Chapter 14 Fundamentals of Electrochemistry Chapter 15 Electrodes and PotentiometryChapter 16: Redox Titration I. Oxidation/Reduction Reactions Reactions in which electrons are transferred from one reactant to another. Example I-1 Cu2+ + Zn (s) <====> Cu (s) + Zn2+ Cu2+ has undergone reduction and is reduced. Oxidizing reagent: Cu2+. Reducing agent: Zn. Zn has undergone oxidation and is oxidized. Oxidizing reagent A reagent which tends to remove electrons from another reactant. Its oxidation number becomes less positive after it reacts. In the above reactant Cu2+ is the oxidizing agent. It has undergone reduction. +2e Cu2+ + Zn (s) <=> Cu (s) + Zn2+ -2e +2e Cu2+ + Zn (s) <=> Cu (s) + Zn2+ -2e .Reducing agent A reagent which tends to give up electrons to another reactant. In the above reaction Zn is the reducing agent. It has undergone oxidation. Its oxidation number becomes more positive after it reacts. Relativity of Oxidation and Reduction Agents: Reagents can be oxidizing agents in one reaction and reducing agent in another depending on the ability of the other reactants of remove or give up electrons. Example I-2: (a) Cu2+ + Zn (s) <=> Cu (s) + Zn2+ (b) Cu(s) + 2 Ag+ <=> Cu2+ + 2 Ag(s) . Example I-3: Cu2+ + 2 e ------> Cu (s) Zn (s) --------> Zn2+ + 2 e reduction of Cu2+ oxidation of Zn .Half-reactions Any oxidation/reduction reaction can be split into two half-reactions: one gains electrons and the other loses electrons. The number of atoms of each element and the net charge must be the same on both sides of the equation. The number of atoms of each element in each side (Mass Balance) Net charge of each side (Charge Balance) .The rules for balancing half-reactions The same as those for balancing ordinary reactions. Problem 15-7 -3e (b) Cr (s) + Ag+ => Cr3+ + Ag +e Cr (s) + 3 Ag+ Cr3+ + 3Ag The number of atoms of each element in each side (Mass Balance) Net charge of each side (Charge Balance) . MnO4- +7 + H2SO3 5e +4 Mn2+ -2e + SO42- +6 2MnO4.+ 4H++3H2O The number of atoms of each element in each side (Mass Balance) Net charge of each side (Charge Balance) .+ 5H2SO3<=> 2Mn2+ + 5SO42.+ 5H2SO3 2Mn2+ + 5SO42- + 3H2O 2MnO4. Example I-4: balance the following reaction IO3. H2O as needed to balance the equation because these species will always be present in an aqueous solution.+ 10e +12 H+ ----> I2 + 6 H2O [Eq.-----> I2 + H2O Step-1: Write and balance two half-reactions (a) Reduction: IO3. 1] We can add H+. OH-. .+ 5e -----> I2 2IO3.+ I. (b) Oxidation: 2I.+ 6H+ <---> 3 I2 + 3 H2O Step-3: Check the reaction is both mass.+ 5 I. 1] + 5x [Eq.+ 10 I. 2] 2IO3.and charge-balanced. 2] Step-2: Combine two half-reactions and balance charge and mass: [Eq..2e -----> I2 [Eq. .+12 H+ <--> 6 I2 + 6 H2O IO3. Any Questions? . Comparison of Oxidation/Reduction (Redox) Reactions with Acid/Base Reactions Just as with reactions of weak acids with base to form a conjugate base and acid. the reaction of an oxidant with a reductant forms a 'conjugate reductant' and 'conjugate oxidant': Example I-4: 2Fe3+(ox1) + Sn2+(red2)<-->2Fe2+(red1) + Sn4+ (ox2) .II. .III. Electrochemical Cells Electrochemical cells consist of electrodes immersed in electrolyte solution and frequently connected by a salt bridge. the cathode (or anode) electrode is usually made of the metal otherwise an inert material. Metals: Inert materials Pt Au Hg Glassy Carbon . is normally used for the electrode. such as Pt.Electrodes: When the reaction involves the metallic form of the metal. (b)Electrolytic cells: (Fig.Types of electrochemical cells (a)Galvanic or Voltaic cells: (Fig. there electrode reactions proceed spontaneously and are accompanied by current flow when connected via an external conductor. 14-a) store electrical energy. . 14-b) require external sources of electrical energy for operation. (a) Galvanic or Voltaic cells Figure 14-a . Currents in Electrochemical Cells . Example: Cu (s) -----> Cu2+ + 2e2 Ag+ + Cu (s) <===> 2 Ag (s) + Cu2+ .Electrodes (Figure 14-a) Cathode: reduction occurs here.----> Ag (s) Anode: oxidation occurs here. Example: Ag+ + e. (b)Electrolytic cells: Figure 14-b . ----> Cu (s) Anode: oxidation occurs here. Example: Cu2+ + 2e. + Example: Ag (s) -----> Ag + e 2 Ag (s) + Cu2+ <===> 2 Ag+ + Cu (s) .Electrodes (Figure 14-b) Cathode: reduction occurs here. 14-a): Galvanic Cell 2 Ag+ + Cu (s) <===> 2 Ag (s) + Cu2+ For the cell described (Fig.For the cell described (Fig. 14-b): Electrolytic Cell 2 Ag (s) + Cu2+ <===> 2 Ag+ + Cu (s) . Chemical Reversible Reactions are those which when forced to proceed in the opposite direction (add electrical energy by reversing current) results in the original reactants (no new reactions appear) (e. nonrechargeable batteries). do not result in the original reactants but form new products (e.. .. rechargeable batteries) Chemical irreversible Reaction are those which when forced to proceed in the opposite (nonspontaneous) direction.g.g. Any Questions? . The Vertical line represents the interface.0200M) Ag (0. The anode is always displayed on the left.0200M) Ag 2+ + By convention. The cathode is always displayed on the right.IV. Schematic Representation of Cells Cu Cu (0. . (a)Galvanic or Voltaic cells Figure 14-a . ----> Ag (s) Anode: oxidation occurs here. Example: Cu (s) -----> Cu2+ + 2e2 Ag+ + Cu (s) <===> 2 Ag (s) + Cu2+ Cu Cu 2+ (0.Electrodes (Figure 14-a) Cathode: reduction occurs here. Example: Ag+ + e.0200 M) Ag .0200 M) Ag + (0. (b)Electrolytic cells: Figure 14-b . 0200 M) Cu + 2+ . Example: Ag (s) -----> Ag+ + e2 Ag (s) + Cu2+ <===> 2 Ag+ + Cu (s) Ag Ag (0.Electrodes (Figure 14-b) Cathode: reduction occurs here.----> Cu (s) Anode: oxidation occurs here.0200 M) Cu (0. Example: Cu2+ + 2e. Any Questions? . K+. Anions and cations carry electricity within the cell (Ag+. Ionic conduction of the electrolyte solution is coupled to the electronic conduction in the electrode by the electrode reactions at the interface. and the corresponding anions). Currents Within a Cell: 1.VA. Currents carry electricity within the electrodes and external conductor. Cu2+. . 3. 2. Currents in Electrochemical Cells . Cell Potential & Electrode Potential . Cell Potential & Electrode Potential . VB. Electrode Potential This parameter is a measure of the tendency of gaining or losing electrons. . Standard Electrode Potential (E0) of the half-reaction is defined as its electrode potential when the activities of all reactants and products are unity.↔ cC + dD + … . aA + bB + … + ne.VC. H 2 (p = 1.Standard Hydrogen Reference Electrode (SHE) Pt.-----> H2 (g) Define this Electrode potential (SHE) = 0 Electrode potential measurements are comparative (relative) to SHE. + .00 atm) H (a H + = 1.00 M) 2 H+ + 2 e. H2 (P=1.00 E0 SHE = 0.Standard Hydrogen Reference Electrode (SHE) Pt.00 atm) aH+ = 1.000 V . Electrode Potential Measurements E0 SHE = 0.799 V .000 V E0 Ag+/Ag = 0. 000 V E0 AgCl/Ag = 0.222 V .Electrode Potential Measurements E0 SHE = 0. Electrode Potential Measurements E0 SHE = 0.000 V E0 Cd2+/Cd = -0.403 V . I . Thus. when Eo of a half-cell is negative with respect to the SHE. .By convention. we treat the SHE as the anode. SHE acts as cathode. we write all half-cell reactions as reduction potentials vs the SHE (Table 1). that electrode is a better anode than the SHE. Hence. When written as the reduction potential. Any Questions? . .314 J/ (K mol) T : temperature in kelvins n : number of moles of electrons that appear in the half-reaction for the electrode process as it has been written F : the faraday constant = 96.<=> cC + dD + … 0. which is characteristic constant for each half-reaction R : the gas constant 8... The Effect of Concentration on Electrode Potential: The Nernst Equation.0592 [C]c [D]d ..485 C (coulombs) Ln : the nature logarithm = 2.303 log . Ln a b = E0 − Log a b E = E0 − nF [A] [B] ..VI. RT [C]c [D]d ... aA + bB + … + ne. E0 : the Standard electrode potential. n [A] [B] .. .. 0.. n Example I-5: .[C]c [D]d .0592 Log E = E0 − [A]a [B]b .. 0592 1 0 =E − E=E − Log Log 2+ n [Zn ] 2 [Zn 2+ ] 1 0.799V 2 0.0600 M Zn2+ + 2e.0600 M Zn(NO3)2 at 25 0C Zn(NO3)2 is a strong electrolyte and complete dissociate.0600 .0592 = −0.763 V 0.<==> Zn 0 E0 = -0.Problem 15-14: Calculate the potential of a Example I-6a: zinc electrode immersed in (a) 0.0592 1 0. Hence: [Zn2+] = 0.763 − Log = −0. 0592 1 0 E=E − Log =E − Log 2+ n [Zn ] 2 [Zn 2+ ] 0 1 0.0x10-16 [Zn2+] = 3. Ksp = 3.0592 1 0.01000 M NaOH and Saturated with Zn(OH)2 at 25 0C Zn(OH)2 has very limit solubility.0592 = −0.0x10 .0x10-16 Ksp = [Zn2+][OH-]2 =[Zn2+] (0.0x10-12 0.Example I-6b: Problem 15-14: Calculate the potential of a zinc electrode immersed in (b) in 0.104V -12 2 3.01000)2 = 3.763 − Log = −1. 0592 1 0.Example 15-3: calculate the electrode I-7: Example potential of a silver electrode immersed in a 0.82 x10 −10 [ Ag + ] = = = 3 .64 x10 − 9 M [ Cl .0592 1 Log = E0 + − Log Ag n [Ag + ] 1 3.299 V .799V Ag 1 .64x10 -9 = 0 .0500 M solution of NaCl using (a) E 0 + / Ag = 0.0500 E = E0 + − Ag 0.] 0 .799V Ag at 25 0C (b) E 0 = 0.82x10 -10 K sp E 0 + = 0.222V AgCl + - (a) Ag + e ----> Ag (s) K sp = [Ag + ][Cl -] = 1. (b) AgCl (s) + e ----> Ag (s) + Cl - - E 0 = 0.222V AgCl [Cl-] = 0.0500 M 0.0592 [Cl - ] 0.0592 = E0 − Log (0.0500) = 0.299V Log E = E0 − AgCl AgCl 1 1 n Any Questions? VII. The Standard Electrode Potential (Eo) Definition: The Standard Electrode Potential (Eo) of a half-cell is defined as its electrode potential when the activities of all reactants and products are unity. Examples: Pt, H 2 (p = 1.00 atm) H + (a H + = 1.00 M) Ag + (a Ag+ = 1.00 M) Ag( s ) E 0 + = +0.799V Ag 2 Ag+ + H2 (g) <==> 2 Ag (s) + Pt, H 2 (p = 1.00 atm) H + (a H + = 1.00 M) Cd 2+ (a Cd 2+ = 1.00 M) Cd (s) 2H+ E 0 2+ = −0.403V Cd Cd (s) + 2H+ <==> H2 (g) + Cd2+ 0592 [C]c [D]d ... includes the standard potential and some activity coefficients. Definition of Formal Potentials Formal potentials (E0’) is the potential of the half-cell (vs SHE) measured under conditions (a) the ratio of analytical concentrations of reactants and products as they appear in the Nernst equation is exactly unity... .VIII. (b) (c) 0. E = E0 − Log a b n [A] [B] . the concentrations of other species in the system are all carefully specified. Formal Potentials IX. The Thermodynamic Potential of Electrochemical Cells Ecell = Ecathode - Eanode aA + bB + … <====> cC + dD + … E cell 0.0592 [C]c [D]d ... Log a b = E0 − cell n [A] [B] ... Cell Potential & Electrode Potential Cell Potential & Electrode Potential . 0592 [C]c [D]d ..X.. Oxidation/Reduction Reactions in Electrochemical Cells For the react with Ag+ with Cu metal 2 Ag+ + Cu (s) <====> 2 Ag (s) + Cu2+ The equilibrium constant expression is [Cu ] K= = 4. Log a b = E0 − cell n [A] [B] . 2+ .1x1015 + 2 [Ag ] E cell 0... 0200 M) Ag + (0.0592 1 − Log = 0.0592 1 − Log = 0.799 − Log = 0.337 V 0.Example 16-1: Example I-8: Calculate the thermodynamics potential of the following cell Cu Cu 2+ (0.6984 V + 2 + 2 n [Ag ] 2 [Ag ] Anode: E anode = E 0 Cu/Cu 2 + Cu (s) <==> Cu2+ + 2e- E0 Cu2+/Cu = 0.6984 − 0.<=> 2 Ag (s) E0 Ag+/Ag = 0.0200 M) Ag Cathode: E cathode = E 0 Ag/Ag + 2 Ag+ + 2e.412V .799 V 0.0592 1 0.337 − Log = 0.2867 V 2+ 2+ n [Cu ] 2 [Cu ] E cell = E cathode − E anode = 0.2867 = 0.0592 1 0. 0592 [Cu 2+ ] = E0 − = E 0 + − E 0 2+ − Log Log cell + 2 Ag/Ag Cu/cu [Ag + ]2 n [Ag ] n [Cu 2+ ] 0.0592 [Cu 2+ ] 0.462 − = +0.<====> 2 Ag (s) Cu (s) <==> Cu2+ + 2e- Or: 2 Ag+ + Cu (s) <====> 2 Ag (s) + Cu2+ g E cell ( ) g( ) 0.Cathode: Anode: 2 Ag+ + 2e.412V Log + 2 [Ag ] 2 .0592 = 0. 0200 M) Cu Cathode: Cu2+ + 2e.0200 M) Cu 2+ (0.412V .0592 1 Log = 0.0592 1 0.6984V Log Log + 2 + 2 [Ag ] 2 [Ag ] n E cell = E cathode − E anode = 0.2867 − 0.2867 V 2+ 2+ [Cu ] n [Cu ] 2 Anode: 2Ag (s) <==> 2Ag+ + 2e E anode = E 0 + /Ag − Ag E0 Ag+/Ag = 0.0592 1 0.0592 = 0.337 − Log = 0.6984 = −0.<==> Cu (s) E0 Cu2+/Cu = 0.Example 16-2: I-9: Example Calculate the thermodynamics potential of the following cell Ag Ag + (0.799 − = 0.799 V 1 0.337 V E cathode = E 0 2+ /Cu − Cu 0. Cathode: Anode: Cu2+ + 2e.0592 [Ag + ]2 Log = −0.412V 2+ n [Cu ] .<==> Cu (s) 2Ag (s) <==> 2Ag+ + 2e Or: 2 Ag (s) + Cu2+ <====> 2 Ag+ + Cu (s) E cell 0.0592 [Ag + ]2 Log Log = E0 − = E 0 2+ /cu − E 0 + /Ag − cell 2+ Cu Ag n [Cu ] n [Cu 2+ ] 0.462 − = −0.0592 [Ag + ]2 0. Hence.Conclusion of Example I-8 & 9: If thermodynamic potential of Electrochemical Cell is positive. If thermodynamic potential of Electrochemical Cell is negative. the cell is a galvanic cell. It requires external energy to force the reactions to take place. Hence. the tendency to proceed reactions spontaneously. . the reaction will not proceed spontaneously. the cell is an electrolytic cell. Any Questions? . Eanode = 0 Ecathode = Eanode .XI. the thermodynamic potential of the electrochemical cell becomes zero: Ecell = Ecathode . Calculation of Redox Equilibrium Constants At the equilibrium. XIA. the electrode potentials for all halfreactions in an oxidation/reduction system are equal. Electrode Potentials at Equilibrium At equilibrium. Eox1 = Eox2 = Eox3 = Eox4 =… =Ered1 =Ered2=… . 0592 = = Log log K eq b a ab [A red ] [Box ] ab .XIB.0592 0.<==> aBred At the equilibrium: EA = EB [A red ]b [Bred ]a 0.0592 0 E − = EB − Log Log b ab [A ox ] ab [Box ]a 0 A [A red ]b 0.0592 + E0 − E0 = − Log Log B A b [A ox ] ab [Box ]a ab [A ox ]b [Bred ]a 0. The Calculation of Equilibrium Constants A General Expression bAred + aBox <===> bAox + aBred By convention: bAox + ba e.<==> bAred aBox + ba e.0592 0.0592 [Bred ]a 0. XIB.0592 0 B 0 A [Eq1] . The Calculation of Equilibrium Constants A General Expression bAred + aBox <===> bAox + aBred ab(E − E ) log K eq = 0. 337 − 2+ [Cu ] 2 [Cu 2+ ] n 0 Cu 2+ /Cu At equilibrium: E 0 + /Ag − E 0 2+ /cu Ag Cu E0 Ecathode = Eanode 0.0592 = Log = Log K eq + 2 n [Ag ] n 0 + /Ag − E Cu 2 + /cu Ag K eq = 10 0.0592 [Cu 2+ ] 0.Example 16-8: Calculate the Keq for the reaction 2 Ag+ + Cu <===> 2 Ag + Cu2+ 0 Ag + /Ag E cathode = E E anode = E 1 0.1x1015 .0592 Log Log − = 0.0592 1 0.0592 − = 0.0592 1 0.0592 n = 4.799 − Log Log + 2 2 [Ag + ]2 n [Ag ] 1 0. Note: When calculating equilibrium expressions. . apply the Nernst Equation to the balanced half-cell reactions. Summary of Thermodynamic potential of Electrochemical Thermodynamic potential of Electrochemical Cell is a parameter of a measure of the tendency of the reaction to occur.00 V.412 V. . Replacement of Cu with Cd results in a higher cell potential (Ecell = 1. Ecell = 0.150 V) than that for Cu (Ecell = 0. As the reaction progresses.7 X 10-9 M.0300 M and [Ag+] = 2.412 V) because Cd is a better reducing agent than Cu. At original concentrations of [Cu2+] and [Ag+] = 0. the greater the tendency to proceed reactions spontaneously.0200 M. At equilibrium [Cu+2] = 0. Ecell --> 0. The more positive the potential. This sigmoidal-shaped curve similar to those obtained when p-functions are plotted against volume (mL) of titrant. Redox Titration Curves The plot of the electrode potential (Esystem) on yaxis versus the volume (VmL) of added titrant on x-axis.XII. Definition: . Redox Titration Curves . 0 M H2SO4 Fe2+ + Ce4+ <=> Fe3+ + Ce3+ This reaction is rapid and reversible and in equilibrium at all times during the titration Hence: E Ce 4+ /Ce 3+ = E Fe3+ /Fe 2+ = E system Esystem is the potential of the titration and is the value plotted on the y-axis. .Example XII-1: Derive the titration curve of 50.100 M Ce4+ in a medium that is 1.0500 M Fe2+ with 0.00 mL of 0. Potentiometric Titration . 100 .00 = = 25.0mL 0.Esystem represents the potential of the electrochemical cell: SHE||Ce4+.68V Fe 3+ 2+ We can follow either Equivalence point: E Ce 4+ /Ce3+ or E Fe 3+ /Fe 2+ CCe4+ VCe4+ = CFe2+ VFe2+ VCe4+ = CFe2+ VFe2+ CCe4+ 0. Fe3+. Ce3+.0500x50. Fe2+|Pt Ce Fe 4+ +e +e - <=> Ce <=> Fe 3+ E 0' 4+ /Ce3+ = 1.44V Ce E 0 ' 3+ /Fe2+ = 0. (a) At VCe = 0 mL . Before the equivalence point: absence of Fe3+ Hence: no sufficient information for Esystem 4+ . 00x 0.1000 2.64V 3+ [Fe ] n 50.00 55.00 + [Ce 4+ ] ≈ = M 50.0592 [Fe 2+ ] − Log = 0.00 x 0.500 4+ [Fe ] = − [Ce ] ≈ = M 50.00 x 0.00x 0. (b) At VCe = 5mL . pre-ep = E Fe 0' Fe3+ /Fe2 + 3+ /Fe 2+ E system = E Fe3+ /Fe2+ = E [Fe 2+ ] = 0.00 x 0.1000 50.00 + 5.00 55.1000 0.0500 − 5.00 + 5.00 5.excess of Fe2+ concentrations of Fe3+/Fe2+ are measurable.0500 − 5.00 50.00 + 5.00 50. Veq > VCe > 0mL : Before the equivalence point: 4+ 4+ Hence: Esystem.00 + 5.1000 5.00 3+ .00 x 0. 00mL .(c) VCe4+ = 25.0592 [Fe 2+ ] E eq = E Fe3+ /Fe2+ = E 0 ' 3+ /Fe2+ − Log Fe n [Fe3+ ] [Ce 3+ ] 0. at the equivalence point: no excess of Fe2+ [Ce4 + ]Vtotal = [Fe2 + ]Vtotal [Ce3+ ]Vtotal = [Fe3+ ]Vtotal Reaction takes place in one vessel.0592 E eq = E Ce4+ /Ce3+ = E 0 ' 4+ /Ce3+ − Log Ce n [Ce 4+ ] [Eq-1] + [Eq-2]: 2E eq = E 0' Fe3+ /Fe 2+ Eq − 1 Eq − 2 +E 0' Ce 4+ /Ce 3+ 0.0592 [Fe2+ ] [Ce3+ ] Log − = E 0 ' 3+ /Fe 2+ + E 0 ' 4+ /Ce3+ Fe Ce n [Fe3+ ] [Ce 4+ ] . Hence [Ce4+] = [Fe2+] [Ce3+] = [Fe3+] 0. 68 = = 1.E eq = E 0' Fe3+ /Fe 2+ +E 2 0' Ce 4+ /Ce 3+ 1.44 + 0.06V 2 Note: Eeq is affected only by the number of electrons transferred for each species and the standard potentials for the titrant and analyte. . 30V 4+ 1 0.010 2+ [Ce ] = + [Fe ] ≈ M 50.00x 0.44 − log = +1.10mL > Veq = 25.10 25.10 75. post-ep = 0' Ce 4+ /Ce 3+ E Ce 4+ /Ce3+ 0.00 + 25.10 75.1000 − 50.10x 0.00x 0.10 3+ .010 n [Ce ] 25.00 + 25.500 0.0592 [Ce3+ ] − Log = 1.500 2+ [Ce ] = − [Fe ] ≈ M 50. Hence E system = E Ce 4+ /Ce3+ = E 4+ Esystem.(d) VCe4+ = 25.1000 2.0592 2.0500 0.00mL After the equivalence point: excess of Ce4+ and concentration of Ce4+/Ce3+ are measurable. . Completeness of Reaction: The change in Esystem at the equivalence point becomes larger when the reaction is more complete (Keq is larger). See figure 16-4. Reactant Concentration: Relatively independent with concentration of reduced and oxidized species. 2.B. The Effect of System Variables on Redox Titration Curves 1. Completeness of Reaction . <==> Inred The general rule for visualization of an indicator holds: See color of Inox [In red ] 1 ≤ [In ox ] 10 [In red ] ≥ 10 [In ox ] See color of Inred . Inox + ne. Oxidation/Reduction Indicators A.XIII. General Redox Indicators: These indicator change color upon being oxidized or reduced. 0592 E=E ± n 0 In 0. T o o b s e r v e t h e c h a n g e . t h e t r a n s it io n p o t e n t ia l m u s t b e w it h in ± 0 . 0592 n o f E eq.118 n V in the equivalence point range (Table 16-3). color change occurs over a range of 0. T h e in d ic a to r is o x id ize d /re d u c e d a t th e e q u iv a le n c e p o in t p o te n tia l Thus. .Oxidation/Reduction Indicators The indicator chemistry is largely independent of the chemical nature of the analyte and titrant. . the starch is used as a specific indicator. Specific Indicators: These indicators change color based on the presence of specific species that are present in adequate concentrations at the equivalence point.B.as the reductant. Example: The titration using I2 as the oxidant or I. C. Potentiometric End Points We can directly observe Esystem by measuring the analyte solution as part of the electrochemical cell: reference electrode||analyte solution|Pt . Potentiometric Titration . Redox Titration Curves . cathode. standard potential. electrode potential Galvanic and electrolytic cells Chemical reversible and irreversible reactions Schematic representation of cells Thermodynamic potential of electrochemical cells SHE.Summary Basic concepts and terms: Oxidation and reduction agents and reactions Anode. electrode Half-cell reactions. formal standard potential Indicator and factors of titration curves Calculations Electrode potential (Nernst Equation) Thermodynamic potentials of electrochemical cell Keq Titration Curves . 2. C. 9. 10. 21. 8. 13 Before working on Homework. 2. 1. 26 16-A. D. 25 15-4.Homework 14-B. Practice with all examples that we discussed in the class and examples in the textbook!! . 6.