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pb has avalency of 4 and iodine has a valency of 1- so the formulae for lead iodide is pbI2.
at the cathode molten lead is formed:
pb2+ + 2e- gives pb
and at the anode:
2I- gives I2 + 2e-
hope it helps !!!
Though lead has a +2 and a +4 for its oxidation states, iodine is too weak an oxidizer to form the second lead-iodine compound.hence it is pbI2
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