General Chemistry 2 Math Question 2

Question

A voltaic cell was constructed using the following half-reactions: Cu2+(aq) + 2 e-1 " Cu(s) Sn2+(aq) + 2 e-1 " Sn(s) Determine the net cell reaction, the standard cell potential, E0 cell, and ΔG° for this cell. (Use the table of standard reduction potentials in your text.) Does ΔGo indicate a product-favored or a reactant-favored reaction? Is this consistent with a voltaic/galvanic cell? Explain your answer. Calculate the cell potential at 25 o C if the concentration of Cu2+ is 0.100 M and the concentration of Sn2+ is 0.010M

Answer 1

Cu²⁺ + 2e- ==> Cus(s) Eº = +0.34 V cathode - reduction

Sn2+ + 2e- ==> Sn(s) Eº = -0.14 V anode - oxidation

Net cell reaction: Cu²⁺(aq) + Sn(s) ==> Cu(s) + Sn²⁺(aq)

Eº_cell = 0.34 V - (0-.14 V ) = +0.48 V

∆Gº = -nFEº_cell where n = moles of electrons; F = Faraday = -96,500 coulombs/mol e-; Eº = E of cell

∆Gº = - (2 mol e-)(96,500 coulombs/mol)(0.48 V) = -92,640 Coulomb Volt

1 C-V = 1 joule

∆Gº = -92,640 J = -92.6 kJ This represents a product favored reaction b/c it is negative (spontaneous) This is consistent with the derived voltage as it is +0.48 V (again a positive value = spontaneous)

Nernst equation:

E = Eº - RT/nF lnQ

At 25ºC and using log10 we have ...

E = Eº - 0.0592 V/n log Q

Cu²⁺(aq) + Sn(s) ==> Cu(s) + Sn²⁺(aq)

Q = [Sn2+]/[Cu2+] = 0.01/0.100 = 0.1

E = 0.48 - (0.0592/2) log 0.1

E = 0.48 - (0.0296 x -1)

E = 0.48 + 0.0296

E = 0.51 V