Slide 1 of E cell, ΔG, and K eq  Cells do electrical work.  Moving electric charge.  Faraday constant, F = 96,485 C mol -1  elec = -nFE ΔG = -nFE ΔG° = -nFE° Michael Faraday

Slide 2 of 54 Spontaneous Change  ΔG < 0 for spontaneous change.  Therefore E° cell > 0 because ΔG cell = -nFE° cell  E° cell > 0  Reaction proceeds spontaneously as written.  E° cell = 0  Reaction is at equilibrium.  E° cell < 0  Reaction proceeds in the reverse direction spontaneously.

Slide 3 of 54 The Behavior or Metals Toward Acids M(s) → M 2+ (aq) + 2 e - E° = -E° M 2+ /M 2 H + (aq) + 2 e - → H 2 (g) E° H + /H 2 = 0 V 2 H + (aq) + M(s) → H 2 (g) + M 2+ (aq) E° cell = E° H + /H 2 - E° M 2+ /M = -E° M 2+ /M When E° M 2+ /M 0. Therefore ΔG° < 0. Metals with negative reduction potentials react with acids.

Slide 4 of 54 Relationship Between E° cell and K eq ΔG° = -RT ln K eq = -nFE° cell E° cell = nF RT ln K eq

Slide 5 of 54 Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

Slide 6 of E cell as a Function of Concentration ΔG = ΔG° -RT ln Q -nFE cell = -nFE cell ° -RT ln Q E cell = E cell ° - ln Q nF RT Convert to log 10 and calculate constants. E cell = E cell ° - log Q n V The Nernst Equation:

Slide 7 of 54 Pt|Fe 2+ (0.10 M),Fe 3+ (0.20 M)||Ag + (1.0 M)|Ag(s) Applying the Nernst Equation for Determining E cell. What is the value of E cell for the voltaic cell pictured below and diagrammed as follows? EXAMPLE 20-8

Slide 8 of 54 E cell = E cell ° - log Q n V Pt|Fe 2+ (0.10 M),Fe 3+ (0.20 M)||Ag + (1.0 M)|Ag(s) E cell = E cell ° - log n V [Fe 3+ ] [Fe 2+ ] [Ag + ] Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) E cell = V – V = V EXAMPLE 20-8

Slide 9 of 54 Concentration Cells Two half cells with identical electrodes but different ion concentrations. 2 H + (1 M) → 2 H + (x M) Pt|H 2 (1 atm)|H + (x M)||H + (1.0 M)|H 2 (1 atm)|Pt(s) 2 H + (1 M) + 2 e - → H 2 (g, 1 atm) H 2 (g, 1 atm) → 2 H + (x M) + 2 e -

Slide 10 of 54 Concentration Cells E cell = E cell ° - log n V x2x E cell = 0 - log V x2x2 1 E cell = V log x E cell = ( V) pH 2 H + (1 M) → 2 H + (x M) E cell = E cell ° - log Q n V