Hi. can some one please explain me the solution for
this question. the correct answer is E. thanks
Homework Answers
from data table:
Eo(Pb2+/Pb(s)) = -0.126 V
Eo(Ag+/Ag(s)) = 0.7996 V
As per given reaction/cell notation,
cathode is (Ag+/Ag(s))
anode is (Pb2+/Pb(s))
Eocell = Eocathode - Eoanode
= (0.7996) - (-0.126)
= 0.9256 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
0.9256 = (0.0592/2)*log Kc
log Kc = 31.2703
Kc = 1.9*10^31
Answer: e
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