Determine the direction of electron flow and label the anode and the cathode. label each electrode as negative or positive. write a balanced equation for the overall reaction

At anode is where oxidation takes place,  that is
Cr(s)→Cr3+(aq) +3e- this is the oxidation
At cathode is where reduction takes place,
Fe3+(aq)+3e-→Fe(s).This is the reduction
Cell reaction is
Cr(s)→Cr3+(aq)+3e- where E°=0.73V
Fe3+(aq)+3e-→Fe(s) where E°= -0.036V
Cr(s) +Fe3+(aq)→Cr3(aq)+Fe(s) E°cell=0.69V
Cell reaction for Cr(s) +Fe3+(aq)→ Cr3+(aq) + Fe(s) is
E°cell is 0.69V

H tricky one let me think and get back to u