Slide 1 of 54 20-3 E cell, ΔG, and K eq Cells do electrical work. Moving electric charge. ...
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Slide 1 of 54 20-3 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 1791-1867
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Transcript of Slide 1 of 54 20-3 E cell, ΔG, and K eq Cells do electrical work. Moving electric charge. ...
Slide 1 of 54
20-3 Ecell, ΔG, and Keq
Cells do electrical work. Moving electric charge.
Faraday constant, F = 96,485 C mol-1
elec = -nFE
ΔG = -nFE
ΔG° = -nFE°
Michael Faraday 1791-1867
Slide 2 of 54
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.
Slide 3 of 54
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.
Slide 4 of 54
Relationship Between E°cell and Keq
ΔG° = -RT ln Keq = -nFE°cell
E°cell = nF
RTln Keq
Slide 5 of 54
Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.
Slide 6 of 54
20-4 Ecell as a Function of Concentration
ΔG = ΔG° -RT ln Q
-nFEcell = -nFEcell° -RT ln Q
Ecell = Ecell° - ln QnF
RT
Convert to log10 and calculate constants.
Ecell = Ecell° - log Qn
0.0592 VThe Nernst Equation:
Slide 7 of 54
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?
EXAMPLE 20-8
Slide 8 of 54
Ecell = Ecell° - log Qn
0.0592 V
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Ecell = Ecell° - logn
0.0592 V [Fe3+][Fe2+] [Ag+]
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag (s)
Ecell = 0.029 V – 0.018 V = 0.011 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|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-
Slide 10 of 54
Concentration Cells
Ecell = Ecell° - logn
0.0592 V x2
12
Ecell = 0 - log2
0.0592 V x2
1
Ecell = - 0.0592 V log x
Ecell = (0.0592 V) pH
2 H+(1 M) → 2 H+(x M)Ecell = Ecell° - log Qn
0.0592 V
