nernst equation 3

8/8/2019 Nernst Equation 3

1/20

Cell EMFCell EMF

Oxidizing and Reducing AgentsOxidizing and Reducing Agents


2/20

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


3/20

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)


4/20

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


5/20

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


6/20

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


7/20

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 =


8/20

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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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


10/20

Silver

electrodesconcentration

cell

Spontaneous cell

reaction?Which is anode

and which is

cathode?

Line diagram?

Always Start Here


11/20

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


12/20

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


13/20


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


14/20

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)


15/20


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

=+=

=

=

=

+

+

-

-


16/20

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


17/20

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


18/20

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


19/20

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!


20/20

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

nernst equation 3

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