Lecture 3-1-11

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Oxidation-Reduction (Redox)
Reactions
1
Redox: Introduction
• Electrons (e-) are transferred from one
compound to another
– e--donors (lose electrons)
– e--acceptors (gain electrons)
• Loss of e- by e--donor = oxidation
• Gain of e- by e--acceptor = reduction
• Oxidation and reduction always
accompany one another
– Electrons cannot exist freely in solution
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Balancing Redox Reactions
• Reaction needs to be charge balanced; eneed to be removed
– To balance charge, cross multiply reactants
• Redox reactions can also be written as
half reactions, with e- included
– Multiply so that e- are equal (LCM) and
combine half reactions
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Redox Reactions
• Ability of elements to act as e--donors or
acceptors arises from extent to which orbitals
are filled with electrons
– Property depends on decrease in energy of atoms
resulting from having only incompletely filled orbitals
• Some similarities to acid-base reactions
– Transferring electrons (e-) instead of H+
– Come in pairs (oxidation-reduction)
– In acid-base reactions, we measure changes in [H+]
(pH); in redox reactions, we measure voltage
changes
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Measuring voltage
• Standard potential tables have been
created for how much voltage (potential) a
reaction is capable of producing or
consuming
– Standard conditions: P =1 atm, T = 298°K,
concentration of 1.0 M for each product
– Defined as Standard Potential (E°)
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Non-Standard Redox Conditions
(real life)
• For non-standard conditions, E° needs to
be corrected
– Temperature
– Number of electrons transferred
– Concentrations of redox reactants and
products
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Nernst Equation
• Nernst Equation
–
•
•
•
•
•
•
E = electrical potential of a redox reaction
R = gas constant
T = temperature (K)
n = number of electrons transferred
F = Faraday constant = 9.65 x 104 J / V∙mole
Q = ion activity product
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Nernst Equation (example)
• Fe(s) + Cd2+  Fe2+ + Cd(s)
– Fe oxidized, Cd reduced
• Need to find standard potentials of half
reactions
– Fe2+ + 2e-  Fe(s)
– Cd2+ + 2e-  Cd(s)
• Look up in Table
8
Standard Potentials
Written as reductions
Strong Reducing Agents
Half-Reaction
Li+(aq) + e- → Li(s)
K+(aq) + e- → K(s)
Ba2+(aq) + 2 e- → Ba(s)
Sr2+(aq) + 2 e- → Sr(s)
Ca2+(aq) + 2 e- → Ca(s)
Na+(aq) + e- → Na(s)
Mg2+(aq) + 2 e- → Mg(s)
Be2+(aq) + 2 e- → Be(s)
Al3+(aq) + 3 e- → Al(s)
Mn2+(aq) + 2 e- → Mn(s)
2 H2O + 2 e- → H2(g) + 2 OH-(aq)
Zn2+(aq) + 2 e- → Zn(s)
Cr3+(aq) + 3 e- → Cr(s)
Fe2+(aq) + 2 e- → Fe(s)
Cd2+(aq) + 2 e- → Cd(s)
PbSO4(s) + 2 e- → Pb(s) + SO42-(aq)
Co2+(aq) + 2 e- → Co(s)
Ni2+(aq) + 2 e- → Ni(s)
Sn2+(aq) + 2 e- → Sn(s)
Pb2+(aq) + 2 e- → Pb(s)
2 H+(aq) + 2 e- → H2(g)
E0 (V)
-3.05
-2.93
-2.90
-2.89
-2.87
-2.71
-2.37
-1.85
-1.66
-1.18
-0.83
-0.76
-0.74
-0.44
-0.40
-0.31
-0.28
-0.25
-0.14
-0.13
0
The greater the E°, the more
easily the substance reduced
Half-Reaction
E0 (V)
2 H+(aq) + 2 e- → H2(g)
0
4+
2+
Sn (aq) + 2 e → Sn (aq)
0.13
2+
+
Cu (aq) + e → Cu (aq)
0.13
2+
SO4 (aq) + 4 H (aq) + 2 e → SO2(g) + 2 H2O
0.20
AgCl(s) + e → Ag(s) + Cl (aq)
0.22
Cu2+(aq) + 2 e- → Cu(s)
0.34
O2(g) + 2 H2 + 4 e → 4 OH (aq)
0.40
I2(s) + 2 e → 2 I (aq)
0.53
MnO4 (aq) + 2 H2O + 3 e → MnO2(s) + 4 OH (aq)
0.59
+
O2(g) + 2 H (aq) + 2 e → H2O2(aq)
0.68
Fe3+(aq) + e- → Fe2+(aq)
0.77
+
Ag (aq) + e → Ag(s)
0.80
2+
Hg2 (aq) + 2 e → 2 Hg(l)
0.85
2+
2+
2 Hg (aq) + 2 e → Hg2 (aq)
0.92
+
NO3 (aq) + 4 H (aq) + 3 e → NO(g) + 2 H2O
0.96
Br2(l) + 2 e- → 2 Br-(aq)
1.07
+
O2(g) + 4 H (aq) + 4 e → 2 H2O
1.23
+
2+
MnO2(s) + 4 H (aq) + 2 e → Mn (aq) + 2 H2O
1.23
2+
3+
Cr2O7 (aq) + 14 H (aq) + 6 e → 2 Cr (aq) + 7 H2O
1.33
Cl2(g) + 2 e → 2 Cl (aq)
1.36
Au3+(aq) + 3 e- → Au(s)
1.50
+
2+
MnO4 (aq) + 8 H (aq) + 5 e → Mn (aq) + 4 H2O
1.51
4+
3+
Ce (aq) + e → Ce (aq)
1.61
+
2PbO2(s) + 4H (aq) + SO4 (aq) + 2e → PbSO4(s) + 2H2O
1.70
+
H2O2(aq) + 2 H (aq) + 2 e → 2 H2O
1.77
Co3+(aq) + e- → Co2+(aq)
1.82
+
O3(g) + 2 H (aq) + 2 e → O2(g) + H2O
2.07
F2(g) + 2 e -----> F (aq)
2.87
Strong Oxidizing Agents
Nernst Equation (example)
• By convention, half-reactions are shown
as reduction reactions
• Fe(s) + Cd2+  Fe2+ + Cd(s)
– Fe2+ + 2e-  Fe(s), E° = -0.44 V
– Cd2+ + 2e-  Cd(s), E° = -0.40 V
– For oxidation, use negative value (+0.44 V)
– Added together, get +0.04 V as standard
potential for complete reaction
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Nernst Equation (example)
• Assume [Fe2+] = 0.0100 M and [Cd2+] =
0.005 M (instead of the standard 1.0 M)
–
• For Fe2+
– E = -0.44 – [(8.314)(298)]/[(2)(9.65x10-4)] ln(1/0.001)
= -0.50
• Fe Cd2+
– E = -0.40 – (0.128) ln (1/0.005) = -0.47
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Redox and Thermodynamics
• Energy released in a redox reaction can
be used to perform electrical work using
an electrochemical cell
– A device where electron transfer is forced to
take an external pathway instead of going
directly between the reactants
– A battery is an example
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Electrochemical Cell
Negative
Electrode
(e- removed)
Positive
Electrode
(e- added)
Zn(s)  Zn2+ + 2e-
Cu2+ + 2e-  Cu(s)
Zn(s) + Cu2+  Zn2+ + Cu(s)
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Redox Cell using Platinum
• Platinum is a good inert means of
transferring electrons to/from solution
• Consider the half-reaction in the presence
of a Pt electrode:
– Fe3+ + e- ↔ Fe2+
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Redox Cell
Pt wire
electrode
Salt bridge
H2 gas (1 atm)
Fe2+
and
Fe3+
[H+] = 1
Fe3+ + e- ↔ Fe2+
←: Pt wire removes electrons from half cell A
→: Pt wire provides electrons to the solution
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Redox Cell using Platinum
• If Pt wire not connected to source/sink of
electrons, there is no net reaction
– But wire acquires an electrical potential
reflecting tendency of electrons to leave
solution
– Defined as activity of electrons [e-]
• pe = -log [e-]
– Can be used in equilibrium expressions
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Redox Cell using Platinum
• Fe3+ + e- ↔ Fe2+
–
–
• [e-] proportional to the ratio of activities of the
reduced to the oxidized species
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Redox Cell
Pt wire
electrode
Salt bridge
H2 gas (1 atm)
Fe2+
and
Fe3+
[H+] = 1
H+ + e- ↔ ½ H2(g)
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Redox Cell using Platinum
• Other half-cell (B)
• H+ + e- ↔ ½ H2(g)
• Can write an equilibrium expression:
•
– SHE = standard hydrogen electrode
– By convention, [e-] =1 in SHE
19
Redox Cell using Platinum
• If switch is closed, electrons will move from
solution with higher activity of e- to the solution
with lower activity of e• Energy is released (heat)
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Redox Cell using Platinum
• Combine half reactions:
• Fe3+ + ½ H2(g) ↔ Fe2+ + H+
• Direction of reaction depends on which half-cell has
higher activity of electrons
– Now open switch: no transfer of e• Voltage meter registers difference in potential (E)
between the 2 electrodes
– Potential of SHE = 0, so E = potential of electrode in half-cell A
– Defined as Eh
– Measured in volts
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Eh
• Eh is positive when:
– [e-] in solution A less than [e-] in SHE
• Eh is negative when:
– [e-] in solution A greater than [e-] in SHE
•
– At 25°C, pe = 16.9
– Eh = 0.059 pe
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Eh as Master Variable
• Fe3+ + ½ H2(g) ↔ Fe2+ + H+
• Recall: G° = -RT ln Keq
• Standard state
• At non-standard state: GR = G° + RT ln Keq
–
–
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Eh as Master Variable
• From electrochemistry: GR = -nF Eh
– n = number of electrons
– By convention, sign of Eh set for half-reaction
written with e- on left side of equation
– Divide through by –nf
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Eh as Master Variable
•
• Rewrite to put oxidized species on top
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Eh (example)
• SO42- + 8e- + 10H+ ↔ H2S + 4 H2O
– 8 electrons transferred
–
•
• E° = ( -1 / (8 x 96.5)) x (-231.82) = +0.30 V
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