Transcript Document

Prof . Dr. Hisham Ezzat Abdellatef
Professor of pharmaceutical analytical chemistry
Electrochemical methods of
analysis
Electrochemistry is the
 relationship between electrical properties
and chemical reactions.
 measured involve either
voltage,
current or
resistance or
combination of these.
Classification of electrochemical methods
1.
POTENTIOMETRY
Measure electrical potential developed by an electrode in an electrolyte solution at zero current
flow. Use NERNST EQUATION relating potential to concentration of some ion in solution.
2.
VOLTAMMETRY
Determine concentration of ion in dilute solutions from current flow as a function of voltage when
POLARIZATION of ion occurs around the electrode.
POLARIZATION = depletion of concentration caused by electrolysis.
If using a dropping mercury electrode, method is termed POLAROGRAPHY.
3.
COULOMETRY
Electrolysis of a solution and use of Faraday's law* relating quantity of electrical charge to amount
of chemical change.
[* essentially states that it takes 9.65 x 104 Coulombs of electrical charge to cause electrolysis of 1
mole of a univalent electrolyte species.]
4.
CONDUCTIMETRY
Measure conductance of a solution, using INERT ELECTRODES, ALTERNATING CURRENT, AND
AN ELECTRICAL NULL CIRCUIT - thereby ensure no net current flow and no electrolysis. The
concentration of ions in the solution is estimated from the conductance.
NOTE:
Methods 1 and 4,
NO ELECTROLYSIS of solution. Sample recoverable, unaltered by analysis.
Methods 2 and 3
must cause ELECTROLYSIS OF THE SAMPLE.
http://www.science.uts.edu.au/subjects/91326/Section12/section12.html
Electrochemistry
Electrochemistry
• Electrochemistry
– deals with interconversion between chemical
and electrical energy
– involves redox reactions
• electron transfer reactions
•Oh No! They’re back!
Nomenclature
• “Redox” Chemistry: Reduction and Oxidation
• Oxidation: Loss of electrons
• Reduction: Gain of electrons
(a reduction in oxidation number)
LEO goes GER
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
Reduction form – electrons = oxidized form
The sum of two half – reactions constitutes the redox
reaction, e.g. the oxidation of ferrous ion by ceric ion:
Fe2+ – e = Fe3+ (oxidation half – reaction)
Ce4+ + e = Ce3+ (reduction half – reaction)
Adding
Fe2+ + Ce4+ = Fe3+ + Ce3+ (redox reaction)
Redox Reaction
Rules of Oxidation State Assignment
•
1. Ox # = 0: Element in its free state
combine with different element) (e.g.
•
(not
Na, H2)
2. Ox # = Charge of ion:
Grp1 = +1, Grp2 = +2, Grp7 = -1, ...
•
3. O = -2:
•
4. H = +1:

5  Ox. # = charge of molecule or ion.
Except with Na2O2 = -1…
Except with electropositive element
(i.e., Na, K) H = -1.
Highest and lowest oxidation numbers
of reactive main-group elements. The
A group number shows the highest
possible oxidation number (Ox.#) for a
main-group element. (Two important
exception are O, which never has an
Ox# of +6 and F, which never has an
Ox# of +7.) For nonmetals, (brown)
and metalloids (green) the A group
number minus 8 gives the lowest
possible oxidation number
Types of Chemical Equations
• Half-Cell Equation
• Net Ionic Equation
• Molecular Equation
7/17/2015
Half-Cell Equation
atom balanced
charge balanced with free electrons
C2O42–(aq)  2 CO2(aq) + 2 e
also called a half-reaction
7/17/2015
–
Net Ionic Equation
–
5 C2O42–(aq) + 2 MnO4 (aq) + 16 H+(aq) 
10 CO2(aq) + 2 Mn2+(aq) + 8 H2O(l)
A balanced, complete (not half-cell) equation in
which
strong electrolytes are ionized
spectator ions have been eliminated
7/17/2015
Net Ionic & Molecular Equations
5 C2O42–(aq) + 2 MnO4– (aq) + 16 H+(aq) 
10 CO2(aq) + 2 Mn2+(aq) + 8 H2O(l)
5 Na2C2O4 + 2 KMnO4 + 16 HCl 
10 CO2 + 2 MnCl2 + 8 H2O + 10 NaCl + 2 KCl
7/17/2015
Redox Reactions - Ion electron method.
Under Acidic conditions
1. Identify oxidized and reduced species
Write the half reaction for each.
2. Balance the half rxn separately except H & O’s.
Balance: Oxygen by H2O
Balance: Hydrogen by H+
Balance: Charge by e 3. Multiply each half reaction by a coefficient.
There should be the same # of e- in both half-rxn.
4. Add the half-rxn together, the e - should cancel.
Redox Reactions - Ion electron method.
Under Basic conditions
1, 2. Procedure identical to that under acidic conditions
Balance the half reaction separately except H & O’s.
Balance Oxygen by H2O
Balance Hydrogen by H+
Balance charge by e-
3. Mult each half rxn such that both half- rxn have same
number of electrons
4. Add the half-rxn together, the e- should cancel.
5. Eliminate H+ by adding:
H+ + OH- H2O
Example: Basic Conditions
H2O2 (aq) + Cr2O7-2(aq )  Cr 2+ (aq) + O2 (g)
Half Rxn (oxid): 6e- + 14H+ + Cr2O7-2 (aq)  2Cr3+ + 7 H2O
Half Rxn (red): ( H2O2 (aq)  O2 + 2H+ + 2e- ) x 3
8 H+ + 3H2O2 + Cr2O72-  2Cr+3 + 3O2 + 7H2O
add:
Net Rxn:
 8 H+ + 8 OH8 H+ + 3H2O2 + Cr2O72-  2Cr+3 + 3O2 + 7H2O
8H2O  8 H+ + 8 OH8H2O
3H2O2 + Cr2O72 - + H2O  2Cr+3 + 3O2 + 8 OH-
basic concepts of electrochemistry
Electrochemistry:
a science that studies the relationship between
electric and chemical phenomena and the conversion
disciplines between electric and chemical energy
Differences between the ordinary oxidation-reduction
reaction (redox reaction) occurring in solution and in the
electrochemical cell.
2Fe3+ + Sn2+  2Fe2+ + Sn4+
To harvest useful energy, the oxidizing and reducing agent has
to be separated physically in two different compartments so
as to make the electron passing through an external circuit
half-reactions:
oxidation / anode reaction:
Sn2+ - 2e-  Sn4+
reduction / cathode reaction:
2Fe3+ + 2e-  2Fe2+
Reaction takes place at
electrode/solution interface
Electrochemical apparatus:
chemical  electric:
primary cell (Galvanic cell)
electric  chemical:
electrolytic cell
Electrode: anode, cathode
positive electrode; negative electrode
We are talking about electricity
So we have
to talk
about
Luigi
Galvani
Galvani ran the experiment
Galvanic Cells (cont.)
8H+ + MnO4- + 5eFe2+
Mn2+ + 4H2O
Fe3+ + e-
x5
Galvanic Cells (cont.)
• In turns out that we still will not get electron flow
in the example cell. This is because charge buildup results in truncation of the electron flow.
• We need to “complete the circuit” by allowing
positive ions to flow as well.
• We do this using a “salt bridge” which will allow
charge neutrality in each cell to be maintained.
Galvanic Cells (cont.)
Salt bridge/porous disk: allows for ion migration such
that the solutions will remain neutral.
Galvanic Cells (cont.)
• Galvanic Cell: Electrochemical cell in which chemical
reactions are used to create spontaneous current (electron)
flow
Voltmeter
Zn (–)
Zn2+
Salt bridge
Na+
SO42–
(+) Cu
Cu2+
Voltmeter
e–
Anode Salt bridge
+
Zn (–) Na
SO 2–
(+) Cu
4
Zn2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Cu2+
e–
2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e–
Anode Salt bridge
+
Zn (–) Na
SO 2–
(+) Cu
4
Zn2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Cu2+
e–
2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e–
e–
Anode Salt bridge Cathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized
Zn
Cu2+
Zn2+
–
Cu e
Voltmeter
e–
e–
Anode Salt bridge Cathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized
Zn
Cu2+
Zn2+
–
Cu e
Voltmeter
e–
e–
Anode Salt bridge Cathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
Overall (cell) reaction
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
Electrode potentials
•
1.
2.
If a metal plate is
dipped into water,
or solution of its salt
an equilibrium
potential difference
established called
electrode potential.
There are two major
factors that
determine the
electrode potential.
electrolytic
solution pressure
ionic pressure
The Nernst Equation
– Why would concentration matter in
electrochemistry?
– The Nernst equation
– Applications
The Nernst Equation
E t  the oxidation potential of the redox electrode at T

 o

E  a constant value, characteri stic for every redox system


called "standard electrode potential"


R

gas
constant

8.314




o
T  absolute temperature of the system which at 25 C

  25  273  298 o C



F  Farady  96500 coulombs

log  natural logarithm, that is to the base 2.718, and is

 e

convertabl e to common logarithms , that is to the




base 10, by multiplying by 2.303


n  valency of the ions.

 n

A M  activity of the metal ions in solution.

A  activity of the metal.


 M
• The potential between
a metal and its ions
can be calculated from
the equation
formulated by Nernst
in 1889 as follows:
•
n
RT
Et  E 
nF
o
log e
AM
AM
25o C
0.0591
n
 E 
log (M )
n
25o C
0.0591
n
 E 
log (M )
n
E
E
o
o
Reduction Half-Reaction
E(V)
F2(g) + 2e-  2F-(aq)
2.87
Au3+(aq) + 3e-  Au(s)
1.50
Cl2(g) + 2 e-  2Cl-(aq)
1.36
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O
1.33
O2(g) + 4H+ + 4e-  2H2O(l)
1.23
Ag+(aq) + e-  Ag(s)
0.80
Fe3+(aq) + e-  Fe2+(aq)
0.77
Cu2+(aq) + 2e-  Cu(s)
0.34
Sn4+(aq) + 2e-  Sn2+(aq)
0.15
2H+(aq) + 2e-  H2(g)
0.00
Sn2+(aq) + 2e-  Sn(s)
-0.14
Ni2+(aq) + 2e-  Ni(s)
-0.23
Fe2+(aq) + 2e-  Fe(s)
-0.44
Zn2+(aq) + 2e-  Zn(s)
-0.76
Al3+(aq) + 3e-  Al(s)
-1.66
Mg2+(aq) + 2e-  Mg(s)
-2.37
Li+(aq) + e-  Li(s)
-3.04
Red. agent strength increases
Ox. agent strength increases
Standard Reduction Potentials
Cell Potential
Electromotive Force (emf)
Cell Potential
• Cell Potential (electromotive force), Ecell
(V)
– electrical potential difference between the two
electrodes or half-cells
• Depends on specific half-reactions, concentrations,
and temperature
• Under standard state conditions ([solutes] = 1 M,
Psolutes = 1 atm), emf = standard cell potential, Ecell
• 1 V = 1 J/C
– driving force of the redox reaction
Cell Potential
high electrical
potential
low electrical
potential
Cell Potential
Ecell = Ecathode - Eanode = Eredn - Eox
E°cell = E°cathode - E°anode = E°redn - E°ox
(Ecathode and Eanode are reduction potentials by definition.)
Cell Potential
• E°cell = E°cathode - E°anode = E°redn - E°ox
– Ecell can be measured
• Absolute Ecathode and Eanode values cannot
• Reference electrode
– has arbitrarily assigned E
– used to measure relative Ecathode and Eanode for
half-cell reactions
• Standard hydrogen electrode (S.H.E.)
– conventional reference electrode
Standard Hydrogen Electrode
• E = 0 V (by
definition;
arbitrarily
selected)
• 2H+ + 2e-  H2
eH2 (1 atm)
ePt
1 M H+
H2  2H+ + 2e-
Cu
1 M Cu2+
Cu2+ + 2e- Cu
Example 1
A voltaic cell is made by connecting a
standard Cu/Cu2+ electrode to a S.H.E. The
cell potential is 0.34 V. The Cu electrode is
the cathode. What is the standard reduction
potential of the Cu/Cu2+ electrode?
eH2 (1 atm)
ePt
1 M H+
2H+ + 2e-  H2
Zn
1 M Zn2+
Zn  Zn2+ + 2e-
Example 2
A voltaic cell is made by connecting a
standard Zn/Zn2+ electrode to a S.H.E. The
cell potential is 0.76 V. The Zn electrode is
the anode of the cell. What is the standard
reduction potential of the Zn/Zn2+
electrode?
Concentration and Ecell
• With the Nernst Eq., we can determine the effect
of concentration on cell potentials.
Ecell = E°cell - (0.0591/n)log(Q)
• Example. Calculate the cell potential for the
following:
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
Where [Cu2+] = 0.3 M and [Fe2+] = 0.1 M
Concentration and Ecell (cont.)
• First, need to identify the 1/2 cells
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
Cu2+(aq) + 2e-
Cu(s)
E°1/2 = 0.34 V
Fe2+(aq) + 2e-
Fe(s)
E°1/2 = -0.44 V
Fe(s)
Fe(s) + Cu2+(aq)
Fe 2+(aq) + 2e-
E°1/2 = +0.44 V
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Concentration and Ecell (cont.)
• Now, calculate Ecell
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Ecell = E°cell - (0.0591/n)log(Q)
2
Fe


(0.1)
Q

 0.33
2
Cu  (0.3)
Ecell = 0.78 V - (0.0591 /2)log(0.33)

Ecell = 0.78 V - (-0.014 V) = 0.794 V
Concentration and Ecell (cont.)
• If [Cu2+] = 0.3 M, what [Fe2+] is needed so that Ecell
= 0.76 V?
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Ecell = E°cell - (0.0591/n)log(Q)
0.76 V = 0.78 V - (0.0591/2)log(Q)
0.02 V = (0.0591/2)log(Q)
0.676 = log(Q)
4.7 = Q
Concentration and Ecell (cont.)
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
4.7 = Q
Q
2
Fe
 
 4.7
Cu 
Fe 

Q
 4.7
2
2

0.3
[Fe2+] = 1.4 M

Concentration Cells
• Consider the cell
presented on the left.
• The 1/2 cell reactions
are the same, it is just
the concentrations that
differ.
• Will there be electron
flow?
Concentration Cells (cont.)
Ag+ + e-
Ag
E°1/2 = 0.80 V
• What if both sides had 1 M
concentrations of Ag+?
• E°1/2 would be the same;
therefore, E°cell = 0.
Concentration Cells (cont.)
Ag+ + e-
Anode: Ag
Cathode: Ag+ + eAg 

Q
Ag 
Ag

anode

cathode

E1/2 = ? V
E1/2 = 0.80 V
0.1
 0.1
1
Ecell = E°cell - (0.0591/n)log(Q)
 0 V
1
Ecell = - (0.0591)log(0.1) = 0.0591 V
Concentration Cells (cont.)
Another Example:
What is Ecell?
Concentration Cells (cont.)
Ecell = E°cell - (0.0591/n)log(Q)
e-
2
0
Fe2+ + 2e-
Fe
2 e- transferred…n = 2
Fe 

Q
Fe 
2
anode
2

cathode
anode
cathode

Ecell = -(0.0296)log(.1) = 0.0296 V
0.01
 0.1
.1