Prof Dr Hisham Ezzat Abdellatef Professor of pharmaceutical

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

Electrochemical methods of analysis

Electrochemistry is the q relationship between electrical properties and chemical reactions. q 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/Section 12/section 12. 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: Fe 2+ – e = Fe 3+ (oxidation half – reaction) Ce 4+ + e = Ce 3+ (reduction half – reaction) Adding Fe 2+ + Ce 4+ = Fe 3+ + Ce 3+ (redox reaction)

Redox Reaction

Rules of Oxidation State Assignment • 1. Ox # = 0: Element in its free state • 2. Ox # = Charge of ion: • 3. O = -2: • 4. H = +1: • 5 Ox. # = charge of molecule or ion. combine with different element) (e. g. Na, H 2) (not Grp 1 = +1, Grp 2 = +2, Grp 7 = -1, . . . Except with Na 2 O 2 = -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 10/7/2020

Half-Cell Equation atom balanced charge balanced with free electrons C 2 O 42–(aq) 2 CO 2(aq) + 2 e– also called a half-reaction 10/7/2020

Net Ionic Equation 5 C 2 O 4 2–(aq) – + 2 Mn. O 4 (aq) + 16 H+(aq) 10 CO 2(aq) + 2 Mn 2+(aq) + 8 H 2 O(l) A balanced, complete (not half-cell) equation in which strong electrolytes are ionized spectator ions have been eliminated 10/7/2020

Net Ionic & Molecular Equations 5 C 2 O 42–(aq) + 2 Mn. O 4– (aq) + 16 H+(aq) 10 CO 2(aq) + 2 Mn 2+(aq) + 8 H 2 O(l) 5 Na 2 C 2 O 4 + 2 KMn. O 4 + 16 HCl 10 CO 2 + 2 Mn. Cl 2 + 8 H 2 O + 10 Na. Cl + 2 KCl 10/7/2020

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 H 2 O 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 H 2 O 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- H 2 O

Example: Basic Conditions H 2 O 2 (aq) + Cr 2 O 7 -2(aq ) Cr 2+ (aq) + O 2 (g) Half Rxn (oxid): 6 e- + 14 H+ + Cr 2 O 7 -2 (aq) 2 Cr 3+ + 7 H 2 O Half Rxn (red): ( H 2 O 2 (aq) O 2 + 2 H+ + 2 e- ) x 3 8 H+ + 3 H 2 O 2 + Cr 2 O 72 - 2 Cr+3 + 3 O 2 + 7 H 2 O add: 8 H 2 O 8 H+ + 8 OH- 8 H+ + 3 H 2 O 2 + Cr 2 O 72 - 2 Cr+3 + 3 O 2 + 7 H 2 O 8 H 2 O 8 H+ + 8 OH- Net Rxn: 3 H 2 O 2 + Cr 2 O 72 - + H 2 O 2 Cr+3 + 3 O 2 + 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. 2 Fe 3+ + Sn 2+ 2 Fe 2+ + Sn 4+

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: Sn 2+ - 2 e- Sn 4+ reduction / cathode reaction: 2 Fe 3+ + 2 e- 2 Fe 2+ 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. ) 8 H+ + Mn. O 4 - + 5 e. Fe 2+ Mn 2+ + 4 H 2 O Fe 3+ + e- x 5

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

Voltmeter Zn (–) Zn 2+ Salt bridge Na+ SO 42– (+) Cu Cu 2+

Voltmeter e– Anode Salt bridge + Zn (–) Na SO 2– 4 Zn 2+ Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e– (+) Cu Cu 2+

e– 2 e– lost per Zn atom oxidized Zn Zn 2+ Voltmeter e– Anode Salt bridge + Zn (–) Na SO 2– 4 Zn 2+ Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e– (+) Cu Cu 2+

e– 2 e– lost per Zn atom oxidized Zn Zn 2+ Voltmeter e– e– Anode Salt bridge Cathode + Zn (–) Na SO 2– (+) Cu 4 Zn 2+ Cu 2+ Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e– Reduction half-reaction Cu 2+(aq) + 2 e– Cu(s)

e– 2 e– gained per Cu 2+ ion reduced 2 e– lost per Zn atom oxidized Zn Cu 2+ Zn 2+ – Cu e Voltmeter e– e– Anode Salt bridge Cathode + Zn (–) Na SO 2– (+) Cu 4 Zn 2+ Cu 2+ Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e– Reduction half-reaction Cu 2+(aq) + 2 e– Cu(s)

e– 2 e– gained per Cu 2+ ion reduced 2 e– lost per Zn atom oxidized Zn Cu 2+ Zn 2+ – Cu e Voltmeter e– e– Anode Salt bridge Cathode + Zn (–) Na SO 2– (+) Cu 4 Zn 2+ Cu 2+ Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e– Reduction half-reaction Cu 2+(aq) + 2 e– Cu(s) Overall (cell) reaction Zn 2+(aq) + Cu(s) Zn(s) + Cu 2+(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 • The potential between a metal and its ions can be calculated from the equation formulated by Nernst in 1889 as follows: •

Reduction Half-Reaction E (V) F 2(g) + 2 e- 2 F-(aq) 2. 87 Au 3+(aq) + 3 e- Au(s) 1. 50 Cl 2(g) + 2 e- 2 Cl-(aq) 1. 36 Cr 2 O 72 -(aq) + 14 H+(aq) + 6 e- 2 Cr 3+(aq) + 7 H 2 O 1. 33 O 2(g) + 4 H+ + 4 e- 2 H 2 O(l) 1. 23 Ag+(aq) + e- Ag(s) 0. 80 Fe 3+(aq) + e- Fe 2+(aq) 0. 77 Cu 2+(aq) + 2 e- Cu(s) 0. 34 Sn 4+(aq) + 2 e- Sn 2+(aq) 0. 15 2 H+(aq) + 2 e- H 2(g) 0. 00 Sn 2+(aq) + 2 e- Sn(s) -0. 14 Ni 2+(aq) + 2 e- Ni(s) -0. 23 Fe 2+(aq) + 2 e- Fe(s) -0. 44 Zn 2+(aq) + 2 e- Zn(s) -0. 76 Al 3+(aq) + 3 e- Al(s) -1. 66 Mg 2+(aq) + 2 e- 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) • 2 H+ + 2 e- H 2

e. H 2 (1 atm) e. Pt 1 M H+ H 2 2 H+ + 2 e- Cu 1 M Cu 2+ + 2 e- Cu

Example 1 A voltaic cell is made by connecting a standard Cu/Cu 2+ 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/Cu 2+ electrode?

e. H 2 (1 atm) e. Pt 1 M H+ 2 H+ + 2 e- H 2 Zn 1 M Zn 2+ + 2 e-

Example 2 A voltaic cell is made by connecting a standard Zn/Zn 2+ 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/Zn 2+ 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) + Cu 2+(aq) Fe 2+(aq) + Cu(s) Where [Cu 2+] = 0. 3 M and [Fe 2+] = 0. 1 M

Concentration and Ecell (cont. ) • First, need to identify the 1/2 cells Fe(s) + Cu 2+(aq) Fe 2+(aq) + Cu(s) Cu 2+(aq) + 2 e- Cu(s) E° 1/2 = 0. 34 V Fe 2+(aq) + 2 e- Fe(s) E° 1/2 = -0. 44 V Fe(s) + Cu 2+(aq) Fe 2+(aq) + 2 e- E° 1/2 = +0. 44 V Fe 2+(aq) + Cu(s) E°cell = +0. 78 V

Concentration and Ecell (cont. ) • Now, calculate Ecell Fe(s) + Cu 2+(aq) Fe 2+(aq) + Cu(s) E°cell = +0. 78 V Ecell = E°cell - (0. 0591/n)log(Q) 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 [Cu 2+] = 0. 3 M, what [Fe 2+] is needed so that Ecell = 0. 76 V? Fe(s) + Cu 2+(aq) Fe 2+(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) + Cu 2+(aq) Fe 2+(aq) + Cu(s) 4. 7 = Q [Fe 2+] = 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. ) Anode: Ag Ag+ + e- E 1/2 = ? V Cathode: Ag+ + e- Ag E 1/2 = 0. 80 V 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- 0 2 Fe 2+ + 2 e- Fe 2 e- transferred…n = 2 anode cathode Ecell = -(0. 0296)log(. 1) = 0. 0296 V