Chemistry 232

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Chemistry 232
Electrochemistry Notes
Electrochemical Cells

Galvanic cells –



an electrochemical cell that drives
electrons through an external circuit
spontaneous redox reaction occurring
inside cell.
The Zn/Cu Galvanic Cell

half-reactions
Cu2+ (aq) + 2 e-  Cu (s) (cathode, RHS)
Zn2+ (aq) + 2 e-  Zn (s) (anode, LHS)
A Schematic Galvanic Cell
e-
Porous Disk
e-
eReducing Agent
Anode
Oxidizing Agent
Cathode
The Zinc/Copper galvanic cell.
e-
Zn(s)
1.10 V
Porous Disk or
Salt Bridge
e-
Cu(s)
e-
ea(Zn2+) = 1.00
Anode
a(Cu2+) = 1.00
Cathode
Cell Reactions
The difference in the RHS and the LHS
reaction
Cu2+ (aq) + Zn (s)  Cu (s) + Zn2+ (aq)
 For each half reaction, we can write the
reaction quotient as follows
Cu2+ (aq) + 2 e-  Cu (s)
Q = 1/ a(Cu2+)
Zn2+ (aq) + 2 e-  Zn (s)
Q = 1/ a(Zn2+)
Overall  Qcell = a(Zn2+) / a(Cu2+)

Cell Diagrams


A shorthand way of expressing what
takes place in an electrochemical
cell.
For the above electrochemical cell.
Pt Cu (s) Cu2+ (aq) Zn2+ (aq) Zn (s) Pt
Note
phase boundary
salt bridges
liquid junction
Another Example

The cell reaction
H2 (g) + Cu2+ (aq)  2 H+ (aq) + Cu
(s)
Pt H2 (g) H+ (aq) Cu2+ (aq) Cu (s) Pt

Electrochemical cells

a cell that has not reached equilibrium can
do electrical work by driving electrons
through an external wire.
Reversible Electrochemical Cells

In order for us to make measurements
on an electrochemical cell, it must be
operating reversibly.


Place an opposing source of potential in
the external circuit
Cell operates reversibly and at a constant
composition.
we,max = G
The Measurement of Cell
Potentials

Measure the potential of an electrochemical
cell when the cell is at equilibrium, i.e., the
state between the galvanic and the
electrolytic cell.
Counter potential
(load)
ePorous Disk
e-
eReducing Agent
Anode
Oxidizing Agent
Cathode
Derivation of the Nernst Equation

Consider an electrochemical cell that
approaches the equilibrium state by
an infinitesimal amount d
dG   J  J d  rxn G d
J
Reminder
 dG
r G   J  J  
J
 d


T ,P
The Work in Transporting Charge

The maximum work
dw e ,max  rxn G d

For the passage d electrons from the
anode (LHS) to the cathode (RHS)
 eN A d  F d
F = Faraday’s constant = e NA = 96485 C/mole
The Cell Potential

The work to transport charge
dw e ,max   F d E cell
dw e ,max
 rxn G
d
rxn G
 - E cell
F
Standard Cell Potentials

From the reaction Gibbs energy
r G  r G o  RT lnQcell   F E cell
o
RT lnQcell
r G
r G


 E cell
 F
 F
 F
We define
E cell

r G

F
o
The Nernst Equation
E cell

RT
E 
lnQcell
F

E represents the standard cell potential,
the potential of the cell when all cell
components are under standard
conditions.




f (all gases) = 1
a (solutes) = 1
T = 298.15 K
P = 1.00 bar pressure
Cells at Equilibrium

When the electrochemical cell has
reached equilibrium
E cell  0 V  Qcell  K cell
Kcell = the equilibrium constant for the cell reaction.
RT
 FE
E 
ln K cell  ln K cell 
F
RT


Knowing the E° value for the cell, we can estimate
the equilibrium constant for the cell reaction.
Equilibrium Constant Calculations
from Cell Potentials

Examine the following cell.
Pt Sn2+ (aq), Sn4+ (aq) Fe3+ (aq) Fe2+ (aq)

Pt
Half-cell reactions.
Sn4+ (aq) + 2 e-  Sn2+ (aq) E(Sn4+/Sn2+) = 0.15 V
Fe3+ (aq) + e-  Fe2+ (aq)
E (Fe3+/Fe2+) = 0.771 V

Cell Reaction
Sn2+ (aq) + 2 Fe3+ (aq)  Sn4+ (aq) + 2 Fe2+ (aq)
Ecell = (0.771 - 0.15 V) = 0.62 V
Standard Reduction Potentials




Standard reduction potentials are
intensive properties.
We cannot measure the potential of
an individual half-cell!
We assign a particular cell as being
our reference cell
Assign values to other electrodes on
that basis.
The Standard Hydrogen Electrode

Eo (H+/H2) half-cell = 0.000 V
e-
f{H2(g)} = 1.00
H2 (g)
a (H+) = 1.00
Pt gauze
A Galvanic Cell With Zinc and the
Standard Hydrogen Electrode.
e-
0.763 V
e-
Porous Disk or
Salt Bridge
Zn(s)
a(Zn2+) = 1.00
Zn2+, SO42-
Anode
a (H+) = 1.00
Source of H+ (e.g.,
HCl (aq), H2SO4 (aq))
Cathode
H2 (g)
Pt gauze
The Cell Equation for the ZincStandard Hydrogen Electrode.
The cell reaction
2 H+ (aq) + Zn (s)  H2 (g) + Zn2+ (aq)

Pt Zn (s) Zn2+ (aq),a=1 H+ (aq), a=1 H2 (g), f=1 Pt

When we measure the potential of this cell
Ecell = ERHS - ELHS
but ERHS = E(H+/H2) = 0.000 V
 Ecell = E(Zn2+/Zn) = 0.763 V
The Spontaneous Direction of
a Cell Reaction

Examine the magnitude the of the
standard cell potential!
E cell


rxn G

F
o
If the standard cell potential is positive, the
rG is negative!
The Composition Dependence of the
Cell Potential

Nonstandard cell potential (Ecell) will be
a function of the activities of the
species in the cell reaction.
E cell

RT
E 
lnQcell
F

To calculate Ecell, we must know the cell
reaction and the value of Qcell.
Example

For the following system
Pt H2 (g) H+ (aq) Cu2+ (aq) Cu (s) Pt

Calculate the value of the cell potential when
the f (H2) = 0.50, a(Cu2+) = 0.20, and a(H+) =
0.40.
Concentration Cells

Electrolyte concentration cell

the electrodes are identical; they simply
differ in the concentration of electrolyte in
the half-cells.
Concentration Cells (II)

Electrode concentration cells

the electrodes themselves have
different compositions. This may be
due to.
Different fugacities of gases involved in
electrode reactions (e.g., The H+ (aq)/H2
(g) electrode).
 Different compositions of metal amalgams
in electrode materials.

Applications of Electrochemistry

Measurement of activities and activity
coefficients.

Electrochemical series.

Equilibrium constants and
thermodynamic functions of cell
reactions
Obtaining Standard Cell Potentials

Look at the following cell
Pt H2 (g) HCl (aq) AgCl (s) Ag (s) Pt
E cell  E

cell
   



RT
a H a Cl

ln
F
f (H 2 )



Ecell = E(AgCl/Ag) - E (H+/H2)
= E(AgCl/Ag)
Ecell Values and Activity Coefficients

E cell
In dilute solution, using the DHLL
1
4.606 RT
2.344 RT


log m  E (AgCl Ag ) 
m 2
F
F
Plot LHS vs. m1/2

Once Ecell is known, we can obtain
experimental estimates of the mean activity
coefficients.
The Calculation of Standard Cell
Potentials
0.236
0.234
0.232
0.230
0.228
y = 0.0582x + 0.2222
R2 = 0.9836
0.226
0.224
0.222
0.220
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
Electrochemical Series

Look at the following series of reactions
Cu2+ (aq) + 2 e-  Cu (s)
Zn2+ (aq) + 2 e-  Zn (s)

Zn has a thermodynamic tendency to
reduce Cu2+ (aq)
Pb2+ (aq) + 2 e-  Pb (s)
Fe2+ (aq) + 2 e-  Fe (s)

E(Cu2+/Cu) = 0.337 V
E(Zn2+/Zn) = -0.763 V
E(Pb2+/Pb) = -0.13 V
E(-Fe2+/Fe) = -0.44 V
Fe has a thermodynamic tendency to
reduce Pb2+ (aq)
Thermodynamic Information

Note
 G

 


 FE r G
T ,P
 And
 FE r G


Entropy Changes

To obtain the entropy change for the
cell reaction
 S
 rxn S  
 
 


 
 T
T ,P


 

rxn S  
rxn G 
 T
 G

 

T ,P


T ,P

  F  E 

 T P





E


  F 

 T P
Enthalpy Changes

To obtain the enthalpy change for the cell reaction
 H
rxn H  
 

 G

 
T ,P  


T ,P
 E
 FE  FT 
 T
rxn H  FE


 S
 T 
 


P
 E 
 FT 
 T


P


T ,P
The Liquid Junction Potential



Examine the following electrochemical
cell
Activity difference of the HCl between
compartment 1 and compartment 2
There should be a transport of matter
from one cell compartment to the
other!
A Concentration Cell
e-
0.0592V
e-
Porous Disk or
Salt Bridge
Ag(s)
a(Cl-) = 0.010
Left
a (Cl -) = 0.0010
Right
Ag(s)
The Development of Liquid Junction
Potentials

The cell
compartments
are identical
except for the
activities of
the electrolyte
solutions.
HCl (a2)
HCl (a1)
Ag/AgCl electrode

Note that we
now have the
migration of both
cations and
anions through
the liquid
junction.
H+
Cl-
Ag/AgCl electrode

After a
period of
time
+++++
- - - - -----------
Ag/AgCl electrode

Choose the lower compartment as our
LHS electrode.
Ag AgCl
Cl- (aq) a1 Cl- (aq), a2 AgCl (s) Ag (s)
Note: liquid junction
 For
the passage of one mole of charge
through the cell
-F Ecell =  GJ
The Cell Reactions

For the LHS and RHS electrodes
AgCl (s) + e-  Ag (s) + Cl- (a1) LHS
AgCl (s) + e-  Ag (s) + Cl- (a2) RHS

Net change
Cl- (a1)  Cl- (a2)

Note that the charge at the interface is
transported by the anions and cations
in the cell reaction!
The Transport Numbers


How is the charge carried at the interface
of the cells?
 t+ moles of charge carried by the H+
(cation).
 t- moles of charge carried by the Cl(anion).
Passage of one mole of “+” charge through
the interface

requires the passage of t+ moles of H+ (aq) from
the LHS  RHS, and the passage of t- mole of Clcharge from the RHS  LHS.
At the boundary
t+ H+(a1) + t- Cl-(a2)  t+ H+(a2) + tCl-(a1)
 For the entire cell
Cl- (a1) t+ H+(a1) + t- Cl-(a2)  Cl- (a2)
t+ H+(a2) +
t- Cl-(a1)
 The cell reaction involves the
transport of t+ moles of HCl from the
LHS to the RHs of the cell.

The Gibbs Energy Changes

For the above cell reaction, we can
write the Gibbs energy expressions as
follows
G  t    (H  )  RT lna (H  )2     (Cl  )  RT lna (Cl  )2 




 t    (H  )  RT ln a (H  )1    (Cl  )  RT ln a (Cl  )1
G  t 

a (H
RT ln
a (H


)a (Cl  ) 2


)a (Cl ) 1


Cells With Transference

Note a(H+) a (Cl-) = {a (HCl)}2
G

a  (HCl )2
 2 t RT ln
a  (HCl )1
Note that the cell potential with
transference, Ewt is determined as follows
E wt
RT a  (HCl )2
 2 t 
ln
a  (HCl )1
F
Cells without Transference


What if we were able to set up a cell so that
the transport at the interface did not
contribute to the overall G?
The potential of this cell would be the cell
potential without transference, Ewot.
Cl- (a1)  Cl- (a2)
G  RT
E wot

a  (HCl )2
ln
a  (HCl )1
RT a  (HCl )2

ln
a  (HCl )1
F
The Liquid Junction Potential

The liquid junction potential is the
difference in the cell potentials with
and without transference!
E LJ  E wt  E wot
E LJ
RT a  (HCl )2
 1  2 t  
ln
a  (HCl )1
F
L.J. Potentials Depend on
Transport Numbers


What is the following were true?
t+  t-  0.5  ELJ would be very small
and would only make a small
contribution to the overall cell
potential !
L.J. Potentials Depend on
Transport Numbers


ELJ a potential problem any time we
measure the cell potential whose
electrodes have different electrolytes
How does the salt bridge help?

e.g., for species with t+  t-  0.5, the ELJ
values are small and are readily
established!
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