Homework Answers
from data table:
Eo(Ba2+/Ba(s)) = -2.90 V
Eo(Cu2+/Cu(s)) = 0.337 V
As per given reaction/cell notation,
cathode is (Cu2+/Cu(s))
anode is (Ba2+/Ba(s))
Eocell = Eocathode - Eoanode
= (0.337) - (-2.90)
= 3.237 V
number of electrons being transferred, n = 2
F = 96500.0 C
use:
ΔGo = -n*F*Eo
= -2*96500.0*3.237
= -625000 J/mol
= -625 KJ/mol
Answer: option 3
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