Homework Answers
from data table:
Eo(Cu2+/Cu(s)) = 0.50 V
Eo(Ag+/Ag(s)) = 0.76 V
As per given reaction/cell notation,
cathode is (Ag+/Ag(s))
anode is (Cu2+/Cu(s))
Eocell = Eocathode - Eoanode
= (0.76) - (0.50)
= 0.26 V
Answer: 0.26 V
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