chapter-14.ppt

Fundamentals of Electrochemistry
Introduction
1.) Electrical Measurements of Chemical Processes
 Redox Reaction involves transfer of electrons from one species to another.
- Chemicals are separated
 Can monitor redox reaction when electrons flow through an electric current
- Electric current is proportional to rate of reaction
- Cell voltage is proportional to free-energy change
 Batteries produce a direct current by converting chemical energy to electrical
energy.
- Common applications run the gamut from cars to ipods to laptops

Basic Concepts
1.) A Redox titration is an analytical technique based on the transfer of
electrons between analyte and titrant
 Reduction-oxidation reaction
 A substance is reduced when it gains electrons from another substance
- gain of e- net decrease in charge of species
- Oxidizing agent (oxidant)
 A substance is oxidized when it loses electrons to another substance
- loss of e- net increase in charge of species
- Reducing agent (reductant)
(Reduction)
(Oxidation)
Oxidizing
Agent
Reducing
Agent

Basic Concepts
2.) The first two reactions are known as “1/2 cell reactions”
 Include electrons in their equation
3.) The net reaction is known as the total cell reaction
 No free electrons in its equation
4.) In order for a redox reaction to occur, both reduction of one compound
and oxidation of another must take place simultaneously
 Total number of electrons is constant
½ cell reactions:
Net Reaction:

Basic Concepts
5.) Electric Charge (q)
 Measured in coulombs (C)
 Charge of a single electron is 1.602x10-19C
 Faraday constant (F) – 9.649x104C is the charge of a mole of
electrons
6.) Electric current
 Quantity of charge flowing each second
through a circuit
- Ampere: unit of current (C/sec)
F
n
q 

Relation between
charge and moles:
Coulombs moles

e
mol
Coulombs

Basic Concepts
7.) Electric Potential (E)
 Measured in volts (V)
 Work (energy) needed when moving an electric
charge from one point to another
- Measure of force pushing on electrons
q
E
work
G 





Relation between
free energy, work
and voltage:
Joules Volts Coulombs
Higher potential
difference
Higher potential difference requires
more work to lift water (electrons) to
higher trough

Basic Concepts
7.) Electric Potential (E)
 Combining definition of electrical charge and potential
q
E
work
G 




 F
n
q 

nFE
G 


Relation between free energy
difference and electric potential
difference:
Describes the voltage that can be generated by a chemical reaction

Basic Concepts
8.) Ohm’s Law
 Current (I) is directly proportional to the potential difference (voltage)
across a circuit and inversely proportional to the resistance (R)
- Ohms (W) - units of resistance
9.) Power (P)
 Work done per unit time
- Units: joules per second J/sec or watts (W)
R
E
I 
I
E
s
q
E
P 




Galvanic Cells
1.) Galvanic or Voltaic cell
 Spontaneous chemical reaction to generate electricity
- One reagent oxidized the other reduced
- two reagents cannot be in contact
 Electrons flow from reducing agent to oxidizing agent
- Flow through external circuit to go from one reagent to the other
Net Reaction:
Reduction:
Oxidation:
AgCl(s) is reduced to Ag(s)
Ag deposited on electrode and Cl-
goes into solution
Electrons travel from Cd
electrode to Ag electrode
Cd(s) is oxidized to Cd2+
Cd2+ goes into solution

Galvanic Cells
1.) Galvanic or Voltaic cell
 Example: Calculate the voltage for the following chemical reaction
G = -150kJ/mol of Cd
V
.
C
J
.
mol
C
.
)
mol
(
J
E
nF
G
E
nFE
G
777
0
777
0
10
649
9
2
10
150
4
3






















Solution: n – number of moles of electrons

Galvanic Cells
2.) Cell Potentials vs. G
 Reaction is spontaneous if it does not require external energy

Galvanic Cells
2.) Cell Potentials vs. G
 Reaction is spontaneous if it does not require external energy
Potential of overall cell = measure of the tendency of this reaction to proceed to
equilibrium
 Larger the potential, the further the reaction is from equilibrium
and the greater the driving force that exists
Similar in concept to balls
sitting at different heights
along a hill

Galvanic Cells
3.) Electrodes
Cathode: electrode
where reduction takes
place
Anode: electrode
where oxidation takes
place

Galvanic Cells
4.) Salt Bridge
 Connects & separates two half-cell reactions
 Prevents charge build-up and allows counter-ion migration
Two half-cell reactions
Salt Bridge
 Contains electrolytes not
involved in redox reaction.
 K+ (and Cd2+) moves to cathode with
e- through salt bridge (counter
balances –charge build-up
 NO3
- moves to anode (counter
balances +charge build-up)
 Completes circuit

Zn|ZnSO4(aZN2+ = 0.0100)||CuSO4(aCu2+ = 0.0100)|Cu
anode
Phase boundary
Electrode/solution interface
Solution in contact with
anode & its concentration
Solution in contact with
cathode & its concentration
2 liquid junctions
due to salt bridge
cathode
Galvanic Cells
5.) Short-Hand Notation
 Representation of Cells: by convention start with anode on left

Ag+ + e-  Ag(s) Eo = +0.799V
Standard Hydrogen Electrode (S.H.E)
Hydrogen gas is bubbled over a Pt electrode
Pt(s)|H2(g)(aH
2
= 1)|H+(aq)(aH+ = 1)||
Standard Potentials
1.) Predict voltage observed when two half-cells are connected
 Standard reduction potential (Eo) the measured potential of a half-cell
reduction reaction relative to a standard oxidation reaction
- Potential arbitrary set to 0 for standard electrode
- Potential of cell = Potential of ½ reaction
 Potentials measured at standard conditions
- All concentrations (or activities) = 1M
- 25oC, 1 atm pressure

Standard Potentials
1.) Predict voltage observed when two half-cells are connected
As Eo increases, the more
favorable the reaction and the
more easily the compound is
reduced (better oxidizing
agent).
Reactions always written as
reduction
Appendix H contains a more extensive list

Standard Potentials
2.) When combining two ½ cell reaction together to get a complete net
reaction, the total cell potential (Ecell) is given by:

 
 E
E
Ecell
Where: E+ = the reduction potential for the ½ cell reaction at the positive electrode
E+ = electrode where reduction occurs (cathode)
E- = the reduction potential for the ½ cell reaction at the negative electrode
E- = electrode where oxidation occurs (anode)
Electrons always flow towards
more positive potential
Place values on number line to
determine the potential difference

Standard Potentials
3.) Example: Calculate Eo, and Go for the following reaction:

Nernst Equation
1.) Reduction Potential under Non-standard Conditions
 E determined using Nernst Equation
 Concentrations not-equal to 1M
aA + ne-  bB Eo
For the given reaction:
The ½ cell reduction potential is given by:
a
b
o
a
A
b
B
o
]
A
[
]
B
[
log
n
V
E
E
A
A
ln
nF
RT
E
E
0.05916





Where: E = actual ½ cell reduction potential
Eo = standard ½ cell reduction potential
n = number of electrons in reaction
T = temperature (K)
R = ideal gas law constant (8.314J/(K-mol)
F = Faraday’s constant (9.649x104 C/mol)
A = activity of A or B
at 25oC

Nernst Equation
2.) Example:
 Calculate the cell voltage if the concentration of NaF and KCl were each
0.10 M in the following cell:
Pb(s) | PbF2(s) | F- (aq) || Cl- (aq) | AgCl(s) | Ag(s)

Eo and the Equilibrium Constant
1.) A Galvanic Cell Produces Electricity because the Cell Reaction is
NOT at Equilibrium
 Concentration in two cells change with current
 Concentration will continue to change until Equilibrium is reached
- E = 0V at equilibrium
- Battery is “dead”













 


 d
b
o
a
c
o
cell
]
D
[
]
B
[
log
n
.
E
]
A
[
]
C
[
log
n
.
E
E
E
E
05916
0
05916
0
aA + ne-  cC E+
o
dD + ne-  bB E-
o
Consider the following ½ cell reactions:
Cell potential in terms of Nernst Equation is:
b
a
d
c
o
o
cell
]
B
[
]
A
[
]
D
[
]
C
[
log
n
.
)
E
E
(
E
05916
0


 

Simplify:

b
a
d
c
o
cell
]
B
[
]
A
[
]
D
[
]
C
[
log
n
.
E
E
05916
0


1.) A Galvanic Cell Produces Electricity because the Cell Reaction is
NOT at Equilibrium
Since Eo=E+
o- E-
o:
At equilibrium Ecell =0:
K
log
n
.
Eo 05916
0

Definition of
equilibrium constant
05916
0
10 .
nEo
K 
at 25oC
at 25oC

2.) Example:
 Calculate the equilibrium constant (K) for the following reaction:

Cells as Chemical Probes
1.) Two Types of Equilibrium in Galvanic Cells
 Equilibrium between the two half-cells
 Equilibrium within each half-cell
If a Galvanic Cell has a nonzero voltage then
the net cell reaction is not at equilibrium
For a potential to exist, electrons
must flow from one cell to the
other which requires the reaction
to proceed  not at equilibrium.
Conversely, a chemical reaction within a ½
cell will reach and remain at equilibrium.

Ni(s)|NiSO4(0.0025M)||KIO3(0.10 M)|Cu(IO3)2(s)|Cu(s)
Cells as Chemical Probes
2.) Example:
 If the voltage for the following cell is 0.512V, find Ksp for Cu(IO3)2:

Biochemists Use Eo´
1.) Redox Potentials Containing Acids or Bases are pH Dependent
 Standard potential  all concentrations = 1 M
 pH=0 for [H+] = 1M
2.) pH Inside of a Plant or Animal Cell is ~ 7
 Standard potentials at pH =0 not appropriate for biological systems
- Reduction or oxidation strength may be reversed at pH 0 compared to pH 7
Metabolic Pathways

Biochemists Use Eo´
3.) Formal Potential
 Reduction potential that applies
under a specified set of
conditions
 Formal potential at pH 7 is Eo´
b
a
d
c
o
cell
]
B
[
]
A
[
]
D
[
]
C
[
log
n
.
E
E
05916
0


Need to express concentrations as
function of Ka and [H+].
Cannot use formal concentrations!
Eo´ (V)

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