Chapter 17, Electrochemistry

Transcription

1 Chapter 17, Electrochemistry 1) Electrochemistry has tremendous commercial importance. 2) You will rely heavily on your understanding of redox rxns. to describe electrochemical cells (2 types): a. Galvanic (a.k.a.: voltaic; rxn. produces energy) b. Electrolytic (a.k.a.: rxn. requires/consumes energy) 3) The chemical (redox) rxns involve e transfer; we can relate these rxns to a common scale, E, standard reduction potential. 4) Relationship of E to ÄG? Background, Fig. 17.1, p

2 1) This clearly energetically downhill. (Rxn done in lab?) Cu 2+ + Zn Cu + Zn 2+ 2

3 2) Is there a way to use/capture/store some of the energy that is released when this rxn. occurs? Comparisons to biological systems? 3) Can we force a normally non-spontaneous rxn. to occur by adding energy? (See electrolytic cells.) Comparisons to biological systems? 3

4 I. Galvanic Cells A. The rxn in a galvanic cell is spontaneous. 1. Example: Fig. 17.1, p. 698: 2. Net ionic equation for the above: Zn (s) + Cu 2+( aq) Zn 2+( aq) + Cu (s) 3. Half-rxns for the above are: Oxidation: + Reduction: + 4

5 Summed half-reactions give the overall balanced rxn. 4. This is not a homogeneous system with respect to phases of the reactants & products. How does that influence the way you think about this rxn? B. It s possible to get this rxn to go even when the reactants are not in direct physical contact if: 1. a wire that conducts e is placed in direct contact with the appropriate electrodes and 2. an ion conducting channel is present so redox inert ions can move to minimize charge accumulation. See Fig a) p

6 3. A design variation is the Daniell cell Fig b). 6

7 C. Nomenclature 1. Electrode where the oxidation half-rxn occurs is called the anode. Anions move toward the anode. 2. Electrode where the reduction half-rxn occurs is the cathode. Cations move toward the cathode. Read the paragraph at the bottom of p. 699 (-700) to get a better feel for the charge designations assigned to the anode and cathode. From the perspective of the liquid phase, they may seem illogical. View the Zn 2+ ions given off by the anode as attracting the anions. Try Prob. 17.1, p

8 II. Shorthand Notation for Galvanic Cells A. As in many other areas of human communication, efficiency is useful in representing a galvanic cell. 1. The cell that we looked at above employing the rxn: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 8

9 can be described by: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) where: a) represents a phase boundary b) represents a salt bridge c) anode half-cell on left, cathode on right d) electrodes are on extreme left and right (s) e) reactants within each half-cell come first 9

10 2. So anode half-cell cathode half-cell bridge Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) phase boundary e flow phase boundary Try prob. 17.2, p

11 III. Cell Potentials & ÄG? for Cell Reactions A. What makes the e move through the wire? 1. This force is an electrical potential called the electromotive force (emf), a.k.a. cell potential (E). Galvanic E defined to be +. Unit = volt. 2. More on units: a) 1 J = 1 C 1 V joules, coulombs, volts b) 1 C = 1 amp for 1 sec c) a) means 1 C moving across 1 V yields 1 J work 11

12 B. Relationship between free energy (G) and cell potential (E): ÄG = nfe note the sign 1. n = number of moles of e transferred 2. F = Faraday constant (96,486 C/mol e ) You will confirm this value in the last lab. 3. E = cell potential C. As before, it is useful in comparing reactions to define a specific set of standard conditions: ÄG = nfe Try prob. 17.5, p

13 IV. Standard Reduction Potentials (E ) A. Sum of the standard half-cell potentials for anode and cathode is equal to the standard potential for the cell: E cell = E ox + E red 1. Example: Cu 2+ /H 2 cell (comment re. H 2 rxn) E cell = E H 2 H + + E Cu 2+ 2 Cu = V V = 0.34 V Why is E H 2 H + expressed as ? 13

14 B. Think back to previous quantitative treatment of G & H. 1. We can t define these in absolute terms (contrast with ). 2. Therefore, we approached quantitative treatment of G and H by examining changes: ÄG and ÄH. 3. The approach is similar for quantitative problems with E. In this case we define one of the half-reactions as having a numerical value of V, and then analyze all other rxns relative to that one. 14

15 4. Reference rxn is the standard H electrode (S.H.E.): 2 H + (aq, 1 M) + 2 e H 2 (g, 1 atm) E = 0 V H 2 (g, 1 atm) 2 H + (aq, 1 M) + 2 e E = 0 V Try prob. 17.6, p Al(s) Al 3+ (aq) Cr 3+ (aq) Cr(s) If E = 0.92 V, look up E for the Al 3+ Al half-cell in Table 17.1 and then calculate E for the Cr 3+ Cr half-cell. Scan table a bit. 15

16 16

17 V. Using E Values A. Because redox rxns all use e (common currency) we can relate these types of rxns through a common scale. 1. For a redox table with 100 half-cell entries, we can calculate = 9,900 cell potentials 17

18 2. Example: Zn (s) + 2 Ag +( aq) Zn 2+ (aq) + 2 Ag (s) E cell = E ox + E red E Oxidation half-cell: Reduction half-cell: Cell: 18

19 B. Remember: E is an intensive property (doesn t depend on how many electrons), therefore, you don t multiply the Ag E value by 2. Derive logic from ÄG = nfe ÄG nf = E ÄG (J/mol) [n (mol) F(C mol)] = E (J/mol) C [(V C J)] = V mol = E You can see that the potential E is in terms of volts per mole. It does not matter if you have X# or Y# of moles, because it is the quantity per mole. Try prob. 17.8, p

20 VI. E cell & Composition of Rxn Mixture: Nernst A. From Chap 16: amount of energy derived from a system is related both to the energies of the reactants/products & their concentrations: ÄG = ÄG + RT ln Q Recall that: 1. ÄG free energy change under standard conditions. 2. Q is the rxn quotient: [products] [reactants], which you can clearly control by how you set it up. 3. ÄG therefore can vary quite a bit depending on the way the system s initial concentrations are set. 20

21 B. By direct substitution from V. B. above (ÄG = nfe): 1. nfe = nfe + RT ln Q divide by nf 2. E = E (RT nf) ln Q the Nernst Equation 3. Because ph is based on the common log (base 10), the following version is often used: E = E ( V n) log Q applies at 25 C 21

22 Try prob , p Cu(s) + 2 Fe 3+ (aq) Cu 2+ (aq) + 2 Fe 2+ (aq) What is E for a cell at 25 C that has the [following]: [Fe 3+ ] = M [Cu 2+ ] = 0.25 M [Fe 2+ ] = 0.20 M 22

23 Try Conceptual prob , p

24 VII. Electrochemical Determination of ph What a ph meter does. A. Cell description: Pt H (1 atm) H + (?M) ref cathode 2 1. Overall cell potential: E cell = EH 2 H + + E ref 2. Calculate E for the hydrogen half-cell rxn via Nernst: H 2(g) 2 H + (aq) + 2 e EH 2 H + = E H 2 H + ( V n) (log [H + ] 2 PH 2 ) 24

25 Because the standard hydrogen cell potential is defined to be zero and PH 2 = 1 atm: EH 2 H + = ( V 2) (log [H + ] 2 ) Extracting the 2 terms & using the definition of ph: EH 2 H + = ( V 2) 2 (log [H + ]) EH 2 H + = ( V) (ph) 3. Now substitute this term back into: E cell = EH 2 H + + E ref 25

26 E cell = ( V) (ph) + E ref solve for ph: ph = ( E cell E ref ) V B. In practice, a glass electrode is used in place of the standard hydrogen electrode & the reference electrode used is the calomel electrode. Half-cell rxns are: 1. glass: 2 [Ag (s) + Cl (aq) AgCl (s) + e ] E = 0.22 V Glass electrode dips into a dilute HCl soln. inside glass membrane separating ph electrode from soln whose ph is to be measured. See Fig. 17.7, p. 715, below. 26

27 2. calomel: Hg 2 Cl 2(s) + 2 e 2Hg(l) + 2Cl (aq) E = 0.28 V D. Based on the above two half-cell rxns, you might (should?) ask, How does ph fit in to the E cell? H + outside of the thin glass membrane develops a potential relative to the H + of the dilute HCl solution. Try prob , p

28 VIII. E & K eq A. We now want to examine the relationship between standard cell potentials and equilibrium constants. 28

29 B. Derivation of relationship: 1. As noted in III above: ÄG = nfe 2. From Chap 17: ÄG = RT ln K 3. Combining 1 & 2: nfe = RT ln K 4. As before, we can convert to log and set T = 25 C: E = ( V n) log K 29

30 Graphically: See Fig. 17.8, p Note that small voltage differences give large K differences. 30

31 C. We have now added a third way to determine K values. The different approaches are: 1. from measuring [solute] values: K = [prod] [react] 2. from thermodynamic data: K = e (ÄG RT) 3. from electrochemical data: ln K = nfe RT Try prob , p

32 IX. Batteries (...most important practical application of galvanic cells. ) A. Physics element. In a multicell battery, a number of batteries are wired together in series (as opposed to in parallel). The voltage for the overall battery is the sum of the voltages for the individual cells. Some examples of batteries: 32

33 B. Pb Storage Battery Anode rxn: oxidation of Pb (s) 1. Single cell has a voltage 2 V. Six cells yield 12 V. 2. This battery can be readily recharged (run reactions in non-spontaneous direction) because the rxn product, PbSO 4 (s), adheres to the electrodes. 3. What part of a car normally recharges the battery and what is its energy source? 33

34 C. Dry-Cell Battery Anode rxn: oxidation of Zn (s) 1. A number of types of these are commercially in use. 2. The major differences are in the cathodic rxn. 3. Which type did the bunny (Eveready) peddle (originally)? D. Ni-Cadmium Anode rxn: oxidation of Cd (s) 1. A.k.a. Ni-cad 2. Can be recharge efficiently because the rxn products adhere to the electrodes. 34

35 E. Lithium Battery Anode rxn: oxidation of Li (s) 1. Rechargeable 2. Relatively high voltage per cell (3 V) 3. Light weight (Have we seen this before in a different context???) 35

36 F. Fuel cell (rxns below) 1. Anode: 2 H 2 (g) + 4 OH (aq) 4 H 2 O(l) + 4 e 2. Cathode: O 2 (g) + 2 H 2 O(l) + 4 e 4 OH (aq) 3. Overall: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) 4. What are advantages and disadvantages for this rxn? 5. Have we seen this before in a different context? 36

37 X. Corrosion A. In our society, probably the most important corrosion rxn is for Fe. See Fig , p. 725: B. Corrosion half-cell rxns: 1. Anode region: Fe (s) Fe 2+ (aq) + 2 e E = 0.45 V 2. Cathode region: O 2(g) + 4 H + (aq) + 4 e 2 H 2 O (l) E = 1.23 V 37

38 a) Note that under normal environmental conditions, [H + ] is much less than 1 M. b) Under normal conditions cathodic E ~ 0.81 V. c) This means E cell is still quite positive. d) What does c) say about spontaneity and whether the rxn shown above goes toward products? 38

39 C. Why don t metals such as Al and Ti, which are electrochemically as vulnerable as Fe corrode? 39

40 D. Preventing Corrosion 1. Shielding with paint, but what if paint chips? 2. Galvanizing with Zn. (See Fig , p. 715.) a) The metal you want to preserve (normally Fe) is coated with Zn. b) From the reduction potentials, you can see that any Fe that is oxidized will be rereduced by the Zn, yielding Zn 2+, which eventually forms ZnCO 3, which adheres relatively tightly to the metal surface. 3. Cathodic protection (the sacrificial anode). a) Involves long range e movement in the metal. b) Anode material must eventually be replenished. 40

41 We won t cover the Chapter 17 topics shown below during spring

42 XI. Electrolysis and Electrolytic Cells A. Electrolysis of molten NaCl Products obtained are? and B. E lectrolysis of water. Products obtained are? and XII. Commercial Applications of Electrolysis A. Production of Na(s) from NaCl B. Production of Cl 2 and NaOH C. Production of Al(s) 1. Large current requirement: 1 mol of e produces only 9 g of Al(s). (Can you derive that number?) 2. Electrolytic production of Al is the largest single process consumer of electricity in U.S.A. today. 3. Comment re. Al recycling. D. Electrorefining and Electroplating Examples of: 42

43 1. Chrome plating of auto components 2. Ag/Au plating of jewelry, dining implements, musical instruments, etc. XIII. Quantitative Aspects of Electrolysis Interesting, but we don t have enough time to cover. 43