J.R. S. answered 01/22/22
Ph.D. University Professor with 10+ years Tutoring Experience
a) I2 is the potential oxidizing agent and Ni is the potential reducing agent
b) Eºcell = 0.536 - (-0.257) = 0.793 V
c) I2 + 2e- ==> 2I- + 2e- Eº = 0.536 V (cathode)
Ni ==> Ni2+ + 2e- Eº = -0.257 V (anode)
c) I2 + Ni ==> 2I- + Ni2+