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CHEM 114 Fundamental Chemistry

Free energy and EMF. Let’s say we have a chemical process that involves 1 mole of electrons (e.g. 1 mol Fe 2+ -> 1 mol Fe 3+ + 1 mol e – ). How much charge is involved? Charge on 1 electron is e = –1.60217649 × 10 –19 C. A mole is N A particles:. N A = 6.0221418 × 10 23 mol –1.

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CHEM 114 Fundamental Chemistry

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  1. Free energy and EMF Let’s say we have a chemical process that involves 1 mole of electrons (e.g. 1 mol Fe2+ -> 1 mol Fe3+ + 1 mol e– ) How much charge is involved? Charge on 1 electron is e = –1.60217649 × 10–19 C A mole is NA particles: NA= 6.0221418 × 1023 mol–1 q = NAe = 6.0221418 × 1023 mol–1 × –1.60217649 × 10–19 C = 96485.3 C mol–1 = 1 F (Faraday) ΔGE = wE= qV = –nFE ΔGE,m = –FE for a 1 electron/mole process ΔGE,m = –NFE for a N electron/mole process Zn(s.) + 2Ag+(aq.) → 2Ag(s.) + Zn2+(aq.) EMF = 1.563 V What is ΔGE,m? As written, we have a 2 electron/mole Zn process. CHEM 114 Fundamental Chemistry ΔGE,m = –2 × 96485.3 C/mol × 1.563 V = –301,613 J/mol = –301.6 kJ/mol

  2. The Nernst equation ΔGE,m = ΔG°E,m + RT ln Q –NFE = –NFE° + RT ln Q E = E° – (RT/NF) ln Q This is the Nernst Equation: it shows the dependence of cell potential on concentration Walther Nernst, 1864 - 1941. Nobel Prize in Chemistry, 1920. Forced to retire by the Nazis, 1933. Problem: find the half-cell potential of a zinc cell in equilibrium with 0.01 M ZnSO4: E0 (Zn2+/Zn) = –0.7628 V at 298 K half-cell reaction: Zn2+ + 2e– → Zn: E° = –0.7628 V Q = aZn/aZn2+ ~ 1/cZn2+ = 1/0.01 = 100 CHEM 114 Fundamental Chemistry = –0.8219 V Useful numbers: RT/F = 0.025680 V at 298 K (RT/F) ln 10 = 0.059130 V at 298 K

  3. Non-standard conditions: example 20-9 Half-cell standard potentials Fe3+ (aq) + e– → Fe2+ (aq): E° = 0.771 V Ag+ (aq) + e– → Ag (s): E° = 0.800 V Under standard conditions, reduction of Ag+ will occur. E° = –0.771 V Fe2+ (aq)→ Fe3+ (aq) + e– E° = 0.800 V Ag+ (aq) + e– → Ag (s) Ag+ (aq) + Fe2+ (aq)→ Fe3+ (aq) + Ag (s) E° = 0.029 V Let’s assume aFe2+ = 0.10, aFe3+ = 0.20 RT/F ln (10) = 0.0591 V N = 1 E = E° – (RT/NF) ln Q CHEM 114 Fundamental Chemistry = 0.029 V – (0.0591 V) log10 [0.20/0.10) = 0.011 V

  4. Concentration cells Same reaction, different cell concentrations! E = 0 – (RT/NF) ln (aH+/1) E = –0.0591 log10aH+ Crude pH meter! Ksp= 8.5 × 10–17 [Ag+] = [I–] = 9.2 × 10–9 Q = 9.2 × 10–8 so log10Q = –7.035 CHEM 114 Fundamental Chemistry E = –0.0591 × –7.035 = 0.417 V

  5. Concentration cells AgCl/HCl reference Modern pH electrode Silver/silver chloride Calomel CHEM 114 Fundamental Chemistry 1 M Cl– 1 M Cl– AgCl (s) + e– → Ag (s) + Cl– (aq) ½ Hg2Cl2 (s) + e– → Hg (l) + Cl– (aq) Ag (s)| AgCl (s)|1 M Cl–,1 M H+|unknown H+ |1 M Cl– | AgCl (s)| Ag (s) E = 0.22233 V E = 0.2680 V

  6. Concentration cells Silver/silver chloride 1 M Cl– AgCl (s) ⇌ Ag+(aq) + Cl– (aq) Break this into two equilibria AgCl (s) + e– → Ag (s) + Cl– (aq) E = 0.22233 V Ag+(aq) + e– → Ag (s) E° = 0.800 V Reverse the second reaction and add to the first Ag (s) → Ag+(aq) + e– E° = –0.800 V AgCl (s) → Ag (aq) + Cl– (aq) E° = –.578 V E° = (RT/NF) ln K so ln K = NFE°/RT = –22.50 Ksp = 1.7 × 10–10 CHEM 114 Lecture 18 2/23/2011

  7. Dry cells Cathode reaction 2 MnO2 (s) + H2O (l) + 2e– → Mn2O3 (s) + 2 OH– (aq) E° = 0.118 V NH4+ (aq) + OH– (aq) → NH3 (g) + H2O (l) The cell has NH4+ and NH3 present ⇒ it is buffered to the pKa of NH4+ (~ 9.24). So log10 [OH–] = – 4.76 E = E° – (RT/NF) ln Q = 0.118 –(0.0591 × – 4.76) = .399 V Anode reaction E° = –0.763 V Zn2+(aq) + 2 e– → Zn (s) E° = 0.763 V Zn (s) → Zn2+(aq) + 2 e– This gives only 1.16 V. The rest of the effect comes from the insolubility of the zinc under alkaline conditions. CHEM 114 Lecture 18 2/23/2011

  8. Lead acid battery Cathode reaction E° = 1.74 V PbO2 (s) + 3 H+ (aq)+HSO4– (aq) + 2 e– → PbSO4 (s) + 2 H2O (l) Anode reaction E° = –0.28 V PbSO4 (s)+ H+ + 2 e– → Pb (s) + HSO4– (aq) But it’s at the anode, so it must be an oxidation: CHEM 114 Lecture 18 2/23/2011 E° = 0.28 V Pb (s) + HSO4– (aq) → PbSO4 (s)+ H+ + 2 e– Total reaction E° = 2.06 V PbO2 (s) + Pb (s) + 2 H+ (aq)+2HSO4– (aq) → 2 PbSO4 (s) + 2 H2O (l)

  9. Silver-zinc button battery Zn (s) | ZnO (s) | KOH(satd) | Ag2O (s)| Ag (s) Cathode reaction Ag2O (s) + H2O (l) + 2 e– → 2 Ag (s) + 2 OH– (aq) E° = 0.604 V Anode reaction E° = –1.246 V ZnO (s)+ H2O (l) + 2 e– → Zn (s) + 2 OH– (aq) E° = 1.246 V Zn (s) + 2 OH– (aq) → ZnO (s)+ H2O (l) + 2 e– Total reaction CHEM 114 Lecture 18 2/23/2011 E° = 1.85 V Ag2O (s) + Zn (s) → 2 Ag (s) + ZnO (s)

  10. Lithium battery Cathode reaction Lithium cobalt oxide is a ‘sponge for Li+; charge balance is taken care of by oxidizing the cobalt. Li(1–x)CoO2 (s) + x Li+ + x e– → LiCoO2 (s) Anode reaction CHEM 114 Lecture 18 2/23/2011 Graphite-intercalated lithium is a source of lithium metal x Li (s, intercalated)→ x Li+ + x e– Total reaction Li(1–x)CoO2 (s) + x Li (s, intercalated) → LiCoO2 (s)

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