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2TABLE OF CONTENTSNO. CHAPTERS PAGE1. Acids and Bases 152. Electrochemistry 323. Reaction Kinetics 604. Alcohols 785. Phenols 856. Synthetic Polymers 917. Nitrogen Compounds 1018. Carbonyl Compounds 1159. Carboxylic Acid and its derivatives 123
3TABLE OF RELATIVE ATOMIC MASSESElement Symbol Atomic Number Relative AtomicHydrogen H 1 Mass 1.0Helium He 2 4.0Lithium Li 3 6.9Berylium Be 4 9.0Boron B 5 10.8Carbon C 6 12.0Nitrogen N 7 14.0Oxygen O 8 16.0Fluorine F 9 19.0Neon Ne 10 20.2Sodium Na 11 23.0Magnesium Mg 12 24.3Aluminium Al 13 27.0Silicon Si 14 28.1Phosphorus P 15 31.0Sulphur S 16 32.1Chlorine Cl 17 35.5Argon Ar 18 39.9Potassium K 19 39.1Calcium Ca 20 40.1Scandium Sc 21 45.0Titanium Ti 22 47.9Vanadium V 23 50.9Chromium Cr 24 52.0Manganese Mn 25 54.9Iron Fe 26 55.8Cobalt Co 27 58.9Nickel Ni 28 58.7Copper Cu 29 63.5Zinc Zn 30 65.4Gallium Ga 31 69.7Germanium Ge 32 72.6Selenium Se 34 79.0Bromine Br 35 79.9Krypton Kr 36 83.8Strontium Sr 38 87.6Silver Ag 47 107.9Cadmium Cd 48 112.4Tin Sn 50 118.7Iodine I 53 126.9Xenon Xe 54 131.3Barium Ba 56 137.3Platinum Pt 78 195.1Gold Au 79 197.0Mercury Hg 80 200.6Lead Pb 82 207.2
4LIST OF SELECTED CONSTANT VALUESIonisation constant for water at 25 oC, Kw =1.00 x 10-14 mol2 dm-6Molar volume of gases, Vm = 22.4 dm3mol-1 at s.t.p.= 24 dm3mol-1at room temperatureSpeed of light in vacuum, cSpecific heat of water cw==3.0 x 108 m s-14.18 kJ kg-1 K-1= 4.18 J g-1 K-1= 4.18 J g-1 oC-1Avogadro’s number NA = 6.02 x 1023 mol-1Faraday constant F = 9.65 x 104 C mol-1Planck’s constant h = 6.63 x 10-34 J sRydberg constant RH = 1.097 x 107 m-12.18 x 10-18 JMolar of gases constant R = 8.314 J K-1 mol-1= 0.0821 L atm mol-1 K-1Density of water ρ = 1 g cm-3Freezing points of water = 0.00 oCVapour pressure of water Pwater = 23.8 torrUNIT AND CONVERSION FACTORVolume 1 liter = 1 dm31 mL = 1 cm3Energy 1 J = 1 kg m2 s-2 = 1 N m1 calorie = 4.184 Joule1 eV / molecular = 96.7 kJ mol-1Pressure 1 atm = 760 mm Hg = 760 torr = 101.325 kPa = 101325N m-2 Others 1 Faraday (F) = 96500 Coulomb1 Newton (N) = 1 kg m s-2
5STANDARD REDUCTION POTENTIALS AT 25OC
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14CHEMISTRY LECTURERS 1. Pn Nurul Elyani Binti Elleas2. Pn Nursafura Binti Abd Manap3. Dr Norashdimah Binti Misdi 4. Pn Siti Zubaidah Binti AzizanREFERENCES1. Atkins, P., & de Paula, J. (2006). Physical chemistry (8th ed.). W.H. Freeman and Company. 2. 2. Brown, T. L., LeMay, H. E., Bursten, B. E., Murphy, C. J., Woodward, P., & Stoltzfus, M. W. (2018). Chemistry: The central science (14th ed.). Pearson. 3. Burns, R. A. (2003). Fundamentals of chemistry (4th ed.). Pearson Education. 4. Christian, G. D. (2004). Analytical chemistry (6th ed.). Wiley. 5. Ebbing, D. D., & Gammon, S. D. (2016). General chemistry (11th ed.). Cengage Learning. 6. Lehninger, A. L., Nelson, D. L., & Cox, M. M. (2017). Principles of biochemistry (7th ed.). W.H. Freeman and Company. 7. Miessler, G. L., Fischer, P. J., & Tarr, D. A. (2013). Inorganic chemistry (5th ed.). Pearson. 8. Solomons, T. W. G., & Fryhle, C. B. (2008). Organic chemistry (10th ed.). Wiley.9. Tan, Y. T. (2012). Chemistry for matriculation semester 1. Oxford Fajar Sdn. Bhd10. Tan, Y. T. (2012). Chemistry for matriculation semester 2. Oxford Fajar Sdn. Bhd
15CHAPTER 1 ACIDS AND BASES
16SUBTOPICS EXPLANATORY NOTES1.1 Theory of acids and basesa) define acid and base according to Arrhenius, Lewis and Bronsted-Lowry Theories (include conjugate acid and conjugate base).1.2 Conjugate acidbase paira) Identify conjugate acid and conjugate base according to Bronsted-Lowry theory1.3 Strengths of acids and basesa) Define strong acid and base, weak acid and base.1.4 The terminology of pH, pOH, Kw, Ka, Kb and application in calculationsa) Define pH, pOH and pKwb) Relate pH and pOH to the ionic product of water, Kw at 25 oCc) calculate the pH values of a strong acid and based) Relate the strength of a weak acid and a weak base to the respective dissociation constants, Ka and Kbe) Perform calculations involving weak acid and weak base involving(i) pH(ii) dissociation constant, Ka,Kb(iii) initial concentration and equilibrium concentration(iv) Degree of dissociation, αf) relate Ka, Kb, Kw, pKa, pKb, pH and pOH1.5 Buffer solution a) Define buffer solutionb) Describe how a buffer solution controls its pHc) Write the Henderson-Hasselbalch equation for a given buffer solutiond) Calculate the pH of buffer solution. 1.1 Theory of Acids and Bases1.1.1 Acid - base Definitiona) Arrhenius TheoryAcid: a substance that contains hydrogen and releases hydrogen ion (H +) as one of the products of ionic dissociation in water.Example:HCl(aq) ⎯⎯→ H+(aq) + Cl−(aq) Base: a substance that produces hydroxide ions in waterExample:NaOH(aq) ⎯⎯→ Na+(aq) + OH−(aq)
17b) Brönsted-Lowry (BL) TheoryAcid: any substance that can donate a proton to another substance (it can be a neutral compound or an ion)Example:HNO3(aq) + H2O(l) ⎯⎯→ NO3−(aq) + H3O+(aq) acidNH4+(aq) + H2O(l) ⎯⎯→ NH3(aq) + H3O+(aq) acidBase: any substance that can accept a proton from another substance (a base can also be a neutral compound or an ion)Example:NH3(aq) + H2O(l) ⎯⎯→ NH4+(aq) + OH−(aq)baseCO32-(aq) + H2O(l) ⎯⎯→ HCO3-(aq) + OH−(aq)baseConjugate Acid-base PairsThe acid - base reaction is reversible. For example, HCO3−(aq) + H2O(l) CO32−(aq) + H3O+(aq) acid base base acid▪ The bicarbonate ion, HCO3− acts as an acid and can transfer a proton to water molecules which accepts the proton and acts as a Brönsted-Lowry base.▪ The hydronium ions however, can transfer a proton to carbonate ion, CO32− to form HCO3−ion and H2O.▪ According to the reaction above, the HCO3− and CO32− are related to one another by the loss or gain of H+ as are H2O and H3O+. A pair of compounds or ions that differ by the presence of one H+ unit is called a conjugate acid-base pair.▪ Thus, CO32− ion is the conjugate base to the acid HCO3- or HCO3-is the conjugate acid of the base CO32−. HCO3−(aq) + H2O(l) CO32−(aq) + H3O+(aq) acid base conjugate base conjugate acid of HCO3− of H2O
18Exercise 1. In the following reaction, identify the acid-base conjugate pairs.a) ClO−(aq) + H2O(l) HOCl(aq) + OH−(aq)b) CH3NH2(aq) + H2PO4-(aq) CH3NH3+(aq) + HPO42−(aq)c) HBr(aq) + NH3(aq) NH4+(aq) + Br−(aq)d) PO43−(aq) + H2O(l) HPO42−(aq) + OH−(aq)e) CH3COOH(aq) + H2O(l) CH3COO−(aq) + H3O+(aq)2. Write the conjugate base formula for the following acids.a) HS− d) H3PO4b) HCN e) H3O+c) N2H5+f) CH3NH3+3. Write the conjugate acid formula for each following base.a) HS− d) NH3b) C6H5COO− e) H2Oc) OH−f) H2PO4−c) Lewis TheoryAcid: a substance that can accept a pair of electrons to form a covalent bondBase: a substance that can donate a pair of electrons to form a covalent bond.Example:H3N: + BF3 ⎯⎯→ H3N: → BF3base acid .. H HH+ + :OH− ⎯⎯→ acid base :O: Exercises 1. Give the formulae of the substances which act as Lewis acids in the following reactionsa. BF3 + F - → BF4- b. CaO + SO3 → CaSO4c. Cl2 + AlCl3 → Cl+ + AlCl4- d. BeF2 + 2NH3 → BeF2(NH3)2
191.2 Strength of acids and bases▪ A strong acid is an acid that is fully dissociates into its ions in water. In contrast, a weak acid is an acid that only partially dissociates into its ions in an aqueous solution or water.▪ A strong base is a base that is fully dissociates into its ions in water. In contrast, a weak base is a base that only partially dissociates into its ions in an aqueous solution or water.1.3 Terminology of pH, pOH, Kw, Ka, Kb and application in calculation1.3.1 Water and pH ScaleThe Water Ionization Constant, Kw▪ Water molecules auto ionizes, transferring a proton from one water molecule to another producing a hydroxonium and hydroxide ion.2H2O(l) H3O+(aq) + OH−(aq)▪ The equilibrium lies far to the left because OH−is a much stronger base than H2O and H3O+is a much stronger acid than water.▪ The equilibrium constant expression can be written as follows:H3O+(aq)OH−(aq)K = H2O (l) 2▪ Because concentration of water is a constant,KH2O 2 = H3O+OH− or Kw = H3O+ OH-Kw is the ionization constant for water at 25C H2O = (1000/18) mole / L = 55.5 MIn pure water, H3O+ = OH− at 25C. H3O+ = OH− = 10-7 M Kw = H3O+ OH− = (1 x 10-7) (1 x 10-7) Kw = 1 x 10-14 M2
20In neutral solution : H3O+ = OH−. = 1 x 10-7MIn acidic solution: H3O+ > OH−or H3O+ > 1 x 10-7M and OH− < 1 x 10-7MIn basic solution: OH− > H3O+or OH− > 1 x 10-7M and H3O+ < 1 x 10-7M1.3.2 pH Scale▪ The pH scale is a widely used method of expressing acidity. The pH of a solution is defined as the negative logarithm (log) of the hydroxonium ion concentration. pH = − log H3O+▪ In similar way, pOH is the negative logarithm of the hydroxide ion concentration.pOH = − log OH−▪ In pure water, H3O+ = OH− = 1 x 10-7M pH = − log H3O+ = − log (1 x 10-7 ) = 7▪ The water ionization constant expression: Kw = H3O+OH− = 1 x 10-14▪ Taking the negative log of both sides of the expression:− log (Kw) = − log ( H3O+OH− )− log (1 x 10-14) = (− log (H3O+ ) + (− log OH− ) 14 = pH + pOH 1.3.3 pH of the Acid and Base SolutionStrong Acid and Strong Base▪ A strong acid and a strong base is an acid or a base that is essentially 100 % dissociated in aqueous solution to produce a high concentration of H3O+ or OH−. ▪ The degree of dissociation, = 1 or 100%▪ In strong acids and bases, the H3O+ or OHconcentration can be obtained from the original concentration of the acid or base used. For example:HCl ⎯⎯→ H+ + Cl−0.5 M 0.5M
21NaOH ⎯⎯→ Na+ + OH−0.1 M 0.1M H2SO4 ⎯⎯→ 2H+ + SO42− 0.2M 0.4MNote: Unless the acid/base solution is extremely dilute, H3O+ / OH− < 1 x 10-7 M, the H3O+ / OH−which come from the dissociation of water is excluded in the calculation of the pH of the solution.Example Calculate the pH of the following solutions:1. An aqueous solution contains 0.700 g NaOH in 485 mL water.NaOH is a strong base:NaOH ⎯⎯→ Na+ + OH− 1 mole NaOH No of moles = 0.700 g NaOH x 40 g NaOH = 0.0175 moles of NaOH 0.0175 moles of NaOH OH− = 485 x 10-3 L = 0.0361 M pOH = −log OH− = − log 0.0361 = 1.44 pH = 14 − pOH = 14 − 1.44 = 12.552. A solution of 5 x 10-3 M H2SO4H2SO4 ⎯⎯→ 2H+ + SO42−5 x 10-3 2(5 x 10-3) pH = −log H+ = −log (1 x 10-2) = 2
22Exercise1. Calculate pH for the following solutions:a) 2.0 x 10-3 M HCl d) 0.025 M KOHb) 5.0 x 10-8 M HNO3 e) 0.02 M H2SO4c) 5.4 x 10-9 M NaOH f) 0.005 M Mg(OH)22. Calculate the H+ and OH−concentration in solutions with pH:a) 2.50b) 8.75c) 11.733. What is the NaOH concentration in a solution with pH = 10.50. [Ans: 3.16 x 10-4M]4. How many grams of NaOH is needed to prepare 250 mL solution with pH = 12.40?Weak Acid and Weak Base▪ Very few acids and bases can donate or accept protons, respectively. The vast majority of acids and bases are weak. Weak acid and weak base only partially ionize in an aqueous solution. ▪ At equilibrium, an aqueous solution of weak acid or weak base contains a mixture of nonionized acid or base molecule and its conjugate base or acid, respectively.▪ One way to specify the extent of the dissociation of weak acid or base is by giving its percent dissociation which is defined as, mole/L acid or base dissociated % dissociation = mole/L acid or base availableExample: The amount of H+ dissociated in a solution of 0.10 M acetic acid is 1.3 x 10-3 M, the percent dissociation is:1.3 x 10-3= x 100 0.1= 1.3 % Consider a general weak acid, HA. The ionization equation in water is given by:HA(aq) + H2O(l) H3O+(aq) + A−(aq)Given,c = initial concentrationx = concentration of H3O+ion dissociated HA(aq) + H2O(l) H3O+(aq) + A-(aq)Initial concentration c 0 0Change -x +x +xEquilibrium conc. c – x x x
23• The ionization expression is given by: H3O+ A−Ka = HAwhere the subscript ‘a’ indicate that it is an equilibrium constant for a weak acid in water. Ka is called an acid-dissociation constant . (x)(x) Ka = (c−x)Assuming, that only a small amount of acid dissociates, i.e. Ka <<< 1, therefore:HA = c − x ~ c x2 Ka = Cx2 = Ka.c x = Ka . c = H3O+ Example:Calculate the pH of 0.10 M acetic acid solution , given Ka for acetic acid is 1.8 x 10 -5 M.The dissociation of acetic acid in water is given by: CH3COOH(aq)+H2O(l) CH3COO-(aq) + H3O+(aq)Initial concentration 0.10 0 0Change −x +x +xEquilibrium concentration 0.10−x x x H3O+ CH3COO−Ka = CH3COOH (x)(x) 1.8 x 10-5 = (0.10−x)Assuming x is small, 0.1 – x ≈ 0.1x2 = 1.8 x 10-5(0.1) x = 1.34 x 10-3 M = H3O+ pH = - log H3O+ = - log 1.34 x 10-3 = 2.87Note : If the value of x that is obtained by making the approximation (initial concentration – x ~ initial concentration ) is less than about 5 % of the initial concentration, it is safe to assume that the approximation is valid.
24 HA(aq) + H2O(l) H3O+(aq) + A−(aq)Initial concentration c 0 0Change c +c +cEquilibrium concentration c(1-) c c H3O+ A−Ka =HA (c)(c)Ka = c(1−) c2= 1−Assuming is very small, i.e. <<< 1, so 1− ~ 1, the equation above becomes: Ka = c2 or = √ Ka / c• Now let us consider the dissociation of a weak base, B.B(aq) + H2O(l) BH+(aq) + OH−(aq)Given,c = Initial concentrationx = concentration of OH−ion dissociated B(aq) + H2O(l) BH+(aq) + OH−(aq)Initial concentration c 0 0Change -x +x +xEquilibrium concentration c-x x x Kb = BH+ OH− B (x)(x) =(c−x)• Assuming, that only a small amount of acid dissociates, i.e. Kb <<< 1, therefore: B = c−x ~ c x2 Kb = c x = √ Kb.c = OH−
25If the degree of dissociation, is given, the same equation can be derived as the weak acid. B(aq) + H2O(l) BH+(aq) + OH-(aq)Initial concentration c 0 0Change c +c +cEquilibrium concentration c(1−) c c BH+OH−Kb =B (c)(c)= c(1− ) c 2= 1−Assuming is very small, i.e. <<< 1, so 1- ≈ 1, the equation above becomes, Kb = c2 = Kb / cExercises 1. The table below list the Ka values for some acids at 298 K.Acid Ka / mol dm-3CH3COOH 1.8 x 10-5HCN 4.9 x 10-10HSO4- 1.2 x 10-2a. Write down the expression of Ka for each of the acids.b. Arrange the acids in order of increasing acid strength.2. The table below lists the Kb values, at 298 K, for some bases.Base KbC2H5NH2 4.7 x 10 - 4N2H4 1.7 x 10 -6NH3 1.8 x 10 -5a. Write down the expression of Kb for each of the bases.b. Arrange the bases in order of increasing strength.3. Calculate the degree of dissociation of 0.10 mol dm-3solution of ethanoic acid.[Ka for CH3COOH = 1.8 x 10 -5 mol dm-3] [Ans: 0.0134]
264. The H+ of an aqueous solution of an acid, HA , of concentration 0.15 mol dm-3, is 1.3 x 10 – 4 mol dm-3. Calculate:a. its degree of dissociationb. its acid dissociations constant, Ka [Ans: a. 0.00087; b. 1.13 x 10 -7 moldm-3]5. The dissociation constant for phenol is 1.2 x 10-10 mol dm-3. Calculate the degree of dissociation of a 0.010 mol dm-3solution of phenol [Ans: 1.01 x 10-6]Relationship Between Ka and Kb for Acid-base Conjugate Pairs• Consider the acid-base conjugate pair of ammonium ion, NH4+ and ammonia, NH3. The dissociation of both species in water is given by, NH4+(aq) NH3(aq) + H+(aq)NH4+(aq) + OH−(aq) NH3(aq) + H2O(l) For ammonium ion, the expression corresponding to Ka is NH3 H+Ka = NH4+For ammonia, the expression corresponding to Kb is: NH4+ OH−Kb = NH3Therefore, the product of Ka x Kb is NH3 H+ NH4+ OH−x = H+ OH− = KwNH4+ NH3• It is clearly seen that the product of the acid dissociation constant for an acid and the base dissociation constant for its conjugate base is the ion product constant for water, Ka x Kb = Kwsince −log Ka, = pKa , −log Kb = pKb and −log Kw = pKw,the equation above can be written as,pKw = pKa + pKbpH + pOH = 14
271.4 Buffer Solution Buffer solution is a type of solution which has the ability to maintain its pH when a small amount of strong acid or base is added to the solution.OrA buffer solution can absorb a small amount of H+ or OH─ without any significant change in pH Buffer solutions have limited capacity to resists pH changes. If too much acid is added such that the weak base (or conjugate base) is used up, no more buffering action against acid is possible. On the other hand, if too much base is added such that the weak acid (or conjugate acid) is used up, no more buffering action against base is possible. The buffer capacity of a buffer solution is the amount of acid or base that can be added to a given volume of the buffer solution before the pH of the buffer solution changes by 1 unit. The buffer capacity depends on the actual amount of acid/conjugate base or base/conjugate acid present in the buffer solution. A buffer solution has two requirements,i) it should have the capacity to control pH after the addition of reasonable small amount of strong acid or strong base. ii) it should control the pH at the desired value. In general, there are two types of buffer solution: i) acidic buffer solutionii) basic buffer solution1.4.1 Acidic Buffer Solution• Prepared by adding a weak acid to a salt which contains its conjugate base in some appropriate amount of water.Example : • The reaction which occurs in the buffer solution is:CH3COOH (aq) CH3COO−(aq) + H+(aq)CH3COONa (s) ⎯⎯→ CH3COO−(aq) + Na+(aq)• The amount of CH3COOis mainly from the complete dissociation of CH3COONa because CH3COOH is a weak acid.• Addition of a small amount of strong acid:CH3COO−(aq) + H3O+ (aq) ⎯⎯→ CH3COOH(aq) + H2O(l)• If a strong base is added to the buffer solution, the OHion will beneutralized by CH3COOHCH3COOH(aq) + OH−(aq) ⎯⎯→ CH3COO−(aq) + H2O(l)
28• Therefore, the pH of the buffer solution can be maintained to an appreciable amount.• Calculate pH of the buffer solution using the CH3COOH/CH3COONa pair.CH3COO−H3O+Ka = CH3COOH CH3COOH H3O+ = x Ka CH3COO− acid H3O+ = x Ka conjugate baseTaking - log at both sides of the equation: acid - log H3O+ = (- log Ka) x (- log ) conjugate baseconjugate basepH = pKa + log acidExampleCalculate the pH of a solution prepared by adding 2.00 g of benzoic acid, C6H5COOH and 2.00 g sodium benzoate, C6H5COONa in enough water to make 1.00 L solution. (Ka (C6H5COOH) = 6.3 x 10-5 M)No of moles of C6H5COOH = 2.00/122.1 = 1.64 x 10-2 molesNo of moles of C6H5COONa = 2.00 / 144 = 1.39 x 10-2 moles C6H5COOH = 1.64 x 10-2/ 1.00 = 1.64 x 10-2 M C6H5COONa = 1.39 x 10-2/ 1.00 = 1.39 x 10-2 M conjugate basepH = pKa + log acid 1.39 x 10-2pH = - log 6.3 x 10-5 + log 1.64 x 10-2= 4.13HendersonHasselbalch equation
291.4.2 Basic Buffer Solution Prepared by adding a weak base with its conjugate acid salt.ExampleNH3 and NH4Cl• The reaction, occurring in the buffer solution are:Partial dissociation of weak base:NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)Complete dissociation of its conjugate salt:NH4Cl(s) ⎯⎯→ NH4+(aq) + Cl−(aq)• According to Le Chatelier`s Principle, the presence of a large amount of NH4+reduces the dissociation of NH3. • Therefore, the species present in large amount is the undissociated NH3 and the NH4+ion which come from the complete dissociation of NH4Cl.Addition of a small amount of strong acidH+ (aq) + NH3(aq) ⎯⎯→ NH4+(aq)Addition a small amount of strong baseOH−(aq) + NH4+(aq) ⎯⎯→ NH3(aq) + H2O(l) There will only be a slight change of pH even when a strong acid or strong base is added to the solution.Calculating pH of the basic buffer solution: NH4+]OH− Kb = NH3 NH3 OH− = x Kb NH4+Taking the − log of both sides of the equation: NH3−logOH− = ( - log Kb ) x − log NH4+ NH4+pOH = pKb + log NH3 salt of conjugate acidpOH = pKb + log weak basepH of the solution is then calculated using: pH = 14 − pOHHenderson-Hasselbalch equation
30Example 1. Calculate the pH of the solution prepared by mixing 500 mL 0.1 M hydrazinium chloride, N2H5Cl with 500 mL 0.2 M hydrazine, N2H4.No. of moles of N2H5Cl = 500_ x 0.1M1000= 0.05 molesN2H5Cl = 0.05 x 1000 ( 500 + 500 )= 0.05 MNo of moles of N2H4 = 500_ x 0.2 1000= 0.1 moles N2H4 = 0.1 x 1000____ ( 500 + 500 ) = 0.1 M N2H5ClpOH = pKb + log N2H4 0.05 pOH = − log (1.7 x 10-7) + log 0.1 = 6.469 pH = 14 - pOH= 14 - 6.469= 7.5312. With the help of an equation, explain what will happen to the pH of the solution when a small amount of strong acid and strong base is added to the solution.(Kb N2H4 = 1.7 x 10-7 M)Addition of a strong acid:H+(aq) + N2H4 (aq) ⎯⎯→ N2H5+(aq)and when a strong base is added:OH−(aq) + N2H5+(aq) ⎯⎯→ N2H4 (aq) + H2O (l)- the pH of the buffer solution does not change to an appreciable amount.
31Exercises 1. a. Calculate the pH of a buffer solution consisting of 0.10 mol dm-3CH3COOH and 0.25 mol dm-3 CH3COONa[Ka for CH3COOH = 1.8 x 10 -5 mol dm-3] [Ans: 5.14]b. Calculate the change in pH when i. 10 cm3 of 1.0 mol dm-3 HCl and ii. 10 cm3 of 1.0 mol dm-3 NaOH is added to 1.0 dm3 of the buffer solution above. [Ans: i. 0.06 , ii. 0.06]2. 4.28 g of ammonium chloride was added to 250 cm3 of 0.50 mol dm-3 ammonia solution. Calculate the pH of resulting solution.[Kb for NH3 = 1.8 x 10 -5 moldm-3] [Ans: 9.49]
32CHAPTER 2 ELECTROCHEMISTRY
33SUBTOPICS EXPLANATORY NOTES2.1 Half-cell and Redox reactiona. Define and explain a redox reaction.b. Determine the oxidation number of element in a redox reaction.c. Balance redox equation using ion-electron method.2.2 Cell Voltage, Standard Hydrogen Electrod (SHE), Standard Electrod Potential (SEP)a. Explain reactions in an electrochemical cell.b. Define cell potential and explain the use of standard reduction potential values.c. Show the use of standard hydrogen electrode in determining the half- cell potentiald. Calculate cell potential and determine spontaneous or non- spontaneous reactionse. Calculate the value of the cell potential, Ecell based on the concentration of solution using Nerst equation at 25o C. Formula : Ecell = Eocell - (0.0592)/n log Q2.3 Electrolysis i. Explain electrolytic cell.ii. Explain the electrolysis of molten NaCl & aqueous NaClusing inert electrodes.2.4 Quantitative Aspect of Electrolysisa. Define the first and the second law of Faraday and its quantitative application
342.1 REDOX REACTIONSRedox reactions are reactions that involve the transfer of electrons from one substance to another. The oxidation and reduction occurred simultaneously.Oxidation is a process of losing electron:Fe2+ → Fe3+ + eReduction is a process of gaining electron:Al3+ + 3e- → AlAn oxidizing agent is an electron acceptor, while a reducing agent is an electron donor.Consider the following reaction:Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+In this reaction:• Fe3+ is reduced to Fe2+2Fe3+ + 2e- → 2Fe2+The electrons are donated by Sn2+. Hence, Sn2+ is a reducing agent.• Sn2+ is oxidized to Sn4+Sn2+ → Sn4+ + 2eThe Fe3+ accepts (or removes) electrons from Sn2+. Hence, Fe3+ is an oxidizing agent.In a redox reaction, the oxidizing agent undergoes reduction, while the reducing agent undergoes oxidation. Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+ 3HNO2 → HNO3 + 2NO + H2OOxidation number: +3 +5 +2In this reaction, nitric (III) acid is simultaneously being oxidized to nitric (V) acid and reduced to nitrogen (II) oxide.oxidationreductionReducing agent Oxidizing agent
352.1.1 How to determine the oxidation number of atomsFor polyatomic substances (ion/molecules), the Rules for Oxidation Numbers is applied to to determine the oxidation number of atoms. 1. The oxidation number of a free element is always 0.Eg: Cl2, Na, O2, Mg ect.2. The oxidation number of a monatomic ion equals the charge of the ion. Eg:Monoatomic ions Oxidation numberCl--1Na+ +1Mg2+ +2Mn7+ +73. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements. Eg:Molecule Element Oxidation numberH2O H O+1-2NaH Na H+1-1MgH2 Mg H+2-14. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides. Eg:Molecule Element Oxidation numberH2O2 H O+1-1NaO Na O+1-25. The oxidation number of a Group 1 element in a compound is +1. Eg:Molecule Element Oxidation numberKCl K Cl+1-1Na2O Na O+1-26. The oxidation number of a Group 2 element in a compound is +2. Eg:Molecule Element Oxidation numberMgCl2 Mg Cl+2-1BaO Ba O+2-27. The oxidation number of a Group 17 element in a binary compound is -1. (A binary compound is a substance composed of exactly two different elements, which are substances that cannot be simplified further by chemical means.) Eg:
36Molecule Element Oxidation numberMgCl2 Mg Cl+2-1AlF3 AlF+3-18. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0. Eg:Molecule Element Oxidation number Sum MgCl2 Mg Cl+2-1(+2) + 2(-1) = 0BaO Ba O+2-29. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. Eg:Polyatomic ion Element Oxidation number Sum SO42- S O+6-2(+6) + 4(-2) = -2NO3- N O+5-2(+5) + 3(-2)=-1PO43- PO+5-2(+5)+4(-2)=-3Exercise :Balance the following chemical equations by applying inspection method.1) NH3 + CuO ⎯→ Cu + N2 + H2O2) C6H6 + O2 ⎯→ CO2 + H2O3) AgNO3 + Na2CrO4 ⎯→ Ag2CrO4 + NaNO34) Fe(OH)3 + H2SO4 ⎯→ Fe2(SO4)3 + H2O 5) N2O5 + H2O ⎯→ HNO3How to balance the redox reaction equationThere are two methods in balancing a chemical equation
37Method 1 Inspection methodThe method is used to balance a simple chemical equation and the equation does not involve redox reaction. Method 2 Ion Electron MethodThis method is used to balance a complex chemical equation involving redox reaction. There are two conditions for this balancing equation.Example:Lets consider this equation:MnO4- + I- → I2 + Mn2+Tips 1: The equation is separated into two half-equations, one for oxidation, and one for reduction.Reduction: MnO4- → Mn2+Oxidation : I- → I2Tips 2: The equation is balanced by adjusting coefficients and adding H2O, H+ , and e-in this order:To balance the Oxygen atoms, add the appropriate number of water (H2O) molecules to the other side and add H+ ions to the other side to balance the added H (H2O)Reduction: MnO4- + 8H+ → Mn2+ + 4H2OOxidation : 2I- → I2
38Tips 3: Add up the charges on each side. They must be made equal by adding enough electrons (e-) to the more positive side. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same. Mn7+ needs 5 electrons to form Mn2+Reduction: 5e + MnO4- + 8H+ → Mn2+ + 4H2OOxidation : 2I- → I2 + 2eTips 3: Cancel out as much as possible.The half-equations are added together, cancelling out the electrons to form one balanced equation. Oxidation : 5e + MnO4- + 8H+ → Mn2+ + 4H2O (2x)Reduction : 2I- → I2 + 2e (5x)_______________________________________________________Reduction : 10e + 2MnO4- + 16H+ → 2Mn2+ + 8H2OOxidation : 10I- → 5I2 + 10eTips 4: Add and balanced both equations. The equation can now be checked to make sure it is balanced.Reduction: 2MnO4- + 16H+ → 2Mn2+ + 8H2OOxidation : 10I- → 5I2 Overall: 10I- + 2MnO4- + 16H+ → 5I2 + 2Mn2+ + 8H2O(in acidic medium)If the equation is being balanced in a basic solution, the appropriate number of OH- must be added to turn the remaining H+ into water moleculesOverall: 10I- + 2MnO4- + 16H+ + 16OH- → 5I2 + 2Mn2+ + 8H2O + 16OH- 16H2O Overall: 10I- + 2MnO4- + 8 H2O → 5I2 + 2Mn2+ + 16OH-(in basic medium)Exercise1. Balance the following redox equation in acidic solution:MnO4- + C2O42- → Mn2+ + CO2 + H2O2. Balance the following redox equation in basic solution:Zn + NO3- → Zn2+ + NH4+
392.2 CELL VOLTAGEElectrochemistry involves TWO main types of processes:Galvanic @ Voltaic cells Electrolytic cells▪ spontaneous chemical reaction (battery)▪ non-spontaneous chemical reaction, require external source (DC source)▪ chemical energy produces electrical energy▪ electrical energy produces chemical energy Galvanic cell2.2.1 Reaction in Electrochemical Voltaic/ Galvanic CellA voltaic cell is an electrochemical cell in which electric current is generated from a spontaneous redox reaction.In this cell, electrons will flow from anode to cathode due to difference in their potential energy. Anode has higher potential energy than the cathode. As a result of it, electrons flow from anode to cathode through the external circuit to lower its potential energy. The anode is the electrode at which oxidation occurs, source of electrons, it is also the negative electrode. The cathode is the electrode at which reduction occurs.Example Eq 1: Zn2+ + 2e- → Zn E° = -0.763V more negative, Eq 2: Cu2+ + 2e- → Cu E° = +0.34V more positive,
40Cu2+ + 2e- → Cu E° = +0.34V (reduction/cathode) Zn → Zn2+ 2e- E° = +0.763V (oxidation/anode)Overall reaction Cu2+(aq) + Zn (s) → Cu(s) + Zn2+(aq) E°cell = E°red + E°ox= +0.34V + (+0.763V) = +1.103V@E°cell = E°cathode - E°anode= +0.34V – (-0.763V) = +1.103VExample Figure above shows the flow of process in galvanic cell:▪ Zn atoms lose electrons and enter the solutions as Zn2+ ions(Zn → Zn2+ + 2e-)▪ The electrons flow through an external circuit from the Zn electrode (anode) to the Cu electrode (cathode).▪ At Cu electrode, Cu2+ ions from copper (II) sulphate solution will gain electrons and deposits as Cu metal on the Cu electrode (Cu2+ + 2e- → Cu)Half-equation:Anode : Zn(s) → Zn2+ (aq) + 2eCathode : Cu2+ (aq) + 2e- → Cu (s)Overall reaction (redox reaction)Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
41▪ Electrons carry the electric current through the external circuit, but they cannot carry electric charges through the solutions.▪ Cations and anions carry electric charge through the solutions▪ The number of Zn2+ ions increases at anode, more positive charge in the solution.▪ The number of Cu2+ ions decreases at cathode, more negative charge in the solution.▪ Salt bridge will complete the electrical circuit by balancing the electrical neutrality (charge).2.2.2 Cell Potential and Standard Reduction PotentialThe electrode potential of an electrode system depends on several factors. Among them are:• Concentration of the aqueous ions.• Pressure• TemperatureHence, conditions for the measurement of the electrode potential have to be standardized. The standard conditions are:• Concentrations of aqueous ions are fixed at 1.0 mol dm-3 @ M.• Temperature is fixed at 25oC or 298 K.• Pressure is fixed at 1 atm or 101 kPa.• Platinum is used as the inert electrode in a system which does not contain a solid electrode.The electrode potential measured under standard conditions is called the standard electrode potential (SEP), E° and has the unit of volt (V).The absolute value of the standard electrode potential of an electrode system cannot be measured. A voltmeter has to be connected to another metal and dipped into the solution to complete the circuit. However, the other metal will also produce an electrode potential. Hence, what is being measured is the difference between the electrode potential of the two electrodes system.A more useful reference electrode for accurate work is the standard hydrogen electrode (SHE). The hydrogen electrode is a platinum sheet, coated with platinum black, immersed in 1 M hydrogen ions. This electrode compares the ability of various reactions to give electrons to aqueous hydrogen ions.H2(g) → 2H+(aq) + 2eWhen the hydrogen electrode is used as the standard of reference, the voltage of the complete cell is called the standard electrode potential (Eo).Cell notation represents a cell in the written form. The Zn-Cu cell above (Figure 2.2) can be written as below:
42 Half-cell Half cellZn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s) Anode Cathode Oxidation Reduction Left Right Negative electrode Positive electrode▪ Place the anode on the left side of the diagram. ▪ Place the cathode on the right side of the diagram ▪ Single vertical line ( | ) represent the boundary between different phases, such as between an electrode and a solution. ▪ Double vertical line ( || ) represent a salt bridge or porous barrier separating two half-cells.2.2.3 STANDARD HYDROGEN ELECTRODEAn electrode potential is calculated from a cell potential where the potential of one electrode is known. SHE is used as the reference electrode in order to calculate unknown electrode potential for certain electrode.It is made up of a platinum electrode, immersed in an aqueous solution of H+ (1 M) and bubbled with hydrogen gas at 1 atm pressure, and temperature at 25oC. The standard reduction potential of SHE is 0 V.Diagram of standard hydrogen electrode (SHE)2.2.3.1 Example determination of standard electrode potential of a metal.The diagram below shows the arrangement of the apparatus used in the determination of the standard electrode potential of zinc half –cell. The reading of the potentiometer or high resistance voltmeter is 0.76V. Zinc is the negative terminal of the cell (undergo oxidation). The standard electrode potential of the zinc half-cell is -0.76V. Connecting,Pt wirePlatinum electrode coated with platinum backHydrogen gas, 1 atmSolution containing H+ion, 1M.
43The arrangement of the apparatus used in the determination of the standard electrode potential of zinc half –cell.The zero point of electrode potentials is taken to be that of the half-cell called the Standard Hydrogen Electrode (SHE) . Hydrogen gas at exactly 1 atm pressure is bubbled over an inert platinum electrode and into an aqueous solution by [H+] = 1 M. The cell voltage called the standard cell potential (E0cell), is the difference between the standard potential of the cathode and the anode.E°cell = E° (cathode) - E° (anode) @ = E°ox + E°redCell notation of the above cell diagram is:Zn (s) | Zn2+(aq, 1 M) || H+(aq, 1 M) | H2(g ,1 atm) | Pt (s) E° cell = 0.763VThis standard cell potential and the standard electrode potentials are still related as followsE°cell = E° (cathode) - E° (anode)0.763V = E°H+/H2 - E°zn2+/Zn0.763V = 0 V - E°zn2+/ZnE°zn2+/Zn = - 0.763VThe value of electrode potential measured depends on what is the other electrode used. For example, the differences between the Zn2+/Zn half cell with Cu2+/Cu half cell, the reading of voltmeter is 1.10V. This problem can be solved if a particular electrode is chosen as the standard or reference electrode and all other electrode systems measured against this standard electrode. Furthermore, if the standard electrode potential of the reference electrode is arbitrarily fixed as 0.00 V, then the potential difference between the half cell and the reference electrode will be the same as the standard electrode potential of that half-cell. The standard hydrogen electrode is chosen as the reference electrode.H2 at 1 atmPlatinum[H+] = 1.0MSalt bridge (KCl)[Zn2+] = 1.0MZn electrode which its electrode potential to be measuredH2 at 1 atm0.76V
44Example The apparatus below (figure 2.5) is used to measure the standard electrode potential of the chlorine half-cell. The results are:The e.m.f of the cell is 1.35VThe chlorine half-cell is the positive terminal (undergo reduction)The hydrogen half cell is the negative terminal (undergo oxidation)Hence, the standard electrode potential for the chlorine half-cell is +1.35 V.Example The apparatus below (figure 2.6) is used to measure the standard electrode potential of the Fe3+/Fe2+ half-cell.The standard electrode potential for the Fe3+/Fe2+ half-cell is measured using the above arrangement. The cell potential or e.m.f of the cell is recorded as 0.77V. The Fe3+/Fe2+ half-cell is the positive terminal. Hence, the standard electrode potential for the Fe3+/Fe2+ half-cell is +0.77V.
45Example Sketch the arrangement of the apparatus used to determine the standard electrode potential of the MnO4-/ Mn2+SolutionThe standard electrode potential of zinc is -.076 V. Zinc is the negative terminal of the cell when connected to the standard hydrogen electrode. The reactions taking place at both the electrodes are:Zn (s) → Zn2+ (aq) + 2e- oxidation 2H+ (aq) + 2e- → H2(g) reductionThe overall cell reaction is : Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2(g)Hence, zinc is a stronger reducing agent than hydrogen. The H+ ion is a stronger oxidizing agent than Zn2+ ion.The standard electrode potential of the silver (Ag) half-cell is + 0.80 V. When connected to the standard hydrogen electrode, silver forms the positive terminal of the cell. The reaction taking place at the respective electrodes are:H2(g) → 2H+ (aq) + 2e- oxidationAg+ (aq) + e- → Ag (s) reductionThe overall cell reaction is:H2(g) + Ag+ (aq) → 2H+ (aq) + Ag (s)Hence, silver is a weaker reducing agent when compared to hydrogen and, Ag+ is a stronger oxidizing agent than H+.The standard electrode potential values can be used to compare the relative strengths of oxidizing agents and reducing agents. When all the electrode systems (half-cells) are arranged in the order of increasing S.R.P. values (from the most negative to the most positive), the standard electrode potential series is obtained.* The convention is to write half cell reaction as reduction:Oxidising agent + electron reducing agent.
46Point to remember when using the SRP series:• All ions are aqueous ions.• All half cells are written in reduction form. For example, the standard electrode potential for the zinc half-cell is - 0.76 V refer to the reaction:Zn2+ (aq) + 2e- Zn (s) E°red = - 0.76V• The more negative the value in SRP, the higher tendency to undergo oxidation.• The more positive the value in SRP, the higher tendency to undergo reduction.• The magnitude of the standard potential of a half-cell does not depend on the stoichiometric equation. Cl2(g) + 2e- 2 Cl-(aq)E°red = + 1.36V½ Cl2(g) + e- Cl-(aq)E°red = + 1.36V2Cl2(g) + 4e- 4 Cl-(aq) E°red = + 1.36V• Reversing a half cell equation, the sign will be reverse as well. Example:Zn2+ (aq) + 2e- Zn (s) E°red = - 0.76VIf it undergoes oxidation, half cell and sign will be reverse.Zn (s) Zn2+ (aq) + 2e- E°oxi = + 0.76V• Oxidizing agents are found on the left-hand side of the respective half-cell.• Reducing agents are found on the right-hand side of the respective half-cells.Sn4+ (aq) + 2e- Sn2+ (aq)Cu2+ (aq) + 2-e Cu (s)I2 (s) + 2e- 2 I-(aq)Fe3+ (aq) + e- Fe2+(aq) [Oxidizing agent] [Reducing agent]
47Half-cell reaction E°/ V (298)Li+ (aq) + e Li(s)K+ (aq) + e K(s)Ca2+ (aq) + 2e Ca(s)Na+ (aq) + e Na (s)Mg2+ (aq) + 2e Mg(s)Al3+ (aq) + 3e Al (s)Zn2+ (aq) + 2e Zn (s)Fe2+(aq) + 2e Fe (s)Cr3+ (aq) + e Cr2+ (aq)Ni2+ (aq) + 2e Ni (s)Sn2+ (aq) + 2e Sn (s)Sb2+(aq) + 2e Sb(s)2H+ (aq) + 2e H2 (g)Sn4+ (aq) + 2e Sn2+ (aq)Cu2+ (aq) + 2e Cu (s)I2 (s) + 2e 2 I-(aq)Fe3+ (aq) + e Fe2+(aq)Ag+ (aq) + e Ag (s)VO2+ (aq) + 2H+ (aq) + e VO2+ (aq) + H2O (l)Br2 (l) + 2e 2 Br-(aq)MnO2 (s) + 4 H+(aq) + 2e Mn2+(aq) + 2 H2O (l)Cr2O72-(aq) + 14H+ (aq) + 6e 2Cr3+(aq) + 7H2O(l)Cl2(g) + 2e 2 Cl-(aq)Ce4+ (aq) + e Ce3+ (aq)PbO2 (s) + 4 H+(aq) + 2e Pb2+(aq) + 2 H2O (l)MnO4-(aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4 H2O (l)H2O2 (aq) + 2 H+ (aq) + 2e 2 H2O (l)F2(g) + 2e 2 F-(aq)- 3.04-2.92-2.87-2.71-2.38-1.66-0.76-0.44-0.41-0.25-0.14-0.130.000.150.340.540.770.801.001.071.231.331.361.451.471.521.772.872.2.4 STANDARD CELL POTENTIAL (E°cell)An electrochemical cell converts chemical energy (produced from redox reaction) to electrical energy.An electrochemical cell can be constructed by combining two half-cells of different E° values. The half-cell with the more positive E ° value (as obtained from SEP table) forms the positive terminal (cathode) of the cell. The half-cell with the more negative value forms the negative terminal (anode) of the cell.Electrons flow from the anode to cathode, while the current flows from the positive terminal to negative terminal.The cell potential @ e.m.f. (E °) of the cell is given by the expression:E°cell = E°cathode - E°anode @ E°cell = E°oxi + E°red
48The reaction taking place at the positive terminal is reduction and at the negative terminal is oxidation.The positive terminal of the cell is called cathode, while the negative terminal is called anode.Example Calculate the standard e.m.f. of the following cells. Write the cell notation and cell reactions of each cell.a) Sn2+ / Sn and Mg2+ / Mgb) Cl2 / Cl− and MnO4−/ Mn2+c) Pt / H2, H+ and AgCl, Cl−/ AgSolutiona) Sn2+ (aq) + 2e- Sn (s) Eo = - 0.14V (less negative, more positive : reduction)Mg2+ (aq) + 2e- Mg(s) Eo = - 2.38V (more negative : oxidation)Method 1, E°cell = E°cathode - E°anode = (- 0.14V) – (- 2.38V)= + 2.24 VMethod 2, E°cell = E°oxi + E°redOxidation : Mg(s) Mg2+(aq) + 2e- Eooxi = + 2.38V (reverse above equation and sign)Reduction : Sn2+(aq) + 2e- Sn(s) Eored = - 0.14V E°cell = E°oxi + E°red = +2.38V + (- 0.14V) = + 2.24 VCell diagram : Mg(s)| Mg2+ (aq) || Sn2+ (aq) | Sn(s)Cell reaction : Mg(s) + Sn2+ (aq) → Mg2+(aq) + Sn (s)b) Cl2 (g) + 2e- 2 Cl-(aq) Eo = + 1.36V (negative pole)MnO4-(aq) + 8 H+(aq) + 5e- Mn2+(aq) + 4H2O(l) Eo=+1.52V (positive pole)E.m.f. = Eo(cathode) - Eo(anode)= (+1.52) – (+ 1.36)= + 0.16V.Cell diagram : Pt(s)| Cl−(aq) | Cl2 (g) // MnO4−(aq) | H+ (aq), Mn2+ (aq) |Pt(s)
49Cell reaction : MnO4− + 8 H+ + 5e- → Mn2+ + 4H2O ………(i) 2 Cl− → Cl2 + 2e- ……………(ii)(i) x 2 + (ii) x5 : 2MnO4−(aq) + 16 H+ (aq) + 10 Cl−(aq) → 2Mn2+(aq) + 8H2O(l) +5Cl2(g)a) 2 H+(aq) + 2e- H2(g) Eo = 0.00V (negative pole)AgCl(aq) + e- Ag(s) + Cl−(aq) Eo = + 0.22V (positive pole)E.m.f., E = Eo(cathode) - Eo(anode) = (+0.22) – (0.00) = + 0.22 V.Cell diagram : Pt(s)| H2 (g) | H+ (aq)|| AgCl (s) | Cl−(aq) | Ag(s)Cell reaction: H2(g) + 2 AgCl(aq) → 2Ag(s) + 2H+(aq) + 2Cl−(aq)Example Determine E° for the reduction half-reaction: Ce4+ (aq) + e- → Ce3+ (aq), given the cell voltage for the following voltaic cell.Co(s) | Co2+ (1 M) (aq) || Ce4+ (1 M) , Ce3+ (1M) | Pt (s) E0cell= 1.887 VSolutionOxidation half reaction: Co(s) → Co2+ + 2e occurs in the anode half-cell on the left. From the table we find that E°Co2+/Co = - 0.227V (Co2+(aq) + 2e- → Co(s)) Reduction half reaction : Ce4+ (aq) + e- → Ce3+ (aq) E° = ?E0cell = E0(cathode) - E0(anode) E0 Ce4+/Ce3+ - E0Co2+/Co = 1.887VE0 Ce4+/Ce3+ - (- 0.227 v ) = 1.887VE0 Ce4+/Ce3+ = 1.610VStandard electrode potentials series can be used i) To Compare the relative strengths of oxidizing agents and reducing agents.Reducing agents The more negative (relatively) the S.E.P.; the stronger the reducing agent. Oxidizing agent The more positive (relatively) the S.E.P.; the stronger the oxidizing agent.
50When comparing the strengths of oxidizing agents and reducing agents, the sign of the S.E.P. remain unchanged. This means, determination of oxidizing and reducing agents must be made in the form of reduction form. For example:Ni2+ (aq) + 2e Ni (s) Eo = - 0.25 VThe oxidizing strength of Ni2+ is equivalent to – 0.25 V (respect to H+)The reducing strength of Ni2+ is equivalent to – 0.25 V(respect to H2)Example Arrange the species in the order of increasing oxidizing strengths and reducing strengths.Zn2+ (aq) + 2e- Zn (s) Eo = - 0.76 VI2 (s) + 2e- 2 I-(aq) Eo = + 0.54 VVO2+ (aq) + 2H+ (aq) + e- VO2+ (aq) + H2O (l) Eo = + 1.00 VFe3+ (aq) + e- Fe2+(aq) Eo = + 0.77 VMnO4-(aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4 H2O (l) Eo = + 1.52 VSolutionOxidising agent: Zn2+ < I2 < Fe3+ < VO2+ < MnO4-Reducing agent: Mn2+ < VO2+ < Fe2+ < I - < Znii) To predict the feasibility of redox reactions.To predict the feasibility of a redox reaction under standard conditions, one has to calculate the e.m.f. (for Eo) of the reaction. If the e.m.f is positive, the reaction might take place spontaneously. If the e.m.f. of the reactions is negative, the reaction cannot take place (non-spontaneous reaction)To predict whether the following reaction is feasible or not, one need to calculate the e.m.f. of the reaction.Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s)Half-cell equation:Oxidation: Zn2+(aq) + 2e → Zn(s) Eored = - 0.76 VReduction: Cu(s) → Cu2+(aq) + 2e Eooxi = - 0.34 VZn2+ + Cu → Cu2+ + Zn e.m.f = E°oxi + E°red = (-0.76 ) + (-0.34)= - 1.10 V.A negative e.m.f. indicates that the reaction is not feasible. So, this is non spontaneous reaction. (However, the reverse reaction will take place)The sign of Eo value of Cu2+/Cu half-cell is reversed because the half-equation is reversed