Use the relationship DeltaG^@= -nF E_(cell)^@ to estimate the minimum voltage required to electrolyse Al_2O_3 in the Half-Heroult process. DeltaG_f^@(Al_2O_3)= -1520 kJ mol^(-1) , DeltaG_f^@(CO_2)= -394 kJ mol^(-1) Show that the oxidation of the graphite anode to CO_2 permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were Al_2O_3to2Al+3O_2

Use the relationship DeltaG^@= -nF E_(cell)^@ to estimate the minimum

1.	Use the relationship DeltaG^@= -nF E_(cell)^@ to estimate the minimum voltage required to electrolyse Al_2O_3 in the Half-Heroult process. DeltaG_f^@(Al_2O_3)= -1520 kJ mol^(-1) , DeltaG_f^@(CO_2)= -394 kJ mol^(-1) Show that the oxidation of the graphite anode to CO_2 permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were Al_2O_3to2Al+3O_2
Answer» Solution :1.6 V in Hall-Heroult process Net reaction in Hall-Heroult process is : `3C+2Al_2O_3 to 4Al +3CO_2` or `4Al^(3+)12E^(-)to 4Al`, number of electrons (n)=12 `DeltaG^@ =3 DeltaG_f^@(CO_2)-2DeltaG_f^@(Al_2O_3)` `=3xx394-2(-1520)=1858 kJ` `DeltaG^@=-nFE_(cell)^@` ltbgt `-E_(cell)^@=(DeltaG^@)/(nF)=(1858xx1000)/(12xx96500)=1.60 V` For the reaction `Al_2O_3 to 2Al+3O_2` `DeltaG^@=`1520 kJ `2Al^(3+)+6e^(-)to2Al, n=6` `-E_(cell)^@=(DeltaG)/(nF)=(1520xx1000)/(6xx96500)=2.62` V Thus Hall-Heroult process takes place at lower voltage.