calculate the [Cu2+]/[Zn2+] ratio at which the following reaction is spontaneous at 25°C.
Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s)
The standard reduction potentials for the half-reactions involved in this reaction are:
Cu2+(aq) + 2e- → Cu(s) E° = +0.34 V
Zn2+(aq) + 2e- → Zn(s) E° = -0.76 V
To determine the equilibrium constant (K) for the reaction, we must find the standard cell potential (E°cell) by subtracting the reduction potential of the anode from the reduction potential of the cathode:
E°cell = E°cathode - E°anode
E°cell = (+0.34 V) - (-0.76 V)
E°cell = +1.10 V
Using the relation between standard cell potential, equilibrium constant, and temperature:
ΔG° = -nFΔE°cell
ΔG° = -RT lnK
We can find the equilibrium constant, K:
ΔG° = -RT lnK
-96485 J/mol·V x (298 K) x lnK = -nFΔE°cell
lnK = (1 x 96485 J/mol·V x 298 K) / (8.314 J/K·mol)
lnK = 33.87
K = 2.89 x 10^14
The reaction is spontaneous when the equilibrium constant (K) is greater than 1. Therefore, the [Cu2+]/[Zn2+] ratio must be greater than 2.89 x 10^14 for the reaction to be spontaneous at 25°C.