1. Electrochemistry Electrochemistry (Bozeman 8:43 min)

2. REDOX REACTIONS  Specify which of the following are redox reactions and identify the oxidizing agent, reducing agent, substance being oxidized and substance being reduced:  Cu (s) + 2Ag + (aq)  Cu 2+ (aq) + 2Ag (s)  Cu: oxidized (loses electrons)  Ag + : reduced (gains electrons)  Cu 2+ : reducing agent  Ag: oxidizing agent

3. Balance the following reactions using the half-reaction method.  For acidic solutions:  Cu (s) + NO 3 - (aq)  Cu 2+ (aq)) + NO (g)  Which one is oxidized and which is reduced?  Cu  Cu 2+ + ? e -  ?e - + NO 3 -  NO  3(Cu  Cu 2+ + 2 e - )  2( 4H + + 3e - + NO 3 -  NO + 2H 2 O)  8H + + 6e - + 3Cu + 2NO 3 -  3Cu 2+ + 6e - + 2NO + 4H 2 O  ∴ 8H + + 3Cu + 2NO 3 -  3Cu 2+ + 2NO + 4H 2 O

4. For basic solutions:  CN - (aq) + MnO 4 - (aq)  CNO - (aq) + MnO 2(s)  3(OH - + CN-  CNO - + 2 e - + H + )  2(2H + + 3e - + MnO 4 -  MnO 2 + 2OH - )  3OH - + 3CN - + 4H + + 6e - + 2 MnO 4 -  3CNO - + 6e - + 3H + + 2MnO 2 + 4OH -  ∴ 3CN - + H + + 2MnO 4 -  3CNO - + 2MnO 2 - + OH -

5.  Br (aq) + MnO 4 - (aq)  Br 2(l) + Mn 2+ (aq)  18H + + 5e - + MnO 4 - + 10Br -  5Br 2 + 5 e - + Mn 2+ + 4H 2 O  Cl 2(g)  Cl - (aq)  OCl - (aq)  2OH - + Cl 2  2Cl - + 2OCl - + 2H +  Al (s) + MnO 4 - (aq)  MnO 2(s) + Al(OH) 4(aq)  2H 2 O + Al + MnO 4  Al(OH) 4 + MnO 2  CN - (aq) + MnO 4 - (aq)  CNO - (aq) + MnO 2(s)  3CN - + H + + 2MnO 4  3CNO - + 2MnO 2 + OH -

6. Galvanic and Electrolytic CellsGalvanic and Electrolytic Cells (Hank 11:00) Electrolytic cells have an external power source They are thermodynamically unfavorable. Used for electroplating stuff like chrome bumpers Galvanic cells are thermodynamically favorable because of the chemical reactions--batteries

7. Anode and Cathode songAnode and Cathode song 3:28 Oxidation occurs at the anode. (mass may decrease) AN OX Reduction occurs at the cathode. (mass may increase) RED CAT FAT CAT : electrons always flow F rom the A node T o the CAT hode. Salt bridge: Maintains electrical neutrality in a galvanic cell. Electrons flow through the wire from anode to cathode. Voltmeters measure the cell potential (emf) in volts.

8. Inert electrodes serve as a source for electrons to travel and are used when a gas is involved OR ion to ion involved such as Fe 3+ rather than Fe 0. These are made of precious metals like Pt, mercury or graphite. Animation (3:26)

9.  Each half-reaction has a cell potential.  Each potential is measured against a standard hydrogen electrode which has an assigned value of 0.00 V.  Standard conditions: 1 atm, 1.0M for solutions and 298K  E cell 0, Emf 0, or ε cell 0 all mean they are at standard conditions.  Tale of Standard Electrode Potentials  Elements that have the most positive reduction potentials are easily reduced (mostly non-metals like carbon)  Elements with least positive reduction potentials are easily oxidized (metals)  Metals with less positive reduction potentials are more active and will replace metals with more positive potentials.

10. Steps to calculating standard cell potential: E o cell, Emf o or ε cell o  Decide which metal is oxidized or reduced. The metal with the more positive reduction potential is reduced.  Write both equations as is from the chart with their associated voltages.  Reverse the equation that will be oxidized and change the sign of its voltage—this is now E o oxidation.  Balance the two half reactions but don’t multiply the voltage values because a volt is equivalent to a J/coulomb which is a ratio.  Add the two half reactions and the voltages together.  E cell o = E o oxidation + E o reduction

11. Consider a galvanic cell based on the reaction: Al 3+ (aq) + Mg (s)  Al (s) + Mg 2+ (aq)  Give the balanced cell reaction and calculate E o for the cell.  Go to the chart--which one is oxidized?  2(Al 3+ + 3e -  Al)-1.66 V  3(Mg  Mg 2+ + 2e -) - 2.37 V (this is being oxidized so change the sign but do not multiply it by 3).)  Balance equation:  6e - + 2Al 3+ + 3Mg  2Al + 3Mg 2+ + 6e - (electrons cancel out)  E o = 2.37 – 1.66 = 0.71V

12. MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) + Mn 2+ (aq) + H 2 O (l) Give balanced cell reaction and calculate E o for the cell. Link for additional standard potentialsLink for additional standard potentials  MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O reduced : E =1.51V  ClO 3 - + H 2 O  ClO 4 - + 2e - + 2H + oxidized : E = 1.19 V (reverse sign)  2(MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O)  5(ClO 3 - + H 2 O  ClO 4 - + 2e - + 2H + )  2MnO 4 - + 5ClO 3 - + 6H +  2Mn 2+ + 5ClO 4 - + 3H 2 O  E = 1.51 V – 1.19 V = 0.32V

13. Balance the reaction and calculate the cell voltage for : Fe 3+ (aq) + Cu (s)  Fe 2+ (aq) + Cu 2+ (aq)  Fe 3+ + e -  Fe 2+ reduced: E = 0.77V  Cu  Cu 2+ + 2e - oxidized: E = -0.34 V  Balanced equation: 2Fe 3+ + Cu  2Fe 2+ + Cu 2+  E cell = 0.77 – 0.34 = 0.43V

14. Draw a diagram of this galvanic cell.  You will need two beakers with a salt bridge and voltmeter connecting the two electrodes. The salt bridge will contain potassium nitrate.  Label everything and explain (in words) what is happening in this diagram.  Electrons flow from the anode to the cathode, down through the Pt (carbon) plate and to the iron (III) solution reducing it to Iron (II). With the loss of a + charge, the potassium cations will enter the solution from the salt bridge to balance out the charge. In the other beaker, as the electrons leave, more copper (II) ions are produced and for every ion produced, two nitrate ions leave the salt bridge to balance out that charge.  Eventually, when the reaction reaches equilibrium the E cell = 0 and the battery no longer works.

15. Calculate the cell voltage for the galvanic cell that would utilize silver metal and involve iron (II) ion and iron (III) ion. Draw a diagram of the galvanic cell for the reaction and label completely.  First write out the half-reaction for iron (III)  iron (II) off of chart:  Fe 3+ + e -  Fe 2+ E = 0.77V  This half reaction is lower on the chart so its more easily oxidized than Ag. Therefore, reverse the sign of E, so E = -0.77V. Since its oxidized, which electrode will it be? AN OX  Ag + + e -  Ag o E = 0.80V  ∴ 0.80V – 0.77V = 0.03 V

16. Cell potential, electrical work and free enegy  Work done when electrons are transferred through a wire depends on the “push” or emf which is the potential difference (in volts) between two points in a circuit.  emf (V) = ε = work(J)/charge (C)  One J of work is produced (or required) to transfer one coulomb of charge between two points in the circuit that differ by one potential of one volt.  If work flows out of the system, it has a MINUS sign (Joules are lost).  When a cell produces a current, the cell potential is positive and the current can be used to do work. So, ε and work have opposite signs.

17. Michael Faraday: English scientist who was a pioneer in electromagnetism and electrochemistry  ε = work(J)/charge(C) = - w/q ∴ -w = εq  Where q = number of moles x F  F = faraday: charge on one mole of electrons = 96 485.3365 coulombs or just 96500 coulombs if you’re working multiple choice problems.

18.  w max = ΔG  ΔG o = - nFE o  G = Gibb’s Free Energy  n = number of moles of electrons  F = Faraday constant 96, 485 J/V· mol  - E o is thermodynamically unfavorable  + E o is thermodynamically favorable

19. Calculate ΔG o for : Cu 2+ (aq) + Fe (s)  Cu (s) + Fe 2+ (aq)  ΔG o = - n FE o  Reduced: Cu 2+ + 2e -  CuE = 0.34 V  Oxidized: Fe  Fe 2+ + 2e - E = + 0.44V (sign reversed)  E cell + 0.34V + 0.44V = 0.74V  ΔG o = - 2 mol e- (96,485 J/V · mol )(0.74V) = -1.5 x 10 5 J  Is it thermodynamically favorable?  Yes, because the E cell is positive and ΔG o is negative.

20.  Using the standard reduction potentials, explain why 1 M HNO 3 will NOT dissolve gold metal to form a 1 M Au 3+ solution.  Au + 3e -  Au 3+ reduced E = -1.50V  NO 3 - + 4 H + + 3e -  NO + 2H 2 O oxidized E = 0.96 V  E cell = -.54V  ΔG = - 3mol (96,485 J/V· mol)(-.54V) = 1.6 x 10 5 J  Not favorable because E is negative and ΔG is positive. Not much work going on there…

21. Voltaic Cells and concentration  If the voltaic cells are at nonstandard conditions. LeChatelier’s principle is used. An increase in the concentration of a reactant will favor the forward reaction and the cell potential will increase. A reduction in concentration will reduce the cell’s potential as well.  2Al (s) + 3Mn 2+ (aq)  2Al 3+ (aq) + 3Mn (s)  Explain why the E cell is larger or smaller than the E o cell for the following cases:  a. [Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M  b. [Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M

22.  2Al (s) + 3Mn 2+ (aq)  2Al 3+ (aq) + 3Mn (s)  2(2Al  2Al 3+ + 3e - ) oxidized E = 1.66 V (reversed sign)  3Mn 2+ + 6e -  3Mn reduced E = -1.18 V  E o cell = 0.48V  4Al + 3Mn 2+  4Al 3+ + 3Mn  When Al 3+ concentration is increased, equilibrium shifts to the left and cell potential is reduced.  When Mn 2+ concentration is increased, equilibrium shifts to the right and the cell potential is increased.

23. Determine E o cell and E cell based on the following half-reactions: VO 2 + + 2H + + e -  VO 2+ + H 2 O Zn 2+ + 2e -  Zn  2(VO 2 + + 2H + + e -  VO 2+ + H 2 O)E = 1.00V  Zn 2+ + 2e -  Zn E = -0.76V  VO 2 + is more positive so it is reduced and Zn is oxidized.  Reverse Zn equation and voltage.  2(VO 2 + + 2H + + e -  VO 2+ + H 2 O)E = 1.00V  Zn  Zn 2+ + 2e - E = +0.76V  2VO 2 + + 4H + + Zn  2VO 2+ + Zn 2+ + 2H 2 O  E o cell = 1.00v + 0.76V = 1.76V

24.  E cell = E o cell – (0.0592/n)( log Q)  Q = [VO 2+ ] 2 [Zn 2+ ]  [VO 2 + ] 2 [H + ] 4  Q = [1x10 -2 ] 2 [1.0 x 10 -1 ]  [2.0] 2 [.50] 4 = 4 x 10 -5  E cell = 1.76V – (0.0592÷ 2) (log Q)  E cell = 1.76V – (0.0592÷ 2) (-4.398)  E cell = 1.76V – (-.1302) = 1.76 +.1302 = 1.89V

25. S 4 O 6 2- (aq) + Cr 2+ (aq)  Cr 3+ (aq) + S 2 O 3 2- (aq) S 4 O 6 2- + 2e -  2S 2 O 3 2- E o = 0.17V Cr 3+ + e -  Cr 2+ E o = -0.50V  S 4 O 6 2- + 2e -  2S 2 O 3 2- E o = 0.17V more positive thus reduced  Cr 3+ + e -  Cr 2+ E o = -0.50V must be oxidized so flip  S 4 O 6 2- + 2e -  2S 2 O 3 2-  2Cr 2+  2Cr 3+ + 2e -  S 4 O 6 2- + 2Cr 2+  2Cr 3+ + 2S 2 O 3 2-  E cell = 0.17 + 0.50 = 0.67V

26. ΔG = - nFE o ΔG = -RTlnK ∴ - nFE o = -RTlnK  - nFE = - RTlnKlnK = nFE/RT  ln K = (2)(98485)(.67) ÷ (8.3145)(298K)  ln K = 52.2  e 52.2 = 4.68 x 10 22

27. 1 volt = 1 J/C 1 amp = 1 C/s 1 Faraday = 96,485 C/mole of e -  How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g of metal?  10.5 g Ag x 1mole/107.868=.0973 mol Ag needed  0.0973 mol x 96485 C/mole e - = 9.39 x 10 3 C 1 amp = 1 C/s ∴ s = C/amp 9.39 x 10 3 C/5.00A = 1878 s = 31.3 min

28. Electrolysis with mixtures of ionsElectrolysis with mixtures of ions (2:30) Easy nickel plating Easy nickel plating  An acidic solution contains the ions Ce 4+, VO 2 +, Fe 3+. Using the E o values listed in Table 17.1 (Zumdahl) give the order of oxidizing ability of these species and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage.  Ce 4+ + e -  Ce 3+ E o = 1.70V  VO 2 + + 2H + + e -  VO 2+ + H 2 O E o = 1.00V  Fe 3+ + e -  Fe 2+ E o = 0.77 V  The more positive the E o value, the more the reaction has a tendency to proceed in the direction indicated. Of these choices, the Ce 4+ occurs most easily followed by VO 2 + and then Fe 3+  Why do we care? You can predict which metal will plate out onto the cathode first and in this case, second and third.

29. How to use a multimeter