We are asked to find the **equilibrium ****constant K** for the disproportionation of the copper(I) ion at room temperature:

2Cu^{+} (aq) → Cu^{2 + } (aq) + Cu(s).

We will use the **Gibbs Free Energy**** ****Equation** to calculate the equilibrium constant.

The Gibbs Free relates the potential of compounds and cell equilibrium.

The Gibbs free energy is related to the potential by:

$\overline{){\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{{\mathbf{nFE}}}_{{\mathbf{cell}}}^{{\mathbf{o}}}{\mathbf{}}}$

If the E^{o}_{cell }> 0, then the process is spontaneous (galvanic cell)

If the E^{o}_{cell }< 0, then the process is nonspontaneous (electrolytic cell)

ΔG^{o} can also be expressed by the equilibrium constant Keq of the reaction.

$\overline{){\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{-}}{\mathbf{}}{\mathbf{RT}}{\mathbf{}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{eq}}}{\mathbf{}}}$

Combining the two equations since both are equal to ΔG^{o} gives:

$\overline{){{\mathbf{nFE}}}_{{\mathbf{cell}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{RT}}{\mathbf{}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{eq}}}{\mathbf{}}}$

where:

n = # of e^{-} transferred

F = Faraday’s constant = 96485 J/(mol e^{-})

E°_{cell} = standard cell potential, V

R = gas constant, 8.314 J/mol e^{- }K

T = temperature in K

K_{eq} = equilibrium constant of the reaction

In the Gibbs Equation, we have to determine the E°_{cell} first, as well as the **anode (oxidation) **and the **cathode (reduction)** in the concentration cell and the **number of electrons transferred (n).**

Using data from Appendix E in the textbook, calculate the equilibrium constant for the disproportionation of the copper(I) ion at room temperature: 2Cu^{+} (aq) → Cu^{2 + } (aq) + Cu(s).

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