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20-3 E cell , Δ G , and K eq

20-3 E cell , Δ G , and K eq. Cells do electrical work. Moving electric charge. Faraday constant, F = 96,485 C mol -1.  elec = - nFE. Michael Faraday 1791-1867. Δ G = - nFE. Δ G° = - nFE °. Spontaneous Change. Δ G < 0 for spontaneous change.

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20-3 E cell , Δ G , and K eq

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  1. 20-3 Ecell, ΔG, and Keq • Cells do electrical work. • Moving electric charge. • Faraday constant, F = 96,485 C mol-1 elec = -nFE Michael Faraday 1791-1867 ΔG = -nFE ΔG° = -nFE°

  2. Spontaneous Change • ΔG < 0 for spontaneous change. • Therefore E°cell > 0 because ΔGcell = -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.

  3. The Behavior or Metals Toward Acids M(s) → M2+(aq) + 2 e-E° = -E°M2+/M 2 H+(aq) + 2 e-→ H2(g) E°H+/H2 = 0 V 2 H+(aq) + M(s) → H2(g) + M2+(aq) E°cell = E°H+/H2 - E°M2+/M = -E°M2+/M When E°M2+/M < 0, E°cell > 0. Therefore ΔG° < 0. Metals with negative reduction potentials react with acids.

  4. RT E°cell = ln Keq nF Relationship Between E°cell and Keq ΔG° = -RT ln Keq = -nFE°cell

  5. Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

  6. RT Ecell = Ecell° -ln Q nF 0.0592 V Ecell = Ecell° - log Q The Nernst Equation: n 20-4 Ecell as a Function of Concentration ΔG = ΔG° -RT ln Q -nFEcell = -nFEcell° -RT ln Q Convert to log10 and calculate constants.

  7. EXAMPLE 20-8 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)

  8. EXAMPLE 20-8 0.0592 V Ecell = Ecell° - log Q 0.0592 V [Fe3+] n Ecell = Ecell° - log [Fe2+] [Ag+] n Ecell = 0.029 V – 0.018 V = 0.011 V Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

  9. Concentration Cells Two half cells with identical electrodes but different ion concentrations. 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- 2 H+(1 M) → 2 H+(x M)

  10. 0.0592 V Ecell = Ecell° - log Q 0.0592 V x2 n Ecell = Ecell° - log 12 n 0.0592 V x2 Ecell = 0- log 1 2 Concentration Cells 2 H+(1 M) → 2 H+(x M) Ecell = - 0.0592 V log x Ecell = (0.0592 V) pH

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