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CHEM 114 Lecture 18 2/23/2011

Dry cells. Cathode reaction. 2 MnO 2 ( s ) + H 2 O (l) + 2e – → Mn 2 O 3 ( s ) + 2 OH – ( aq ). E° = 0.118 V. NH 4 + ( aq ) + OH – ( aq ) → NH 3 ( g ) + H 2 O (l).

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CHEM 114 Lecture 18 2/23/2011

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  1. 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

  2. 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)

  3. 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)

  4. 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)

  5. Galvanization The more electropositive metal dissolves, leaving electrons (and a slight negative charge) on the less electropositive metal. This prevents it from dissolving. E° = 0.763 V Zn (s) → Zn2+(aq) + 2 e– So we can build up a voltage of > –0.8 V on the metal, and the reaction still goes forward CHEM 114 Lecture 18 2/23/2011 E° = 0.440 V Fe (s) → Fe2+(aq) + 2 e– With –0.8 V on the Fe, this reaction will not go forward!

  6. Sacrificial anodes CHEM 114 Lecture 18 2/23/2011

  7. Electrolysis We ain’t talking about hair removal! If we put a more negative voltage than 1.103 V across the cell, the process will reverse! CHEM 114 Lecture 18 2/23/2011 Zn(s.) + Cu2+(aq.) → Cu(s.) + Zn2+(aq.) EMF = 1.103 V

  8. Chloralkali process Na+(aq.) + e– → Na(Hg. amalgam) E° = –2.713 V CHEM 114 Lecture 18 2/23/2011 2 Cl– (aq) → Cl2 (g) + 2 e– E° = –1.358 V 2 Na+(aq.) + 2 Cl– (aq) → 2 Na(Hg. amalgam) + Cl2 (g) E° = –4.071 V 2 Na(Hg. amalgam) + 2 H2O → 2 Na+(aq) + 2 OH– (aq) + H2 (g)

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