Problem: 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)  →  Cu2 +  (aq) + Cu(s).

FREE Expert Solution

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

2Cu+ (aq)  →  Cu2 +  (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:

Go = -nFEcello 

     If the Eocell > 0, then the process is spontaneous (galvanic cell)

     If the Eocell < 0, then the process is nonspontaneous (electrolytic cell)


ΔGo can also be expressed by the equilibrium constant Keq of the reaction.

Go =- RT ln Keq 


Combining the two equations since both are equal to ΔGo gives:

nFEcello = RT ln Keq 

where:

n = # of e- transferred
F = Faraday’s constant = 96485 J/(mol e-)
cell = standard cell potential, V

R = gas constant, 8.314 J/mol e-  K

T = temperature in K

Keq = 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).

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Problem Details

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)  →  Cu2 +  (aq) + Cu(s).

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Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.