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Lecture 12Lecture 12
Oxidation & Reduction II
Suggested reading: Chapter 5
Recall from Last time: Redox Potentials
QRTEE l0The Nernst equation: Q
FEEcell ln0
Cell Potential and pH
For the H+/H couple at 1 bar and 25oC:For the H /H2 couple at 1 bar and 25oC:
pHpHRTE -59 15mV10ln pHpHF
Ecell -59.15mV
In general for aOx+ν e-+ν H+a’RedIn general, for aOx+νee +νHH a Red
pHEpHRT
EE HH )V0590(
10ln '' pHEpH
FEE
eecell )V059.0(
aRT ]Red[ '
Th i l d (b i i ) h
ae xF
RTEE]O[]Red[ln0'
The potential decreases (becoming more negative) as the pH increases and the solution becomes more basic
Redox stabilities
When assessing the thermodynamic stability of a i i l i id ll ibl species in solution, we must consider all possible
reactants:
• the species itself• the solvent
• other solutes• dissolved O2
Redox Chemistry of Water
What is in pure water? )(),(),( 2 lOHaqOHaqH
Water can act as an oxidizing agent, when it is reduced to H2:
(aq)OH(g)H21eO(l)H 22
For the equivalent reduction of hydronium ions in water:
1 (59 V) HE
Water’s oxidation power increases with decreasing pH
(g)H21e(aq)H 2 (59mV)pH
/2HH
E
Water s oxidation power increases with decreasing pH.
Redox Chemistry of Water
Water can also act as a reducing agent, when it is oxidized to O2:
O(l)2H4e(aq)4H(g)O O(l)2H4e(aq)4H(g)O 22
When the partial pressure of O2 is 1 bar, the Nernst equation gives:p p 2 b , q g
pH0.059V1.23V]H[
1ln4 4
0 FRTEE
]H[4F
Oxidation by Water
For metals with large negative standard potentials reaction For metals with large, negative standard potentials, reaction with aqueous acid will occur
Oxidation of the metal by water or hydrogen ions
The overall reaction occurs via one of the following two processes:
(aq)OH(g)H21(aq)MO(l)HM(s) 22
(g)H21(aq)M(aq)HM(s) 2
These reactions are thermodynamically favorable when M is an s-block metal of a 3d series metal (i.e., Ti, Cr, Al, Ni, etc)
(g)H3(aq)2M(aq)H62Sc(s) 23
Oxidation by atmospheric oxygen
•Reactions of Al, Ti metal with moist air are spontaneous but moist air are spontaneous, but they can be used for many years in the presence of water y pand oxygen.
h h f b•Why?: Their surfaces become passivated by an oxide film
•ie: aluminum oxide, titanium oxide, also oxides of Cu, Fe, Zn, , ,
Copper carbonate on the roofs of the Chateau Frontenac in Quebec
Cu Oxidation
R d h lf iRedox half reactions:
O(l)2H4e(aq)H4(g)O 2-
2 pH0.059V1.23V E( )( q)(g) 22
Cu(s)e2(aq)Cu -2
pH0.059V1.23VE
V34.0E
Full Reaction:
O(l)2H(aq)2Cu(aq)H4(g)O2Cu(s) 22
2
pH0.059VV89.0 E p
Since E>0 in neutral and acidic environments atmospheric oxidation Since E>0 in neutral and acidic environments, atmospheric oxidation is spontaneous (E=0.48V at pH=7)
Reduction by water
The positive potential of the O2, H+/H2 couple is large:
pH0.059V1.23V E pH0.059V1.23VE
Acidified water is a poor reducing agent (i.e., a poor electron li ) d idi i supplier) except toward even stronger oxidizing agents
23 (aq)4H(g)O4CoO(l)2H(aq)4Co 22
23
1.92VE .9 V
(Lowering acidity favors the reaction)
Standard potentials at 298KOnly a few redox couples have standard potentials greater than
h 1 23V greater than 1.23V.
Water splitting & Photosynthesis
(g)O(g)2H2eO(l)2H 222
Easy with metals
(aq)OH(g)H2eO(l)2H 22
4eO(l)2H(g)O4OH 22
Hard in general: barrier is kinetic• transfer 4 electrons, form O=O
PhotosynthesisNature uses a metal-oxide cluster containing 4 Mn atoms and 1 Ca g
atom that is located in photosystem II for O2 evolution
The Porphyrin Porphine Chlorophyll-a (located in giant proteins known as “light
harvesting antennae”harvesting antennae
Photosynthesis
A li h A light harvesting complexp
Reduction of CO to
hvP* CO2CH2O
e- (in 1 ps)
Reduction of CO2 to carbohydrates via
reduction of NADP+ to NADPH
Ch d ti t P+
Charged reaction center chlorophyll, reduced
back to it’s ground state P
In PSII, this electron comes from oxidation of water into O and by accepting an electron
e-
of water into O2 and H+
O2 evolution catalyst
Various hypotheses for O2 evolution:) l d d d d l l1) As Mn sites are progressively oxidized, coordinated H2O molecules
become increasingly polarized and lose protonsH2OOH-O2
2-2 2
2) Since Mn(IV) and Mn(V)=O are equivalent to Mn(II)-[O] and Mn(III)-[O], if two [O] atoms are close together, and O2 molecule
may formmay form3) Presence of Ca2+ is essential: remains in 2+ state and provides a fast
binding site for H2O
Dis & co-proportionation
DisproportionationDisproportionationA redox reaction in which the oxidation number of element is simultaneously raised and lowered.
Example: Cu+VsCueaqCua
sCuaqCuaqCuba520)()(:)
)()()(2:)) 2
VEVaqCueaqCub
VsCueaqCua
36.016.052.016.0),()(:)
52.0),()(:)
0
2
FFF )(3)()(2 23
VFFVaqFeeaqFe
aqFesFeaqFe
450)(2)(77.0),()()(3)()(2
2
23
23
CoproportionationThe reverse of disproportionation.
VEVsFeeaqFe
22.1)45.0(77.045.0),(2)(
0
2
Example Fe3+ and Fe
Redox Reactions in Biology: Neuroscience
http://en.wikipedia.org/wiki/Neuron
The Soma Membrane
Cl-
http://www.nicerweb.com/bio1152/Locked/media/ch48/resting_potential.html
Re-deriving the Nernst Equation
The Nernst equation specifies the voltage that will result from maintenance of a given concentration difference.
Diff i fl Th b f h (i ) i
Flux due to drift and diffusion
“Fick’s First Law”xdCDJ )(
Diffusion flux: The net number of charges (ions) crossing some position per unit time per unit area, Γ is:
“Fick’s First Law”dx
DJdiff
D = diffusion coefficient of charges = l2/τ
dC/dx = charge concentration gradient
Drift flux: Ions will drift with a velocity proportional to their mobility mu, charge, and the strength of the E field:
xdrift xCqJ E)(xdrift q Ev
(Assuming single charged ions, so coefficient of q=1)
xdC )(Total Flux
dxxdCDxCqJ xtot)()( E
kTD Einstein relation
In equilibrium, J=0
CdCkTdxx
EC
out
V din
V dxxE
Membrane Potential
))](ln())([ln(Vout
i
outCinCqkTdxx
E
in q
)()( inCinCkT)(
)(ln25mV)(
)(lnVoutCinC
outCinC
qkT
(Recall: R=Nak, F=qNa)
[Ki+]:[K +]=10:1
ΔV=-58mV resting l f
[Kin ]:[Kout ] 10:1potential of neuron
Action potentialsIn the resting state of a nerve cell, sodium mobility across the membrane is much lower than
potassium and the cell maintains a negative voltage When a nerve or muscle cell is potassium, and the cell maintains a negative voltage. When a nerve or muscle cell is stimulated by synaptic transmission, the mobility (or channel conductance, or permeability)
for sodium transiently increases to a value greater than that for potassium and the cell internal voltage "spikes" above zero volts for about a millisecond.
Clark, Nature Neuroscience 8 (2005)http://www.brown.edu/Departments/Engineering/Courses/En123//Lectures/electroN.htm
Electrode : Solution Interactions
Cations
Outer
Cationssurrounded by solvent particles Specifically Primary solvent Outer
Helmholtz plane
p(solvated cations)
Specifically adsorbed
anionsElectrolytic
solution
ylayer (acting as a
dielectric)
Inner H l h l
Metal planeMetal (with negative
h )Helmholtz plane
charge)
Electrical Double Layer
T d i h To derive the potential ψ, start
with Poisson’s
Negatively charged particle
equation:
/2
ρ: charge density = ρ(x)
Boundary conditions:
“slipping ψ
ψ(0)=ψo
ψ(∞)=0pp g
plane”
Pot
entia
l, P
Electrical Double Layer
A h i b B l di ib i ( b bili i Assume the ions obey a Boltzmann distribution (probability an ion having a local potential ψ ):
kTqz
i
i ienn /
#ions/vol valencei
iiqnz
#ions/vol
i
iiq
kTqd /2
2
i
kTqzii
ienzqdxd /
22
Poisson-Boltzman equation
Electrical Double Layer: Huckel approximation
P i B l i h l l iPoisson-Boltzman equation has no general solution.
kTAssume:
kTqzi
mV7.25~kT T=300KmV7.25q
(Low surface potential approximation)
d 2
Expand exponential into a power series
i
iii kT
qznzqdxd )1(2
2
At infinity, electrical neutrality – all charges cancel0
Electrical Double Layer: Huckel approximation
d 222 1
i
ii
kTnqz
dxd 2
2
2 1
ii nzkTq 2
22
ikT
22
d xe 2 dx oe
κ: Double layer thickness (typically 1-3 nm)y ( yp y )
Beyond the Huckle approximation: Gouy-Chapman-Stern:
kTzqkTni
o 2sinh8