1. Electrochemistry Why Do Chemicals Trade Electrons?

2. Galvanic (Voltaic) Cells Remember oxidation reduction reactions Good, you can do half reactions no problem Because you practiced them, over and over Until they were easy Lucky, because otherwise you would have to learn them now in addition to all the rest of the stuff in this chapter

3. Electrochem Oxidation involves loss of electrons (OIL) –Increase of oxidation number Reduction involved gain of electrons (RIG) –Reduction of oxidation number Oxidation agent is the one that’s reduced Reducing agent is the one that’s oxidized

4. Potassium permanganate is added to an acidified solution of iron (II) chloride. Write the net ionic reaction MnO 4 - +H + + Fe +2  Mn +2 + H 2 O + Fe +3 Write the half reactions 8H + + MnO 4 - + 5e -  Mn +2 + 4 H 2 O 5(Fe +2  Fe +3 + e - ) 8H + + MnO 4 - + 5Fe +2  Mn +2 + 4 H 2 O + 5Fe +3

5. Galvanic Cell A device that changes chemical energy to electrical energy. The difference between a redox reaction and a cell –Instead of mixing the chemicals in a flask and letting the electrons move to a new substance –Separate the chemicals –Use moving electrons to do work –Need to allow ions to flow! Or will only work for a moment

6. Voltaic Cells (Galvanic Cells) A voltaic cell is a chemical concept that led to the manufacture of batteries. There are three major components to a voltaic cell: –Anode – oxidation reaction –Cathode – reduction reaction –Salt bridge – replaces ions to keep solutions neutral

7. Salt bridge or porous disk Allows ions to move Oxidation occurs Reduction Occurs Red Cat An OX

8. Cell Potential Between the anode and the cathode, electrons have a difference in energy – the cathode is a lower energy state for the electrons, so they will naturally “flow” there. This is a form of electrical potential energy. –There is a “force” behind electrons moving –Electromotive force (emf) or cell potential This potential can be measured with a potentiometer Like all chemical reactions there is a driving force. –Which thermodynamic concept do you think would work here?

10. Zn (s) + Cu +2 => Zn +2 + Cu (s) What is the anode half reaction? Zn  Zn +2 + 2 e - What is the cathode half reaction? Cu +2 + 2e -  Cu

11. Cells The maximum amount of electrons that can move (Current in Amps) is based on: –Voltage – Volts - Pond on top of hill, water flows downhill. –Resistance – Ohms - Things in its way slow the water down. Galvanic cell consists of two half reactions –Trading electrons –The force driving them is called “voltage” Or potential

12. Water in a tank has a force That pushes it out a pipe Which pipe would have a higher water pressure? Analogous to a cell with more electromotive force. It has a higher voltage

13. Electromotive Force (EMF) The potential difference between anode and cathode is called the electromotive force (the force “pushing” electrons”), or emf. E cell = emf EMF is measured in volts, where –1 Volt = 1 Joule per Coulomb –1 e - = 1.60 x 10 -19 C

14. To compute the value of the cell potential –The maximum voltage –Need a standard for all half cells –Use hydrogen at [H + ] = 1M, P H 2 = 1 atm –All half cells are compared to hydrogen. Hydrogen half cell is assigned a value of zero. Zero

16. Determining Relative Agent Strengths The strongest oxidizers (or oxidizing agents) are those which have the most positive E o red values. (they are the most easily reduced) The strongest reducing agents have the most negative E o red values. F 2 is the strongest oxidizing agent, while Li is the strongest reducing agent.

17. Zn (s) + Cu +2 => Zn +2 + Cu (s) What is the anode half reaction? (An Ox) Zn  Zn +2 + 2 e - E° = +0.76 What is the cathode half reaction? (Red Cat) Cu  Cu +2 + 2e - E ° = -0.34 E o cell = E o red (cathode) – E o red (anode) We subtract the anode’s E o red because oxidation happens at the anode, and we are technically using the reverse reaction. The cell potential is 0.76 + 0.34 = 1.10V What kind of reaction is this? –We called this a single replacement reaction! –Remember the activity series? Why are some substances more active then others?

18. Standard Reduction Potentials Calculate the standard emf for the voltaic cell based on the reaction: Cr 2 O 7 2- + 14 H + + 6 I -  2 Cr 3+ + 3 I 2 + 7 H 2 O –E o red (Cr 2 O 7 2- + 14 H + + 6 e -  Cr 3+ + 7 H 2 O) = +1.33 V –E o red (I 2 + 2 e -  2 I - ) = +0.54 V Note: even though there are coefficients, you do NOT multiply emfs by them. EMFs are an intensive property of matter and are not subject to the same rules as joules, grams, etc.

19. Line Notation Anode   Cathode Mg (s)  Mg +2 (aq)   Al +2 (aq)  Al (s) Mg + Al +2  Mg +2 + Al For reactions producing/using gases or ions a platinum electrode is used Pt (s)  ClO 3 -, ClO 4 -, H +   H +, MnO 4 -, Mn+2  Pt (s)

20. Two Cents Worth of Physics A galvanic cell consists of two half reactions The “difference” in cell potentials is the electromotive force E This force is related to the ability to do work and ideally equal to the free energy change

21. Equations emf = potential difference (V) = Work (J) charge (C) Work is viewed from the point of view of the system! E = -w q Where E = electromotive force (voltage) -w = work done (system lost energy) q = charge (number of electrons x charge)

22. Free Energy Without the derivation! ∆G = -nF E n = moles of electrons F = 96,485 C/mol (the constant FARADAY) E = emf = electromotive force Under standard conditions ∆G° = -nF E ° (25 °C, 1 atm, [1M])

23. Units in Electrochem F = Faraday = 1 mole e - charge on mole of electrons F = 96,485 Coulombs / mol e - Coulomb = C (unit of charge) q = amount of charge in coulombs E = electromotive force = EMF E ° = standard EMF E = joule / C

24. Predicting Spontaneity ∆G for spontaneous reactions < 0 (negative) ∆G = -nF E Any process with a positive E  Will have a negative ∆G  Will be spontaneous, ∆G < 0, negative  K>1

25. Calculate ∆G° for the reaction Cu +2 + Fe → Fe +2 + Cu The half reactions are: Cu +2 + 2e - → Cu E ° = 0.34 V Fe → Fe +2 + 2 e - E ° = 0.44 V The E ° is 0.78 V so the delta G is ∆G ° = -nF E ° = -2 * 96500 * 0.78 = -15.1 kJ

26. What is the Voltage of a Cell? There are several different types of cells Each has its uses Some are batteries Some are electrodes (pH meter) Fuel cells use electrochemistry –Instead of combustion

27. Concentration Cells A galvanic cell can be made with the anode and cathode with the same half reaction. Porous disk 0.1M Ag + 1.0 M Ag + Which direction will Electrons flow? Nature will try to make the concentration On both sides The same Ag Which is the cathode? Which side plates silver metal? Cathode

28. Concentration Cells If the Eº of both the cells is the same, how can a voltage be formed? Eº are at 1M concentrations So there is a potential difference between the half cells 0.1M Ag + 1.0 M Ag +

29. The Nernst Equation The cell potential is dependent on concentration Free energy is dependent on concentration Remember  G =  Gº + RT lnQ Q and the concentration affects  G

30. Nernst Again  G = -nF E Replacing the  G’s with –nF E’s E = E  - RT lnQ nF At 25º C E = E  - 0.0591 log (Q) n

31. 2Al + 3Mn +2  2Al +3 + 3Mn [Mn +2 ] = 0.50M [Al +3 ] = 1.50M –Remember heterogeneous equilibria –and half reactions! E = E  - 0.0591 log (Q) n Al  Al +3 +3e - Mn +2 +2e -  Mn n = 6 (3 x 2) E = 0.48V – 0.0591 log (1.50) 2 = 0.48 – 0.01 = 0.47V 6 (0.50) 3

32. What is a Dead Battery? The cell will spontaneously discharge When the Q = K, the reaction is at equilibrium There is no longer a driving force The two cells have the same free energy E cell =  G = 0

33. Lets Consider The Reaction VO 2 +1 and Zn react. 2VO 2 + + Zn  2VO +2 + Zn +2 Use half reaction method to get equation 2VO 2 + +4H + + Zn  2VO +2 +2H 2 O + Zn +2

34. What is the cell potential? VO 2 + +2H + + e -  VO +2 +H 2 O E  = 1.00V Zn  2 e - + Zn +2 E  = 0.76V 2VO 2 + +4H + + Zn  2VO +2 +2H 2 O + Zn +2 Where T = 25  C, [VO 2 + ] = 2.0M, [H + ] = 0.50M [VO +2 ] = 1.0x 10 -2 M [Zn +2 ]= 1.0 x 10 -1 M Not at standard conditions, n = ? (number of electrons) E = 1.76 - 0.0591 log ([VO +2 ] 2 [Zn +2 ]) = 1.76 + 0.13 = 1.89V 2 [VO 2 + ] 2 [H + ] 4

35. Ion Selective Electrodes pH electrodes are an example –Sensitive to the H + ion –Others available Cd +2, Ca +2,Cu +2, K +1. Ag +1, Na +1 Cl -1, Br -1, CN -1, F -1, NO3 -1, S -2 Works like a concentration cell –Inside electrode is a half cell of known concentration –Measures the potential difference

36. Equilibrium Constants E = E  - 0.0591 log (Q) n When Q = K, E = 0 E  = +0.0591 logK n Solve for K. K is typically large for a redox reaction. (The useful ones anyway)

37. Electrolytic Cells An electrolytic cell consists of two electrodes in a molten salt or a solution. –Electrolysis is the decomposition of a molecule using electricity. Example: –2 NaCl(l)  2 Na(l) + Cl 2 (g) Unlike a voltaic cell, electrolytic cells require a voltage source (battery) to function.

38. Electroplating Electrolysis can be used for the industrial practice of electroplating – that is, plating one metal on another. Example: Copper plating of a quarter.

40. The Ampere An ampere is a coulomb per second –It is the unit of electric current. 1 A = 1 C/s We can calculate the extent of electrolysis given the electric current and time.

41. Moles of Electrons When you have total charge in coulombs, you can use Faraday’s constant to convert to moles of electrons. Faraday’s constant = 96,483 C/mol e - This will allow us to then use stoichiometry to solve for the amount of substance oxidized/reduced.

42. Practice Problems Calculate the number of grams of aluminum produced in 1.00 hour by the electrolysis of molten AlCl 3 if the electrical current is 10.0 A (amperes).

43. Electrical Work Gibbs’ free energy is the extra energy that can be used to power other devices. ΔG = -nF E cell Therefore, the maximum amount of useful electrical work (energy) from a voltaic cell is –w max = -nF E cell Or, if you are applying an external potential (like in an electrolytic cell)… –w = nF E ext (positive because you are putting energy into the system) q=nF w max = q E

44. Electrical Work and Efficiency The unit of electrical work is the kilowatt-hour (kWh). Based on the watt (1 J/s), a unit of electrical power. 1 kWh = 3.6 x 10 6 J Efficiency = w x 100% w max

45. Practice Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 10 3 kg of aluminum by electrolysis of Al 3+ if the applied voltage is 4.50 V.