In a fume cupboard molten lead(ii) bromide is electrolysed using graphite electrodes. Write half-equations for the reaction at the (i).cathode (ii). Anode (iii). Overall equation (states required).

Answer 1

PbBr2 ==> Pb^2+ + 2Br^- are the ions you have.
Pb^2+ + 2e ==> Pb a the cathode (you can add the states).
2Br^- ==> Br2 + 2e at the anode (again you add the states).

Add the two half equations to obtain the overall equation.