Free Energy and Cell Potential

 G o = - nF  o Faraday (F) = the charge of 1 mole of electrons q = nFw = -q  heat work

 G o = - nF  o Calculate  G o for Cu 2+ (aq) + Fe (s)  Cu(s) + Fe 2+ (aq) Is the rxn spon? Cu 2+ (aq) + 2 e -  Cu(s)  o = 0.34 V Fe(s)  Fe 2+ (aq) + 2 e-  o = 0.44 V  o cell = 0.78 V  G o = - nF  o = -(2 mol)(96,485 C/mol)(0.78 J/C)  G o = x 10 5 J Neg means spon

Dependence of cell potential on [ ] 2Al(s) + 3Mn 2+ (aq)  2Al 3+ (aq) + 3Mn(s)  o cell =0.48V Predict  cell larger or smaller than  o cell a)[Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M product [ ] > 1 M therefore  cell >  o cell b) [Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M Reactant [ ] > 1 M therfore  o cell >  cell

Nerst Equation mol e- [products] [reatants] All at 1.0 M Q = 1, log(Q) = 0  cell =  o cell Q = K at eq  cell = 0 V Dead Battery