2. Electrochemistry The study of the interchange of chemical and electrical energy. Sample electrochemical processes: 1) Corrosion 4 Fe (s) + 3 O 2(g) ⇌ 2 Fe 2 O 3(s) 2) Biological processes C 6 H 12 O 6 + 6 O 2 ⇌ 6 CO 2 + 6 H 2 O 3) Batteries (Galvanic or Voltaic cells) Electrochemical cells that produce a current (flow of electrons) as a result of a redox reaction 4) Electrolytic cells Electrical energy is used to produce chemical change Used to prepare or purify metals (such as sodium, aluminum, copper)

3. Chemical Change  Electron Flow Copper: Cu (s), Cu 2+ (aq) Cu (s)  Cu 2+ (aq) + 2e - ΔG° rxn = ΔG° f (Cu 2+ ) = 65.6 kJ Silver: Ag (s), Ag + (aq) Ag (s)  Ag + (aq) + e - ΔG° rxn = ΔG° f (Ag + ) = 77.2 kJ Cu (s)  Cu 2+ (aq) + 2e - ΔG° = +65.6 kJ Ag + (aq) + e-  Ag (s) ΔG° = -77.2 kJ Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s) ΔG° = -88.8 kJ Spontaneous w max = -88.8 kJ Cu 2+ in solution 2( ) 2( ) Ag(s)

4. Harnessing the Energy Separate the half-reactions –Creates a galvanic or voltaic cell CuAg 1 M CuSO 4 Cu (s)  Cu 2+ (aq) + 2e - 1 M AgNO 3 Ag + (aq) + e -  Ag (s) Luigi Galvani Alessandro Volta Cu 2+ SO 4 2- Ag + NO 3 - KNO 3(aq) K+K+ NO 3 - e-e- salt bridge Oxidation Reduction AnodeCathode “cathode” and “reduction” begin with consonants “anode” and “oxidation” begin with vowels − (produces electrons) + (attracts electrons) Red Cat

5. Line Notation for Galvanic Cells CuAg 1 M CuSO 4 Cu (s)  Cu 2+ (aq) + 2e - 1 M AgNO 3 Ag + (aq) + e -  Ag (s) Cu 2+ SO 4 2- Ag + NO 3 - K+K+ e-e- Oxidation Reduction Anode (−) Cathode (+) Anode always on the left Cu (s)  Cu 2+ (1 M)  Ag + (1 M)  Ag (s) Cathode always on the right

6. Chemical Change  Electrical Work Chemical change produces electrical energy Electrical energy can be used to do work! ΔG = w max Electrical work: w = -nF ℰ n = # of moles e - transferred F = charge on a mole of e - ℰ = electrical potential (electromotive force) Cell Potential ( ℰ) or Electromotive Force (emf): The driving force pushing the electrons from the anode to the cathode. Units = Volts1 Volt = 1 joule/coulomb

7. Standard Reduction Potentials The cell potential ℰ ° cell can be determined from the standard reduction potentials ( ℰ °red) for the half-reactions: –Reduction potential = tendency for reduction to happen Positive ℰ °red  spontaneous reduction reaction Negative ℰ °red  non-spontaneous reduction or spontaneous oxidation (reverse reaction) –Standard ( o ) = standard conditions (1 M solutions, 1 atm gases)

8. Standard Reduction Potentials Half-Reaction ℰ ° (V) F 2 + 2e -  2F - 2.87 Au 3+ + 3 e -  Au 1.50 Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 2H + + 2e -  H 2 0.00 Ni 2+ + 2e -  Ni-0.23 Zn 2+ + 2e -  Zn-0.76 Al 3+ + 3e -  Al-1.66 Li + + e -  Li-3.05  ℰ ° = 0 (SHE) Standard Hydrogen Electrode ℰ ° > 0 Spontaneous reduction ℰ ° < 0 Non-Spontaneous reduction Reduction potential = tendency for reduction to happen Standard = standard conditions (1 M solutions, 1 atm gases) Spontaneous oxidation (reverse rxn)

9. Standard Reduction Potentials Half-Reaction ℰ ° (V) F 2 + 2e -  2F - 2.87 Au 3+ + 3 e -  Au 1.50 Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 2H + + 2e -  H 2 0.00 Ni  Ni 2+ + 2e - +0.23 Zn  Zn 2+ + 2e - +0.76 Al  Al 3+ + 3e - +1.66 Li  Li + + e - +3.05 ℰ ° > 0 Spontaneous reduction But remember, an oxidation CANNOT happen without a reduction Spontaneous oxidation

10. Standard Reduction Potentials Half-Reaction ℰ ° (V) F 2 + 2e -  2F - 2.87 Au 3+ + 3 e -  Au 1.50 Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 2H + + 2e -  H 2 0.00 Ni 2+ + 2e -  Ni-0.23 Zn 2+ + 2e -  Zn-0.76 Al 3+ + 3e -  Al-1.66 Li + + e -  Li-3.05  Strongest Oxidizing Agent (most easily reduced) Reduction potential = tendency for reduction to happen Standard = standard conditions (1 M solutions, 1 atm gases)  Strongest Reducing Agent (most easily oxidized)

11. Cell Potential ℰ ° cell = ℰ ° reduction + ℰ ° oxidation Ag + (aq) + e -  Ag (s) ℰ ° = 0.80 V Cu 2+ (aq) + 2 e -  Cu (s) ℰ ° = 0.34 V Reduction reaction:2(Ag + (aq) + e -  Ag (s) ) ℰ ° = +0.80 V Oxidation reaction:Cu (s)  Cu 2+ (aq) + 2 e - ℰ ° = - 0.34 V Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s) ℰ ° cell = +0.46 V ℰ ° is intensive, unlike ΔG o Cu (s)  Cu 2+ (1 M)  Ag + (1 M)  Ag (s) The ℰ °cell MUST be + and thus spontaneous for Galvanic cells

12. Free Energy and Cell Potential  G  = w max =  nF ℰ  n= number of moles of electrons transferred F= Faraday’s constant = 96,485 coulombs per mole of electrons (C/mol e - ) ℰ °= standard cell potential (V or J/C) Michael Faraday Cu (s)  Cu 2+ (1 M)  Ag + (1 M)  Ag (s) ℰ ° cell = +0.46 V ΔG° = -nF ℰ ° cell ΔG° = -(2 mol e - )(96485 C/mol e - )(0.46 V) ΔG° = -88,800 J or -88.8 kJ

13. Practice Time Given the following information, draw a galvanic cell. Fe(s)  Fe 2+ (1 M)  Au 3+ (1 M)  Au(s) Be sure to include the following: Anode/Cathode reactions Balanced overall reaction Complete circuit (external wire with e- flow direction, salt bridge) Label all parts of the cell (solution, electrode, etc.)

14. Fe(s)  Fe 2+ (1 M)  Au 3+ (1 M)  Au(s) FeAu 1 M Fe 2+ Fe (s)  Fe 2+ (aq) + 2e - 1 M Au 3+ Au 3+ (aq) + 3e -  Au (s) Fe 2+ Au 3+ anions e-e- Oxidation Reduction AnodeCathode 3Fe(s) + 2Au 3+ (aq) → 3Fe 2+ (aq) + 2Au(s) cations ℰ °cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 V

15. Reaction Quotient The reaction quotient (Q) sets up a ratio of products and reactants For a reaction, A + 2B → 3C + 4D [C] 3 [D] 4 [A] 1 [B] 2 Only concentrations (aq) or pressures (g) are used to solve for Q Solids (s) and liquids (l) are not included in the expression Q =

16. Reaction Quotient practice Write the Q expression for the following reaction CH 4 (g) + O 2 (g) → CO 2 (g) + H 2 O(g) Reaction must be balanced first CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g ) (CO 2 )(H 2 O) 2 (CH 4 )(O 2 ) 2 Q =

17. Reaction Quotient practice Write the Q expression for the following reaction Cu(s) + 2Ag + (aq)  Cu 2+ (aq) + 2Ag(s) (Cu 2+ )(Ag) 2 (Cu)(Ag + ) 2 Is this correct? NO: Solids aren’t included in the equation! (Cu 2+ ) (Ag + ) 2 Q =

18. Non-standard conditions: The Nernst Equation We can calculate the potential of a cell in which some or all of the components are not in their standard states (not 1 M concentration or 1 atm pressure). ΔG = ΔG° + RT lnQ ΔG = -nF ℰ ΔG° = -nF ℰ ° -nF ℰ = -nF ℰ ° + RT lnQ Walther Nernst ℰ = ℰ ° - R= 8.3415 J/mol K T= temperature n= moles of e- F= Faraday’s constant 96,485 C/mol e -

19. Practice with the Nernst Equation What will be the cell potential ℰ of a Cu/Ag cell using 0.10 M Cu 2+ and 1.0 M Ag + solutions at 25°C? Cu (s)  Cu 2+ (aq) + 2e - Ag + (aq) + e -  Ag (s) CuAg Cu 2+ SO 4 2- Ag + NO 3 - Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s) ℰ= 0.46 V – (-0.03 V) Cu (s)  Cu 2+ (0.10 M)  Ag + (1.0 M)  Ag (s) ℰ ℰ= 0.49 V ℰ = ℰ ° -

20. Brain Warmup Half-Reaction ℰ ° (V) Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 Zn 2+ + 2e -  Zn-0.76 Al 3+ + 3e -  Al-1.66 What is ℰ ° for each of the following reactions? Which reaction(s) are spontaneous? 3 Ag + (aq) + Al (s)  3 Ag (s) + Al 3+ (aq) Cu 2+ (aq) + Zn (s)  Cu (s) + Zn 2+ (aq) 2 Al 3+ (aq) + 3 Zn (s)  2 Al (s) + 3 Zn 2+ (aq) ℰ°ℰ° 2.46 V 1.10 V -0.90 V Spontaneous? Y Y N Zn can reduce Cu 2+, but not Al 3+