Thermochemistry

Question

Calculate the non-standard cell potential (in V) for the following reaction at 86.51 ℃ with the following concentrations: [Mg+2] = 7.00×10−7 M; [Cu+2] = 2.158 M

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

Answer 1

For the problem, we will use the Nernst Equation:

E = Eº - RT/nF ln Q

E = non standard cell potential = ?

Eº = standard cell potential (see below)

R = 8.314 J/mol-K

T = temp = 86.51 + 273 = 360K

n = mols electrons transferred = 2

F = 96,500 C/mol e-

Q = reaction quotient (see below)

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First, we find the standard cell potential, Eº:

Cu²⁺ + 2e- ==> Cu Eº = 0.34 V

Mg²⁺ + 2e- ==> Mg Eº = -2.38

Reaction is Cu²⁺ + Mg ==> Cu + Mg²⁺

Eº = 0.34 + 2.38 = 2.72 V

Q = [Mg²⁺]/[Cu²⁺] = 7x10^-7 / 2.158 = 3.23x10^-7

Plugging values into the Nernst equation we have...

E = Eº - RT/nF ln Q

E = 2.72 - (8.314 x 360)/(2)(96,500) ln 3.23x10^-7

E = 2.72 - (0.0155 )( -14.9) = 2.72 + 0.23

E = 2.95 V

Answer 2

You must use the Nernst equation:

E_cell = E^o_cell - RT/nF ln Q

Find the E^o_cel using standard reduction potentials.

R = 8.314 J/mole

F = 96487 coulombs /mole

T = absolute temperature

Q is reaction quotient 7 x 10^-7 M / 2.158 M

n = 2

Plug these values into the Nernst equation.

Sorry I didn't have time to do it all.