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FUNDAMENTALS OF ELECTROCHEMISTRY ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT DEALS WITH THE RELATIONSHIP BETWEEN EL

REDOX REACTIONS. FUNDAMENTALS OF ELECTROCHEMISTRY ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT DEALS WITH THE RELATIONSHIP BETWEEN ELECTRICITY AND CHEMICAL REACTIONS. PRODUCTION OF ELECTRICAL CURRENT BY CHEMICAL REACTIONS (Batteries, Fuel cells )

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FUNDAMENTALS OF ELECTROCHEMISTRY ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT DEALS WITH THE RELATIONSHIP BETWEEN EL

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  1. REDOX REACTIONS • FUNDAMENTALS OF ELECTROCHEMISTRY • ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT • DEALS WITH THE RELATIONSHIP BETWEEN ELECTRICITY AND • CHEMICAL REACTIONS. • PRODUCTION OF ELECTRICAL CURRENT BY CHEMICAL REACTIONS (Batteries, Fuel cells) • CHEMICAL CHANGES PRODUCED BY ELECTRIC CURRENT (Electrolysis, Electroplating and refining of metals) • ELECTRIC CURRENT = Transfer of chargeper unit time • BIOELECTROCHEMISTRY – study of electron transfer in biological regulations of organisms

  2. BASIC CONCEPTS • A redox reaction involves transfer of electrons from one species to another. • Oxidation: loss electrons - Reducing agent • Reduction : gain electrons - Oxidizing agent

  3. Electric charge (q) = n x F • = Coulombs • F – Faraday constant = 96 485.3415 C/mol of e- • Electric current (I) – is the quantity of charge flowing each second through the circuit. Unit – Amperes (A) • Eg. Calculate the mass of aluminum produced in 1 hour by electrolysis of molten AlCl3 if the electrical current is 20 A. • (Answer: 33 g)

  4. Potential difference (E) between 2 points is the work needed when moving an electric charge from 1 point to another, unit is Volt • Work = E. q OR J = CV • 1 Joule is the energy gained or lost when 1 coulomb of charge moves between points whose potentials differ by 1V • The greater the potential difference between 2 points the stronger will be the “push” on a charged particle travelling between those points. A 12V battery pushes e- 8X harder than a 1.5V dry cell

  5. The free energy of change, ∆G, represents the maximum electrical work that can be done by the reaction on its surroundings. • Work done on surroundings = ∆G = - work = -E.q • ∆G = - nFE (-ve ∆G = spontaneous rxn) • Ohm’s Law, I = E/R • Unit of resistance is ohms or Greek symbol Ω (omega) • Power , P = work/time • = E.q = E. q = E. I = I2. R • s s

  6. BALANCING REDOX REACTIONS • In an acidic medium • In a basic medium Split the reaction into 2 components or half reactions by identifying which species are oxidised and which ones are reduced. Add electrons appropriately to match the change in oxidation state of each element • Introduce H2O to balance the oxygen atoms • Introduce H+ to balance the charges as well as the hydrogen atoms • Multiply the half reactions so as to have an equal no. of e- and add the half reactions Whichever case, balance for the acid (H+) and then for basic media add OH- to the side where there is H+ to eliminate it as a H2O molecule (this is a 1:1 reaction)

  7. STANDARD POTENTIALS The voltmeter tells how much work is done by e- flowing from one side to the other, +ve V means e- flow into negative terminal LHS – negative terminal Reference electrode RHS connected to the positive terminal Line notation for cell

  8. STANDARD REDUCTION POTENTIAL (Eo) for each half cell is measured or setup by the above experiment. • ‘Standard’ means the activities ( A ) of all species are unity. • The half reaction of interest is : 2Ag+ + 2e- 2Ag(s) • AAg+ = 1 by definition activity of Ag (s) = unity • Standard Hydrogen Electrode (SHE), consists of a catalytic Pt surface in contact with an acid solution H+ (aq,1M), AH+ =1 • By convention the LHS electrode(Pt) is attached to the negative terminal of the potentiometer • equilibrium rxn at SHE : 2H+ (aq, A = 1) + 2e- H2 (g, A = 1) • half reaction – always written as reduction reactions

  9. BY INTERNATIONAL AGREEMENT THE SHE IS ARBITARILY ASSIGNED Eo = 0.00V at 25oC. • E0cell = + 0.799V • E0cell = Eored (cathode) – Eored (anode) or • E0cell = RHS electrode potential – LHS electrode potential E0cell = E0red (reduction process) – E0red (oxidation process) • A positive E0 = spontaneous process • A negative E0 = non spontaneous process • Sketch the cell construction : SHE ll Cd2+(aq, A=1) l Cd(s) The half reaction with a more positive E0 is more reduced and that with a less positive E0 is less reduced or more oxidised

  10. b RT A = - o B E E ln a nF A A NERNST EQUATION Le Chatelier`s principle tells us that conc. of reactants or products drives rxn to the right or left respectively. The net driving force is expressed by the NERNST EQUATION For the half reaction : a A + ne-↔ b B Nernst Equation: for the half cell potential, E Q = Reaction quotient

  11. where Eo = standard reduction potential ( AA = AB = 1) Ai = activity of species i R = gas constant ( 8.314 J/K.mol) T = Temperature (K) N = number of electrons in the half reaction F = Faraday constant (9.6485 x 104C/mol) [concentration inKc can be replaced by activities to account for ionic strength ] Ac = [C] γc γ = activity co efficient – measures the deviation from ideality ,γ = 1 behavior is ideal, low ionic strength

  12. Pure solids and liquids are omitted from Q, because their activities are (close to) unity. Concentration mol/L and pressure in bars When all activities are unity Q = 1, ln Q = 0 then E = Eo Log form of the Nernst equation, at 25oC T = 298K

  13. NERNST EQUATION FOR A COMPLETE CELL E = E+ - E- Find the voltage for the cell if the right half cell contains 0.50M AgNO3 (aq) and the left half cell 0.010M Cd(NO3)2 (aq).

  14. STEP 1 Right electrode : 2 Ag+ + 2e-↔ 2Ag(s) Eo = 0.799V Left electrode : Cd2+ + 2e-↔ Cd(s) Eo = - 0.402V STEP 2 Nernst equation for the right electrode = 0.781V STEP 3 Nernst equation for the left electrode = - 0.461V STEP 4 Cell Voltage E = E+ - E- = 0.781 - (-0.461) = + 1.242V STEP 5 Net cell reaction: Eqn Right electrode – Eqn Left electrode Cd (s) + 2 Ag+↔ Cd2+ + 2 Ag (s)

  15. Note : Multiplying a half reaction by a number does not change E0 nor E To figure half cell reactions look for the element in two different oxidations states Eo and the Equilibrium Constant A galvanic cell produces electricity because the cell is not at equilibrium. Relating E to the reaction quotient Q : Consider: Right electrode: aA + ne-↔ cC Eo+ Left electrode: dD + ne- ↔ bB Eo-

  16. The Nernst equation will be: E = E+ - E- = = E0 Q = When the cell is at equilibrium E = 0 and Q = K Eo can be calculated and used to find K for 2 half reactions!

  17. Question 10 in Tutorial on Complexation rxns Use the following standard-state cell potentials to calculate the complex formation equilibrium constant for the Zn(NH3)42+ complex ion. Zn(NH3)42+ + 2e- ⇌ Zn + 4NH3Eored = -1.04 V Zn2+ + 2e- ⇌ Zn Eored = - 0.7628 V SOLUTION: E0cell = E0red (reduction process) – E0red (oxidation process) (i) Zn + 4NH3⇌ Zn(NH3)42+ + 2e-E0 = + 1.04 V (ii) Zn2+ + 2e-⇌Zn E0 = - 0.7628 V (i) + (ii): Zn2+ + 4NH3⇌Zn(NH3)42+E0 = + 0.28 V =10 (2)(0.28)/0.05916= 2.92 x 109

  18. REDOX TITRATIONS • Theory of redox titrations and common titrants. • A redox titration is based on an oxidation- reduction reaction between an analyte and titrant. • Environmental and biological analytes can be measured by redox titrations. Other analytes include laser and superconductor materials. • REDOX TITRATION CURVE • Consider the potentiometric titration of Fe(II) and cerium (IV). • Titration rxn: Ce4+ + Fe2+ Ce3+ + Fe3+ (1) • Each mole of Ce4+ ion oxidizes 1 mole of ferrous ion, the titration creates a mixture of Ce4+, Fe2+, Ce3+ and Fe3+ions.

  19. e- flow from the anode to the cathode, the circuit measures the potential for Fe3+ /Ce4+ reduction at the Pt surface by e- from the calomel electrode

  20. Calomel reference electrode : 2Hg(l) + 2 Cl- Hg2Cl2 (s) + 2e- Pt indicator electrode: 2 rxns coming to equilibrium: Fe3+ + e-↔ Fe2+ Eo = + 0.767V (2) Ce4+ + e- ↔ Ce3+ Eo = + 1.70V (3) Cell rxns 2Fe3+ + 2Hg(l) + 2Cl- ↔ Fe2+ + Hg2Cl2 (s)(4) 2Ce4+ + 2Hg(l) + 2Cl- ↔ Ce3+ + Hg2Cl2 (s) (5) At equilibrium the potential driving rxns 1 and 2 must be the same. THE CELL RXNS ARE NOT THE SAME AS THE TITRATION RXN! The titration reaction goes to completion and is an oxidation of Fe2+ and reduction of Ce4+ Cell rxns proceed to negligible extent. The cell is used to measure activities, not to change them.

  21. HOW CELL VOLTAGE CHANGES AS Fe2+ IS TITRATED WITH Ce4+ REGION 1 : BEFORE THE EQUIVALENCE POINT As each aliquot of Ce4+ is added , it is consumed (eqn 1) and creates an equal number of moles of Ce3+ and Fe3+ Prior to the equivalence point excess unreacted Fe remains in solution Since the amounts of Fe2+ and Fe3+ are known, cell voltage can be calculated from eqn 2 rather than 3 E = E+ - E- (calomel) When volume titrant is half the equivalence point ( V = 1/2Ve) the concentration of Fe2+ and Fe3+ are equal, thus E+ = Eo For an acid-base titration pH = pKa when V = 1/2Ve

  22. SHAPES OF TITRATION CURVES E ~ Eo(Ce4+I Ce3+) - 0.241V = 1.46V Notsymmetric about the equivalence point – 2:1 1:1stoichiometry symmetric about equivalence point & same curve for diluted sample

  23. REGION 2 : AT THE EQUIVALENCE POINT All cerium is in the form of Ce3+ [Ce3+] = [Fe3+] Thus the equilibrium form of eqn 1 Ce 4+ + Fe2+↔Ce3+ + Fe3+ If a little Fe3+ goes back to Fe2+, an equal no. of moles Ce4+ must be made and [Ce4+] = [Fe2+] Eqns 2 and 3 are in equilibrium at the Pt electrode, it is convenient to use both these eqns to calculate the cell voltage At equilibrium in a redox titration, Ecell = (Eo1 + Eo2)/2

  24. REGION 3 : AFTER THE EQUIVALENCE POINT Almost all the iron atoms are Fe3+, the moles of Ce3+ = moles Fe3+, and there is a known excess of unreacted Ce4+. We know [Ce3+] and [Ce4+] and so we can use eqn 3 to calc E .

  25. FINDING THE ENDPOINT As in acid-base titration, indicators and electrodes are commonly used to find the endpoints of a redox titration. REDOX INDICATORS A redox indicator is a compound that changes colour when it goes from oxidized to a reduced state, eg. Ferroin changes from pale blue to red. By writing the Nernst equation we can predict the potential range over which the indicator will change : In (oxidised) + ne- ↔ In (reduced) (6)

  26. As with acid base indicators, the colour of In (reduced) will be observed when: [In( reduced)] 10 [In(oxidised)] 1 And the colour of the In(oxidised) will be observed when: [In( reduced)] 1 [In(oxidised)] 10 Using these 2 quotients in eqn 6, tells us the colour range will occur over the range Eo= 1.147V, we expect the colour change to occur ~ 1.088 – 1.206V wrt SHE

  27. STARCH IODINE COMPLEX Many analytical procedures use redox titrations involving iodine. Starch is used as the indicator, since it forms an intense blue complex with iodine. Starch is not a redox indicator because it responds to I2, not to a change in potential. Starch is readily biodegradable and must be freshly prepared. ADJUSTMENT OF ANALYTE OXIDATION STATES Before titration we adjust the oxidation state of the analyte, eg Mn2+ can be pre-oxidised to MnO4- and then titrated with Fe2+ Pre-adjustment must be quantitative and all excess reagent must be destroyed. Pre-oxidation : Persulphate S2O82- is a powerful oxidant that requires Ag+ as a catalyst: S2 O82-+ Ag+ SO42- + SO4- + Ag2+ Two powerful oxidants

  28. Excess reagent is destroyed after by boiling the solution after oxidation is complete 2S2O82- + 2 H2O boiling 4SO42- + O2 + 4H+ H2O2 is a good oxidant in basic solution and reductant in acidic solution. The excess spontaneously disproportionate in boiling water. H2O2 H2O + O2 PRE-REDUCTION Stannous & chromous chloride, SO2 , H2S are used to pre-reduce analytes to a lower oxidation state. An important pre-reduction technique uses a packed column to pre-reduce analyte to a lower oxidation state (analyte is drawn by suction).

  29. Jones reductor, which contains Zn coated with Zn amalgam. Zn is a powerful reducing agent (Eo = -0.764V) making the Jones reductor unselective, other species eg, Cr3+ are reduced and may interfere with the titration analysis. OXIDATION WITH POTASSIUM PERMANGANATE KMnO4 is a strong oxidant with an intense violet colour. In strongly acidic solutions (pH< 1) it is reduced to colourless Mn2+ (manganous). In neutral or alkaline the product is a brown solid – MnO2 In strongly alkaline solution, green manganate MnO42- is produced.

  30. KMnO4 is not a primary standard, and can contain traces of MnO2, thus it must be standardized with pure Fe wire or sodium oxalate (pink end-point) for greater accuracy, The KMnO4 solution is unstable : 4MnO4- + 2H2O 4MnO2 + 3O2 + 4OH-

  31. OXIDATION WITH Ce4+ Reduction of Ce4+ (yellow) to Ce3+ (colorless) can be used in place of KMnO4. Ce4+ is used for the quantitative determination of malonic acid as well as alcohol, ketones and carboxylic acids. The primary standard is prepared by dissolving the salt in 1M H2SO4 and is stable indefinitely. OXIDATION WITH K2Cr2O7 Powerful oxidant – in acidic solution orange dichromate iron is reduced to green chromic ion. In 1M HCl the formal potential is 1.00V and 2M H2SO4 it is 1.11V, thus less powerful than Ce 4+ and MnO4-

  32. Stable primary standard that is employed to determine Fe2+ Also used in environmental analysis of oxygen demand. COD or chemical oxygen demand is defined as the oxygen that is equivalent to the Cr2O72- consumed by the oxidation of organics in water. METHODS INVOLVING IODINE

  33. When a reducing analyte is titrated with iodine to produce I- the method is called Iodimetry (titration with I3-). In the iodimetric determination of vitamin C-starch is added to give an intense blue end-point. Iodometry – oxidizing analyte added to I- to produce I2 which is then titrated with thiosulfate standard, starch is added only before the endpoint. When we speak of using iodine as a titrant we mean a solution of I2 plus excess I- I2 (aq) + I- I3- K= 7x102 A 0.05M solution of I3- is prepared by dissolving0.12M KI plus 0.05M I2 in water. Reducing agent + I3- 3I- Oxidizing agent+ 3I- I3-

  34. Precipitation Reactions Gravimetric Analysis: Solid product formed Relatively insoluble Easy to filter High purity Known Chemical composition Precipitation Conditions: Particle Size Small Particles: Clog & pass through filter paper Large Particles: Less surface area for attachment of foreign particles.

  35. 2. Particle Growth 1. Nucleation Crystallization Molecules form small Aggregates randomly Addition of more molecules to a nucleus. More solute than should be present at equilibrium. Supersaturated Solution: Supersaturated Solution: Nucleation faster; Suspension (colloid) Formed. Nucleation slower, larger particles formed. Less Supersaturated Solution:

  36. How to promote Crystal Growth 1. Raise the temperature Increase solubility Decrease supersaturation 2. Precipitant added slowly with vigorous stirring. 3. Keep low concentrations of precipitant and analyte (large solution volume).

  37. Homogeneous Precipitation Precipitant generated slowly by a chemical reaction pH gradually increases Large particle size

  38. Net +ve charge on colloidal particle because of adsorbedAg+

  39. Precipitation in the Presence of an Electrolyte Consider titration of Ag+ with Cl- in the presence of 0.1 M HNO3. Colloidal particles of ppt: Surface is +vely charged Adsorption of excess Ag+ on surface (exposed Cl-) Colloidal particles need enough kinetic energy to collide and coagulate. Addition of electrolyte (0.1 M HNO3) causes neutralisation of the surface charges. Decrease in ionic atmosphere (less electrostatic repulsion)

  40. Digestion and Purity Digestion: Period of standing in hot mother liquor. Promotion of recrystallisation Crystal particle size increases and expulsion of impurities. Purity: Adsorbed impurities: Surface-bound Inclusions & Occlusions Within the crystal Absorbed impurities: Inclusion: Impurity ions occupying crystal lattice sites. Occlusion: Pockets of impurities trapped within a growing crystal.

  41. Coprecipitation: Adsorption, Inclusion and Occlusion Colloidal precipitates: Large surface area BaSO4; Al(OH)3; and Fe(OH)3 How to Minimise Coprecipitation: 1.Wash mother liquor, redissolve, and reprecipitate. 2. Addition of a masking agent: Gravimetric analysis of Be2+, Mg2+, Ca2+, or Ba2+ with N-p-chlorophenylcinnamohydroxamic acid. Impurities are Ag+, Mn2+, Zn2+, Cd2+, Hg2+, Fe2+, and Ga2+. Add complexing KCN.

  42. Ca 2+ + 2RH  CaR2(s) + 2H+ Analyte Precipitate Mn2+ + 6CN- Mn(CN)64- Impurity Masking agent Stays in solution Postprecipitation: Collection of impurities on ppt during Digestion: A supersaturated impurity e.g., MgC2O4 on CaC2O4. Peptization: Breaking up of charged solid particles when ppt is washed with water. AgCl is washed with volatile electrolyte (0.1 M HNO3). Other electrolytes: HCl; NH4NO3; and (NH4)2CO3.

  43. Product Composition Difficult to weigh accurately Hygroscopic substances: Some ppts: Variable water quantity as water of Crystallisation. Drying Change final composition by ignition:

  44. Thermogravimetric Analysis Heating a substance and measuring its mass as a function of temperature.

  45. Example In the determination of magnesium in a sample, 0.352 g of this sample is dissolved and precipitated as Mg(NH4)PO4.6H2O. The precipitate is washed and filtered. The precipitate is then ignited at for 1 hour 1100 oC and weighed as Mg2P2O7.The mass of Mg2P2O7 is 0.2168 g. Calculate the percentage of magnesium in the sample.

  46. Solution: Relative atomic mass Of Mg The gravimetric factor is: FM of Mg2P2O7 Note: 2 mol Mg2+ in 1 mol Mg2P2O7.

  47. Mass of Mg 2+ = 0.0471 g = 13.45 %

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