J.R. S. answered 11/02/22
Ph.D. University Professor with 10+ years Tutoring Experience
From the table:
Cu2+ + 2e- ==> Cu ... Eº = 0.34 V
Cr3+ + e- ==> Cr2+ ... Eº = -0.407 V
Reduction half reaction will be Cu2+ + 2e- ==> Cu @ the cathode
Oxidation half reaction will be Cr2+ ==> Cr3+ + e- @ the anode
Eºcell = 0.34 V + 0.407 V
Eºcell = 3.807 V