The EMF of the cell No/Ni2+ (0.01M) // cl-(0.01M)/ Cl2 PT it ? . If SRP of nickel and chlorine electrode are -0.25V and +1.36 respectively . Share with your friends Share 0 Vartika Jain answered this Dear Student, At anode:Ni2+ + 2e- → Ni E°=-0.25 VAt cathode :2Cl- + 2e- → Cl2 E°=1.36 VNow, overall reaction :2Cl- + Ni → Ni2+ + Cl2E°cell=E°cathode-E°anode =1.36-(-0.25) =1.61 VNow, Ecell=E°cell-0.0591nlog[Ni2+][Cl-]2 =1.61-0.05912log(0.01)(0.01)2 =1.61 -0.05912log (102) =1.61-0.05912×2 =1.61-0.0591 =1.55 V 11 View Full Answer Rohit Singh answered this E=E cation subtracted by E anion E=1.36-(-0.25) E=1.61 1
