20-3 Ecell, G, and KeqCells do electrical work.Moving electric charge.Faraday constant, F = 96,485 C mol-1

elec = -nFEG = -nFEG = -nFEMichael Faraday 1791-1867

Spontaneous ChangeG < 0 for spontaneous change.Therefore Ecell > 0 because Gcell = -nFEcell Ecell > 0Reaction proceeds spontaneously as written.Ecell = 0Reaction is at equilibrium.Ecell < 0Reaction proceeds in the reverse direction spontaneously.

The Behavior or Metals Toward AcidsM(s) M2+(aq) + 2 e- E = -EM2+/M2 H+(aq) + 2 e- H2(g) EH+/H2 = 0 V2 H+(aq) + M(s) H2(g) + M2+(aq)Ecell = EH+/H2 - EM2+/M = -EM2+/MWhen EM2+/M < 0, Ecell > 0. Therefore G < 0.Metals with negative reduction potentials react with acids.

Relationship Between Ecell and KeqG = -RT ln Keq = -nFEcell

Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

20-4 Ecell as a Function of ConcentrationG = G -RT ln Q-nFEcell = -nFEcell -RT ln QConvert to log10 and calculate constants.

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)Applying the Nernst Equation for Determining Ecell. What is the value of Ecell for the voltaic cell pictured below and diagrammed as follows?

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag (s)Ecell = 0.029 V 0.018 V = 0.011 V

Concentration CellsTwo half cells with identical electrodes but different ion concentrations.2 H+(1 M) 2 H+(x M)Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)2 H+(1 M) + 2 e- H2(g, 1 atm)H2(g, 1 atm) 2 H+(x M) + 2 e-

Concentration CellsEcell = - 0.0592 V log xEcell = (0.0592 V) pH2 H+(1 M) 2 H+(x M)

Chemistry 140 Fall 2002