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Zn + Cu 2 +  Zn 2 + + Cu PowerPoint Presentation
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Zn + Cu 2 +  Zn 2 + + Cu

Zn + Cu 2 +  Zn 2 + + Cu

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Zn + Cu 2 +  Zn 2 + + Cu

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  1. Zn + Cu2+ Zn2+ + Cu

  2. Applied Electrochemistry • Voltaic (or galvanic) cells: spontaneous • redox reaction  electricity (or electrical work) • Electrolytic cells: non-spontaneous • electricity  redox • To produce electricity, we must direct the electron flow through an external circuit. We cannot have direct redox. • Daniell cell Zn – Cu rxn:

  3. Lemon Battery

  4. 17.1 Galvanic Cells 3:25

  5. Galvanic Cells • To produce electricity, we need: • Isolated half-reactions, using half-cells • Conductive solids (electrodes) connected by external circuits • May consist of a reactant/product or be an inert substance such as platinum or graphite • Anode: oxidation half-reaction • Cathode: reduction half-reaction

  6. Galvanic Cells

  7. Galvanic Cells • Externally, the anode has the negative charge; internally, it has a positive charge • Anions flow towards the anode; cations move away from it and towards the cathode. • Two half-cells must be connected to pass ions • Salt bridge or porous glass

  8. Molecular View of Electrode Processes

  9. 17.2 Cell Potentials • If the half-reactions are carried out separately (but coupled), we find they generate an electrical current characterized by a voltage

  10. Cell Potentials • The voltage produced by a voltaic cell is called the cell potential, Eocell (also the reaction potential, Eorxn, when the half-reactions are not separated) • Under standard conditions, the voltage is also called the standard electromotive force (emf), Eo. • Unit = Volt (V) • Volt = 1 Joule of energy / coulomb of charge transferred = 1 J/C • J = Joule C = coulomb • 1 mol e- = 96,500 coulombs

  11. Cell Potentials • Reference for Eo is the standard hydrogen electrode, using the reaction: • 2H+(aq) + 2e- H2(g) Eo = 0 V = 0 J/C 1 M 1 atm (std conditions)

  12. Cell Potentials • Then get other half-reaction potentials from measured Eocell values. • Zn + 2H+ Zn2+ + H2 Eocell = 0.76 V • Zn  Zn2+ + 2e- Eoox = 0.76 V • 2H+ + 2e- H2 Eored = 0.00 V • How do these reactions relate?

  13. Cell Potentials • Cell potential values can thus be determined relative to the standard hydrogen electrode, with Eo = 0 V • Eored = cell potential for the reduction half-rxn • Eoox = cell potential for the oxidation half-rxn • Thus, Eocell = Eored + Eoox • We can measure Eocell, using the standard reference of 0 V, we can measure Eoox and Eored for half reactions paired with the H2 half reaction.

  14. Determining Cell Potentials Values • If we reverse a half-reaction, what happens to the sign of Eo. Zn  Zn2+ + 2e- Eoox = 0.76 V Zn2+ + 2e- Zn Eored = ? • Now consider data for Eocell = 0.63 V for the following reaction. Zn + Pb2+ Zn2+ + Pb • What is the Eored of Pb2+ + 2e- Pb ?

  15. Cell Potential Values • Zn  Zn2+ + 2e- Eoox = 0.76 V • Pb2+ + 2e- Pb Eored = ? • Zn + Pb2+ Zn2+ + Pb Eocell = 0.63 V • Eocell = Eored + Eoox • 0.63 V = Eored + 0.76 V • Eored = 0.63 V - 0.76 V = -0.13 V • Values determined in this way are listed in Table 17.1 and Appendix 5.5

  16. Reduction Potentials

  17. Relative Strengths of Oxidizing and Reducing Agents

  18. Cell Potential • Reduction or oxidation values can also be measured from Eocell with other known half-cells. • AgCl + e- Ag + Cl- Eredo = 0.22 V • Hg2Cl2 + 2e- 2Hg + 2Cl- Eredo = 0.2802 V • Half-cells such as these are used as reference electrodes. The Ag/AgCl electrode, along with a glass electrode, is used in a pH meter.

  19. 17.2 Describing Galvanic Cells • What will happen if we place a piece of Zn and a piece of Cu in a solution that contains a mixture of Zn2+ and Cu2+? • Two possibilities: • Zn + Cu2+ Zn2+ + Cu • Cu + Zn2+ Cu2+ + Zn

  20. Spontaneous Redox • Two possible reduction half-reactions: • Zn2+ + 2e- Zn Eored = -0.76 V • Cu2+ + 2e- Cu Eored = 0.34 V • Two possible oxidation half-reactions: • Zn  Zn2+ + 2e- Eoox = 0.76 V • Cu  Cu2+ + 2e- Eoox = -0.34 V • One way to combine them: Zn2+ + 2e- Zn Eored = -0.76 V Cu  Cu2+ + 2e- Eoox = -0.34 V ———————— ———————— Cu + Zn2+ Cu2+ + Zn Eorxn = -1.10 V

  21. Spontaneous Redox • Other way to combine them: Cu2+ + 2e- Cu Eored = 0.34 V Zn  Zn2+ + 2e- Eoox = 0.76 V ———————— ——————— Zn + Cu2+ Zn2+ + Cu Eorxn = 1.10 V • Which combination is observed to be spontaneous. skip

  22. Stability in Aqueous Systems • Reaction with Water • Reduce hydronium ion to release hydrogen gas: • 2H+(aq) + 2e-H2(g) E°red = 0.000 V • Any substance with Eoox > 0 will reduce H+ to H2 • Examples are V, V2+, Cr, Cr2+, Mn • The ions will react, but tend to react only very slowly. There seems to be a kinetic factor that results in a fast reaction only if Eorxn > 0.4-0.5 V (called an overvoltage).

  23. Stability in Aqueous Systems • Reaction with Water • Oxidize water to release oxygen gas: • 2H2O(l) O2(g) + 4H+(aq) + 4e- E°ox = -1.23 V • Any Eored > 1.23 V will result in production of O2. Generally need Eorxn > 0.4-0.5 V for fast reaction. • Examples are Cr2O72- (very slow), MnO42- (disproportionates faster), Mn3+ (disproportionates faster)

  24. Stability in Aqueous Systems • Oxidation by O2 in Air • O2(g) + 4H+(aq) + 4e-2H2O(l) E°red = 1.23 V • Any Eoox > -1.23 V will result in oxidation by air. Many substances fall into this category (Eorxn > 0.4-0.5 V for fast reaction). • V Cr Mn • V2+ Cr2+ not Mn2+ • V3+ not Cr3+ Mn3+ • VO2+ (very slow) not MnO2 • MnO42- (disproportionates faster)

  25. Eocell and Spontaneity • We have seen three criteria for spontaneity: • Eo > 0 • DGo < 0 1 V = 1 J/C, so 1 J = 1 C x 1 V • These criteria are related: • K >> 1 • DGo = - RT lnK • DGo = - n F Eo, DG = - nFE where n = number of e- transferred and F = Faraday constant (charge on 1 mole e-) 1 F = 96,500 coul/mol e- = 96,500 J/V mol e-

  26. Thermodynamics • These relationships work for half-reactions or complete redox reactions. • Zn + Cu2+ Zn2+ + Cu Eo = 1.10 n = 2 • DGo = -2mol e- x 96500 J/V mol e- x 1.10 V • DGo = -212,300 J = -212.3 kJ • DGo depends on the number of moles, but Eo does not

  27. Voltage and Moles • Note that different size alkaline cells all deliver the same voltage, in spite of different number of moles of reactants.

  28. Thermodynamics • We can add Eoox to Eored to give Eocell or Eorxn in the same way that we can add half-reactions to give an overall reaction. • Fe  Fe2+ + 2e- Eoox = +0.44 V DGoox = - 2 x 96500 x 0.44 = -84900 J • Cl2 + 2e- 2Cl- Eored = 1.36 V DGored = - 2 x 96500 x 1.36 = -262500 J • Fe + Cl2 Fe2+ + 2Cl- Eorxn = 1.80 V DGo = - 2 x 96500 x 1.80 = -347400 J • DGorxn = DGoox + DGored = -84900 + -262500 = -347400 J

  29. Thermodynamics • From Chapter 16, we know that DGo values are additive when we add reactions. • Eos are additive when we add half-reactions to give a complete reaction because the value of n is the same for the half-reactions and the complete reaction. • Eos are not additive when adding two half-reactions to give a third half-reaction because the value of n is not constant.

  30. Thermodynamics • We can add DGo under all circumstances: DGo3 = DGo1 + DGo2 -n3FEo3 = -n1FEo1 - n2FEo2 n3Eo3 = n1Eo1 + n2Eo2 Eo3 = (n1Eo1 + n2Eo2)/n3 • V  V2+ + 2e- Eo1 = 1.20 V V2+V3+ + e- Eo2 = 0.26 V V V3+ + 3e- Eo3 < 1.20 + 0.26 • Eo3 = (2 x 1.20 + 1 x 0.26)/3 = 0.887 V

  31. Thermodynamics • We can calculate Keq from Eo: • DGo = - nFEo = - RT ln K • Eo = (RT/nF) ln K = 2.303 (RT/nF) log K • At 25oC, 2.303 RT/F = 0.05916 • Eo = (0.05916/n) log K at 25oC • Thus, we can measure Eo for a redox reaction and then calculate the equilibrium constant for that reaction.

  32. 17.4 Effect of Concentration on Cell EMF • So far, we have been using standard state conditions, but we don’t always have 1 M solutions. We can correct Eo to E by using the Nernst equation. • DG = DGo + RT ln Q • But, DG = - nFE and DGo = -nFEo, so • - nFE = -nFEo + 2.303 RT log Q • E = Eo - (2.303 RT/nF) log Q • At 25oC, E = Eo - (0.05916/n) log Q

  33. Nernst Equation • At 25oC, E = Eo - (0.05916/n) log Q • E = Eo if Q = 1 • When the system reaches equilibrium, Q = K, and E = 0, because Eo = (0.05916/n) log K, and the cell has “run down”. • Consider the Zn/Cu2+ reaction if more Cu2+ is added to the cell. The voltage becomes greater than 1.10 V. 21m07an1

  34. Nernst Equation • What is E of the Zn/Cu2+ reaction if [Cu2+] = 0.010 M and [Zn2+] =1.99 M? Note that this corresponds to starting with standard conditions and changing to 99% completion of reaction. Eo = 1.10 V (with [Cu2+] = [Zn2+] = 1.00 M) • Zn + Cu2+ Zn2+ + Cu • For this reaction, n = 2. • Q = [Zn2+]/[Cu2+]

  35. Nernst Equation • E = Eo - (0.05916/n) log Q • E = 1.10 V - (0.05916/2) log (1.99/0.010) • E = 1.10 V - (0.05916/2) log 199 • E = 1.10 - 0.068 = 1.03 V • For 99.9% reaction (1.999 M Zn2+, 0.001 M Cu2+), E = 1.10 - 0.098 = 1.00 V • For 99.99% reaction, 1.9999 M Zn2+, 0.0001 M Cu2+), E = 1.10 - 0.127 = 0.97 V

  36. Concentration Cells • We can generate a voltage with a cell that contains the same materials in the cathode and anode compartments, but at different concentrations.