OXIDATIONREDUCTION

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OXIDATION-REDUCTION
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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.

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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.

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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)

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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-

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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.

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• Another example - balance the redox reaction
• FeS2 O2 ? Fe(OH)3 SO42-
• Fe2S20 O20 ? Fe3(OH)3 S6O42-
• ? ?
• 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!

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• Final example
• C2H6 NO3- ? HCO3- NH4
• C-32H6 N5O3- ? 2HC4O3- N-3H4
• ? ?
• 14e- 8e-
• 8C2H6 14NO3- ? 2HCO3- NH4
• 8C2H6 14NO3- ? 16HCO3- 14NH4
• 8C2H6 14NO3- 12H 6H2O? 16HCO3- 14NH4

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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.

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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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Figure 16.1 from Faure. Schematic diagram of a
Zn-Cu electro-chemical cell. The two metal
electrodes are immersed in solutions of ZnSO4 and
CuSO4, which are prevented from mixing by a
porous partition.
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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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? (Faraday Constant) 96,489 C mol-1
23.06 kcal/V-1
mol-1 Eo ?Gro/(2?) -50.8 kcal/(2
mol23.06 kcal V-1 mol-1) -1.1 V
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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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If we always write half-cell reactions with the
electrons on the right hand side, E0 tells us the
position of the reaction in the electromotive
series relative to SHE.
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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?

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• In this particular case
• IAP Zn2 and n 2
• so

At 25C we have in this particular case
Or in the most general case at 25C
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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

However, for the SHE by definition, pH2 1 bar
and H 1 mol L-1, so
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STABILITY LIMITS OF WATER IN Eh-pH SPACE

• Upper limit
• H2O(l) ? 2H 1/2O2(g) 2e-
• Eh E0 0.0295 log (pO21/2H2)
• E0 ?Gro/(n?) -56.687/(223.06) 1.23 V
• Eh 1.23 0.0148 log pO2 - 0.0592 pH
• At the Earths surface, pO2 can be no greater
  than 1 bar so
• Eh 1.23 - 0.0592 pH

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• 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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Eh-pH diagram depicting the limits of stability
of liquid water.
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Eh-pH diagram showing the redox conditions of
various natural environments.
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Eh-pH DIAGRAM FOR THE SYSTEM Cd-H2O-CO2
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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/(2.30251.987298.15) -6.42

By definition H2CO30 HCO3- so K H and
pH -log K 6.42.
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• HCO3-/CO32- boundary
• HCO3- ? CO32- H
• ?Gro -126.15 - (-140.24 ) - (0.00) 14.09 kcal
  mol-1
• log K -?Gro/(2.3025RT)
• -14090/(2.30251.987298.15) -10.33

By definition HCO3- CO32- so K H and
pH -log K 10.33.
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• 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/(423.06) 0.229 volts

We must assign a value to ?C. Here we assign ?C
10-3 M. Because H2CO30 is predominant, H2CO30 ?
?C.
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• 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/(423.06) 0.324 volts

Because HCO3- is predominant, HCO3- ? ?C 10-3
M
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• 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/(423.06) 0.476 volts

Because CO32- is predominant, CO32- ? ?C 10-3
M
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• CH4(aq)/C(graphite) boundary
• CH4(aq) ? C 4H 4e-
• ?Gro -(-8.28) 8.28 kcal mol-1
• E0 ?Gro/(n?) 8.28/(423.06) 0.0898 volts

Because CH4(aq) is predominant, CH4 ? ?C 10-3
M
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• 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/(823.06) 0.207 volts

By definition CH4(aq) HCO3-, so
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• 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/(823.06) 0.283 volts

By definition CH4(aq) CO32-, so
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Eh-pH diagram depicting fields of predominance
and stability in the system C-O-H.
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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/(2.30251.987298.15) 5.53

We assign ?C 10-3 M and ?Cd 10-3 M. Now log K
2pH log H2CO30 log Cd2 5.53 2pH - 3
- 3 pH 5.77
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• 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/(423.06) 0.311 volts

We assign ?Cd 10-3 M.
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• 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/(823.06) 0.200 volts

We assign ?Cd 10-3 M and ?C 10-3 M.
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• 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/(2.30251.987298.15) 24.97

We assign ?C 10-3 M. Now log K 24.97 2pH -
log CO32- 2pH 3.0 pH 10.98
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• 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/(823.06) 0.098 volts

We assign ?C 10-3 M.
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• 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/(2.30251.987298.15) 13.75

We assign ?Cd 10-3 M. Now log K 13.75 log
Cd2 2pH -3 2pH pH 8.38
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• 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/(2.30251.987298.15) -33.15

We assign ?Cd 10-3 M. Now log K -33.15 log
CdO22- - 2pH -3 2pH pH 15.08 This
boundary does not plot on our diagram.
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Eh-pH diagram depicting fields of predominance
and stability in the system Cd-C-O-H.
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Eh-pH diagram depicting fields of predominance
and stability in the system Cd-C-S-O-H. The
conditions are ?C 10-3 M, ?S 10-3 M, ?Cd
10-8 M