The correct option is C Cathode:Pb2+→Pb++2e−; Anode:Cl−→Cl+e−
We know oxidation happens at anode and reduction at cathode
In PbCl2 ions are Pb2+ and Cl− , so Pb2+ reduces and Cl− oxidizes.
So at anode reaction is
Cl−→Cl+e− (oxidation)
Pb2++2e−→Pb (reduction at cathode)