= 0. 52 V for the reduction Cu+(aq) + e− → Cu(s), calculate E
, and K for the following reaction at 25°C:
2Cu+(aq) ⇌ Cu2+(aq) + Cu(s)
The equilibrium constant of the reaction is 1.21 * 10^6 while the change in free energy is -34.7 kJ.
What is equilirium constant?
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
Learn more about equilibrium constant:https://brainacademy.pro/question/10038290