In: Chemistry
1)
from data table:
Eo(Fe2+/Fe(s)) = -0.44 V
Eo(Cu2+/Cu(s)) = 0.337 V
the electrode with the greater Eo value will be reduced and it will be cathode
here:
cathode is (Cu2+/Cu(s))
anode is (Fe2+/Fe(s))
The chemical reaction taking place is
Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
Eocell = Eocathode - Eoanode
= (0.337) - (-0.44)
= 0.777 V
Since Eocell is positive, iron can reduce copper.
2)
from data table:
Eo(Fe2+/Fe(s)) = -0.44 V
Eo(Na+/Na(s)) = -2.71 V
for given cell reaction to occur
cathode is (Na+/Na(s))
anode is (Fe2+/Fe(s))
Eocell = Eocathode - Eoanode
= (-2.71) - (-0.44)
= -2.27 V
Since Eocell is negative, iron can’t reduce copper.