Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox...

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C.

Pb 2+(aq) + Cu(s) → Pb(s) + Cu2+(aq)

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Answer #1

from data table:

Eo(Cu2+/Cu(s)) = 0.337 V

Eo(Pb2+/Pb(s)) = -0.126 V

As per given reaction,

cathode is (Pb2+/Pb(s))

anode is (Cu2+/Cu(s))

Eocell = Eocathode - Eoanode

= (-0.126) - (0.337)

= -0.463 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

-0.463 = (0.0592/2)*log Kc

log Kc = -15.6419

Kc = 2.28*10^-16

Answer: 2.28*10^-16

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