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Ch17 z5e electrochem

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  • 1. Electrochemistry pp Applications of Redox
  • 2. 17.1 Galvanic Cells Oxidation reduction reactions involve a transfer of electrons. OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction
  • 3. Applications Moving electrons = electric current. 8H + +MnO 4 - + 5Fe +2  Mn +2 + 5Fe +3 +4H 2 O It helps to break the reactions into half reactions. 8H + +MnO 4 - +5e -  Mn +2 +4H 2 O 5(Fe +2  Fe +3 + e - ) In the same mixture this happens without doing useful work, but if separate . . .
  • 4. Connected this way the reaction starts. Stops immediately because charge builds up. H + MnO 4 - Fe +2
  • 5. Galvanic Cell H + MnO 4 - Fe +2 Salt Bridge allows current to flow
  • 6. Galvanic Cell H + MnO 4 - Fe +2 Electrons flow in the wire, ions flow through the salt bridge
  • 7. Electricity travels in a complete circuit Instead of a salt bridge . . . H + MnO 4 - Fe +2 e -
  • 8. H + MnO 4 - Fe +2 Porous Disk
  • 9. Reducing Agent Oxidizing Agent e - e - e - e - e - e - Oxidation at Anode Reduction at Cathode
  • 10. Cell Potential Oxidizing agent pulls the electron. Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential E cell Also called the electromotive force (emf). Unit is the volt (V). = 1 joule of work per coulomb of charge transferred (1 V = 1 J/C). Measured with a voltmeter.
  • 11. 17.2 Standard Reduction Potentials The reaction in a galvanic cell is a redox reaction. So, break it down into two half-reactions. Assign a potential to each. Sum the half-cell potentials to get the overall cell potential. “Active anodes” - the more active metal is always the anode (good for multiple choice questions).
  • 12. Zn +2 SO 4 -2 1 M HCl Anode 0.76 1 M ZnSO 4 H + Cl - H 2 in Cathode
  • 13. Standard Hydrogen Electrode This is the reference all other oxidations are compared to. E º = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. 1 M HCl H + Cl - H 2 in
  • 14. Z5e 841 Figure 17.5: Zn/H Galvanic Cell. Notice the electron flow also.
  • 15. Cell Potential Zn(s) + Cu +2 (aq)  Zn +2 (aq) + Cu(s) The total cell potential is the sum of the potential at each electrode. E º cell = E º Zn  Zn +2 + E º Cu +2  Cu We can look up reduction potentials in a table (see, p. 796). For Br, s/b 1.0 7 (not 1.09) need for online HW!!) Since one of the 1/2 reactions is oxidation its table value must be reversed , so change its sign .
  • 16. Z5e 842 Fig 17.6 Zn/Cu Galvanic Cell
  • 17. Cell Potential pp Determine the cell potential for a galvanic cell based on the redox reaction . . . Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) steps follow Fe +3 (aq) + e -  Fe +2 (aq) E º = 0.77 V Cu +2 (aq)+ 2 e -  Cu(s) E º = 0.34 V Since one of these must be oxidation, one of them needs to be reversed. Which one?
  • 18. Cell Potential pp Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) This is spontaneous only if Eº cell = (+), so reverse the copper half-reaction. Fe +3 (aq) + e -  Fe +2 (aq) E º = 0.77 V Cu(s)  Cu +2 (aq)+ 2 e - E º = - 0.34 V Must balance the e - s, so multiply the Fe 1/2-reaction by 2, BUT do not multiply E º. Why? Cell potential is an intensive property (doesn’t depend on number of times the reaction occurs).
  • 19. Cell Potential pp Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) 2 Fe +3 (aq) + 2 e -  2 Fe +2 (aq) E º = 0.77 V Cu(s)  Cu +2 (aq)+ 2 e - E º = - 0.34 V Cu(s) + 2Fe +3 (aq)  Cu +2 (aq) + 2Fe +2 (aq) E º = 0.43 V
  • 20. Cell Potential Be sure to use the correct 1/2-reactions! For example, table 17.1, p. 796 lists three 1/2-reactions for MnO 4 1- Find them. Answers next slide.
  • 21. Cell Potential Three 1/2-reactions for MnO 4 1- : MnO 4 1- + 4H 1+ 3e -  MnO 2 + 2H 2 O Eº = 1.68 MnO 4 1- + 8H 1+ 5e -  Mn 2+ + 4H 2 O Eº = 1.51 MnO 4 1- + e -  MnO 4 2- Eº = 0.56 Pick the reaction that “works” with your overall reaction (look at reactants and products).
  • 22. Line Notation pp solid  Aqueous  Aqueous  solid Anode on the left  Cathode on the right Single line to show different phases. Double line  porous disk or salt bridge. If all the substances on one side are aqueous , a platinum electrode is used. For: Cu(s) + Fe +3 (aq)  Cu +2 ( aq ) + Fe +2 ( aq ) Cu( s )  Cu +2 (aq)  Fe +3 ( aq ),Fe +2 ( aq )  Pt (s) Remember to show the electrodes!!
  • 23. Complete Galvanic Cell Description (AP Test) pp The reaction always runs spontaneously in the direction that produced a positive cell potential. Four things for a complete description: Cell Potential and balanced reaction Direction of flow Designation of anode and cathode Nature of all components -- electrodes & ions (plus inert conductor like Pt if needed). Use line notation.
  • 24. Practice pp Completely describe the galvanic cell based on the following half-reactions under standard conditions. MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º = 1.51 V Fe +2 +2e -  Fe(s) E º = - 0.44 V
  • 25. Practice - Item 1 pp Determine cell potential & balanced reaction MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º = 1.51 V Fe +2 +2e -  Fe(s) E º = - 0.44 V Since (+) E required, reverse the 2nd reaction E º cell =1.51 + ( + 0.44) = 1.95 V Complete, balanced reaction is 2MnO 4 - + 5Fe(s) + 16 H +  2Mn +2 + 5Fe 2+ (aq) + 8H 2 O(l) Note: multiplying the half reactions to balance the reaction does NOT multiply the E º values!!! (intensive property).
  • 26. Practice - Item 2 pp Determine e - flow by inspecting 1/2 rxns & using the direction that gives a (+) E º cell MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º = 1.51 V Fe (s)  Fe +2 +2e - E º = + 0.44 V E º cell =1.51 + ( + 0.44) = 1.95 V So, electrons flow from Fe(s) to MnO 4 - (aq)
  • 27. Practice - Item 3 pp Designate the anode and cathode MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º = 1.51 V Fe (s)  Fe +2 +2e - E º = + 0.44 V E º cell =1.51 - (0.44) = 1.95 V Elections flow from Fe( s ) to MnO 4 - So, oxidation occurs in the compartment containing Fe(s) -- the anode Reduction occurs in the compartment containing MnO 4 - -- Use Pt for the cathode Note: e - always flow from anode to cathode ” Red cat ate an ox". Red/cat = reduct/cathode
  • 28. Practice - Item 4 pp Describe nature of each electrode & ions present (use line notation) Complete, balanced reaction is 2MnO 4 - + 5Fe(s) + 16 H +  2Mn +2 + 5Fe 2+ (aq) + 8H 2 O(l) Electrode in Fe/Fe 2+ compartment is iron metal An inert conductor like Pt must be used in MnO 4 -1 / Mn +2 compartment (don’t forget). Line notation is: Fe(s)  Fe +2 (aq)  MnO 4 -1 (aq),Mn +2 (aq)  Pt(s)
  • 29. pp Figure 17.7: A Schematic of the previous Galvanic Cell Eº = 1.95 v Be able to draw this as well as write the line notation for the AP exam.
  • 30. 17.3 Cell Potential, Work &  G emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q (emf & work have opposite signs) Use (-)w because it is flowing out from system. Charge is measured in coulombs. -w = q E (where q = the charge) Faraday = 96 485 C/mol e - q = nF = moles of e - x charge per mole e - w = -q E = -nF E =  G
  • 31. Potential, Work and  G pp  Gº = -nF E º (at standard conditions) if E º > 0, then  Gº < 0 spontaneous if E º < 0, then  Gº > 0 nonspontaneous In fact, reverse is spontaneous.
  • 32. Potential, Work and  G Calculate  Gº for the following reaction: Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq) Fe +2 (aq) + 2e -  Fe(s) E º = - 0.44 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V  Gº = -nF E º Answer? -1.5 x 10 5 J Calculation with units is . . . -(2 mol e - )(96 485 C / mol e - )(0.78 J / C )
  • 33. Putting It Together pp Using Table 17.1, predict if 1 M HNO 3 will dissolve gold to form a 1 M Au 3+ solution? What are the half reactions? . . . Gold needs to be oxidized so HNO 3 must be reduced. Look for a half-reaction with HNO 3 where NO 3 1- is being reduced . . . NO 3 1- + 4H 1+ + 3e -  NO + 2H 2 O Eº = 0.96 v Au  Au 3+ + 3e - Eº = -1.50 v
  • 34. Putting It Together pp Using Table 17.1, predict if 1 M HNO 3 will dissolve gold to form a 1 M Au 3+ solution? NO 3 1- + 4H 1+ + 3e -  NO + 2H 2 O Eº = 0.96 v Au  Au 3+ + 3e - Eº = -1.50 v Au + NO 3 1- + 4H 1+  Au 3+ + NO + 2H 2 O Eº cell = -0.54 v Since Eº cell = negative, this cannot be spontaneous because . . .  Gº = -nF E º = (-)(3)(96 485)(-0.54) = +156kJ Since  Gº is (+), not spontaneous.
  • 35. 17.4 Cell Potential and Concentration pp Notes for online HW 1 lb = 453.6 g 1 Faraday = 96 485 c/s (not 96 500 c/s) Use 0.059 2 in Nernst equation (not 0.0591)
  • 36. 17.4 Cell Potential and Concentration pp Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell of 0.48 v if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5 M if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M Steps . . .
  • 37. Cell Potential pp 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell of 0.48 v if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M Answer . . . Since a product [ ] has been raised above 1.0 M , Le Chatelier predicts a shift left and E cell < Eº cell
  • 38. Cell Potential pp 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell of 0.48 v if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5 M Answer . . . Since a reactant [ ] has been raised above 1.0 M , Le Chatelier predicts a shift right and E cell > Eº cell
  • 39. Cell Potential pp 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell of 0.48 v if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M Answer . . . Since both [ ]s have been raised above 1.0 M , cannot use Le Chatelier! Must use the Nernst Equation (coming to your community soon!)
  • 40. Le Chatelier, ∆G, & Concentration Cells Figure 17.9 A Concentration Cell That Contains a Sliver Electrode and Aqueous Silver Nitrate in Both Compartments Since the right compartment has higher [Ag 1+ ] there is a shift right of e - So, Ag metal is deposited on the right side while [Ag 1+ ] decreases on right side and increases on the left.
  • 41. Le Chatelier, ∆G, & Concentration Cells pp So, Ag metal is deposited on the right side while [Ag 1+ ] decreases on right side and increases on the left. You will need to recognize this concept to solve for ∆G on the AP exam. Hint: the electrode with the largest [ ] will always be the cathode (where reduction occurs).
  • 42. The Nernst Equation  G =  Gº + RT ln(Q), since  G = -nF E . . . -nF E = -nF E º + RT ln(Q) E = E º - RT ln(Q) nF What is n in Al (s) + Mn 2+  Al 3+ + Mn (s) ? Always have to figure out “n” by balancing. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E º = 0.48 V n = 6. Why? . . . n = mole of e - , not mole of compound.
  • 43. The Nernst Equation continued  G =  Gº + RT ln(Q) -nF E = -nF E º + RT ln(Q) E = E º - RT ln (Q) = E º - 2.303 RT log (Q) nF nF Since we know R and F, at 25 o C, above is aka E = E º - 0.059 2 log(Q) n Textbook has typo: 0.0591 should be 0.059 2 Use 0.0592 for all your calculations!
  • 44. The Nernst Equation pp E = E º - 0.0592 log(Q) n For concentration cells ( i.e. , not at 1 M ) this equation must be done separately for each 1/2 cell, then subtract the results (or flip one and add) to get the cell potential. See the following problem . . .
  • 45. We’ll do “a,” “b,” & “c”. pp
  • 46. E cell when [Ag 1+ ] on the right = 1.0 M pp Since [Ag 1+ ] is same on both sides E cell = Eº cell which is 0 because . . . Ag 1+ + e -  Ag Eº = .80v Ag  Ag 1+ + e - Eº = -.80v Eº cell = 0 E = E º - 0.0592 log(Q) n Since log(1) = 0, E cell = Eº cell and Eº = 0 from above, so E cell also = 0 .
  • 47. E cell when [Ag 1+ ] on the right = 2.0 M pp Cathode always has the higher [ ] and e - always flow from the anode to the cathode. Q = [ ] anode ÷ [ ] cathode Here, [ cathode ] is on the right ( 2.0 M ) , & in the denominator for Q E = E º - 0.0592 log(Q) n E = 0 - 0.0592 log 1.0 = . 018 v 1 2.0
  • 48. E cell when [Ag 1+ ] on the right = 0.10 M pp Cathode always has the higher [ ] and e - always flow from the anode to the cathode. So, [cathode] is on the left & in the denominator for Q E = E º - 0.0592 log(Q) n E = 0 - 0.0592 log 0.1 = . 059 v 1 1.0 You will have a test question on this and it is NOT on the pre-test, so . . . Do p. 832 #53!!
  • 49. E cell when [Ag 1+ ] on the right = 0.10 M pp Notes for all of these: Eº is always 0 for concentration cells because the 1/2 reactions cancel. E = E º - 0.0592 log(Q) n E = 0 - 0.0592 log anode . n cathode Memorize this equation! You will have a test question on this and it is NOT on the pre-test, so . . . Do p. 832 #53!!
  • 50. The Nernst Equation As reactions proceed concentrations of products increase and reactants decrease. Reaches equilibrium where Q = K and E cell = 0 Since at equilibrium E cell = 0 = E º - RT ln( K ) nF E º = RT ln(K) nF nF E º = ln(K) at 25 º C aka log(K) = n E º RT 0.0592 Use ln(K) version in online HW even if at 25ºC!!
  • 51. Nernst Equation & K pp Calculate K sp of silver iodide at 298 K AgI(s)  Ag + + I - where E o AgI(s) + e -  Ag(s) + I - -0.15 v I 2 (s) + 2e -  2I - +0.54 v Ag + + e -  Ag(s) +0.80 v logK = nEº/0.0592 1st, get Eº cell for overall reaction. Steps. Find overall reaction, then Eº cell Only need 1st & 3rd equations to get . . . AgI(s)  Ag + + I - where E o cell = -0.95 v
  • 52. Nernst Equation & K pp Calculate K sp of silver iodide at 298 K AgI(s)  Ag + + I - where E o cell = -0.95 v at 25º C log(K) = n E º = (1)(-0.95) = -16.05 0.0592 0.0592 K sp = 10 -16.05 = 9.0 x 10 -17
  • 53. 17.5 Batteries are Galvanic Cells Car batteries are lead storage batteries. Pb +PbO 2 +H 2 SO 4  PbSO 4 (s) +H 2 O Be able to recognize the anode & cathode from the half reactions.
  • 54. Figure 17.13 One of the Six Cells in Storage Battery a 12-V Lead Storage Battery
  • 55. Batteries are Galvanic Cells Dry Cell - acid battery Zn + 2NH 4 + + 2MnO 2  Zn +2 + 2NH 3 + Mn 2 O 3 + H 2 O Dry Cell - alkaline battery Zn + 2MnO 2  ZnO + Mn 2 O 3 (in base) NiCad - can be re-charged indefinitely NiO 2 + Cd + 2H 2 O  Cd(OH) 2 +Ni(OH) 2 Fuel Cell - reactants are continuously supplied CH 4 + 2O 2  CO 2 + 2H 2 O + energy
  • 56. Figure 17.14 A Common Dry Cell Battery
  • 57. 17.6 Corrosion Rusting - spontaneous oxidation . Most structural metals have reduction potentials that are less positive than O 2 . So, they oxidize while O 2 is reduced. Fe +2 +2e -  Fe E º= - 0.44 V O 2 + 2H 2 O + 4e -  4OH - E º= + 0.40 V Reverse top half reaction, then add both Fe + O 2 + H 2 O  Fe 2 O 3 (rust) + H + Eº cell = 0.84 v Reaction happens in two places . . .
  • 58. Water Rust Iron Dissolves - Fe  Fe +2 e - Salt speeds up process by increasing conductivity
  • 59. Figure 17.17 The Electrochemical Corrosion of Iron
  • 60. Preventing Corrosion Coating - to keep out air and water. Galvanizing - Putting on a zinc coat Zinc has a lower reduction potential than iron, so it is more easily oxidized. So, zinc is a more active metal than iron. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of a more active metal like magnesium that get oxidized instead of iron (iron stays reduced ).
  • 61. Preventing Corrosion Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead of iron. Attach Mg wire to iron pipe (and replace periodically). Attach titanium bars to ships’ hulls. In salt water the Ti acts as the anode and is oxidized instead of the steel hull, which now acts as the cathode.
  • 62. Figure 17.18 Cathodic Protection
  • 63. Running a galvanic cell backwards . Put a voltage whose magnitude is bigger than the potential which reverses the direction of the redox reaction. Produces a chemical change which would not normally happen because the potential is negative . Remember: 1 A = 1 C/s and 1 F = 96 485 C Used for electroplating -- depositing the neutral metal onto the electrode by reducing the metal ions in solution. 17.7 Electrolysis
  • 64. 1.0 M Zn +2 e - e - Anode Cathode 1.10 Zn Cu 1.0 M Cu +2 Galvanic Cell - spontaneous
  • 65. 1.0 M Zn +2 e - e - Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2 Electrolytic Cell Forces the opposite reaction.
  • 66. Figure 17.19 (a) A Standard Galvanic Cell (b) A Standard Electrolytic Cell
  • 67. Calculating plating Have to include the charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second 1 A = 1 C/s q = I x t = the charge q/nF = moles of metal Mass of plated metal
  • 68. Calculating plating pp How many minutes must a 5.00 amp current be applied to produce 10.5 g of Ag from Ag + Steps follow . . . Set up a picket fence that includes Current & time Quantity of charge (in coulombs) Moles of electrons Moles of metal (may be different) Grams of metal Arrange the picket fence so the units give you what you’re looking for.
  • 69. Calculating plating pp How many minutes must a 5.00 amp current be applied to produce 10.5 g of Ag from Ag + Steps follow . . . The pieces of the picket fence are: 5.00 amp (rewrite as 5.00 C/s) 10.5 g Ag 107.868 g Ag/1mol Ag 1 mol e - /mol Ag (Ag  Ag 1+ + 1e - ) 96 485 C/mol e - 60 sec/min Your answer? . . .
  • 70. Calculating plating pp How many minutes must a 5.00 amp current be applied to produce 10.5 g of Ag from Ag + (amp = C/s) 31.3 minutes ( 10.5 g Ag )( 1 mol Ag / 107.868 g Ag )( 1 mol e - / 1 mol Ag )( 96 485 C / 1 mol e - )( 1 s / 5.00C )( 1 min / 60 s )
  • 71. Calculating plating pp An antique automobile bumper is to be chrome plated by dipping into an acidic Cr 2 O 7 2- solution serving as the cathode of an electrolytic cell. M Cr = 51.996 Using 10.0 amperes, how long to deposit 1.00 x 10 2 grams of Cr(s)? Steps . . . Find overall reaction from 1/2-reactions (to get moles of e-). Steps follow. The only substances you start with are Cr 2 O 7 2- and H 2 O. Which is oxidized? Reduced? . . .
  • 72. Calculating plating pp Using 10.0 amperes, how long to deposit 1.00 x 10 2 grams of Cr(s) by dipping into acidic Cr 2 O 7 2- (aq) , M Cr = 51.996? The only substances you start with are Cr 2 O 7 2- and H 2 O. Which is oxidized? Which is reduced? . . . Cr 2 O 7 2- must be reduced to get to Cr (s) so H 2 O must be oxidized . Use Table 17.1 p. 796 to find the 1/2-reaction of H 2 O . . .
  • 73. Calculating plating pp Use Table 17.1 p. 843 to find the 1/2-reaction of H 2 O. Which is it? H 2 O 2 + 2H 1+ + 2e -  2H 2 0 2H 2 0  O 2 + 4H 1+ + 4e - O 2 + 2H 2 0 + 4e -  4OH 1- 2H 2 0 + 2e -  H 2 + 2OH 1- ***** #2 above is the only one that starts with water and is oxidized.
  • 74. Calculating plating pp Using 10.0 amperes, how long to deposit 1.00 x 10 2 grams of Cr(s) by dipping into acidic Cr 2 O 7 2- (aq) ? , M Cr = 51.996 So, the 1/2-reactions needed are . . . 6e - + 14H 1+ + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 2H 2 0  O 2 + 4H 1+ + 4e - and one more! (Why?) You only got to Cr 3+ , but you need Cr (s) Also need Cr 3+ + 3e -  Cr(s) Now, write the balanced equation . . .
  • 75. Calculating plating pp Using 10.0 amperes, how long to deposit 1.00 x 10 2 grams of Cr(s) by dipping into acidic Cr 2 O 7 2- (aq) ? , M Cr = 51.996 Now, write the balanced equation . . . 2H 1+ + Cr 2 O 7 2-  H 2 O + 3O 2 + 2Cr(s) How many mol e - changed in this reaction (comprised of three 1/2-reactions)? 12 mol e - Now, you can start to do the problem. Your answer in days ? . . .
  • 76. Calculating plating pp Using 10.0 amperes, how long to deposit 1.00 x 10 2 grams of Cr(s) by dipping into acidic Cr 2 O 7 2- (aq)? , M Cr = 51.996 with 2H 1+ + Cr 2 O 7 2-  H 2 O + 3O 2 + 2Cr(s) and 12 mol e - moving. Your answer is . . . 1.29 days . . . 100.g Cr(s) • 1mol Cr/51.996g Cr(s) • 12 mol e - / 2 mol Cr(s) • 96 486 C/1 mol e - • 1s/10.0C • 1 h/3600 s • 1 d/24 h = 1.29 days
  • 77. Calculating plating Electrolysis of a molten salt , MCl, using 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal? Use picket fence to get moles of metal. Since g/mol = M, use calculated moles of metal and 1.52 g to get M. Answer . . . Potassium. The solution is . . .
  • 78. Calculating plating Electrolysis of a molten salt , MCl, using 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal? 3.86 C/s • 16.2 min • 60s/1m • 1mol e-/96 485 C • 1 mol M 1+ /1mol e- = 0.039 = moles of metal. 1.52g/0.039 mol = 39.1 g/mol = potassium
  • 79. Other uses pp Electrolysis of water. Separating mixtures of ions A more positive reduction potential means that reaction proceeds forward. The metal with the most positive reduction potential is easiest to plate out of solution (and is the best oxidizer).
  • 80. Relative Oxidizing Abilities pp An acidic solution has Ce 4+ , VO 2 1+ , & Fe 3+ . Use Table 17.1 p. 796 to predict the order of oxidizing ability. Answer . . . Order of oxidizing ability is the same as the order for being reduced, So. . . Ce 4+ + e -  Ce 3+ E = 1.70 v VO 2 1+ + 2H 1+ + e -  VO 2+ + H 2 O E = 1.00 v Fe 3+ + e -  Fe 2+ E = 0.77 v Also, predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage. Answer . . . Since Ce 4+ is the greatest oxidizer it is the most easily reduced (needs least voltage). Do section 17.8 on your own.