Homework Answers
from data table:
Eo(Sn2+/Sn(s)) = -0.14 V
Eo(Cl2(g)/2Cl-) = 1.36 V
As per given reaction/cell notation,
cathode is (Cl2(g)/2Cl-)
anode is (Sn2+/Sn(s))
Eocell = Eocathode - Eoanode
= (1.36) - (-0.14)
= 1.50 V
here, number of electrons being transferred, n = 2
Eo = (2.303*R*T)/(n*F) log Kc
At 25 oC or 298 K, R*T/F = 0.0592
So, Eo = (0.0592/n)*log Kc
1.5 = (0.0592/2)*log Kc
log Kc = 50.6757
Kc = 4.739*10^50
Answer: 4.7*10^50
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