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}Pb2++2e−→Pb
Pb2++e−→Pb\text{Pb}^{2+} + \text{e}^- \rightarrow \text{Pb}Pb2++e−→Pb
Pb2+−2e−→Pb\text{Pb}^{2+} - 2\text{e}^- \rightarrow \text{Pb}Pb2+−2e−→Pb