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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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Chem 114 fundamental chemistry

Your road map for the third hour exam

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

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


Chem 114 fundamental chemistry

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


Chem 114 fundamental chemistry

Properties of Group 1 elements

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

Solar system abundances of the elements

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

Abundances of Group 1 elements

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

How sodium compounds are made

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

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


Chem 114 fundamental chemistry

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


Chem 114 fundamental chemistry

Properties of Group 2 elements

CHEM 114 Fundamental Chemistry


Chem 114 fundamental chemistry

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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