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8/8/2019 Nernst Equation 3
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Cell EMFCell EMF
Oxidizing and Reducing AgentsOxidizing and Reducing Agents
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Cell Potential Determine the cell potential and thespontaneous reaction for a voltaic cell with
electrodes Cu2+/Cu and Fe3+/Fe2+
Fe+3(aq) + e- Fe+2(aq) E= 0.77 V
Cu+2(aq)+2e- Cu(s) E= 0.34 V
Cu(s) Cu+2(aq)+2e- E= -0.34 V
2Fe+3(aq) + 2e- 2Fe+2(aq) E= 0.77 V
Cu(s) + 2Fe3+ Cu2+ + 2Fe2+ Eocell= 0.43V
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Line Notation
Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
solidAqueousAqueoussolid
Anode on the leftCathode on the right
Single line different phases.
Double line porous disk or salt bridge. If all the substances on one side are
aqueous, a platinum electrode is indicated.
Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)
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Line Notation Galvanic Cell
Zn Zn2+
(anode)
porous
plate
Cu2+ Zn2+
Cu2+ Cu(cathode)
e-
1.10 V
Start HereEnd Here
Tell the story. What you see during your trip from the anode to the cathode?
Zn(s) Zn2+ (aq, 1M) Cu2+ (aq, 1M) Cu(s)
1.00 M1.00 M
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Practice
Completely describe the voltaic cell based on
the following half-reactions under standardconditions.
MnO4-
+ 8 H
+
+5e
-
Mn+2
+ 4H2O E= 1.51
Fe+3 +3e- Fe(s) E= 0.036V
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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions
Consider a silver-nickel voltaic cell
E1/2(Ag+/Ag) = 0.80 V; E1/2(Ni
2+/Ni) = -0.28 V
E(cell) = E1/2(Ag+/Ag) - E1/2(Ni2+/Ni)= (0.80 V) - (-0.28 V) = 1.08 V
Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s)
The voltaic cell reaction is spontaneous
EMF and FreeEMF and Free--Energy ChangeEnergy Change
G= -nFE
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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions
EMF and FreeEMF and Free--Energy ChangeEnergy Change
Gis the change in free-energy, nis the
number of moles of electrons transferred, FisFaradays constant, and Eis the emf of the
cell.
Since nand Fare positive, ifG< 0 then E>0.
mol-J/V96,500C/mol500,961 =
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Effect of Concentration on Cell EMFEffect of Concentration on Cell EMF
A voltaic cell is functional until E= 0 at which
point equilibrium has been reached:G = 0.
The point at which E= 0 is determined by theconcentrations of the species involved in the
redox reaction.
The Nernst EquationThe Nernst Equation
The Nernst equation relates emf to
concentration usingQRTGG ln
QRTnFEnFE ln
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Effect of Concentration on Cell EMFEffect of Concentration on Cell EMF
The Nernst EquationThe Nernst Equation This rearranges to give the Nernst equation:
The Nernst equation can be simplified by collecting
all the constants together using a temperature of298 K:
(Note the change from natural logarithm to base-10
log.)
Remember that nis number of moles ofelectrons.
QnFRTEE ln
Qn
EE log0592.0
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Silver
electrodesconcentration
cell
Spontaneous cell
reaction?Which is anode
and which is
cathode?
Line diagram?
Always Start Here
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The Spontaneous Process in a
Concentration Cell - Mixing
0.1 M
1.0 L
1.0 M
1.0 L
Mix SolutionsSpon
taneous
0.55 M
2.0 L
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Silver Concentration Cell Anode Reaction Oxidation - dilute solution
becomes more concentrated.
Ag(s) Ag+
(0.1 M) + e-
Cathode Reaction Reduction-concentrated
solution becomes more dilute.
Ag+ (1.0 M) + e- Ag (s) Net Cell Reaction
Ag+
(1.0 M) Ag+
(0.1 M)
Ag(s)| Ag+ (0.1 M)||Ag+ (1.0 M)|Ag (s)
Construct the cell line diagram
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Effect of Concentration on Cell EMFEffect of Concentration on Cell EMF
Concentration CellsConcentration Cells We can use the Nernst equation to describe a cell
that has an emf based solely on difference in
concentration.
One compartment will consist of a concentrated
solution, while the other has a dilute solution.
Example: 1.00 MNi2+(aq) and 1.00 10-3 MNi2+(aq).
The cell tends to equalize the concentrations of
Ni2+(aq) in each compartment.
The concentrated solution must convert Ni2+(aq) (to
Ni(s)), so it must be the cathode.
Qlogn
0592.0EE
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Nickel electrodes concentration cell
Ni Ni
0.001 M Ni2+ 1.00 M Ni2+
Ni2+ (aq, 1 M) + 2e- Ni(s)Ni(s) 2e- + Ni2+ (aq, .001 M)
Ni(s) + Ni2+ (aq, 1 M)Ni2+ (aq, .001 M) + Ni (s)
e- e-
anodecathode
0.1
10x1Q
3-
=
Ni(s)Ni2+(.001 M, aq)Ni2+(1.0 M, aq)Ni(s)
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Effect of Concentration on Cell EMFEffect of Concentration on Cell EMF
Concentration CellsConcentration Cells Since the two half-reactions are the same, E
will be zero.
mV88.8V0888.0
M1.00
M101.00log
2
0592.0V0
]Ni[
]Ni[log
2
0592.0-V0
Qlogn
0592.0EE
3-
edconcentrat2
dilute2
=+=
=
=
=
+
+
-
-
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Origins of Membrane Potentials
Intracellular Fluids
K+
(140 mM)
K+
(2.5 mM)
Cl-
(120 mM)
Cell
Membrane
Cl-
(4 mM)
Na+
(9.2 mM)
Na+
(120 mM)
Extracellular Fluids
Concentration
differences of ions
suggests a Nernst
Potential
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Nernst Equation for Membrane
Potentials
[ ][ ]insideoutsidemem xxlogmVZ62)x(E The constant, 62, is applicable to tissue at 370 C.
Z represents the charge on the ion
This equation calculates the potential that the cell must
generate to prevent transfer of ions between the
intracellular and extracellular fluids.
The potential to prevent transfer in and out of the cell is
minus the potential for the spontaneous process
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Calculation of Membrane Potentials
1204Cl-2.5140K+
1209.2Na+
[x] mM, Outside[x] mM, Insidex
Consider the potential for transfer of K+
out.
There is a positive potential for this (G negative).
This means cell must generate a potential of 108 mV to
prevent transfer out.
( ) ( )( ) mV1081405.2logmV162KEmem
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Electrostatics of Membrane Potential
Negative Ions are attached to large proteins that
cannot migrate through cell membranes at the wall
No redox process or wire to conduct electrons!
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Calculation ofE, Non-standard Cell
Zn(s)|Zn2+ (1.00M)Cu2+ (0.05M)|Cu(s)
Write the balanced cell reaction
Zn(s) + Cu2+(0.05M) Zn2+ (1.00M) + Cu(s) Is E> or < Eo?
Concentrations indicate E less than EoLook for concentration effect
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