Chapter 18 - Sections 18.1-18.9 - Exercises - Problems by Topic - Page 905: 53b

$E^{o}_{Cat} = -0.13$ $E^{o}_{An} = -0.23$ $E^{o}_{cell} = E^{o}_{Cat} - E^{o}_{An} = -0.13 - (-0.23)$ = +0.10 Since the value is positive, the redox reaction occurs spontaneously in the forward direction.

Here Ni is acting as anode and Pb is acting as cathode.