Question

1B: Consider the cell Cd(s)+Cu²⁺(aq)-->Cu(s)+Cd²⁺. Calculate K for this reaction.

Answer 1

Cd(s) + Cu2+(aq) ------> Cd2+(aq) + Cu(s)

E°cell = E°cathode - E°anode

E°cell = E°Cu2+/Cu - E°Cd2+/Cd

E°cell = 0.34 V - (-0.40V)

E°cell = 0.74 V

The relationship between E°cell and K is as follows:

∆G° = -nFE° ------>(1)

∆G° = -RTlnK ----->(2)

From 1 & 2

nFE° = RTlnK

lnK = RT /nFE°

Here , K is equilibrium constant, R is gas constant, T temperature, F Faraday constant, E° cell potential, n number of electrons transferred

By substitution of R, T and F values then we will get

lnK = n × E° / (0.0257)

Here n = 2

ln K = 2 × 0.74 / 0.0257

lnK = 57.59

K = e^57.59

K = 1.02×10^25