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Electrochemistry

the study of the interchange of chemical and electrical energy. Electrochemistry. Electrochemistry Terminology #1. Oxidation – the loss of electrons, increase in oxidation number Na(s)  Na + + e - Reduction – the gain of electrons, decrease of oxidation number

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Electrochemistry

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  1. the study of the interchange of chemical and electrical energy Electrochemistry

  2. Electrochemistry Terminology #1 • Oxidation – the loss of electrons, increase in oxidation number Na(s)  Na+ + e- • Reduction – the gain of electrons, decrease of oxidation number Cl2 + 2e- 2Cl-

  3. Electrochemistry Terminology #2 An old memory device for oxidation and reduction goes like this… LEOsaysGER LoseElectrons =Oxidation GainElectrons=Reduction

  4. Electrochemistry Terminology #3 • Oxidizing agent The substance that is reduced is the oxidizing agent • Reducing agent The substance that is oxidized is the reducing agent

  5. Electrochemistry Terminology #4 • Voltaic or Galvanic cell is a battery but not a dry cell; generates useful electrical energy • Electrolytic cell Requires useful electrical energy to drive a thermodynamically unfavorable reaction.

  6. Electrochemistry Terminology #5 • Anode The electrode where oxidation occurs • Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode

  7. Table of Reduction Potentials Measured against the StandardHydrogenElectrode

  8. Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

  9. Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E0 A negative free energy change, (-G)

  10. Zn - Cu Galvanic Cell Zn2+ + 2e- Zn E = -0.76V Cu2+ + 2e-  Cu E = +0.34V From a table of reduction potentials:

  11. Zn - Cu Galvanic Cell Cu2+ + 2e-  Cu E = +0.34V Zn  Zn2+ + 2e- E = +0.76V Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

  12. Line Notation An abbreviated representation of an electrochemical cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Anode material Cathode material Anode solution Cathode solution | || |

  13. Calculate cell potential Eocell = EoCathode – Eo Anode

  14. Calculating G0 for a Cell G0 = -nFE0 n= moles of electrons in balanced redox equation F=Faraday constant = 96,485 coulombs/mol e- Zn + Cu2+  Zn2+ + Cu E0= + 1.10 V

  15. The Nernst Equation Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R= 8.31 J/(molK) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e-

  16. Nernst Equation Simplified At 25 C (298 K) the Nernst Equation is simplified this way:

  17. Equilibrium Constants and Cell Potential At equilibrium, forward and reverse reactions occur at equal rates, therefore: • The battery is “dead” • The cell potential, E, is zero volts Modifying the Nernst Equation (at 25 C):

  18. Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu E0= + 1.10 V

  19. ??? Concentration Cell Both sides have the same components but at different concentrations. Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

  20. ??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration Zn2+ (1.0M) + 2e- Zn (reduction) Zn  Zn2+ (0.10M) + 2e- (oxidation) Zn2+ (1.0M)  Zn2+ (0.10M)

  21. Concentration Cell ??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C). Zn2+ (1.0M)  Zn2+ (0.10M)

  22. Nernst Calculations Zn2+ (1.0M)  Zn2+ (0.10M)

  23. Electrolytic Processes Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0) A positive free energy change, (+G)

  24. Electrolysis of Water In acidic solution Anode rxn: -1.23 V Cathode rxn: -0.83 V -2.06 V

  25. Electroplating of Silver Anode reaction: Ag  Ag+ + e- Cathode reaction: Ag+ + e- Ag Electroplating requirements: 1. Solution of the plating metal 2. Anode made of the plating metal 3. Cathode with the object to be plated 4. Source of current

  26. Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current? Ag+ + e-  Ag 5.0 g 1 mol Ag 1 mol e- 96 485C 1 s 20.0 C 1 mol e- 107.87 g 1 mol Ag = 2.2 x 102 s

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