1. CH. 20 ELECTROCHEMISTR Y MAIN CONCEPTS Galvanic Cell Electolytic Cell -electroplating Cell Potentials Nernst Equation Balance 1 / 2-Rxns -Acidic / Basic Solns Electrochemical Cells: classified as Galvanic Cells *spontaneous chem rxn that produce E Electolytic Cell Commercial Importance Galvanic Zn | MnO 2 & Zn | Ag 2 O cells in watches Fuel Cells in space crafts H 2 - O 2 Electrolytic NaOH purify metals electroplating

2. Conversion E elec to E chem ELECTROLYTIC CELLS conversion E chem to E elec GALVANIC - VOLTAIC CELLS ELECTROLYTIC CELLS 1. process of electrolysis 2. pass electricity thru soln w / E to cause nonspont redox 3. commercial importance-- NaOH purify ores electroplate GALVANIC CELLS 1. provides source electricity thru spont redox 2. batteries ZnC Alkaline Ag 2 O Pb Storage NiCd NiMH Fuel Cells

3. REVIEW REDOX RXN REDUCED H +1 ---> 0 oxidizing agent OXIDIZED Fe 0 ---> +2 reducing agent OXIDATION REDUCTION gain of O atoms loss of O atoms loss of H gain of H lose e - gain e - LEO GER OIL RIG incr ox # decr ox #

4. SPONTANEOUS REDOX RXN Two ½-Reactions: Cu +2 Zn +2 Cu 0 Zn 0 Oxidation: Zn 0 (s) ---> Zn +2 (aq) + 2 e - lose 2 e -, OX, ox # incr Reduction: Cu +2 (aq) + 2 e - ---> Cu 0 (s) gain 2 e -, RED, ox # decr No Electrical Work produced heat E released

5. BALANCING 1 / 2 -REACTIONS Write balanced molecular, add # moles of spectator ions to get neutral cmpds

6. Potassium permanganate & potassium iodide in basic soln Notice: K +1 spectator Step 1- step 2 not needed as balanced Step 3- step 4 Step 5 Step 6

7. molecular Step 7 Step 8

8. ENERGY CAPTURE E released in spont REDOX rxn is captured to perform electrical work E of water wheel depends on 1) vol water 2) PE of H 2 O Work from electrochem cell w = (vol H 2 O)(E released / unit vol)

9. R ight is R eduction - Cathode (+) ox - + ne - ----> Red e - flows into cathode reacts w / oxidized species forms reduced species L eft is O xidation - Anode (-) Red’ ---> ox -’ + ne -’ e - flow out of anode GALVANIC CELL Anode (-) OXIDATION Cathode (+) REDUCTION e-e- e-e-

10. MAKE UP OF CELL consists of: 1. 2. 2 0.5-cells; anode(-) & cathode(+) Salt bridge, electrolyte (Na + NO 3 - ) allows slow mixing of ions - - - - - - - Anode (-) Zn +2 Spont Rxn continuous e - flow thru external wire Ions flow thru soln of redox rxn @ eletrodes Zn 0 Cu +2 + 2 e - ---> Cu Zn ---> Zn +2 + 2 e - Cu +2 (aq) + Zn (s) -----> Cu (s) + Zn -2 (aq) Cathode (+) + + + + + + + Cu +2 Cu 0 Volt Meter e-e- e-e- Zn (s) | Zn +2, 1M (aq) || Cu +2, 1M (aq) | Cu (s) NO 3 -1

11. “LINE” Notation of Electrochemical Cell Zn (s) | Zn +2, 1M (aq) || Cu +2, 1M (aq) | Cu (s) - - - - - - - Anode (-) Zn +2 Zn 0 Cu +2 + 2 e - ---> Cu Zn ---> Zn +2 + 2 e - Cathode (+) + + + + + + + Cu +2 Cu 0 Volt Meter Anode Ox electrode Cathode Red electrode Salt Bridge Phase Boundary Aqueous Solutions

12. Voltmeter shows diff in electrical potential bet 2-1 / 2 cells known as “cell potential”; electromotive force (emf) E o cell = E o ox + E o red Spont Rxn: E o cell > 0 Std H 2 Electrode (SHE) reference electrode E o = 0 V 2 H +1 (aq) + 2 e - H 2 0 (g) What measures cell potential E o ox ??? E o red ??? - - - - - - - Volt Meter E 0 cell = 0.76 V Zn 0 Zn +2 Cathode (+) H 2 1 atm 1 M H + Pt Anode (-)

13. Std Electrode Potential in Water @ 25 o C Std Red. Pot. V Reduction 1 / 2-rxn 2.87 F 2 (g) + 2 e - -----> 2 F -1 (aq) 0.80 Ag +1 (aq) + 1 e - -----> Ag (s) 0.34 Cu +2 (aq) + 2 e - -----> Cu (s) 0 2 H +1 (aq) + 2 e - -----> H 2 (g) -0.28 Ni +2 (aq) + 2 e - -----> Ni (s) -0.76 Zn +2 (aq) + 2 e - -----> Zn (s) -3.05 Li +1 (aq) + 1 e - -----> Li (s)

14. EX: Build galvanic cell --- Ag & Zn What is reduced? Cathode? What is oxidized? Anode? What is the cell potential, emf? Highest 0.080 V Ag +1 (aq) + 1 e - -----> Ag (s) Lowest -0.76 Zn (s) -----> Zn +2 (aq) + 2 e - E o cell = E o ox,Zn + E o red,Ag = +0.76 V + 0.80 V = +1.56 V Write the “line” notation Zn (s) | Zn +2, 1M (aq) || Ag +1, 1M (aq) | Ag (s)

15. L eft is O xidation - Anode (+) Red ----> ox - + ne - R ight is R eduction - Cathode (-) ox -’ + ne -’ = RED’ ELECTROLYTIC CELL drive a nonspontaneous reaction E o cell < 0 Anode (+) OXIDATION Cathode (-) REDUCTION e-e- e-e- “e - pump” e - being push by external power source + -

16. ELECTROLYSIS Driven by outside source E (nonspont) MAKE UP OF CELL consists of: 1. 2. 2 electrodes (in molten salt) driven by DC source; e - pump DC source + + + + + + + - - - - - - - Anode (+)Cathode (-) Na + Cl - Na 0 Na + ions gain e- @ cathode; reduce Cl - ions lose e- @ anode; oxidized Cl 2 0 Na + + e - ---> Na 2 Cl - ---> Cl 2 + 2 e - e-e- e-e-

17. Electrical E (V) needed to drive NaCl rxn Oxidize Reduce 2 Cl - (aq) ---> Cl 2 (g) + 2 e - E ox = -1.36 Na + (aq) + e - ---> Na (s) E red = -2.71 V E cell = -4.07 V Is rxn spontaneous as written? Write balanced spont. rxn. Fe +2 (aq) -----> Fe (s) + Fe +3 (aq) 2 e - + Fe +2 (aq) -----> Fe (s) E o = Fe +2 (aq) -----> Fe +3 (aq) + 1 e - E o = -0.44 V -0.77 V 2( ) Fe +2 + 2 Fe +2 (aq) -----> Fe (s) + 2 Fe +3 (aq) 3 Fe +2 (aq) -----> Fe (s) + 2 Fe +3 (aq) E o = -1.21 V Fe (s) + 2 Fe +3 (aq) -----> 3 Fe +2 (aq) E o = +1.21 V

18. SET #1 Using STD reduction potentials, which rxns are spontaneous? 1) I 2 (s) + 5 Cu +2 (aq) + 6 H 2 O (l) ------> 2 IO 3 - (aq) + 5 Cu (s) + 12 H + (aq) 2) Hg 2 +2 (aq) + 2 I - (aq) ------> 2 Hg (l) + I 2 (s) 3) H 2 SO 3 (aq) + 2 Mn (s) + 4 H + (aq) ------> S (s) + 2 Mn +2 (aq) + 3 H 2 O (l) 1) OX: I 2 (s) + 6 H 2 O (l) ------> 2 IO 3 - (aq) + 12 H + (aq) + 10 e - RED: 5 [ Cu +2 (aq) + 2 e - ------> Cu (s) ] I 2 (s) + 5 Cu +2 (aq) + 6 H 2 O (l) ------> 2 IO 3 - (aq) + 5 Cu (s) + 12 H + (aq) E OX = -1.195 E RED = 0.337 E cell = 0.337 + (-1.195) = -0.858 V NONSPONT

19. 2) OX: 2 I - (aq) ------> I 2 (s) + 2 e - RED: Hg 2 +2 (aq) + 2 e - ------> 2 Hg (l) 2 I - (aq) + Hg +2 (aq) ------> I 2 (aq) + Hg (l) E OX = -0.536 E RED = 0.789 E cell = 0.789 + (-0.536) = 0.253 V SPONT 3) OX: 2 [ Mn (s) ------> Mn +2 (aq) + 2 e - ] RED: H 2 SO 3 (aq) + 4 H + (aq) + 4 e - ------> S (s) + 3 H 2 O (l) H 2 SO 3 (aq) + 2 Mn (s) + 4 H + (aq) ------> S (s) + 2 Mn +2 (aq) + 3 H 2 O (l) E OX = 1.18 E RED = 0.45 E cell = 0.45 + 1.18 = 1.63 V SPONT

20. EMF &  G Change in Free E is measure of spontaneity @ constant T & P relationship bet emf & G is:  G o = -nFE o n = # e- pushed F: Faraday; 1 F = 96,500 C / mol = 96,500 J / V-mol K relationship  G o = -RT LnK R=8.314 J / mol-K

21. 4 Ag (s) + O 2 (g) + 4 H + (aq) ------> 4 Ag + (aq) + 2 H 2 O (l) Need values for E o, G o, & K Use eqn  G o = -RT LnK STEP1: RED: O 2 (g) + 4 H + (aq) + 4 e - ---> 2 H 2 O (l) OX: 4 Ag (s) ---> 4 Ag + (aq) + 4 e -  G o very large, so very favored, expect K to be very large also STEP 2:  G o = -nFE = -(4 e - )(96,500 J / V-mol)(+0.43 V) = -170,000 J / mol What can be deduced from this? E red = +1.23 V E ox = -0.80 V E cell = +0.43 V At 25 o C, find standard  G o and K for:

22. STEP 3: equilib constant K  G o = -RT LnK ===> Ln K =  G o / -(RT) K = e 69 = 9.3*10 29 Notice example pg. 864 part b rxn written 1 / 2 of original All values remain the same, even though half of quantities K is 1 / 2, why????

23. ELECTROLYTIC CELL ELECTROPLATING + + + + + + + Anode (+) plating material Cathode (-) object to be plated Ag + Ag 0 e-e- e-e- DC source NO 3 -

24. Coulumb (C) amt of charge that passes thru a pt w / current 1 ampere @ 1 sec 1 C = 1 As 1 F = 96,500 C / mol e - ELECTRICAL WORK -  G = w max w max = nFE w max = n * F * E J = (mol) * (C/mol)*(J/C)

25. Steps 1. Rxn Ag deposited, gain e -, red Ag +1 (aq) + 1 e - ---> Ag (s) 2. Relationship 1 mol Ag ~ 1 mol e - = 96,500 C 3. Current & time, find C 1.5 Ah*[2*(3600 s / h)] = 10,800 As = 10,800 C 10,800 C*(1 mol e - / 96,500 C)*(1 mol Ag / 1 mol e - ) *(107.8 g Ag / 1 mol Ag) = 12.1 g Ag EX. How many grams Ag deposited from AgNO 3 soln by current 1.5 Ah over 2 hrs.

26. QUANTITATIVE Michael Faraday 1 st to describe extent of current used to chem change @ electrodes Faraday (F) amt of electricity supplied to deliver 1 mol of e - ; “a mol of e - ” 1 mol Ag = 107.9 g, know 1 mol e - passed Amt Change - related to amt electricity passed amt moles e - lost / gain in redox rxn Ag + (aq) + e - ---> Ag (s)

27. Coulumb (C) amt of charge that passes thru a pt w / current 1 ampere @ 1 sec Steps 1. Rxn Cu deposited, gain e -, red Cu +2 (aq) + 2 e - ---> Cu (s) 2. Relationship 1 mol Cu ~ 2 mol e - = 2 F 3. Current & time, find C 1.5 A*[2*(3600 s / hr)] = 11,000 As = 11,000 C 11,000 C*(2 mol Cu /2 mol e- )*( 1 F / 96,500 C)*(63.6 g / 1 mol) = 3.5 g Cu EX. How many grams Cu deposited from CuSO 4 soln by current 1.5 A over 2hrs. 1 C = 1 As 1 F = 96,500 C = 1 mol e -

28. CELL POTENTIAL emf 1 V = 1 J / C E cell = E ox + E red Zn ox = -0.76 = +0.76 V Cu red = +0.34 V Zn more diff to reduce E cell = 0.76 + 0.34 = 1.10 V STEPS 1. E 1 = -0.74 V E 2 = +1.28 V Red 2 > Red 1 #1 must be oxidized, reverse 2. Rewrite, balance e -, & sum EX. What is the rxn and E cell from the following: 1) Cr +3 (aq) + 3 e - -----> Cr (s) 2) MnO 2 (s) + 4 H + (aq) + 2 e - ----> Mn +2 (aq) + 2 H 2 O (l) 2 [Cr (s) -----> Cr +3 (aq) + 3 e - ] 3[MnO 2 (s) + 4 H + (aq) + 2 e - ----> Mn +2 (aq) + 2 H 2 O (l) ] 3 MnO 2 (s) + 12 H + (aq) + 2 Cr (s) ----> 3 Mn +2 (aq) + 2 Cr +3 (aq) + 6 H 2 O (l) E cell = 1.28 + 0.74 = 2.02 V

29. Zn | Cu Cell move 2 e - @ 25 o C 1 V = 1J / C (2.303RT / F) is const = 0.0592 J / C E = (0.0592 / n) Log Kc  G = (2 mol e - )(96,500 C / mol)(1.10 J / C) =  G = -2.303RT Log K c Combine Eqns nFE = 2.303 Log K c E = (2.303RT / nF) Log K c

30. QUANTITATIVE, NERST EQN aA + bB ----> cC + dD @ 25 o C (2.303RT / F) = 0.0592 / n Zn | Cu Cell E o = 1.10 V n = 2 Application Nernst 1) w / varying concentrations 2) K sp, solubility product constant 3) pH R: gas constant, 8.314 T: temp, K n: # e- transferred F: Faraday constant, 96,500 Q: rxn quotient

31. Find the standard potential (E o ) for the rxn: Cd (s) + Cu +2 (aq) Cd +2 (aq) + Cu (s) [Cu +2 ] = 0.80 [Cd +2 ] = 0.20 1/2-rxns Anode (ox) Cd (s) Cd +2 (aq) + 2 e - Cathode (red) Cu +2 (aq) + 2 e - Cu (s) E o 0.40 V 0.34 V Cd +2 (aq) + Cu (s) Cu +2 (aq) + Cd (s) E o = 0.40 + 0.34 = 0.74 V Nernst Eqn: E = 0.74 - 0.0592 / 2*Log[0.20] / [0.80] = 0.74 - 0.0296*(-0.602) = 0.75 V

32. Nernst Eqn E = E o cell – (0.0592 V / n) Log ( [Fe +2 ] dil / [Fe +2 ] conc ) E = 0.0 – (0.0592 V / 2) Log ( [0.003] / [1.5] ) = -(0.0296 V) Log(0.002) = -(0.0296 V)(-2.70) = 0.080 V Concen Cell: system consists of 2 half-cells; cell #1: strip Fe metal in 1.5 M Fe +3 soln cell #2: strip Fe metal in 0.003 M Fe +3 soln What is the emf? Anode: Fe (s) ---  Fe +2 (aq) + 2 e - E o = +0.44 Cathode: Fe +2 (aq) + 2 e - ---  Fe (s) E o = -0.44 Fe +2 (aq, conc) ----  Fe +2 (aq,dilute) E o cell = 0.44 + (-0.44) = 0.0 V

33. E cell = 0.00 - (0.0592 / 2) Log([3.73*10 -4 ] / [1.35]) = -(0.0296)Log(2.76*10 -4 ) = -0.0296(-3.56) Oxidize Zn (s) ---> Zn +2 (aq) + 2 e - E ox = 0.763 V Reduced Zn +2 (aq) +2 e- ----> Zn (s) E red = -0.763V Concentration cell with 2 Zn(s)-Zn +2 (aq) half-cells. 1 st half-cell [Zn +2 ]= 1.35 M 2 nd cell [Zn +2 ] = 3.73*10 -4 M a) which half-cell is anode? b) What is emf? 1 st Cell; more concentrated E o = 0.00 V Nernst Eqn E = E o cell – (0.0592 V / n) Log ( [Zn +2 ] dil / [Zn +2 ] conc ) E cell = 0.105 V

34. emf - Electromotive Forces (V) E cell = E ox (Cl - ) + E red (H 2 O) = (-1.36) + (-.083) + -2.19 V Soln: CuCl 2 (s) ----> Cu (s) & Cl 2 (g) EX. Explain why CuCl 2 produces Cu (s) & Cl 2 (g). Min emf Ox: Anode 2 Cl - (l) ---> Cl 2 (g) + 2 e - Red: Cathode 2 Na + (l) + 2 e - ---> Na (l) Cell Rxn 2 Na + (l) + 2 Cl - (l) ---> Cl 2 (g) + Na (l)

35. E cell = (0.34) + (-1.36) = -1.02 V Oxidize 2 Cl - (aq) ---> Cl 2 (g) + 2 e - E ox = -1.36 V anode 2 H 2 O (l) ---> O 2 (g) + 4 H + (aq) + 4 e - E ox = -1.23 V Reduce 2 H 2 O (l) + 2 e - ---> H 2 (g) + 2 OH - (aq) E red = -0.83 V cathode Cu (aq) + 2 e - ---> Cu (s) E red = +0.34 V H 2 produced @ cathode

36. Half Reactions Ox - Anode 2 Cl - (l) ---> Cl 2 (g) + 2 e - Red -Cathode 2 Na + (l) + 2 e - ---> Na (l) Electro- Cathode: e- forced unto (-) Anode: e- withdrawn (+) more complex due to ability H 2 O to RED & OX Possible red-ox solvent & ions solute; whether the solute anion or H 2 O, or the solute cation or H 2 O Cell Rxn 2 Na + (l) + 2 Cl - (l) ---> Cl 2 (g) + Na (l) AQUEOUS SOLN

37. Reduce 2 H 2 O (l) + 2 e - ---> H 2 (g) + 2 OH - (aq) E red = -0.83 V Na + (aq) + e - ---> Na (s) E red = -2.71 V H 2 produced @ cathode Acidic Soln 2 H + (aq) + 2 e - ---> H 2 (g) not major dil. aq soln Oxidize 2 Cl - (aq) ---> Cl 2 (g) + 2 e - E ox = -1.36 V 2 H 2 O (l) ---> O 2 (g) + 4 H + (aq) + 4 e - E ox = -1.23 V Overcharge?? O 2 formation is high to permit Cl - oxid than H 2 O “BRINES” produce H 2 & Cl 2

38. Anode: E = 0.356 V Cathode: E = 1.685 V E cell = 0.356 + 1.685 = 2.041 V Pb Storage Battery - car 12 V Consists of: 6 cells (2 V) anode: Pb Cathode: PbO 2 H 2 SO 4 (aq) Discharge Anode Pb (s) + SO 4 -2 (aq) -----> PbSO 4 (s) + 2 e - Cathode PbO 2 (s) + SO 4 -2 (aq) + 4 H + (aq) + 2 e - ----> PbSO 4 (s) + 2 H 2 O (l) ] Pb (s) + PbO 2 (s) + 4 H + (aq) + 2 SO 4 -2 (aq) ----> 2 PbSO 4 (s) + 2 H 2 O (l)

39. Reactants: Pb; PbO 2 D H2SO4 = 1.8 Charge D = 1.25 - 1.30 Recharge D < 1.20 - + Pb PbO 2 H 2 SO 4

40. Find the equilibrium constant for the rxn @ 25 o C of Ni (s) & Ag +1 (aq) Anode: Ag 0.80 Cathode: Ni 0.25 n = 2 Log K = (2 * 1.05 V) / 0.0592 = 35.473 K = 10 35.473 = 3*10 35 E cell = 0.80 + 0.25 = 1.05 V

41. Find E cell and  G o @ 25 o C for n = 2 & K = 5.0*10 -6  G o = -RT Ln K = -(8.314 J / molK)*(298.15 K) Ln(5.0*10 -6 ) = -(2478.8)*(-12.2) = 30200 J / mol = -0.16 V E cell = -(30200 J/mol) / [ (2 mol e - )*(96,500 C / mol e - ) ] *(V / JC -1 ) E cell = -  G o / nF

42. Ex. MnO 4 -2 is stable in strong basic soln. (frames 6 & 7 basic example) In acidic soln, reacts to form permanganate and MnO 2 (s). Wrtie balanced overall rxn from the two half-rxns.

43. Ex. Write balance eqn of CuS in 3M HNO 3. Cu (s) + H + (aq) + NO 3 -1 (aq) -----> Cu +2 (aq) + S (s) + NO (g) + H 2 O (l)