A concentration cell at 298K consists of two Cu/Cu^2+ electrodes in two separate

Question

A concentration cell at 298K consists of two Cu/Cu^2+ electrodes in two separate compartments connected by a salt bridge. you know that the solution in one compartment is 1.0 M CuSO4, but you do not know the concentration of CuSO4 at present in the other. However, you do know that the volatage of the concentration cell is at present 0.1V. What is the concentraion of CuSO4 in the other compartment at present?

The answer is 0.4 mM but I dont how that came about. If someone can explain this to me step by step it will be really helpful. Thanks!

Explanation / Answer

Overall cell reaction is:

Cu2+ (1.0 M) => Cu2+ (? M)


Eo(cell) = 0.0 V

E(cell) = 0.1 V

Faraday constant F = 96485 C/mol

Molar gas constant R = 8.314 J/mol.K

Temperature T = 298 K

Moles of electrons transferred n = 2


Nernst equation:

E(cell) = Eo(cell) - RT/nF ln([Cu2+(? M)/Cu2+(1.0 M)]


0.1 = 0.0 - 8.314 x 298/(2 x 96485) x ln([Cu2+(? M)]/1.0)

ln([Cu2+(? M)]/1.0) = -7.7887

[Cu2+(? M)] = exp(-7.7887) = 4 x 10^(-4) M = 0.4 mM

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