1. Galvanic (or Voltaic) Cells Electrochemistry = the interchange of chemical and electrical energy = used constantly in batteries, chemical instruments, etc… I.Galvanic Cells A.Definitions 1)Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced) 2)Oxidation = loss of electron(s) to become more positively charged 3)Reduction = gain of electron(s) to become more negatively charged B.Using Redox Reactions to generate electric current (moving electrons) 1)8H + (aq) + MnO 4 - (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) a)Fe 2+ is oxidized and MnO 4 - is reduced b)Half Reaction = oxidation or reduction process only Reduction: 8H + + MnO 4 - + 5e - Mn 2+ + 4H 2 O Oxidation: 5(Fe 2+ Fe 3+ + e-) Sum = Redox Rxn

2. 2)In solution: a)Fe 2+ and MnO 4 - collide and electrons are transferred b)No work can be obtained; only heat is generated 3)In separate compartments, electrons must go through a wire = Galvanic Cell a)Generates a current = moving electrons from Fe 2+ side to MnO 4 - side b)Current can produce work in a motor c)Salt Bridge = allows ion flow without mixing solutions (Jello-like matrix)

3. d)Chemical reactions occur at the Electrodes = conducting solid dipped into the solution i)Anode = electrode where oxidation occurs (production of e-) ii)Cathode = electrode where reduction occurs (using up e-) C.Cell Potential 1)Think of the Galvanic Cell as an oxidizing agent “pulling” electrons off of the reducing agent. The “pull” = Cell Potential a)  cell = Cell Potential = Electromotive Force = emf b)Units for  cell = Volt = V 1 V = 1 Joule/1 Coulomb 2)Voltmeter = instrument drawing current through a known resistance to find V Potentiometer = voltmeter that doesn’t effect V by measuring it

4. II.Standard Reduction Potentials A.Standard Hydrogen Electrode 1)When measuring a value, you must have a standard to compare it to 2)Cathode = Pt electrode in 1 M H + and 1 atm of H 2 (g) Half Reaction: 2H + + 2e - H 2 (g)  1/2 = 0 3)We will use this cathode to find  cell of other Half Reactions

5. 4)Standard Reduction Potentials can be found in your text appendices a)Always given as a reduction process b)All solutes are 1M, gases = 1 atm 5)Combining Half Reactions to find Cell Potentials a)Reverse one of the half reactions to an oxidation; this reverses the sign of  1/2 b)Don’t need to multiply for coefficients = Intensive Property (color, flavor) c)Example: 2Fe 3+ (aq) + Cu o 2Fe 2+ (aq) + Cu 2+ (aq) i.Fe 3+ + e - Fe 2+  1/2 = +0.77 V ii.Cu 2+ + 2e - Cu o  1/2 = +0.34 V iii.Reverse of (ii) added to (i) = -0.34 V + +0.77 V = +0.43 V =  cell

6. B.Direction of electron flow in a cell 1)Cell always runs in a direction to produce a positive  cell 2)Fe 2+ + 2e - Fe o  1/2 = -0.44 V MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O  1/2 = +1.51 V 3)We put the cell together to get a positive potential a)5(Fe o Fe 2+ + 2e - )  1/2 = +0.44 V b)2(MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O)  1/2 = +1.51 V 16H + (aq) + 2MnO 4 - (aq) + 5Fe o (s) 2Mn 2+ (aq) + 5Fe 2+ (aq) + 8H 2 O(l)  cell = 1.95V

7. C.The complete description of a Galvanic Cell 1)Items to include in the description a)Cell potential (always +) and the balanced overall reaction b)Direction of electron flow c)Designate the Anode and the Cathode d)Identity of the electrode materials and the ions present with concentration 2)Example: Completely describe the Galvanic Cell based on these reactions Ag + + e- Ag  o = +0.80 V Fe 3+ + e- Fe 2+  o = +0.77 V Ag + (aq) + Fe 2+ (aq) Ag o (s) + Fe 3+ (aq)  o cell = +0.03 V

8. Electrochemical Potential, Work, and Energy III.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

9. 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 10 5 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)

10. 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 E o cell = +0.78 V

11. C.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:

12. IV.Electrolysis = using electric energy to produce chemical change (opposite of cell) A.Example 1)Consider the Cu/Zn Galvanic Cell a)Anode: Zn Zn 2+ + 2e - b)Cathode: Cu 2+ + 2e - Cu  o cell = +1.10 V 2)If we attach a power source of  o > +1.10 V, we can force e- to go the other way a)Anode: Cu Cu 2+ + 2e - b)Cathode: Zn 2+ + 2e - Zn c)Called an Electrolytic Cell

13. B.Calculations with Electrolytic Cells 1)How much Chemical Change? Is usually the question. 2)Find mass Cu o plated out passing 10 amps (10 C/s) through Cu 2+ solution 30 min. a) Cu 2+ + 2e - Cu o (s) b)Steps: current/time, charge (C), moles e-, moles Cu, grams Cu 3)Example: How long must a current of 5.00 amps be applied to a Ag + solution to produce 10.5 g of silver metal? C.Electrolysis of Water 1)Galvanic: 2H 2 + O 2 2H 2 O (Fuel Cell) 2)Electrolytic Cell: a)Anode: 2H 2 O O 2 + 4H + + 4e - -  o = -1.23 V b)Cathode: 4H 2 O + 4e - 2H 2 + 4OH -  o = -0.83 V c)Overall: 6H 2 O 2H 2 + O 2 + 4(H + + OH - ) 2H 2 O 2H 2 + O 2  o cell = -2.06 V 3)We must add a salt to increase the conductance of pure water [H + ] = [OH - ] = 10 -7

15. D.Electrolysis of Mixtures 1)Mixture of Cu 2+, Ag +, Zn 2+ ; What is the order of plating out? a)Ag + + e - Ag  o 1/2 = +0.80 V b)Cu 2+ + 2e - Cu  o 1/2 = +0.34 V c)Zn 2+ + 2e - Zn  o 1/2 = -0.76 V 2)Reduction of Ag + is easiest (  o = most positive) followed by Cu, then Zn 3)Example: Ce 4+ (  o 1/2 = +1.70 V), VO 2+ (  o 1/2 = +1.00 V), Fe 3+ (  o 1/2 = +0.77 V)