1. Electrochemical Potential, Work, and Energy I.Potential, Work, and Energy A.Units 1)Joule (J) = unit of energy, heat, or work (w) = kgm 2 /s 2 2)Coulomb (C) = unit of electrical charge (q). 1 e - = 1.6 x 10 -19 C 3) = electrical potential (  ) 4)1 J of work is produced when 1 C of charge is transferred between two points differing by 1 V of electrical potential 5)Work flowing out of a system (Galvanic Cell) is taken to be negative work 6)Cell Potential is always positive 7)From last chapter, w max =  G

2. B.Electrochemical Problems 1)When current flows, we always waste some of the energy as heat instead of work w < w max 2)We can, however, measure  max with a potentiometer, so we can find the hypothetical value of w max 3)Example:  o cell = 2.50 V 1.33 mole e - pass through the wire.  actual = 2.10 V a)1 Faraday (F) = the charge on 1 mole of electrons = 96,485 C (6.022 x 10 23 e-/mol)(1.6 x 10 -19 C/e-) = 96,485 C/mol b)w = -q  = -(1.33 mol e-)(96,485 C/mole e-)(2.10 J/C) = -2.69 x 10 5 J c)w max = -q  max = -(1.33 mol e-)(96,485 C/mole e-)(2.50 J/C) = -3.21 x 10 5 J d)Efficiency = w/w max = -2.69 x10-5 J/-3.21 x 105 J = 0.838 or 83.8% 4)Free Energy (  G) a)q = nF where n = number of moles, F = 96,485 C/mole b)  G = -nF  (assuming the maximum  ) c)Maximum cell potential is directly related to  G between reactants and products in the Galvanic Cell (This lets us directly measure  G)

3. 5)Example: Calculate  G o for the reaction Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq) a)Half Reactions: Cu 2+ + 2e- Cu o  o = 0.34 V Fe o Fe 2+ + 2e-  o = 0.44 V b)  G o = -nF  o = -(2 mol e-)(96,485 C/mol e-)(0.78 J/C) = -1.5 x 10 5 J 6)Example: Will 1 M HNO 3 dissolve metallic gold to make 1 M Au 3+ ? a)Half Reaction: NO 3 - + 4H + + 3e - NO + 2H 2 O  o = +0.96 V Au o Au 3+ + 3e -  o = -1.50 V Au(s) + NO 3 - (aq) + 4H + (aq) Au 3+ (aq) + NO(g) + 2H 2 O(l)  o cell = -0.54V b)Since  is negative (  G = +) the reaction will not occur spontaneously II.Cell Potential and Concentration A.Concentration Cells 1)Up until now, concentration for all Galvanic solutions = 1 M (Gives  o ) 2)What happens if we change these concentrations? E o cell = +0.78 V

4. 3)Le Chatelier’s Principle a)Cu(s) + 2Ce 4+ (aq) Cu 2+ (aq) + 2Ce 3+ (aq)  o cell = 1.36 V b)Increase Ce 4+ concentration, (  >  o ) c)Increase Cu 2+ concentration, (  <  o ) d)Example 4)Concentration Cell = Galvanic Cell driven by the fact that concentrations of the same reactants are different on the two sides of the cell. 5)Example: Ag + + e - Ag o  o 1/2 = +0.80 V a)If both sides had [Ag + ] = 1 M, then  o cell = +0.80 V + (-0.80 V) = 0.00 V b)If [Ag + ] right = 1 M and [Ag + ] left = 0.1 M then we should have a potential i.Diffusion would try to equalize Ag + on the right side and the left side (Entropy favors even distribution, like gas particles in two chambers) i.Electrons would flow from left to right to even out [Ag + ] ii.A very small voltage would be generated iii.Example

6. B.The Nernst Equation 1)Derivation a)  G =  G o + RTlnQ = -nF  b)  G o = -nF  o c)-nF  = -nF  o + RTlnQ 2)At 25 o C, this simplifies to 3)Example: 2Al(s) + 3Mn 2+ (aq) 2Al 3+ (aq) + 3Mn(s)  o cell = 0.48 V a)Oxidation: 2Al(s) 2Al 3+ (aq) + 6e- b)Reduction: 3Mn 2+ (aq) + 6e- 3Mn(s) c)[Mn 2+ ] = 0.5 M, [Al 3+ ] = 1.5 M d)Q = [Al 3+ ] 2 / [Mn 2+ ] 3 = (1.5) 2 / (0.5) 3 = 18 e)As the reaction proceeds,  cell 0 (Q K) = Dead Battery! f)Calculating K:

8. g)Example: [VO 2 + ] = 2M, [H+] = 0.5M, [VO 2+ ] = 0.01M, [Zn 2+ ] = 0.1M C.Ion-Selective Electrodes 1)Cell potential depends on concentration of an ion 2)pH meter a)Standard electrode of known potential b)Glass electrode filled with known [HCl] whose potential changes based on external [H + ] c)Potentiometer measures the potential difference 3)You can make similar Na +, K +, or NH 4 +, Cl -, F -, etc…selective electrodes a)Glass “senses” the presence of H + in open sites (pH meter) b)Change the type of glass for sensing other ions VO 2 + + 2H + + e - VO 2+ + H 2 O  o = 1.00 V Zn 2+ + 2e - Zn  o = -0.76 V Line Notation for a typical pH electrode: Ag | AgCl | Cl - || H + outside | H + inside, Cl - | AgCl | Ag Outer ref elec.sample H + sensing glass membrane Known H + Inner ref elec. Find E cell

9. III.Batteries A.Battery Basics 1)Battery = galvanic cells used as a portable source of electrical potential 2)Batteries are a source of direct current only; not suitable for providing alternating current like permanent outlets do B.Lead Storage Batteries 1)Highly rechargeable, durable batteries that can operate between –30 and 120 o F 2)Lead anode, Lead oxide cathode, Sulfuric Acid electrolyte Anode: Pb + H 2 SO 4 PbSO 4 + H + + 2e - Cathode: PbO 2 + HSO 4 - + 3H + + 2e - PbSO 4 + 2H 2 O Cell: Pb(s) + PbO 2 (s) + 2H + (aq) + 2HSO 4 - (aq) 2PbSO 4 (s) + 2H 2 O  o = 2.0V 3)For cars: 6 of these cells in series with grid electrodes provides 12 V (2 V each) 4)Sulfuric Acid is consumed; so density of the acid drops over its life 5)Water is also consumed; can “top off” the battery with water. New Ca/Pb electrodes no longer use up water (sealed batteries) 6)Alternator recharges battery by forcing current in opposite direction 7)Physical Damage, not chemical depletion, usually “kills” the battery

10. C.Other Batteries 1)Dry Cell Batteries = calculators, watches, etc… a)Acid Version: Zn anode, C cathode, MnO 2 /NH 4 Cl/C paste as electrolyte 1.5V Anode: Zn Zn 2+ + 2e - Cathode: 2NH 4 + + 2MnO 2 + 2e - Mn 2 O 3 + 2NH 3 + H 2 O b)Alkaline Version has KOH or NaOH as electrolyte Anode: Zn + 2OH - ZnO + H 2 O + 2e - Cathode: 2MnO 2 + H 2 O + 2e - Mn 2 O 3 + 2OH - c)Rechargable Nickel—Cadmium Batteries Anode: Cd + 2OH - Cd(OH) 2 + 2e - Cathode: NiO 2 + 2H 2 O + 2e - Ni(OH) 2 + 2OH - d)Nickel-Metal Hydride (NiMH) Batteries Anode: M∙H + OH - M + H 2 O + e - Cathode: NiO 2 + 2H 2 O + 2e - Ni(OH) 2 + 2OH - e)Lithium Ion Batteries: flow of Li + inside battery matched by e - in wire

11. D.Fuel cells = galvanic cell with continuous source of reactants 1)The Hydrogen—Oxygen Fuel Cell is used for NASA spaceflights 2)The reactant gases can be stored as liquids in tanks a)Anode: 2H 2 + 4OH - 4H 2 O + 4e -  1/2 = 0.83V b)Cathode: 4e - + O 2 + 2H 2 O 4OH -  1/2 = 1.20V c)Overall: 2H 2 (g) + O 2 (g) + catalyst 2H 2 O(l)  o = 2.03V