Thermochemistry

Question

Calculate the non-standard cell potential (in V) for the following reaction at 110.55 ℃ with the following concentrations: [Mg^+2] = 2.00×10−72.00×10^-7 M; [Cu^+2] = 5.582 M

Cu^+2(aq) + Mg(s) --> Cu(s) + Mg+^2(aq)

Answer 1

We wil have to use the Nernst equation after we determine Eºcell.

Cu²⁺(aq) + Mg(s) ==> Cu(s) + Mg²⁺(aq)

Cathode = reduction = Cu

Anode = oxidation = Mg

From a table of standard reduction potentials, I obtain the following:

Cu²⁺ + 2e- ==> Cu(s) Eº = 0.34 V

Mg²⁺ + 2e- ==> Mg(s) Eº = -2.37 V

Eº_cell = 0.34 V + 2.37 V = 2.71 V

Nernst equation:

E_cell = Eº_cell - RT / nF lnQ

E_cell = ?

Eº_cell = 2.71 V

R = 8.314 J/Kmol

T = 110.55 C + 273 = 383.6K

n = 2 moles electrons

F = 96.500 C / mole e-

Q = [Mg²⁺] / [Cu²⁺] = 2.00x10^-7 / 5.582 = 3.58x10^-8

Plug values into the Nernst equation and solve for E_cell