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Electrochemistry

Electrochemistry. Ch. 17. Electrochemistry. Generate current from a reaction Spontaneous reaction Battery Use current to induce reaction Nonspontaneous reaction Electroplating. Oxidation-Reduction Reaction. aka Redox Transfer of electrons Donor + Acceptor

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Electrochemistry

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  1. Electrochemistry Ch. 17

  2. Electrochemistry • Generate current from a reaction • Spontaneous reaction • Battery • Use current to induce reaction • Nonspontaneous reaction • Electroplating

  3. Oxidation-Reduction Reaction • aka Redox • Transfer of electrons Donor + Acceptor (reducing agent) (oxidizing agent) (is oxidized) (is reduced)

  4. Oxidation Involves Loss of electrons Reduction Involves Gain of electrons Loss of Electrons is Oxidation says Gain of Electrons is Reduction Mnemonics are cool!

  5. Assigning Oxidation States (1) Covalent bond btw identical atoms => Split electrons evenly (2) Covalent bond btw different atoms => All electrons given to more electronegative atom.

  6. (3) For ionic compounds, oxidation states are equal to ionic charge. (4) Oxidation state for an elemental atom is zero.

  7. (5) Oxidation state for monatomic ion is the same as the charge. (6) In compounds, fluorine always has an O.S. of -1.

  8. (7) Oxygen usually has an O.S. of -2, except when in a peroxide or when in OF2. H2O-2 H2O2-1 +2OF2 (8) With a nonmetal, hydrogen has an O.S. of +1. With a metal, H is assigned an O.S. of -1. NH3+1 LiH-1 (9) The sum of the oxidation states must add up to the overall charge.

  9. Examples Assign the oxidation states to each atom of the following compounds. CO2 CH4 K2Cr2O7

  10. Redox Reactions CH4 + O2 CO2 + H2O Which species is oxidized? Which species is reduced? Which species is the oxidizing agent? Which species is the reducing agent?

  11. Balancing Redox Reaction • Balance… …# of atoms …# of electrons transferred …overall charge • Types of reactions • Acidic conditions • Basic conditions

  12. Redox in Acidic Solutions Cr2O72- + C2H5OH  Cr3+ + CO2 • Assign oxidation states • Write half reactions Red: Cr2O72- Cr3+ Ox: C2H5OH  CO2

  13. 3. Balance elements except H and O Cr2O72- 2Cr3+ C2H5OH  2CO2 4. Balance oxygen by adding H2O Cr2O72- 2Cr3+ + 7H2O 3H2O + C2H5OH  2CO2 5. Balance hydrogen by adding H+ 14H+ + Cr2O72- 2Cr3+ + 7H2O 3H2O + C2H5OH  2CO2 + 12H+

  14. 6. Balance charge by adding electrons 6e-+14H+ + Cr2O72- 2Cr3+ + 7H2O 3H2O + C2H5OH  2CO2 + 12H+ + 12e- 7. Equalize the number of electrons 12e- + 28H+ + 2Cr2O72-  4Cr3+ + 14H2O 3H2O + C2H5OH  2CO2 + 12H+ +12e- 8. Cancel like terms and add reactions 16H+ + 2Cr2O72- + C2H5OH  4Cr3+ + 2CO2 + 11H2O 9. Check your answer!

  15. Balancing in basic solution Following the same algorithm used for acidic solutions through step #8 then… 9. Add the same # of OH- to both sides of equation as there are H+ on one side 10. Combine H+ and OH- on same sides of equation to make H2O 11. Cancel any like terms and check

  16. Galvanic Cells • Spontaneous chemistry generating current • Some terms • Reducing agent • Oxidizing agent • Half reactions • Anode • Cathode • Cell potential

  17. Building a Galvanic Cell Overall Reaction 8H+(aq) + MnO4-(aq) + 5Fe2+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Half Reactions Reduction: 8H+ + MnO4- + 5e-→ Mn2+ + 4H2O Oxidation: 5(Fe2+→ Fe3+ + e-)

  18. Reduction: 8H+ + MnO4- + 5e-  Mn2+ + 4H2OOxidation: Fe2+  Fe3+ + e- salt bridge KNO3

  19. Calculating Cell Potential Reduction: 8H+ + MnO4- + 5e-  Mn2+ + 4H2O εº(reduction) = 1.51 V Oxidation: 5(Fe2+  Fe3+ + e-) εº(oxidation) = -0.77 V εº(cell) = εº(red) + εº(ox) = 0.74 V

  20. Comments on Cell Potential • Potential is an intensive property • DO NOT multiply potential by balancing factor • The º indicates standard conditions • 1.0 M and 1 atm • Potentials references to standard H+ red. 2H+ + 2e- → H2εº = 0.00 V

  21. Fe or Pt ←- +→ Cu Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)  e- e- e-  salt bridge Fe3+ Cu2+ Oxidation: Reduction: Cu  Cu2+ + 2e- Fe3+ + e- Fe2+

  22. Demo Cu2+ + Zn  Zn2+ + Cu

  23. Line Notation 2Al3+(aq) + 3Mg(s) → 2Al(s) + 3Mg2+(aq)

  24. Line Notation II MnO4-(aq) + H+(aq) + ClO3-(aq) → ClO4-(aq) + Mn2+(aq) + H2O(l) Pt(s) │ ClO3-(aq), ClO4-(aq) ║ MnO4-(aq), Mn2+(aq) │ Pt(s)

  25. To Review • Full description of galvanic cell requires: • Composition of solutions • Composition of electrodes • Direction of electron flow • Direction of ion flow • Calculation of cell potential • Labels: “anode” and “cathode”

  26. Cell Potential and Free Energy

  27. Reconsidering Cell Potential Given: Al3+ + 3e- Al ΔG1 and ε1 = -1.66V Mg2+ + 2e-  Mg ΔG2 and ε1 = -2.37 V Find ε(cell) for: 2Al3+ + 3Mg  2Al + 3Mg2+

  28. Cell Potential and Spontaneity • Can bromine oxidize iodide to iodine? • Can Cr(II) reduce oxygen gas under acidic conditions to produce water? • Can Ag(I) oxidize chloride to chlorine? • Can hydrogen reduce Fe(II) to elemental iron?

  29. Non-Standard Cell Potential

  30. Practice Problems Determine ε for the following reaction and conditions: 2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn (a) [Al3+] = 2.0 M; [Mn2+] = 1.0 M @ 25°C (b) [Al3+] = 1.0 M; [Mn2+] = 3.0 M @ 25°C

  31. Do Worksheet

  32. Potential and Equilibrium Cell potential at equilibrium = 0.0 V Q = K at equilibrium

  33. Types of Batteries • Lead storage (car battery) • Dry cell battery • Acidic • Alkaline • Rechargeable • Lithium ion battery • Fuel cell

  34. Lead Storage Battery Anode: Pb + HSO4-→ PbSO4 + H+ + 2e- Cathode: PbO2 + HSO4- + 3H+ + 2e- → PbSO4 + 2H2O Overall Potential: εº = 2.04 V

  35. Dry Cell Battery (acidic) Anode: Zn → Zn2+ + 2e- Cathode: 2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O Overall Potential: εº = 1.5 V

  36. Dry Cell Battery (alkaline) Anode: Zn → Zn2+ + 2e- Cathode: 2MnO2 + H2O + 2e-→ Mn2O3 + 2OH- Overall Potential: εº = 1.5 V

  37. Lithium Ion Battery Anode: Li  Li+ + e- Cathode: MnO2 + Li+ + e-  LiMnO2 Cell Potential: εº = 3.6 V

  38. Fuel Cell Anode: 2H2(g) + 4OH-(aq)  4H2O + 4e- Cathode: O2 + 2H2O + 4e-  4OH- Cell Potential: εº = 1.23 V

  39. Corrosion • A significant portion of construction is done to replace corroded materials. Cathode: O2 + 2H2O + 4e- 4OH-ε = 0.40 V Anode: Fe  Fe2+ + 2e-ε = 0.44 V ε(cell) = 0.84 V

  40. Corrosion and Acidic Conditions Cathode: O2 + 4H+ + 4e- 2H2O ε = 1.23 V Anode: Fe  Fe2+ + 2e-ε = 0.44 V ε(cell) = 1.67 V

  41. Electrolysis • Supply current to perform chemistry • Performed in an electrolytic cell • Stoichiometric relationship btw. charge and chemical amount • Factor-label fun! • Current measured in Ampere = 1 coulomb per second

  42. Example Problem I How long will it take to plate out 1.00 kg of aluminum from an aqueous solution of Al3+ using a current of 100.0 A?

  43. Al3+ + 3e- Al

  44. Example Problem II What volume of F2 gas, at 25C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of K metal is produced? At which electrode does each reaction occur?

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