Electrochemical Potential, Work, and Energy

1. Electrochemical Potential, Work, and Energy • Potential, Work, and Energy • Units • Joule (J) = unit of energy, heat, or work (w) = kg•m2/s2 • Coulomb (C) = unit of electrical charge (q). 1 e- = 1.6 x 10-19 C • = electrical potential (e) • 1 J of work is produced when 1 C of charge is transferred between two points differing by 1 V of electrical potential • Work flowing out of a system (Galvanic Cell) is taken to be negative work • Cell Potential is always positive • From last chapter, wmax = DG

2. Electrochemical Problems • When current flows, we always waste some of the energy as heat instead of work w < wmax • We can, however, measure emax with a potentiometer, so we can find the hypothetical value of wmax • Example: eocell = 2.50 V 1.33 mole e- pass through the wire. eactual = 2.10 V • 1 Faraday (F) = the charge on 1 mole of electrons = 96,485 C (6.022 x 1023 e-/mol)(1.6 x 10-19 C/e-) = 96,485 C/mol • w = -qe = -(1.33 mol e-)(96,485 C/mole e-)(2.10 J/C) = -2.69 x 105 J • wmax = -qemax = -(1.33 mol e-)(96,485 C/mole e-)(2.50 J/C) = -3.21 x 105 J • Efficiency = w/wmax = -2.69 x10-5 J/-3.21 x 105 J = 0.838 or 83.8% • Free Energy (DG) • q = nF where n = number of moles, F = 96,485 C/mole • DG = -nFe (assuming the maximum e) • Maximum cell potential is directly related to DG between reactants and products in the Galvanic Cell (This lets us directly measure DG)

3. Example: Calculate DGo for the reaction Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq) • Half Reactions: Cu2+ + 2e- Cuoeo = 0.34 V Feo Fe2+ + 2e- eo = 0.44 V b) DGo = -nFeo = -(2 mol e-)(96,485 C/mol e-)(0.78 J/C) = -1.5 x 105 J • Example: Will 1 M HNO3 dissolve metallic gold to make 1 M Au3+? • Half Reaction: NO3- + 4H+ + 3e- NO + 2H2O eo = +0.96 V Auo Au3+ + 3e-eo = -1.50 V Au(s) + NO3-(aq) + 4H+(aq) Au3+(aq) + NO(g) + 2H2O(l) eocell = -0.54V • Since e is negative (DG = +) the reaction will not occur spontaneously • Cell Potential and Concentration • Concentration Cells • Up until now, concentration for all Galvanic solutions = 1 M (Gives eo) • What happens if we change these concentrations? Eocell = +0.78 V

4. 3) Le Chatelier’s Principle • Cu(s) + 2Ce4+(aq) Cu2+(aq) + 2Ce3+(aq) eocell = 1.36 V • Increase Ce4+ concentration, (e > eo) • Increase Cu2+ concentration, (e < eo) d) Example • Concentration Cell = Galvanic Cell driven by the fact that concentrations of the same reactants are different on the two sides of the cell. • Example: Ag+ + e- Agoeo1/2 = +0.80 V • If both sides had [Ag+] = 1 M, then eocell = +0.80 V + (-0.80 V) = 0.00 V • If [Ag+]right = 1 M and [Ag+]left = 0.1 M then we should have a potential • Diffusion would try to equalize Ag+ on the right side and the left side (Entropy favors even distribution, like gas particles in two chambers) • Electrons would flow from left to right to even out [Ag+] • A very small voltage would be generated • Example

6. The Nernst Equation • Derivation • DG = DGo + RTlnQ = -nFe • DGo = -nFeo • -nFe = -nFeo + RTlnQ • At 25 oC, this simplifies to • Example: 2Al(s) + 3Mn2+(aq) 2Al3+(aq) + 3Mn(s) eocell = 0.48 V • Oxidation: 2Al(s) 2Al3+(aq) + 6e- • Reduction: 3Mn2+(aq) + 6e- 3Mn(s) • [Mn2+] = 0.5 M, [Al3+] = 1.5 M • Q = [Al3+]2 / [Mn2+]3 = (1.5)2 / (0.5)3 = 18 • As the reaction proceeds, ecell 0 (Q K) = Dead Battery! • Calculating K:

8. VO2+ + 2H+ + e- VO2+ + H2O eo = 1.00 V Zn2+ + 2e- Zn eo = -0.76 V Find Ecell • Example: [VO2+] = 2M, [H+] = 0.5M, [VO2+] = 0.01M, [Zn2+] = 0.1M • Ion-Selective Electrodes • Cell potential depends on concentration of an ion • pH meter • Standard electrode of known potential • Glass electrode filled with known [HCl] whose potential changes based on external [H+] • Potentiometer measures the potential difference • You can make similar Na+, K+, or NH4+, Cl-, F-, etc…selective electrodes • Glass “senses” the presence of H+ in open sites (pH meter) • Change the type of glass for sensing other ions Line Notation for a typical pH electrode: Ag | AgCl | Cl- || H+ outside | H+ inside, Cl- | AgCl | Ag Outer ref elec. sample Known H+ Inner ref elec. H+ sensing glass membrane

9. Batteries • Battery Basics • Battery = galvanic cells used as a portable source of electrical potential • Batteries are a source of direct current only; not suitable for providing alternating current like permanent outlets do • Lead Storage Batteries • Highly rechargeable, durable batteries that can operate between –30 and 120 oF • Lead anode, Lead oxide cathode, Sulfuric Acid electrolyte Anode: Pb + H2SO4 PbSO4 + H+ + 2e- Cathode: PbO2 + HSO4- + 3H+ + 2e- PbSO4 + 2H2O Cell: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq) 2PbSO4(s) + 2H2O eo = 2.0V • For cars: 6 of these cells in series with grid electrodes provides 12 V (2 V each) • Sulfuric Acid is consumed; so density of the acid drops over its life • Water is also consumed; can “top off” the battery with water. New Ca/Pb electrodes no longer use up water (sealed batteries) • Alternator recharges battery by forcing current in opposite direction • Physical Damage, not chemical depletion, usually “kills” the battery

10. Other Batteries • Dry Cell Batteries = calculators, watches, etc… • Acid Version: Zn anode, C cathode, MnO2/NH4Cl/C paste as electrolyte 1.5V Anode: Zn Zn2+ + 2e- Cathode: 2NH4+ + 2MnO2 + 2e- Mn2O3 + 2NH3 + H2O • Alkaline Version has KOH or NaOH as electrolyte Anode: Zn + 2OH- ZnO + H2O + 2e- Cathode: 2MnO2 + H2O + 2e- Mn2O3 + 2OH- • Rechargable Nickel—Cadmium Batteries Anode: Cd + 2OH- Cd(OH)2 + 2e- Cathode: NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH- d) Nickel-Metal Hydride (NiMH) Batteries Anode: M∙H + OH- M + H2O + e- Cathode: NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH- e) Lithium Ion Batteries: flow of Li+ inside battery matched by e- in wire

11. Fuel cells = galvanic cell with continuous source of reactants • The Hydrogen—Oxygen Fuel Cell is used for NASA spaceflights • The reactant gases can be stored as liquids in tanks • Anode: 2H2 + 4OH- 4H2O + 4e-e1/2 = 0.83V • Cathode: 4e- + O2 + 2H2O 4OH-e1/2 = 1.20V • Overall: 2H2(g) + O2(g) + catalyst 2H2O(l) eo = 2.03V