Using standard electrode potentials calculate ΔG∘rxn and use its value to estimate the equilibrium constant for each of the reactions at 25 ∘C. Pb2+(aq)+Mg(s)→Pb(s)+Mg2+(aq).
from data table:
Eo(Mg2+/Mg(s)) = -2.372 V
Eo(Pb2+/Pb(s)) = -0.126 V
As per given reaction/cell notation,
cathode is (Pb2+/Pb(s))
anode is (Mg2+/Mg(s))
Eocell = Eocathode - Eoanode
= (-0.126) - (-2.372)
= 2.246 V
a)
number of electrons being transferred, n = 2
F = 96500.0 C
use:
ΔG = -n*F*E
= -2*96500.0*2.246
= -433478 J
= -433.5 KJ
Answer: -433.5 KJ
b)
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
2.246 = (0.0592/2)*log Kc
log Kc = 75.8784
Kc = 7.558*10^75
Answer: 7.56*10^75
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