Respuesta :
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
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
Infotmation about the reactions is needed
Explanation:
Assuming this is a cell (no external electricity added) the process flow will be the following:
In the cathode, the reduction happens meaning that the element takes electrons from the environment.
This electrons come from the oxidation reaction and to be attracted to the cathode it has to be positive (+).
The oxidation reaction happens in the anode and cause of the release of electrons it is negative (-)