Given:
Cr3+(aq)+3e−⇌Cr(s); E°=−0.74 V
Pb2+(aq)+2e−⇌Pb(s); E°=−0.13 V
What is the equilibrium constant (K) at 25°C for the following cell reaction?
2Cr(s)+3Pb2+(aq)→3Pb(s)+2Cr3+(aq)
Select one:
a. 7.4 ×10^61
b. 2.1×10^26
c. 2.9×10^44
d. 8.33x10^56
we have:
Eo(Cr3+/Cr(s)) = -0.74 V
Eo(Pb2+/Pb(s)) = -0.13 V
here:
cathode is (Pb2+/Pb(s))
anode is (Cr3+/Cr(s))
The chemical reaction taking place is
Pb2+(aq) + Cr(s) --> Pb(s) + Cr3+(aq)
Eocell = Eocathode - Eoanode
= (-0.13) - (-0.74)
= 0.61 V
here, number of electrons being transferred, n = 6
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
0.61 = (0.0592/6)*log Kc
log Kc = 61.8243
Kc = 6.673*10^61
Answer: a
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