The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e--- >2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode = 1.07- (-0.13) =1.20V Solution The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e--- >2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode = 1.07- (-0.13) =1.20V.