1. Instrumental Analysis
Prof . Dr. Hisham Ezzat Abdellatef
Professor of pharmaceutical analytical chemistry
Instrumental Analysis

2. Electrochemical methods of analysis

3. Electrochemistry is the
relationship between electrical properties
and chemical reactions.
measured involve either
voltage,
current or
resistance or
combination of these.

4. Classification of electrochemical methods
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.
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.
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.]
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

5. Electrochemistry

6. Oh No! They’re back! Electrochemistry Electrochemistry
deals with interconversion between chemical and electrical energy
involves redox reactions
electron transfer reactions
Oh No! They’re back!

7. Nomenclature LEO goes GER “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

8. 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)
Ce e = Ce3+ (reduction half – reaction)
Adding
Fe Ce4+ = Fe Ce3+ (redox reaction)

9. Redox Reaction

10. Rules of Oxidation State Assignment
1. Ox # = 0: Element in its free state (not combine with different element) (e.g. Na, H2)
2. Ox # = Charge of ion: Grp1 = +1, Grp2 = +2, Grp7 = -1, ...
3. O = -2: Except with Na2O2 = -1…
4. H = +1: Except with electropositive element (i.e., Na, K) H = -1.
5  Ox. # = charge of molecule or ion.
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

11. Types of Chemical Equations
Half-Cell Equation
Net Ionic Equation
Molecular Equation
12/8/2018

12. also called a half-reaction
Half-Cell Equation
atom balanced
charge balanced with free electrons
C2O42–(aq) ® 2 CO2(aq) + 2 e–
also called a half-reaction
12/8/2018

13. A balanced, complete (not half-cell) equation in which
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
12/8/2018

14. 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 KMnO HCl ®
10 CO2 + 2 MnCl2 + 8 H2O + 10 NaCl + 2 KCl
12/8/2018

15. 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.

16. 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

17. Example: Basic Conditions
H2O2 (aq) + Cr2O7-2(aq )  Cr 2+ (aq) + O2 (g)
Half Rxn (oxid): 6e H+ + Cr2O7-2 (aq)  2Cr H2O
Half Rxn (red): ( H2O2 (aq)  O H e- ) x 3
8 H H2O2 + Cr2O72-  2Cr O H2O
add: 8H2O  8 H OH-
8H2O  8 H OH-
Net Rxn: 3H2O2 + Cr2O H2O  2Cr O OH-

18. basic concepts of 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+

19. oxidation / anode reaction: Sn2+ - 2e-  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

20. Electrochemical apparatus:
electric  chemical:
electrolytic cell
chemical  electric:
primary cell (Galvanic cell)
Electrode: anode, cathode
positive electrode; negative electrode

21. We are talking about electricity
So we have to talk about
Luigi Galvani

22. Galvani ran the experiment

23. Galvanic Cells (cont.) 8H+ + MnO4- + 5e- Mn2+ + 4H2O Fe2+ Fe3+ + e-
x 5

24. Galvanic Cells (cont.)
In turns out that we still will not get electron flow in the example cell. This is because charge build-up 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.

26. Galvanic Cells (cont.)
• Galvanic Cell: Electrochemical cell in which chemical
reactions are used to create spontaneous current (electron)
flow

27. Voltmeter
(–)
Salt bridge
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+

28. Voltmeter e–
Anode
(–)
Salt bridge
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–

29. e–
2e– lost
per Zn atom
oxidized Zn
Zn2+
Voltmeter e–
Anode
(–)
Salt bridge
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–

30. e–
2e– lost
per Zn atom
oxidized Zn
Zn2+
Voltmeter e– e–
Anode
(–)
Salt bridge
Cathode
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)

31. e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized Zn
Cu2+ Cu e–
Zn2+
Voltmeter e– e–
Anode
(–)
Salt bridge
Cathode
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)

32. e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized Zn
Cu2+ Cu e–
Zn2+
Voltmeter e– e–
Anode
(–)
Salt bridge
Cathode
(+) Zn
Na+ Cu
SO42–
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)

33. Electrode potentials
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

34. The Nernst Equation
Why would concentration matter in electrochemistry?
The Nernst equation
Applications

35. The Nernst Equation
The potential between a metal and its ions can be calculated from the equation formulated by Nernst in 1889 as follows:

37. Standard Reduction Potentials
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
Ox. agent strength increases
Red. agent strength increases 37

38. Cell Potential Electromotive Force (emf)

39. 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

40. low electrical potential high electrical potential
Cell Potential
low electrical potential
high electrical potential

41. Cell Potential Ecell = Ecathode - Eanode = Eredn - Eox
(Ecathode and Eanode are reduction potentials by definition.)

42. 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

43. Standard Hydrogen Electrode
E = 0 V (by definition; arbitrarily selected)
2H+ + 2e-  H2

44. e-
Cu2+ + 2e- Cu Cu
H2  2H+ + 2e-
H2 (1 atm) Pt
1 M H+
1 M Cu2+

46. 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?

48. e- e-
H2 (1 atm) Pt Zn
1 M H+
1 M Zn2+
2H+ + 2e-  H2
Zn  Zn2+ + 2e-

49. 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?

50. 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

51. 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 = V
Fe(s) Fe 2+(aq) + 2e-
E°1/2 = V
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
E°cell = V

52. Concentration and Ecell (cont.)
• Now, calculate Ecell
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
E°cell = V
Ecell = E°cell - (0.0591/n)log(Q)
Ecell = 0.78 V - ( /2)log(0.33)
Ecell = 0.78 V - ( V) = V

53. 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 = 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

54. Concentration and Ecell (cont.)
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
4.7 = Q
[Fe2+] = 1.4 M

55. 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?

56. 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.

57. Concentration Cells (cont.)
Anode:
Ag Ag+ + e- E1/2 = ? V
Cathode:
Ag+ + e Ag E1/2 = 0.80 V
Ecell = E°cell - (0.0591/n)log(Q)
0 V 1
Ecell = - (0.0591)log(0.1) = V

58. Concentration Cells (cont.)
Another Example:
What is Ecell?

59. Concentration Cells (cont.)
Ecell = E°cell - (0.0591/n)log(Q) e- 2
Fe2+ + 2e Fe
2 e- transferred…n = 2
anode
cathode
Ecell = -(0.0296)log(.1) = V