J.R. S. answered 11/08/23
Ph.D. University Professor with 10+ years Tutoring Experience
In order to address this problem, we must consult a table of standard reduction potentials. Assuming that one half cell is 1.0 M chromium (III) and 1.0 M chromium (II) and not as written where both ions are chromium(II)..
The table gives the following:
Cr3+ + e- ==> Cr2+ ... Eº = -0.407 V
Ce3+ + 3e- ==> Ce ... Eº = -2.34 V
Accordingly, Cr3+/Pt will act as the cathode and Ce3+/Pt will act as the anode
Eºcell = -0.407 V + 2.34 V = 1.93 V