Electrochemistry The study of the interchange of chemical and electrical energy. Sample electrochemical processes: 1) Corrosion 4 Fe(s) + 3 O2(g) ⇌ 2 Fe2O3(s) 2) Biological processes C6H12O6 + 6 O2 ⇌ 6 CO2 + 6 H2O 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)

Chemical Change  Electron Flow Copper: Cu(s), Cu2+(aq) Cu(s)  Cu2+(aq) + 2e- ΔG°rxn = ΔG°f(Cu2+) = 65.6 kJ Silver: Ag(s), Ag+(aq) Ag(s)  Ag+(aq) + e- ΔG°rxn = ΔG°f(Ag+) = 77.2 kJ Ag(s) Cu2+ in solution 2( Ag+(aq) + e-  Ag(s) ΔG° = -77.2 kJ ) 2( ) Cu(s)  Cu2+(aq) + 2e- ΔG° = +65.6 kJ Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ΔG° = -88.8 kJ Spontaneous wmax = -88.8 kJ

Harnessing the Energy − + Separate the half-reactions Red Cat Anode Creates a galvanic or voltaic cell Luigi Galvani Alessandro Volta e- Red Cat salt bridge − (produces electrons) + (attracts electrons) Cu NO3- K+ Ag KNO3(aq) Anode Cathode “anode” and “oxidation” begin with vowels “cathode” and “reduction” begin with consonants Cu2+ SO42- Ag+ NO3- 1 M CuSO4 Cu(s)  Cu2+(aq) + 2e- 1 M AgNO3 Ag+(aq) + e-  Ag(s) Oxidation Reduction

Line Notation for Galvanic Cells Cu NO3- K+ Ag Anode (−) Cathode (+) Cu2+ SO42- Ag+ NO3- 1 M CuSO4 Cu(s)  Cu2+(aq) + 2e- 1 M AgNO3 Ag+(aq) + e-  Ag(s) Oxidation Reduction Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) Anode always on the left Cathode always on the right

Chemical Change  Electrical Work Chemical change produces electrical energy Electrical energy can be used to do work! ΔG = wmax Electrical work: w = -nFℰ 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 = Volts 1 Volt = 1 joule/coulomb

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)

Standard Reduction Potentials Half-Reaction ℰ° (V) F2 + 2e-  2F- 2.87 Au3+ + 3 e-  Au 1.50 Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 2H+ + 2e-  H2 0.00 Ni2+ + 2e-  Ni -0.23 Zn2+ + 2e-  Zn -0.76 Al3+ + 3e-  Al -1.66 Li+ + e-  Li -3.05 ℰ° > 0 Spontaneous reduction  ℰ° = 0 (SHE) Standard Hydrogen Electrode ℰ° < 0 Non-Spontaneous reduction Spontaneous oxidation (reverse rxn) Reduction potential = tendency for reduction to happen Standard = standard conditions (1 M solutions, 1 atm gases)

Standard Reduction Potentials Half-Reaction ℰ° (V) F2 + 2e-  2F- 2.87 Au3+ + 3 e-  Au 1.50 Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 2H+ + 2e-  H2 0.00 Ni  Ni2+ + 2e- +0.23 Zn  Zn2+ + 2e- +0.76 Al  Al3+ + 3e- +1.66 Li  Li+ + e- +3.05 ℰ° > 0 Spontaneous reduction Spontaneous oxidation But remember, an oxidation CANNOT happen without a reduction

Standard Reduction Potentials Half-Reaction ℰ° (V) F2 + 2e-  2F- 2.87 Au3+ + 3 e-  Au 1.50 Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 2H+ + 2e-  H2 0.00 Ni2+ + 2e-  Ni -0.23 Zn2+ + 2e-  Zn -0.76 Al3+ + 3e-  Al -1.66 Li+ + e-  Li -3.05 Strongest Oxidizing Agent (most easily reduced) Strongest Reducing Agent (most easily oxidized) Reduction potential = tendency for reduction to happen Standard = standard conditions (1 M solutions, 1 atm gases)

Cell Potential ℰ°cell = ℰ°reduction + ℰ°oxidation Ag+(aq) + e-  Ag(s) ℰ° = 0.80 V Cu2+(aq) + 2 e-  Cu(s) ℰ° = 0.34 V Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) Reduction reaction: 2(Ag+(aq) + e-  Ag(s)) ℰ° = +0.80 V Oxidation reaction: Cu(s)  Cu2+(aq) + 2 e- ℰ° = - 0.34 V Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ℰ°cell = +0.46 V The ℰ°cell MUST be + and thus spontaneous for Galvanic cells ℰ° is intensive, unlike ΔGo

Free Energy and Cell Potential G = wmax = nFℰ 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) Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) ℰ°cell = +0.46 V ΔG° = -nFℰ°cell ΔG° = -(2 mol e-)(96485 C/mol e-)(0.46 V) ΔG° = -88,800 J or -88.8 kJ Michael Faraday

Practice Time Given the following information, draw a galvanic cell. Fe(s)Fe2+(1 M)Au3+(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.)

Fe(s)Fe2+(1 M)Au3+(1 M)Au(s) anions cations Au Anode Cathode Fe2+ Au3+ 1 M Fe2+ Fe(s)  Fe2+(aq) + 2e- 1 M Au3+ Au3+(aq) + 3e-  Au(s) Oxidation Reduction 3Fe(s) + 2Au3+(aq) → 3Fe2+(aq) + 2Au(s) ℰ°cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 V

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 =

Reaction Quotient practice Write the Q expression for the following reaction CH4(g) + O2(g) → CO2(g) + H2O(g) Reaction must be balanced first CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) (CO2)(H2O)2 (CH4)(O2)2 Q =

Reaction Quotient practice Write the Q expression for the following reaction Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) (Cu2+)(Ag)2 (Cu)(Ag+)2 Is this correct? NO: Solids aren’t included in the equation! (Cu2+) (Ag+)2 Q = Q =

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 = -nFℰ ΔG° = -nFℰ° -nFℰ = -nFℰ° + RT lnQ R = 8.3415 J/mol K T = temperature n = moles of e- F = Faraday’s constant 96,485 C/mol e- ℰ = ℰ° - Walther Nernst

Practice with the Nernst Equation What will be the cell potential ℰ of a Cu/Ag cell using 0.10 M Cu2+ and 1.0 M Ag+ solutions at 25°C? Cu(s)Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s) Cu(s)  Cu2+(aq) + 2e- Ag+(aq) + e-  Ag(s) Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ℰ = ℰ° - ℰ Cu Ag Cu2+ SO42- Ag+ NO3- ℰ = 0.46 V – (-0.03 V) ℰ = 0.49 V

Brain Warmup What is ℰ° for each of the following reactions? Half-Reaction ℰ° (V) Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 Zn2+ + 2e-  Zn -0.76 Al3+ + 3e-  Al -1.66 What is ℰ° for each of the following reactions? Which reaction(s) are spontaneous? ℰ° 2.46 V 1.10 V -0.90 V Spontaneous? Y N 3 Ag+(aq) + Al(s)  3 Ag(s) + Al3+(aq) Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq) 2 Al3+(aq) + 3 Zn(s)  2 Al(s) + 3 Zn2+(aq) Zn can reduce Cu2+, but not Al3+