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CHEM 114 Fundamental Chemistry

Your road map for the third hour exam. CHEM 114 Fundamental Chemistry. Current involved in electrolysis. Na + ( aq. ) + e – → Na( Hg. amalgam ). E° = –2.713 V. 2 Cl – ( aq ) → Cl 2 ( g ) + 2 e –. E° = –1.358 V. 2 Na + ( aq. ) + 2 Cl – ( aq ) → 2 Na( Hg. amalgam ) + Cl 2 ( g ).

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CHEM 114 Fundamental Chemistry

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  1. Your road map for the third hour exam CHEM 114 Fundamental Chemistry

  2. Current involved in electrolysis Na+(aq.) + e– → Na(Hg. amalgam) E° = –2.713 V 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 For each Cl2 (g) we need two electrons. For each mole of Cl2 (g) (71 g) we need two moles of electrons = 2 moles × 1 Faraday = 2 × 96485.3 Coulombs = 192971 C = 192971 Ampere seconds So we could put 10 A through the cell for 19297 seconds (5.4 hours), or 100 A for 32 minutes (etc.) Minimum power to produce 1 mole of Cl2 (g) in 32 minutes is 100 A × –4.071 V = 407.1 W Electrical work needed to produce 1 mole of Cl2 (g) is wE = –nFE = –2 × 96485.3 × 4.071 = 785,583 J = 785,583/3,600,000 kWh = 0.22 kWh CHEM 114 Fundamental Chemistry

  3. Main group elements p-block s-block IE = Ionization energy: ΔE for M → M+ + e– CHEM 114 Fundamental Chemistry EA = Electron affinity: –ΔE for M + e– → M– EN = Electronegativity (hokey made-up quantity) Polarizability α: p = αE, where E is an applied electric field and p the induced dipole moment

  4. Properties of Group 1 elements CHEM 114 Fundamental Chemistry

  5. Solar system abundances of the elements CHEM 114 Fundamental Chemistry

  6. Abundances of Group 1 elements CHEM 114 Fundamental Chemistry

  7. How sodium compounds are made CHEM 114 Fundamental Chemistry

  8. Occurrence of Group 2 elements Beryl: Be3Al2(SiO3)6 varieties: emerald, aquamarine, Bertrandite: Be4Si2O7(OH)2 Chrysoberyl: Al2BeO4 Magnesium: 3rd most abundant element in seawater; 11th in the body. Calcium: 5th most abundant element in seawater; most abundant metal in the body. Strontianite: SrCO3 CHEM 114 Fundamental Chemistry Barite: BaSO4 Celestine: SrSO4

  9. Production of Group 2 elements Magnesium: by electrolysis of molten MgCl2 Mg2+ + 2 e– → Mg 2Cl– → Cl2 + 2 e– Beryllium: by reduction of BeF2 with magnesium Mg + BeF2 → MgF2 + Be Calcium: by electrolysis of molten CaCl2 Ca2+ + 2 e– → Ca 2Cl– → Cl2 + 2 e– Strontium: by electrolysis of molten SrCl2/KCl Sr2+ + 2 e– → Sr 2Cl– → Cl2 + 2 e– Barium: by electrolysis of molten BaCl2 Ba2+ + 2 e– → Ba 2Cl– → Cl2 + 2 e– CHEM 114 Fundamental Chemistry Alternative: reduction of the oxide with aluminum at high temperature 4 BaO + 2 Al → BaAl2O4 + 3 Ba

  10. Properties of Group 2 elements CHEM 114 Fundamental Chemistry

  11. Chemistry of the group 2 metals Halides are ionic, except beryllium Formula MX2 : M = {Be, Mg, Ca, Sr, Ba}; X = {F, Cl, Br, I}; Oxides are ionic, and alkaline, except beryllium. CaO + H2O → Ca(OH)2 (CaO is quicklime; Ca(OH)2 ‘slaked lime’) Beryllium oxide is amphoteric BeO + H2O + 2 H3O+ → [Be(H2O)4]2+ [Be(H2O)4]2+ is a moderately strong Lewis acid BeO + H2O + 2 OH– → [Be(OH)4]2– Barium forms some peroxide when it burns in air. Ba + O2 → BaO2 Ca, Sr, Ba sulfates and carbonates are more-or-less insoluble in water CHEM 114 Fundamental Chemistry

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