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OXIDATION-REDUCTION

2. DEFINITIONS. Oxidation: Loss of electrons.Reduction: Gain of electrons.Reductant: Species that loses electrons.Oxidant: Species that gains electrons.Valence: the electrical charge an atom would acquire if it formed ions in aqueous solution.. 3. RULES FOR ASSIGNMENT OF VALENCES. 1) The valence

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OXIDATION-REDUCTION

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    1. 1 OXIDATION-REDUCTION

    2. 2 DEFINITIONS Oxidation: Loss of electrons. Reduction: Gain of electrons. Reductant: Species that loses electrons. Oxidant: Species that gains electrons. Valence: the electrical charge an atom would acquire if it formed ions in aqueous solution.

    3. 3 RULES FOR ASSIGNMENT OF VALENCES 1) The valence of all pure elements is zero. 2) The valence of H is +1, except in hydrides, where it is -1. 3) The valence of O is -2, except in peroxides, where it is -1. 4) The algebraic sum of valences must equal zero for a neutral molecule or the charge on a complex ion.

    4. 4 VARIABLE VALENCE ELEMENTS Sulfur: SO42-(+6), SO32-(+4), S(0), FeS2(-1), H2S(-2) Carbon: CO2(+4), C(0), CH4(-4) Nitrogen: NO3-(+5), NO2-(+3), NO(+2), N2O(+1), N2(0), NH3(-3) Iron: Fe2O3(+3), FeO(+2), Fe(0) Manganese: MnO4-(+7), MnO2(+4), Mn2O3(+3), MnO(+2), Mn(0) Copper: CuO(+2), Cu2O(+1), Cu(0) Tin: SnO2(+4), Sn2+(+2), Sn(0) Uranium: UO22+(+6), UO2(+4), U(0) Arsenic: H3AsO40(+5), H3AsO30(+3), As(0), AsH3(-1) Chromium: CrO42-(+6), Cr2O3(+3), Cr(0) Gold: AuCl4-(+3), Au(CN)2-(+1), Au(0)

    5. 5 BALANCING OVERALL REDOX REACTIONS Example - balance the redox reaction below: Fe + Cl2 ? Fe3+ + Cl- Step 1: Assign valences, Fe0 + Cl20 ? Fe3+ + Cl- Step 2: Determine number of electrons lost or gained by reactants. Fe0 + Cl20 ? Fe3+ + Cl- ? ? 3e- 2e- Step 3: Cross multiply. 2Fe + 3Cl20 ? 2Fe3+ + 6Cl-

    6. 6 HALF-CELL REACTIONS The overall reaction: 2Fe + 3Cl20 ? 2Fe3+ + 6Cl- may be written as the sum of two half-cell reactions: 2Fe ? 2Fe3+ + 6e- (oxidation) 3Cl20 + 6e- ? 6Cl- (reduction) All overall redox reactions can be expressed as the sum of two half-cell reactions, one a reduction and one an oxidation.

    7. 7 Another example - balance the redox reaction: FeS2 + O2 ? Fe(OH)3 + SO42- Fe+2S20 + O20 ? Fe+3(OH)3 + S+6O42- ? ? 15e- 4e- 4FeS2 + 15O2 ? Fe(OH)3 + SO42- 4FeS2 + 15O2 ? 4Fe(OH)3 + 8SO42- 4FeS2 + 15O2 +14H2O ? 4Fe(OH)3 + 8SO42- + 16H+ This reaction is the main cause of acid generation in drainage from sulfide ore deposits. Note that we get 4 moles of H+ for every mole of pyrite oxidized!

    8. 8 Final example: C2H6 + NO3- ? HCO3- + NH4+ C-32H6 + N+5O3- ? 2HC+4O3- + N-3H4+ ? ? 14e- 8e- 8C2H6 + 14NO3- ? 2HCO3- + NH4+ 8C2H6 + 14NO3- ? 16HCO3- + 14NH4+ 8C2H6 + 14NO3- + 12H+ + 6H2O? 16HCO3- + 14NH4+

    9. 9 STRENGTH OF REDUCING AND OXIDIZING AGENTS Zn + Fe2+ ? Zn2+ + Fe Which way will the reaction go? ?Gr = -16.3 kcal/mole Zn is a stronger reducing agent than Fe. Fe + Cu2+ ? Fe2+ + Cu ?Gr = -34.51 kcal/mole Fe is a stronger reductant than Cu.

    10. 10 ELECTROMOTIVE SERIES Weakest oxidant Strongest reductant Zn ? Zn2+ + 2e- Fe ? Fe2+ + 2e- Cu ? Cu2+ + 2e- Ag ? Ag+ + e- Strongest oxidant Weakest reductant

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    12. 12 ELECTROMOTIVE FORCE Electromotive force (EMF): The electrical potential generated by the half reactions of an electrochemical cell. Consider: Zn + Cu2+ ? Zn2+ + Cu At equilibrium the electrochemical cell has to obey

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    14. 14 STANDARD HYDROGEN ELECTRODE (SHE) In order to assign a ranking of half-cell reactions, we arbitrarily set E = 0.00 V for the reaction: H2(g) ? 2H+ + 2e- when pH2 = 1 bar and pH = 0. In other words, Eo = 0.00 V. This is equivalent to the convention: ?Gfo (H+) = ?Gfo (e-) = 0.00 kcal/mole. We then connect this SHE to any other electrode representing a half-cell reaction and we can obtain Eo for all half-cell reactions. This is called the standard electrode potential.

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    16. 16 THE NERNST EQUATION The value E0 refers to the EMF of a half-cell reaction when all reactants are in the standard state, e.g., for Zn ? Zn2+ + 2e- Eo is the EMF when aZn2+ = 1.0. What if this is not the case?

    17. 17 In this particular case IAP = [Zn2+] and n = 2 so:

    18. 18 DEFINITION OF Eh We define Eh to be the EMF between a half-cell reaction in any state, and the SHE. For example, Eh for the zinc reaction above is given defined by the overall reaction: Zn + 2H+ ? Zn2+ + H2(g) for which:

    19. 19 STABILITY LIMITS OF WATER IN Eh-pH SPACE Upper limit H2O(l) ? 2H+ + 1/2O2(g) + 2e- Eh = E0 + 0.0295 log (pO21/2[H+]2) E0 = ?Gro/(n?) = -56.687/(2·23.06) = 1.23 V Eh = 1.23 + 0.0148 log pO2 - 0.0592 pH At the Earth’s surface, pO2 can be no greater than 1 bar so Eh = 1.23 - 0.0592 pH

    20. 20 Lower limit: 1/2H2(g) ? H+ + e- Eh = 0 + 0.059 log ([H+]/pH21/2) Again, let pH2 = 1 bar. Eh = -0.059 pH

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    23. 23 Eh-pH DIAGRAM FOR THE SYSTEM Cd-H2O-CO2

    24. 24 START WITH H2O-CO2 SYSTEM H2CO30/HCO3- boundary: H2CO30 ? HCO3- + H+ ?Gro = -140.24 - (-149.00) - (0.00) = 8.76 kcal mol-1 log K = -?Gro/(2.3025RT) = -8760/(2.3025·1.987·298.15) = -6.42

    25. 25 HCO3-/CO32- boundary: HCO3- ? CO32- + H+ ?Gro = -126.15 - (-140.24 ) - (0.00) = 14.09 kcal mol-1 log K = -?Gro/(2.3025RT) = -14090/(2.3025·1.987·298.15) = -10.33

    26. 26 H2CO30/C(graphite) boundary: C + 3H2O ? H2CO30 + 4H+ + 4e- ?Gro = -149.00 - 3(-56.7) = 21.10 kcal mol-1 E0 = ?Gro/(n?) = 21.10/(4·23.06) = 0.229 volts

    27. 27 HCO3-/C(graphite) boundary: C + 3H2O ? HCO3- + 5H+ + 4e- ?Gro = -140.24 - 3(-56.7) = 29.86 kcal mol-1 E0 = ?Gro/(n?) = 29.86/(4·23.06) = 0.324 volts

    28. 28 CO32-/C(graphite) boundary: C + 3H2O ? CO32- + 6H+ + 4e- ?Gro = -126.15 - 3(-56.7) = 43.95 kcal mol-1 E0 = ?Gro/(n?) = 43.95/(4·23.06) = 0.476 volts

    29. 29 CH4(aq)/C(graphite) boundary: CH4(aq) ? C + 4H+ + 4e- ?Gro = -(-8.28) = 8.28 kcal mol-1 E0 = ?Gro/(n?) = 8.28/(4·23.06) = 0.0898 volts

    30. 30 CH4(aq)/HCO3- boundary: CH4(aq) + 3H2O(l) ? HCO3- + 9H+ + 8e- ?Gro = -140.24 - (-8.28) - 3(-56.7) = 38.14 kcal mol-1 E0 = ?Gro/(n?) = 38.14/(8·23.06) = 0.207 volts

    31. 31 CH4(aq)/CO32- boundary: CH4(aq) + 3H2O(l) ? CO32- + 10H+ + 8e- ?Gro = -126.15 - (-8.28) - 3(-56.7) = 52.23 kcal mol-1 E0 = ?Gro/(n?) = 52.23/(8·23.06) = 0.283 volts

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    33. 33 NOW ADD Cd Cd2+/CdCO3(s) boundary (H2CO30 field): CdCO3(s) + 2H+ ? Cd2+ + H2CO30 ?Gro = -149.00 - 18.55 - (-160.00) = -7.55 kcal mol-1 log K = -?Gro/(2.3025RT) = 7550/(2.3025·1.987·298.15) = 5.53

    34. 34 Cd2+/CdCO3(s) boundary (graphite field): Cd2+ + C + 3H2O(l) ? CdCO3(s) + 6H+ + 4e- ?Gro = -160.0 - (-18.55) - 3(-56.7) = 28.65 kcal mol-1 E0 = ?Gro/(n?) = 28.65/(4·23.06) = 0.311 volts

    35. 35 Cd2+/CdCO3(s) boundary (CH4 field): Cd2+ + CH4(aq) + 3H2O(l) ? CdCO3(s) + 10H+ + 8e- ?Gro = -160.0 - (-18.55) - 3(-56.7) - (-8.28) = 36.93 kcal mol-1 E0 = ?Gro/(n?) = 36.93/(8·23.06) = 0.200 volts

    36. 36 Cd(OH)2(s)/CdCO3(s) boundary (CO32- field): Cd(OH)2(s) + CO32- + 2H+ ? CdCO3(s) + 2H2O(l) ?Gro = -160.0 + 2(-56.7) - (-113.19) - (-126.15) = -34.06 kcal mol-1 log K = -?Gro/(2.3025RT) = 34060/(2.3025·1.987·298.15) = 24.97

    37. 37 Cd(OH)2(s)/CdCO3(s) boundary (CH4 field): Cd(OH)2(s) + CH4(aq) + H2O(l) ? CdCO3(s) + 8H+ + 8e- ?Gro = -160.0 - (-56.7) - (-113.19) - (-8.28) = 18.17 kcal mol-1 E0 = ?Gro/(n?) = 18.17/(8·23.06) = 0.098 volts

    38. 38 Cd(OH)2(s)/Cd2+ boundary: Cd(OH)2(s) + 2H+ ? Cd2+ + 2H2O(l) ?Gro = -18.55 + 2(-56.7) - (-113.19) = -18.76 kcal mol-1 log K = -?Gro/(2.3025RT) = 18760/(2.3025·1.987·298.15) = 13.75

    39. 39 Cd(OH)2(s)/CdO22- boundary: Cd(OH)2(s) ? CdO22- + 2H+ ?Gro = -67.97 - (-113.19) = 45.22 kcal mol-1 log K = -?Gro/(2.3025RT) = -45220/(2.3025·1.987·298.15) = -33.15

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