Given that E
o
= 0. 52 V for the reduction Cu+(aq) + e− → Cu(s), calculate E
o
, ΔG
o
, and K for the following reaction at 25°C:

2Cu+(aq) ⇌ Cu2+(aq) + Cu(s)

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-04-26T14:36:11+02:00

The equilibrium constant of the reaction is 1.21 * 10^6 while the change in free energy is -34.7 kJ.

The equilibrium constant shows the extent of conversion of reactants to products.

Now we know from the Nernst equation that;

Ecell = E°cell - 0.0592/n logQ

E°cell = 0.52−0.16=0.36 V

Since Ecell = 0 V at equilibrium,

0 = 0.36 - 0.0592/1 log K

0.36 = 0.0592/1 log K

log K = 0.36/ 0.0592

K = antiog (0.36/ 0.0592)

K = 1.21 * 10^6

ΔG = -RT lnK

ΔG =-(8.314 * 298 * ln1.21 * 10^6)

ΔG =-34.7 kJ

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