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Chapter 20

Chapter 20. Electrochemistry. Overview. Oxidation-Reduction reactions Balancing Redox Reactions Half-Reaction method Acidic Solution Basic Solution Voltaic Cells Cell EMF--standard reduction potentials Oxidizing & Reducing reagents Spontaneity of Redox reactions.

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Chapter 20

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  1. Chapter 20 Electrochemistry

  2. Overview • Oxidation-Reduction reactions • Balancing Redox Reactions • Half-Reaction method • Acidic Solution • Basic Solution • Voltaic Cells • Cell EMF--standard reduction potentials • Oxidizing & Reducing reagents • Spontaneity of Redox reactions

  3. Effect of Concentration • Nernst Equation • Equilibrium Constants • Commercial Voltaic Cells • Electrolysis • Quantitative Aspects • Electrical Work

  4. Redox Reactions • Involve a transfer of electrons • Oxidation  loss of one or more electron(s) • oxidation state will increase • Reduction  gain of one or more electron(s) • oxidation state will decrease • Must occur simultaneously Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Zn  Zn2+(aq) + 2e- oxidation ½ rxn oxidation reduction reduction ½ rxn Cu2+(aq) + 2e- Cu(s)

  5. You must know oxidation states:(Review: Section 8.10) • What are the oxidation states of each atom in the following: • H2 • CO • ClO2- • HC2H3O2 H 0 C +2, O -2 Cl +3, O -2 H +1, C1 +3, C2 -3, O -2

  6. Balancing Redox Reactions • Mass balance must be observed • e--transfer must be balanced • Simple reactions: • Sn2+ + Fe3+  Sn4+ + Fe2+ • Sn2+ Sn4+ + 2e- • Fe3+ + e- Fe2+ oxidation ½ rxn x 2 reduction ½ rxn • 2Fe3+ + 2e- 2Fe2+ • Sn2+ + 2Fe3+  Sn4+ + 2Fe2+

  7. Reactions involving H & O in acid: • MnO4- + C2O42- Mn2+ + CO2 • write both ½ reactions • MnO4- Mn2+ • C2O42- CO2 • mass balance (all except H & O) • MnO4- Mn2+ • C2O42- 2CO2 • add H2O & H+ to balance O & H • 8H+ + MnO4- Mn2+ + 4H2O • C2O42- 2CO2

  8. balance charge by adding electrons • 5e- + 8H+ + MnO4- Mn2+ + 4H2O • C2O42- 2CO2 + 2e- • balance electrons transferred • 10e- + 16H+ + 2MnO4- 2Mn2+ + 8H2O • 5C2O42- 10CO2 + 10e- • add half reactions • 16H+ + 2MnO4-+ 5C2O42-  10CO2 + 2Mn2+ + 8H2O • check the balance

  9. H2O + + 2H+ + 2e- 3e- + 4H+ + + 2H2O • Reactions in base: MnO4- + CN- CNO- + MnO2 • use exactly the same process • CN- CNO- • MnO4- MnO2 • since H+ cannot exist in basic solution, add OH- • 2OH- + CN- CNO- + H2O+ 2e- • 3e- + 2H2O + MnO4- MnO2 + 4OH- • balance electrons transferred & sum • 6OH- + 3CN- 3CNO- + 3H2O+ 6e- • 6e- + 4H2O + 2MnO4- 2MnO2 + 8OH- • 3CN- + H2O + 2MnO4- 2MnO2+3CNO- +2OH- • check balance

  10. Voltaic Cells • A spontaneous redox reaction that does work • Anode • electrode at which oxidation occurs • loses mass • electrons released, sign is negative • Cathode • electrode at which reduction occurs • gains mass • electrons consumed, sign is positive

  11. Cell EMF • Difference in potential energy of electrons at the anode and cathode • Diff. in potential energy per electrical charge measured in volts • 1 V = 1 J C • Potential difference = EMF, electromotive force • Ecell = cell potential = cell voltage • Eºcell = cell potential under std. conditions • 1 M, 1 atm, 25 ºC

  12. Standard reduction potentials • E ºred in tables • E ºcell = E ºred (cathode) - E ºred (anode) • Based on “standard hydrogen electrode” • 2H+(aq, 1M) + 2e- H2(g, 1atm) E ºred = 0 V • Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) E ºcell = 0.76 V • 0.76 V = 0 V - E ºred (anode) • Zn2+(aq, 1M) + 2e- Zn(s) E ºred (anode) = -0.76 V

  13. Problem: • Calculate Eºcell for • 2Al(s) + 3I2(s) 2Al3+(aq) + 6I-(aq) • Anode: 2Al  2Al3+ + 6e- • Cathode: 3I2 + 6e-  6I- • Eºcell = E ºred (cathode) - E ºred (anode) • E ºcell = 0.54 V - (-1.66 V) • E ºcell = 2.20 V

  14. Note: stoichiometric coefficient does notaffect the value of the E ºred(it is an intensive property) • E ºox = - E ºred • 2Al(s) + 3I2(s) 2Al3+(aq) + 6I-(aq) • 2Al  2Al3+ + 6e- E ºox = +1.66 V • 3I2 + 6e-  6I- E ºred = +0.54 V • E ºcell = E ºox + E ºred = 2.20V • The more positive the E ºcell the more driving force for the reaction

  15. Oxidizing/Reducing Agents • Oxidizing agents cause oxidation • oxidizing agents are reduced • the more (+) the E ºred the better the ox. agent • Reducing agents causereduction • reducing agents are oxidized • the more (-) the E ºred the better the red. agent

  16. Which is the better oxidizing agent? • NO3- + 4H+ + 3e- NO + 2H2O E ºred 0.96 V • Ag+ + e- Ag E ºred 0.80 V • Cr2O72- + 14H+ + 6e- 2Cr3+ + H2O E ºred 1.33 V • Which is the strongest reducing agent? • I2 + 2e- 2I- Eºred +0.54 V • Fe2+ + 2e- Fe Eºred -0.44 V • MnO4- + 8H+ + 5e-  Mn2+ + 4H2O Eºred +1.51 V

  17. Spontaneity of Redox Reactions • Spontaneous redox rxns have positive potentials • Non-spontaneous redox rxns have negative potentials • Is this rxn spont. or non-spont.? • MnO4- + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H2O • Fe2+ Fe3+ + 1e- Eºox = -0.77 v • MnO4- + 8H+ + 5e-  Mn2+ + 4H2O E ºred = +1.51 v • E ºox + E ºred = + 0.74 v Yes

  18. EMF & Free Energy • If both DG & E are a measure of spontaneity, they must be related • DG = - nFE • F is Faraday’s constant 1 F = 96,500 J/v mol e- • remember: 1 C = 1 J/v • n = mol e- transferred • In the standard state DGº = - nFEº

  19. Calculate the standard free energy change for Hg + 2Fe3+ Hg2+ + 2Fe2+ • n = 2 mol electrons transferred • Hg  Hg2+ + 2e- Eox = - 0.854 v • 2Fe3+ +2e-  2Fe2+ Ered= + 0.771 v • Ecell = - 0.083 v • DG = - (2 mol e-)(-0.083 v)(96,500 J/v mol e-) • = + 16 kJ

  20. Concentration & Cell EMF • Nernst Equation • relationship between DG & concentrations • DG = DGº + RT ln Q Q = [prod]x/[react]y • substitute -nFE for DG • E = Eº - (RT/nF) ln Q or • E = Eº - (2.303 RT/nF) log Q • 2.303 RT/F = 0.0592 v-mol e- at std. temp. • E = Eº - (0.0592/n) log Q

  21. Calculate the emf that the following cell generates when [Mn2+] = 0.10 M & [Al3+] = 1.5 M 2Al + 3Mn2+ 2Al3+ + 3Mn • Eº = + 0.48 v • E = (+ 0.48 v) - (0.0592 v/ 6) log [(1.5)2/(0.10)3] • E = + 0.45 v • when [Mn2+] = 1.5 M & [Al3+] = 0.10 M • E = (+ 0.48 v) - (0.0592 v/ 6) log [(0.10)2/(1.5)3] • E = + 0.51 v

  22. Equilibrium Constants • Remember DG = DGº + RT ln Q, if Q = K, then DG = 0, therefore -nFE = 0 and • 0 = Eº - (RT/nF) ln K or • 0 = Eº - (0.0592/n) log K • K can be calculated from cell potentials • log K = nE º/0.0592

  23. Calculate the equilibrium constant, K, for 2IO3- + 5Cu + 12H+ I2 + 5Cu2+ + 6H2O • Eº = + 0.858 v • n = 10 mol e- transferred • log K = nEº/0.0592 • log K = 145 • K = 1 x 10145

  24. Voltaic Cells • Lead storage battery • PbO2 + SO4-2 + 4H+ + 2e- PbSO4 + H2OPb + SO42- PbSO4 + 2e- • Ecell = + 2.041 v • Dry cell • NH4+ + 2MnO2 + 2e- Mn2O3 + 2NH3 + H2OZn  Zn2+ + 2e- • In an alkaline cell the NH4Cl is replaced with KOH

  25. Ni-Cd • NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH-Cd + 2OH-  Cd(OH)2 + 2e- • Fuel cells • 4e- + O2 + 2H2O  4OH-2H2 + 4OH-  4H2O

  26. Electrolytic Cells • Redox reactions that are not spontaneous • Must be driven by an outside source of electrical energy • Cathode • reduction occurs • by sign convention, is negative • Anode • oxidation occurs • by sign convention, is positive

  27. Quantitative Aspects • Redox reactions occur in stoichiometric relationship to the transfer of electrons • Electrons put into a system through electrical energy, can be quantized • Coulomb = quantity of charge passing through electrical circuit in 1 s at 1 ampere (A) current • Coulomb = (amp) (seconds)

  28. Problem: Calculate the mass of Mg formed upon passage of a current of 60.0 A for a period of 4.00 x 10 3 s. • MgCl2  Mg + Cl2 • Mg2+ + 2e-  Mg 2Cl-  Cl2 + 2e- • we are concerned with the reduction • (60.0 A)(4 x 103s)(1C/1 A-s) = 2.4 x 105 C • (2.4 x 105 C)(1 mol e-/ 96,500 C) = 2.49 mol e- • (2.49 mol e-)(1 mol Mg/2 mol e-) = 1.24 mol Mg • (1.24 mol Mg)(24.3 g/mol) = 30.1 Mg

  29. Electrical Work • DG = wmaxDG = - nFE wmax = - nFE • Max work proportional to potential • wmax = - n F E • J = (mol) (C/mol) (J/C) • Electrical work = (watt) (time) • 1 watt (W) = 1 J/s or watt-s = J • 1 kWh = 3.6 x 106 J

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