CHM 156 Physical Chemistry 1 PDF

CHE 156 Physical Chemistry 1  Atoms, Molecules and Intermolecular Forces Acids, Bases and Salts Oxidation and Reduction/Electrochemistry Thermochemistry, Spontaneous Processes and Equilibria Reaction Kinetics Kinetic Theory Nelson Okpako OBI-EGBEDI Unit 1 Topic Rate of reactio...

CHE 156 Physical Chemistry 1  Atoms, Molecules and Intermolecular Forces Acids, Bases and Salts Oxidation and Reduction/Electrochemistry Thermochemistry, Spontaneous Processes and Equilibria Reaction Kinetics Kinetic Theory Nelson Okpako OBI-EGBEDI Unit 1 Topic Rate of reaction and factors that affect rate.  This topic is about Reaction Kinetics: Rates of Reactions. The study of reaction kinetics is concerned with the measurements of the velocities or rates of reaction together with the factors which affect these rates and reaction mechanism;  The rate of reaction describes how fast the reaction goes; An explanation of how a reaction occurs is mechanism of the reaction; Factors which affect the velocity of a chemical reaction are: concentration of the reactants, temperature of the reacting system, nature of the reacting species and presence of catalysts or inhibitors; There will be two quizzes and one test. Key Concepts Velocities/speed; reactants and products; initial rate; determining/limiting steps Unit 1: Objectives In this Unit, you will learn about factors that affect the rate(speed) of a reaction and how to explain the rate of reactions. The Unit will enable you to understand that some reactions occur very rapidly, whereas others take hours, days, or even years.  Many chemical reactions occur when we prepare food and our recipes require heating to make those reactions occur. You heat an egg so that necessary chemical reactions will happen. The length of time required for the completion of a reaction depends on factors. Expected Outcome(Unit 1) At the end of Unit 1, you should be able to : Define/explain rate or velocity and mechanism of reaction; List the factors that affect the rate of reaction. REACTION KINETICS  It is not enough to understand the stoichiometry and thermodynamics of reaction, We must also understand the FACTORS that govern/affect the rate of reaction. Chemical kinetics/Reaction kinetics is the study of how reaction rates change under varying conditions and what molecular events occur during the OVERALL REACTION.  The study of the rates of chemical reaction is called KINETICS. Question? How fast does a reactions go? And what is the mechanism of the reaction. RATE is (# of events per unit time) WHY MEASURE REACTION RATES? Below are some of the reasons: How fast a chemical product will ‘work’ How quickly, and economically, a product can be made. A knowledge of the reaction mechanism helps the industrial chemist to modify a product, thereby increasing its effectiveness. Learn a great deal about how reactants change into products. FACTORS THAT AFFECT REACTION RATES Chemical nature of the reactants; Ability of the reactants to come in contact with each other; Concentrations of the reactants: increasing the concs.; Temperature of the reaction system and Availability of agents called CATALYSTS.  The rate or velocity with which a chemical reactant is converted under varying conditions into a product and the mechanism by which such a reaction occurs constitute the field of study known as Reaction Kinetics. The rate of such a chemical reaction is determined by measuring the change in the concentration of either the reactant or the product with time. What is Mechanism of a reaction? The mode by which a reactant is transformed into a product is the mechanism of the reaction. Collection of individual kinetic processes/ elementary steps involved in the transformation of reactants into products  Explanation of how a reaction takes place/occurs. The molecular EVENTS which occur during a reaction. The SERIES of individual steps leading to the overall observed reaction. For example: consider the thermal decomposition of dinitrogen pentoxide. The STOICIOMETRIC equation for the reaction is: The reaction can take place in stages, probably as follows: 2 N2O5 4 NO2  O2 N2O5  NO2  NO3 slow NO2  NO3  NO2  NO  O2 fast One Way NO  NO3  2NO2fast 2 N2O5 4 NO2  O2 Rates of Chemical Reaction [Conc] [time]   Equations (1a &1b) indicate that the rate of a reaction may be defined in two major ways, either in terms of the disappearance of the reactants or the appearance of the products, depending upon whether the reactant or the product is being observed or monitored.  What one observes experimentally in determining the rate of a reaction is the change in conc. , of one or more of the reactants or product as a function of time, t. Ci  In practice, the observation of a change in a physical property (spectroscopic, hydrogen ion concentration, etc) of the reactant or product is used to measure the extent of change as a function of time.  Consider a hypothetical homogeneous reaction, which is unidirectional in the direction of the arrow in Equation (2), with reactant concentration as the only monitored variable A(g) + B(g) C(g) + D(g) (undefined) (2)  The conc. of A or B decreases during the time interval, therefore the change in A or B is a negative quantity. The rate of a reaction is a positive quantity, so a minus sign is needed in the rate expression to make the rate positive.  In Equation (2), all the stoichiometric coefficients of the reactants and the products are the same. When they are not, the stoichiometric factors have to be taken into account in the definition of the rate of reaction. Consider a general reaction aA  bB cC  dD (4)  Take an arbitrary reaction A B d [ A] [ A] slope    cons tan t dt Conc. [ B] d [ B] slope   cons tan t dt Time / s linear plot  C D [C ] Conc. [D] Time/s Non-linear plot. The rate depends on reaction conc. and time. Quiz # 1 1. In 300 seconds, the conc of A(g) falls from 3.40 x 10-2 mol dm-3 to 2.50 x 10-2 mol dm-3 by the reaction A B , what is the rate of reaction? 2. Which term describes the measure of the increase in the concentration of a product per unit time? A. activation energy B. kinetics C. reaction rate D. reaction time Express the rate of 2A 3C in terms of change of the concentrations. A. Rate =  - 1 d[C] 3 dt 2 dt B. Rate =  - 1 d[A] d[C] dt 1 d[A] 2 dt C. Rate = -  - 1 d[C] 3 dt 1 d[A] 2 dt D. Rate =  1 d[C] 3 dt 1 d[A] 2 dt  Reference: Fundamental Physical Chemistry(Revised Edition). Edited by Iweibo I., Okonjo K. & Obi-Egbedi N.  CHE 156 Physical Chemistry 1  Nelson Okpako OBI-EGBEDI  Unit2 Topic: Concentration, the rate law and experimental determination of rate expression  Unit2 Objectives In this Unit, you will learn how to write the rate law and use it to predict the speed of reactions. Experimental determination of rate law.  Expected Outcome(Unit 2) At the end of Unit 2, you should be able to : Know how concentration affects rate of reaction; Determine the order of a reaction; Write the rate law/expression; and Do some calculations.  The rate of reaction is a function of factors such as the concentrations of the reactants, the temperature of the reaction system, the identity of the reactants and the catalyst used. Let us consider one of the factors: Concentration  Concentration and the Rate Law The rates of most reactions change when the concentrations of the reactants change. Chemical reactions are usually faster when concs. of the reactants are increased and slower when decreased. In general, the rate is proportional to the concentration of the reactants raised to appropriate powers/indices/exponents. The rate is proportional to the molar concentrations of the reactant raised to an appropriate power. Consider the following chemical equation: xA + yB Product(s) (6) The above equation is like :H2 + 1/2O2 → H2O  In Equation (7), k is the specific reaction rate constant. The exponents n and m are called reaction orders, and the sum (m + n) is the overall reaction order. Equation (7) is known as the rate law.  The rate law is an expression for the rate as a function of concentration at a fixed temperature. Equation (7) is determined experimentally.  Experimental Determination of Rate Laws Most chemical changes consist of several steps called elementary steps. Different reactants can affect the rate of a particular reaction differently. Consequently, Equation (7) cannot be written from the balanced chemical equation but from experimental data. N2O5  NO2  NO3 slow NO2  NO3  fast NO2  NO  O2 One Way NO  NO3  2NO2 fast 2 N2O5 4 NO2  O2  To determine the order with respect to each reactant in expression (7), it is necessary to study the mechanism by which changes in the concentration of each reactant affect the initial rate of the reaction.  A + B → Product(s) Let us conduct a hypothetical experiment for the above-named reaction in order to determine its rate law, using the initial rates method. How the initial rate depends on the starting concs is preferable. Why? It is because …. as the reaction proceeds, the concs of the reactants decrease and it may become difficult to measure the change accurately.  Mix the solutions of the reactants A and B with initial concentrations [A]0 and [B]0, respectively. The temperature of the mixture is fixed.  At convenient time intervals, withdraw some aliquots of this mixture, stop the reaction or slow it down drastically, analyze the withdrawn sample chemically to determine the change in concentration of the reactant A which occurs after reactants have been mixed.  Instead of monitoring the change in concentration of reactant A, the change in its physical property as a function of time could be measured. For example, its optical absorbance, fluorescence, optical activity or refractive index.  Perform other experiments with different values of the initial concentration of A at the same fixed temperature, keeping the initial concentration of reactant B constant. Determine the initial rate, at (time = 0), for each experiment from a plot of concentration against time.  Compare the initial rates obtained from these experiments. If there is no change in rate, the order with respect to A is zero. If the rate of reaction doubles when the initial concentration of A is doubled, the order with respect to A is one.  If the rate quadruples when the initial concentration of A is doubled, then the order with respect to A is two.  Similarly, varying the initial concentration of reactant B and keeping the initial concentration of reactant A and the temperature of the reaction constant, the effect of the concentration of reactant B on the initial rate of the reaction can also be investigated and the order with respect to B determined.  These experimentally determined orders with respect to each reactant will enable us to write Equation (7), the rate law.  For example, consider the reaction: A(g) + B(g) →C(g) The experimental rate law of the above reaction is: Rate = k [A] [B]0 What is the overall order of the reaction?  Since the overall order is the sum of the exponents, we obtain (1 + 0) = 1 as the reaction order.  For a hypothetical reaction in which B and C react to form the product, D, the following data were obtained from three experiments. B+C D Experiments [B] [C] (Rate of formation of D) 1. 0.15M 0.075M 3.5 x 10-3mol dm-3min-1 2. 0.30M 0.15M 1.4 x 10-2mol dm-3 min-1 3. 0.15M 0.15M 7.0 x 10-3mol dm-3 min-1 BrO3 (aq)  5Br  (aq)  6 H  (aq) 3Br2 (aq)  3H 2O(l ) Suppose we have the reaction: o Mixture [Bro3] [Br-] [H+]ions in acid Bromate solution. Relative initial rate of formation of Br2 I 0.05 0.25 0.30 1 II 0.05 0.25 0.60 4 III 0.10 0.25 0.60 8 IV ? ? ? ? R  k[ BrO3 ]l [ Br  ]m [ H  ]n  1  k[0.05] [0.25] [0.30] l m n 4  k[0.05] [0.25] [0.60] l m n 1 [0.30]n  4 [0.60]n 1 1 2  n 2 2 n= 2  The exponents in the rate law appear to be unrelated to the coefficients in the overall chemical equation. See Equation 7 There is, in fact, no way to know for sure what the exponents in the rate law of an overall reaction will be without doing experiments to determine them.  Sometimes, however, the coefficients and the exponents are the same by COINCIDENCE, as in the case of elementary reactions. In general, the order of a reaction must be determined by expt; it cannot be deduced from the overall balanced equation.  Quiz #2 1. A + 2B D The rate of formation of D in the above reaction is found experimentally to be independent of the concentration of B and to quadruple when the concentration of A is doubled. Write the rate law for the reaction.  2. For the general rate law, Rate = K[A][B]2 , what will happen to the rate of reaction if the concentration of A is tripled? [A] The rate will be halved [B] The rate will be doubled [C] The rate will be tripled [D] The rate will remain the same 3. Methanol can be produced by the following reaction: CO(g) + 2H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol? 4. What is the overall reaction order for the reaction represented by the rate law: Rate = k[H2][NO]2? [A] zero [B] first [C] second [D] third  CHE 156 Physical Chemistry 1  Nelson Okpako OBI-EGBEDI  Unit 3 Topic: Molecularity, Mechanism ,Predicted Rate Law and Units of Rate Constant  Expected Outcome: Obtain the units of the specific rate constant for a given order  An overall balanced chemical equation does not tell us much about how a reaction takes place. 2H  O 2H O 2 2 2 The chemical equation represents the sum of a series of simple reactions called the elementary steps/elementary reactions.  They represent the progress of the overall reaction at the molecular level. The sequence of elementary steps that leads to product formation is called the reaction mechanism. What is Mechanism of a reaction? The mode by which a reactant is transformed into a product is the mechanism of the reaction. Collection of individual kinetic processes/ elementary steps involved in the transformation of reactants into products  Explanation of how a reaction takes place/occurs. The molecular EVENTS which occur during a reaction. The SERIES of individual steps leading to the overall observed reaction. For example: consider the thermal decomposition of dinitrogen pentoxide. The STOICIOMETRIC equation for the reaction is: 2 N2O5 4 NO2  O2 The reaction can take place in stages, probably as follows: N2O5  slow  NO2  NO3 NO2  NO3  fast NO2  NO  O2 One Way NO  NO3  fast 2NO2 2 N2O5 4 NO2  O2  Molecularity, Mechanism and Predicted Rate Law Most chemical reactions consist of elementary reactions or steps. One of these steps, which is much slower than the others, determines the rate of the reaction and it is referred to as the rate- determining or rate-limiting step.  An elementary reaction can be characterized by the number of reactant molecules involved in it. This number of molecules or species on the reactant(s) side / in an elementary reaction /step is known as the molecularity of the reaction.  For the elementary process, the order(s) in its rate law are the same as the stoichiometric factors, or equal to the coefficients of the reactants in the chemical equation for the process.  An elementary step in which only one reacting molecule participates is UNIMOLECULAR Two molecules is BIMOLECULAR. TERMOLECULAR reaction involves the participation of three molecules in one elementary step.  The rate of an elementary reaction is proportional to the product of the concentrations of the reactants that are in that step. Therefore, the rate law for an elementary process can be predicted from the chemical equation for the process.  Consider the reaction: PCl5 →PCl3 + Cl2  The above reaction can be explained by a two-step mechanism: Step (1) is the slow (rate limiting) step. PCl5 →PCl4 + Cl. slow Step 2 PCl4 →PCl3 + Cl. fast Net: PCl5 →PCl3 + Cl2 The experimentally determined rate law is: Rate = k [PCl5] and, from the rate-limiting step, the predicted rate law is Rate k [PCl5]  The elementary steps must satisfy two requirements: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step, which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law as is determined experimentally.  The order of multistep reaction must be obtain experimentally. For example, the conversion of A + C D  A B slow, RDS B+ C D very fast A+C D The rate at which D is formed depends on how quickly A gives B, not how quickly B reacts with C. The rate equation would be: Rate = k[A]x  Consider a hypothetical reaction: 2C  D E  F What is the rate law of the rate determining step in terms of the concentrations of the reactants in the overall balanced equation? Fast step: rate1 (forward reaction)  Rate2 (backward reaction) (i) (ii) The rate of the slow step:  Rate (iii) Sub. (ii) into (iii) yields the overall rate as Rate (iv)  UNITS OF RATE CONSTANTS The units of the specific rate constant depend on the overall order of the reaction. The units are worked out from the units of the individual terms in the rate equation. Units of rate constants = Units of initial rate (Units of initial concentration) x (9)  The units of rate are mol dm-3 time-1 and the units of [B] are mol dm-3 Substituting these units into Equations (9)/ (10), and assuming that time is in seconds and x = 1, gives the units of k as s-1 or time-1. [ ] = mol dm-3 M = mol dm-3  Quiz # 3 1. A + 2B D The rate of formation of D in the above reaction is found experimentally to be independent of the concentration of B and to quadruple when the concentration of A is doubled. Write the rate law for the reaction.  2. For the general rate law, Rate = K[A][B]2 , what will happen to the rate of reaction if the concentration of A is tripled? [A] The rate will be halved [B] The rate will be doubled [C] The rate will be tripled [D] The rate will remain the same 3. Methanol can be produced by the following reaction: CO(g) + 2H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol? 4. What is the overall reaction order for the reaction represented by the rate law: Rate = k[H2][NO]2? [A] zero [B] first [C] second [D] third  Reference: Fundamental Physical Chemistry(Revised Edition). Edited by Iweibo I., Okonjo K. & Obi- Egbedi N.  Unit 4 Topic: ORDER, KINETICS EQUATIONS/CONCS-TIME EQUATIONS  Expected Outcome: At the end of this Unit 4, you should be able to: 1. Derive kinetic equations for first and second orders reactions 2. Explain the meaning of half-life of a reaction 3. Do calculations using the kinetic equations  ORDER OF REACTION Reaction can be classified by its order/ we can classify reactions into: Zeroth/Zero order First order Second order Third order(rarely) Fractional order Negative order  KINETICS EQUATIONS/CONCS-TIME EQUATIONS The rate law tells us how the speed/rate of a reaction varies with concentrations of the reactants at a particular moments. The relationship between the concentration of a reactant and TIME can be obtained from the rate law of a reaction.  This can be done by integration of the rate expression/ transform a rate law into a mathematical relationship between conc. and time. x x t t dx  x 0   b  x t 0 kdt  Separating the variables and integrating Equation (13) between the limits x = 0 at t = 0 and x = x at some time t, we have x x t t [ ln(b  x)] x 0  k[t ] t 0 (14) b ln  kt (15) bx 1  xdx  ln x  loge x  We can have the equation as: b ln  kt bx b  exp  kt  bx (b  x)  b exp  kt  [ B]t  [ B]0 exp(kt ) log(b-x) Slope = - k/2.303 Mol dm-3 Time/s Linear plot for a first order reaction  Second-order Reactions let us consider a situation in which the reactants are present with initial equimolar concentrations. Consider a reaction : C+D → Products [C] = [D] i.e., the initial conc. of [C]=[B]  d [C ]  k[C ][ D] OR dt d [C ] d [ D] dx    k2 [c  x]2 OR dt dt dt x t dx  0 (c  x ) 2  k 0 2 dt (17)  x  1  k2t   (18) c  x 0 1 1 k2t   (19) cx c  Equation (19) is the second-order kinetic equation. k2 can be evaluated graphically. Or by substitution of the initial concentration and the concentration of a reactant at time t into Equation (19). 1/c-x Slope = k2 Time/s Linear plot for a second order reaction  Half- life of zero order:  Second-order Half life:  METHODS FOR ORDER The order of a reaction can be determined by: Graphical method Substitution method- constant ks Half-life method. Half-Life Method  Example The rate constant for the first-order decomposition of N2O5 at temperature t0C is 2.2 x 10-1s-1. If the initial concentration of N2O5 is 3.6 x 10-3 mol dm-3, what will the concentration be after four half-lives? For a first-order reaction, the half-life is independent of the initial concentration of the reactant.  After the first half-life, the initial concentration would have decreased by half. Therefore, after the fourth half-life, the concentration will be 2.25 x 10-4 mol dm-3.  CHE 156 ACIDS, BASES AND SALTS NATHANAEL DAMILARE OJO COURSE CONTENT BROAD DIVISION ACIDS BASES pH SALT BUFFER SOLUTION 2 COURSE CONTENT Unit 1: Acid and bases in our environment Unit 2: Properties of acids and bases Unit 3: Concepts of Acids and Bases: Arrhenius, Lowry-Bronsted and Lewis concept Unit 4: Strength of Acids and Bases: Acid and base dissociation constants 3  COURSE CONTENT Unit 5: Acid-Base equilibria in aqueous solutions: The pH concept; calculations of pH of strong/weak acid solution, strong/weak base solution Unit 6: Hydrolysis: Definition and description of salt of strong/weak acid and strong/weak base Unit 7: Buffer Solution: Definition and buffer action: Preparation and uses of buffer solution Unit 8: Calculation of the pH of buffer solutions Unit 9: Acid-Base Indicators: Acid/base strength and chemical structure. 4 Objectives of the Section At the end of the section, you should be able to Define acids, bases and salt List at least two properties each of acids, bases and salt Calculate the pH of strong/weak acid and base Calculate the pH of buffer solutions 5 Expected Learning Outcome By the end of this section, you should be able to Differentiate between an acid and a base using their definitions and properties Explain the concept of neutralization Solve problems related to equilibrium constant, buffer action, pH of acids, bases and buffer solutions 6 Unit 1: Acid and bases in our environment Acids in our environment: Unripe fruits Sour milk Aspirin Car batteries Human stomach acid Acid rain: SO2, CO2, NO2 in rain This Photo by Unknown Author is licensed under CC BY-SA-NC 7 https://creativecommons.org/licenses/by-nc-sa/3.0/ Unit 1: Acid and bases in our environment Bases in our environment: Milk of magnesia Ammonia in cleaning agents Bleach Lye (NaOH) 8  Unit 2 Properties of acids and bases: neutralization; acid salts; monoprotic, diprotic and polyprotic acids 9 Properties of an Acid Physical properties: Sour taste Concentrated acids are corrosive Turn moist blue litmus red pH < 7 Soluble in water Aqueous acidic solutions are electric conductors 10 Properties of an Acid CHEMICAL properties: Releases hydrogen gas when reacted with highly electropositive (active) metals Forms salt with alkali (neutralization) 11 Properties of a Base Physical properties: Bitter taste Slippery to touch Concentrated bases are corrosive Turn moist red litmus blue Good conductor of electricity pH > 7 Bases that are very soluble in water are called ALKALI e.g., NaOH, KOH 12 Properties of a Base/Alkali CHEMICAL properties: Neutralization Reaction: base + acid form salt and water Alkali reacts with ammonium salt to release ammonia gas e.g. Ca(OH)2(s) + (NH4)2SO4(s) Heat CaSO4 + 2NH3(g) + 2H2O(l) Alkali solution precipitates insoluble metal hydroxides from solution of its salt: 2NaOH(aq) + CuSO4(aq) Cu(OH)2(s) + Na2SO4(aq) 13 Properties of a Base/Alkali Reaction with metal: Strong alkalis (NaOH) reacts with aluminium to liberate hydrogen gas (not a general reaction) 2Al(s) + NaOH(aq) +6H2O(l) 2Na3Al(OH)4(s) + 3H2(g) 14  UNIT 3 DEFINITIONS OF ACIDS AND BASES  Only hydrogen atoms involved in polar bonds are ionizable Arrhenius theory of acids  Acids produce H+ (H3O+)ions in aqueous solution +H O  HA 2 H3O+(aq) + A-(aq) 15 DEFINITIONS OF ACIDS AND BASES Arrhenius acids can be grouped into: Non-oxygen acids and oxyacids Monoprotic, diprotic and triprotic acids Strong and weak acids 16 DEFINITIONS OF ACIDS AND BASES Non-oxygen acids: HCN(aq), HI(aq) and HCl(aq) Oxyacids e.g. H2SO4(aq), HClO4(aq) and H3PO4(aq) 17 DEFINITIONS OF ACIDS AND BASES Strong acids: HClO4(aq), HNO3(aq), HCl(aq), HI(aq) and H2SO4(aq) Weak acids: HF(aq), H2S(aq), H3PO4(aq) and HCN(aq) 18 DEFINITIONS OF ACIDS AND BASES Monoprotic acids: HClO4(aq), HNO3(aq), HCl(aq), CH3COOH(aq) Diprotic acids: H2SO4(aq), H2SO3(aq), H2CO3(aq) Triprotic acids: H3PO4(aq), citric acid and arsenic acid NOTE: (aq) means the acid is dissolved in water to form an aqueous solution 19 Arrhenius theory of bases An Arrhenius base releases its OH- ions in water 20 Arrhenius theory of bases Strong Arrhenius base completely ionizes in water e.g. NaOH NaOH(aq) Na+(aq) + OH-(aq) Weak Arrhenius base partially ionizes in water e.g. Al(OH)3(s) Arrhenius definition of acids and bases is limited to aqueous solution 21 BRÖNSTED-LOWRY ACIDS AND BASES An acid is a proton donor A base is a proton acceptor The B-L acid-base definition is broader than Arrhenius definition All Arrhenius acids and bases are Bronsted-Lowry’s acids and bases For example: NH3(g) +HCl(g) NH4Cl(s) Base Acid Salt 22 BRÖNSTED-LOWRY ACIDS AND BASES Every B-L acid has a conjugate base, and vice versa HCl(aq) + H2O(aq) H3O(aq)+ + Cl-(aq) Acid1 + Base2 Acid2 + Base1 Strong acid forms a weak conjugate base; weak base forms a strong conjugate acid B-L theory restricts the definition of acids and bases to proton transfer 23 Lewis Theory of Acids and Bases Lewis acid is an electron pair acceptor Lewis base is an electron pair donor 24 Lewis Theory of Acids and Bases Every B-L base is a Lewis base Every B-L acid is a Lewis acid Electron pair can be donated to other electron deficient species other than H+ Examples of Lewis acid are AlCl3, metal ions, BCl3 Examples of Lewis base are NH3, ethylene diamine, OH- For instance, Na2O(s) + SO3(g) Na2SO4(s) No proton transferred, no hydrogen ion transferred, yet a neutralization reaction: Lewis acid-base reaction 25 Class Quiz Group the following into Arrhenius acids, Bronsted-Lowry and Lewis acids: 1. H2SO4(aq) 7. Cu2+ 2. HCl(g) 8. OH- 3. H2O(l) 4. NH4+(aq) 5. HCl(aq) 6. AlCl3(s) 26  UNIT 4 The Strength of Acids and Bases This can be compared using Degrees of dissociation, α Dissociation constant: Ka and Kb pH and pOH Ka and Kb are temperature dependent; concentration independent The degree of dissociation is based on Ostwald’s dilution law 27 Ostwald’s dilution law This law relates to weak electrolyte; employs the degree of dissociation Consider a weak binary electrolyte MA c: molar concentration α: degree of dissociation MA(aq) + M (aq) + A (aq)- c(1-α) cα cα 28  2 𝑀+ 𝐴− (cα) Dissociation constant, K = = (1) 𝑀𝐴 𝑐(1−α) (cα2 ) K = (1−α) K = cα2 ; α = 𝐾 𝑐 1/2 if α [H+], [OH-] for pure water; Kw = [H+][OH-]; [H+] = [OH-] = 1.00 x 10-7 moldm-3 during full ionization at 25oC Kw : ionic product of water is temperature dependent 30 Dissociation of an acid HA(aq) H+ + A- + − 𝑎𝐻 𝑎𝐴 𝐾𝑎 = 4 𝑎𝐻𝐴 aH+, aA- and aHA are the activities of hydrogen ions, anion and the weak acid Activity ≈ concentration for a dilute solution [𝐻 + ][𝐴− ] Ka = 5 [𝐻𝐴] 31 Dissociation of a base (A-) A-(aq) + H2O HA(aq) + OH- [𝐻𝐴][𝑂𝐻 − ] Kb = 6 [𝐴− ] Equations 5 x 6 become Kw = Ka x Kb = [H+][OH-] 7 pKa = -Log10Ka 8 pKb = -Log10Kb 9 NOTE: 1. Large Ka ≡ small pKa ≡ strong acid 2. Large Kb ≡ small pKb ≡ strong base 32 Table 1: Dissociation constants of acids (Ka) and weak bases (Kb) Adapted from Fundamental Physical Chemistry (Iweibo et al.) 2007 Revised Edition 33  Unit 5 Acid-Base equilibria in aqueous solutions: The pH concept; calculations of pH of strong/weak acid solution, strong/weak base solution 34 Expected Outcomes for Unit 5 At the end of the unit, you should be able to: define pH of a solution calculate the pH of an aqueous solution of a strong acid calculate the pH of an aqueous solution of a weak acid calculate the pH of an aqueous solution of a strong base calculate the pH of an aqueous solution of a weak base 35 The pH concept pH expresses the acidity or basicity of a solution pH = -Log10[H+] 10 [H+]: molar concentration of the hydrogen ion (or H3O+) in solution p : potenz (German word which means power to which 10 is raised) 36 The pH concept When pH < 7: an acidic solution; [H+] > [OH-] When pH >7: a basic/alkaline solution [H+] < [OH-] When pH = 7: a neutral solution [H+] = [OH-] 37 The pH of an aqueous solution of a strong acid A strong acid fully dissociates in solution The pH of a strong acid solution follows equation 10: pH = -Log10[H+] 10 38 Example 1: The pH of an aqueous solution of a strong acid Determine the pH of a 0.20 M HNO3 solution. Solution HNO3(aq) H+(aq) + NO3-(aq) 1 mole 1 mole 1 mole 0.20 moldm-3 0.20 moldm-3 0.20 moldm-3 [H+] = 0.20 moldm-3 pH = -Log10[H+] pH = - Log10(0.20) = 0.70 39 Example 2: The pH of an aqueous solution of a strong acid Determine the pH of a 0.20 M H2SO4 solution. Solution H2SO4(aq) 2H+(aq) + SO42- (aq) 1 mole 2 mole 1 mole 0.20 moldm-3 (2 x 0.20) moldm-3 0.20 moldm-3 [H+] = 0.40 moldm-3 pH = -Log10[H+] pH = - Log10(0.40) = 0.40 40 The pH of an aqueous solution of a weak acid Weak acids dissociate partially in solution The degree of dissociation is less than 1 The concentration is better expressed as activity of the solution 41 The pH of an aqueous solution of a weak acid The pH of a weak acid solution in expressed as: 1 1 𝑝𝐻 = 𝑝𝐾𝑎 − 𝐿𝑜𝑔 𝑐 11 2 2 Where 𝑝𝐾𝑎 = −𝑙𝑜𝑔𝐾𝑎 12 [H+] = 𝐾𝑎 𝑐 13 c = molar concentration of the acid (not the hydrogen ion concentration) Ka = acid dissociation constant 42 Example: The pH of an aqueous solution of a weak acid Calculate the pH of a 0.001 M solution of methanoic acid. Ka =1.8 × 10−4 Solution 1 1 𝑝𝐻 = 𝑝𝐾𝑎 − 𝐿𝑜𝑔 𝑐 11 2 2 1 1 pH = − 𝑙𝑜𝑔 1.8 × 10−4 − 𝑙𝑜𝑔 1.0 × 10−3 2 2 pH = 1.87 + 1.50 pH = 3.37 43 The pH of an aqueous solution of a strong base A strong base dissociates completely in water to give OH- ions A strong base produces a weak conjugate acid 44 The pH of an aqueous solution of a strong base For example: Calculate the pH of 0.05 M NaOH solution. Solution: NaOH(aq) Na+(aq) + OH-(aq) 0.05 M 0.05 M 0.05 M (10−14 ) 𝐻+ = = 2 × 10−13 [𝑂𝐻 − ] 𝑝𝐻 = − log 2 × 10−13 = 12.70 45 The pH of an aqueous solution of a weak base For a weak base, the pH is 1 1 𝑝𝐻 = 𝑝𝐾𝑤 − 𝑝𝐾𝑏 + 𝐿𝑜𝑔 𝑐 12 2 2 𝑝𝐾𝑤 = 14 𝑝𝐾𝑏 = −𝐿𝑜𝑔10 𝐾𝑏 46 Example: The pH of an aqueous solution of a weak base Determine the pH of a 0.05 M aqueous hydroxylamine solution. Kb = 1.1 x 10-8 at 298 K. Solution HONH2(aq) + H2O(l) HONH3+(aq) + OH-(aq) 1 1 𝑝𝐻 = 𝑝𝐾𝑤 − 𝑝𝐾𝑏 + 𝐿𝑜𝑔 𝑐 12 2 2 1 −8 1 𝑝𝐻 = 14 − (−log(1.1 × 10 ) + 𝐿𝑜𝑔 0.05 2 2 𝑝𝐻 = 14 − 3.98 − 0.65 = 9.37 47  Unit 6 Hydrolysis: Definition and description of salt of strong/weak acid and strong/weak base 48 Expected Outcome At the end of this unit, you should be able to Define salt Describe hydrolysis of salt 49 Salt Ionic compound in which H+/OH- ion is totally/partially replaced by a metal ion or positive radical Types: normal, acid, basic, complex and double salt Normal salt: Strong acid + Strong base e.g. NaCl Acid salt: Strong acid + weak base e.g. NH4Cl Basic salt: Weak acid + Strong base e.g. CH3COONa Complex salt e.g. Cu(NH3)42+ Double salt e.g. KAl(SO4)2.12H2O, (NH4)2[Fe(H2O)6](SO4)2 50 Salt of strong acid and strong base Does not hydrolyze in solution It is neutral: pH = 7 MA M+ + A- Examples: NaCl, NaNO3, K2SO4 51 Salt of strong acid and weak base Hydrolyzes to produce acidic solution pH on hydrolysis < 7 Examples: NH4NO3, CuSO4, NH4Cl 52 Salt of weak acid and strong base Hydrolyzes to produce basic solution pH on hydrolysis > 7 Examples: CH3COOK, HCOONa HCOONa HCOO- + Na+(aq) 53 Salt of weak acid and weak base Hydrolyzes to produce acidic, basic or neutral solution Examples: CH3COONH4, HCOONH4, (NH4)2CO3 54  UNITS 7 AND 8 Buffer Solution: Definition and buffer action: Preparation and uses of buffer solution; Calculation of the pH of buffer solutions 55 Expected Outcome At the end of the units, you should be able to Define a buffer Explain the preparation of a buffer solution Determine the pH of a buffer solution 56 Buffer solution Resists sharp change in pH Buffer: helps living organisms to resist drastic change in pH For instance: Human blood (pH = 7.4), enzyme operate at specific pH Blood plasma contains carbonic acid, carbonate and sodium salts of phosphoric acid Citric acid is used to stabilize milk of magnesia Tears possess high buffer capacity 57 Buffer action Buffer contains Weak acid and its salt e.g. ethanoic acid and sodium ethanoate Weak base and its salt e.g. aqueous ammonia and ammonium chloride Salt of a weak acid and a weak base e.g. ammonium ethanoate 58 Buffer action Weak acid and its highly ionizable salt: added hydrogen ions will combine with excess anion e.g. ethanoate ion; Added base (OH-) will react with be removed by the acid part (e.g. ethanoic acid) of the buffer 59 Buffer action Weak base and its salt e.g. aqueous ammonia and ammonium chloride Added hydrogen ions will be removed by aqueous ammonia Added hydroxide ions (base) will be removed by ammonium ions 60 Buffer action Salt of a weak acid and a weak base e.g. ammonium ethanoate Added hydrogen ions will be removed by ethanoate ions Added hydroxide ions (base) will be removed by ammonium ions 61 pH of a buffer solution Buffering capacity and pH of a buffer are very important Buffering capacity: amount of acid/base the buffer can neutralize before pH change becomes appreciable 𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑑𝑑𝑒𝑑 𝑎𝑐𝑖𝑑/𝑏𝑎𝑠𝑒 𝑝𝑒𝑟 𝑙𝑖𝑡𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑢𝑓𝑓𝑒𝑟 Buffer capacity β = 13 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝐻 62 pH of a buffer solution [𝐴− ] pH = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 14 [𝐻𝐴] OR [𝑆𝐴𝐿𝑇] pH = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 [𝐴𝐶𝐼𝐷] [𝐴− ] = concentration of the salt [𝐻𝐴] = concentration of the acid Equation 14 is Henderson-Hasselbalch equation 63 EXERCISE (pH of buffer solution) 0.065 M sodium ethanoate is added to 0.02 M ethanoic acid to prepare a buffer solution. Calculate: 1. the pH of the buffer (assume no change in volume) 2. The change in pH if 1 cm3 of 1 M KOH is added to the buffer Ans: 1. pH = 5.25 2. change in pH = 0.03 64  0.065 M sodium ethanoate is added to 0.02 M ethanoic acid to prepare a buffer solution. Calculate: the pH of the buffer (assume no change in volume) The change in pH if 1 cm3 of 1 M KOH is added to the buffer Determine the new concentration of KOH, CH3COOH and salt [KOH]: C1=1M, V1= 1cm3 (10-3 dm3); V2= 1001 cm3, C2= ? (Ans: 0.000999 M) Since a base was added, [acid] reduces; [salt] increases [Acid] = 0.02 – 0.000999 M = 0.019001 M [Salt] = 0.065 + 0.000999 M = 0.065999 M [𝑆𝐴𝐿𝑇] pH = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 = 5.28 [𝐴𝐶𝐼𝐷] ΔpH = 0.03 NOTE: The buffer experiences a slight change in pH 65 Unit 9: Acid-Base Indicators: Acid/base strength and chemical structure Expected outcome By the end of the unit, you should be able to describe an indicator state appropriate indicator for specific acid-base reactions 66 Indicator Substance to determine the endpoint of an acid-base reaction Each indicator has pH range where it gives a notable colour Phenolphthalein: 8.3 – 10.5 (pink); < 8.3 (colourless) Methyl orange: 3.2 – 4.5 (red); >4.5 (yellow) 67 Acid-base indicator It can be a weak acid or a weak base; partially ionizable Change in concentration of indicator can alter the degree of ionization HIn + H2O H3O+ + In- Indicator Conjugate base The unionized indicator and its conjugate base display different colour 𝐻3 𝑂+ [𝐼𝑛− ] The equilibrium constant, KIn = 15 [𝐻𝐼𝑛] 68 Acid-base indicator Useful relationship between pH and pKIn : pH = 𝑝𝐾𝐼𝑛 ± 1 16 That is, the useful pH range at which the colour of the indicator is perceptible A universal indicator may contain 4-5 different indicators with pH range 4-11 69 Acid-base indicator Table 2: Acid-base indicators and their pH range Adapted from PHYSICAL CHEMISTRY textbook by K.K. Sharma and L.K. Sharma 70 Acid-base indicator in titration Strong acid - Strong base Titration: Methyl orange and phenolphthalein are suitable. Strong acid - Weak base Titration: Methyl orange is suitable Weak acid - Strong base Titration: Phenolphthalein is suitable Weak acid - Weak base Titration: No suitable indicator. No accurate endpoint 71 Practice Questions 1. AlCl3 is an example of ___ (a). Arrhenius base (b). Lewis base (c). Lewis acid (d). Bronsted-Lowry acid 72 2. The pH of 0.01 M H2SO4 is ___ (a). 0.02 (b). 2.0 (c). 1.7 (d). 0.8 73 3. The pH of 0.00023 M citric acid is _______ (Ka = 3.5 x 10-4) (a). 3.55 (b). 1.82 (c). 3.46 (d). 1.73 1 1 𝑝𝐻 = 𝑝𝐾𝑎 − 𝐿𝑜𝑔 𝑐 11 2 2 1 1 pH = − 𝑙𝑜𝑔 3.5 × 10−4 − 𝑙𝑜𝑔 2.3 × 10−4 2 2 pH = 1.73 + 1.82 pH = 3.55 74 4. The Kb of 0.010 M NH3 solution with 0.0424 degree of ionization is _ (a). 1.8 x 10-4 (b). 1.8 x 10-5 (c). 1.8 x 10-8 (d). 1.8 x 10-10 C = 0.010 M, α = 0.042, Kb = ? Kb = cα2 Kb = 0.010 × (0.0424)2 = 1.8 x 10-5 75 5. The pH of a buffer containing 0.100 M lactic acid and 0.100 M sodium lactate is _______ (Ka = 1.4 x 10-4) (a). 1.93 (b). 4.85 (c). 3.85 (d). 3.55 [𝑆𝐴𝐿𝑇] pH = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 14 [𝐴𝐶𝐼𝐷] −4 pH = −𝑙𝑜𝑔 1.4 × 10 + 𝑙𝑜𝑔 1 pH = 3.85 + 0 pH = 3.85 76 77 References Fundamental Physical Chemistry, Revised edition, Edited by Iweibo I., Okonjo K. & Obi-Egbedi N. https://www.pdfdrive.com/physical-chemistry-books.html https://www.pdfdrive.com/download.pdf?id=39587383&h=b2d7b6e2e 394267f626af702fbaff3d0&u=cache&ext=pdf https://www.pdfdrive.com/download.pdf?id=26523760&h=8a81af9a1 729968e5ba2e3d72eb2e984&u=cache&ext=pdf 78  PREMIER DEPARTMENT OF CHEMISTRY 1  Course Outline Atoms and Molecules Acids, Bases and Salts Thermochemistry Oxidation and Reduction`Reactions/ Electrochemistry Rates of Reactions Kinetic Theory of Gases 2 Oxidation and Reductions/ Electrochemistry- Dr. Onyinyechi V. Uhuo (B15) 3 Oxidation and Reductions Definition of Terms Calculation of Oxidation numbers Oxidizing and Reducing Agents Identifying Redox Reactions Balancing Redox Equations 4 Background Knowledge Periodic Table Atomic number Electronic configuration Loss or gain of electron Sign nomenclature of electron loss or gain 5  OXIDATION AND REDUCTION (DEFINITIONS) In terms of 2. REDOX 1. In terms of oxygen oxidation number* 3. In terms of hydrogen *In terms of electron and *In terms of electronegative/electropositive elements 6 1. Definition in term of oxygen Oxidation is the addition of oxygen to a substance while Reduction is the removal of oxygen from a substance Oxidizing agent is the substance that is reduced Reducing agent is the substance that is oxidized An oxidizing agent in one reaction can act as the reducing agent in another reaction Example 1: Extraction/production of metals by reduction of their ores (which exist as metal oxides in most cases) to the metal. E.g production of iron from iron ore (hemitite) 7  reduction (O.A) (R.A) oxidation (reduction) 2. CO → CO2 (addition of oxygen) 8 Example 2: Iron chains when exposed to oxygen from the atmosphere rust because of the addition of oxygen. oxidation 4Fe(s) + 3O2(g) → 2Fe2O3(s) (R.A) (O.A) reduction (Oxidation) O2 is reduced to Fe2O3 (O2 is the O.A) 9 Example 3: oxidation Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) (R.A) (O.A) reduction (Oxidation) CuSO4 is reduced to Cu (CuSO4 is the O.A) 10 2. Definition in terms of hydrogen Oxidation is the removal of hydrogen from a compound. Reduction is the addition of hydrogen to a compound 11 Example 4: Example 5: 12 Example 6: Combustion of Hydrocarbon. oxidation Hydrocarbon CH + O2(g) → CO2(g) + H2O(l) (R.A) (O.A) reduction Oxidation O2 is reduced to H2O (O2 is the O.A) 13 Example 7: Cellular Respiration (oxidation of glucose). oxidation https://micro.magnet.fsu.edu/cells/ mitochondria/mitochondria.html C6H12O6 + 6O2(g) → 6CO2(g) + 6H2O(l) (R.A) (O.A) reduction O2 is reduced to H2O (O2 is the O.A) 14 Exercises CuO(s) + H2(g) → Cu(s) + H2O(l) Cl2(g) + H2(g) HCl(g) 15 16  3). Definition in terms of oxidation number (electron ) Oxidation is the loss of electron Oxidation is the increase in the oxidation number Oxidation number increases Reduction is the gain of electron Reduction is the decrease in the oxidation number Oxidation number decreases 17 The oxidation and reduction half-reactions must always occur together the total number of electron loss in the oxidation reaction(s) must be the total number of electron gained in the reduction reaction(s). In any redox equation, there is at least one oxidation half-reaction and at least one reduction half-reaction. 18  Zn(s) ZnSO4(aq) CuSO4(aq) Cu(s) Addition of metallic zinc to an aqueous CuSO4, which is blue, to give a colourless aqueous solution of ZnSO4 and a red-brownish deposit of Cu metal. Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) This reaction can be divided into two halves, called half-reactions, the oxidation reaction and the reduction reaction. 19 Example 8: oxidation Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) (R.A) (O.A) reduction 20  oxidation Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) reduction Oxidation reaction: Zn(s) Zn2+(aq) + 2e- Reduction reaction: Cu2+(aq) + 2e- Cu(s) Net reaction: Zn(s) + Cu2+(aq) + 2e- Zn2+(aq)+ Cu(s) +2e- Net reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Number of electrons lost during oxidation equals number of electrons gained during reduction 21  3). Definition in terms of oxidation number (electron ) Example 9: Oxidation 2Na(aq) + H2(s) → 2NaH(s) (R.A) (O.A) reduction Na is oxidized to NaH (NaH is the R.A) 22 23 *). Definition in terms of addition and removal of electronegative and electropositive elements Oxidation is the addition of electronegative elements or the removal of electropositive elements Reduction is the removal of electronegative elements or the addition of electropositive elements. Electronegative elements (non-metals): F, O, Cl, N, Br, C, I, P Electropositive elements (metals): K, Na, Ca, Mg, Al, Zn, Pb, Cu, H 24 Oxidation is the addition of electronegative elements or the removal of electropositive elements Reduction is the removal of electronegative elements or the addition of electropositive elements. Oxidation is the addition of Oxygen (an electronegative element) or the removal of Hydrogen (an electropositive elements). Reduction is the removal of Oxygen (an electronegative element) or the addition of Hydrogen (an electropositive elements). 25 Example 10: oxidation 2Na(s) + Cl2(g) → 2NaCl(s) (R.A) (O.A) reduction Cl2 is reduced to NaCl (Cl2 is the O.A) 26  General Rules for Calculating Oxidation Number 1. The oxidation number of an uncombined atom is zero (oxidation of H2, He, Zn, Cu, K = 0; ie H20, Zn0, Cu0, K0; but written as H2, Zn, Cu, K). 2a. The sum of oxidation number of all atoms/ elements in a molecule is equal to zero (ZnSO4)0, (CuSO4)0, (Na2HPO4)0, (KMnO4)0. 2b. For an ionic compound, the sum of oxidation numbers of all atoms/elements is equal to the charge on the compound. Oxidation numbers of PO43- = -3; MnO4- = -1; and NH4+ = +1 27  General Rules for Calculating Oxidation Number 3i. The oxidation number of the alkali metals (Li, Na, K, Rb, Cs; group IA) is +1 while that of the alkaline earth metals (group IIA; Be, Mg, Ca, Ba) is +2. 3ii. Oxidation number of a simple ion (ion comprising of just a single element) is equal to the charge on the ion. E.g. The oxidation number of Na+ = +1; Al3+ = +3; Cl- = -1; F- = -1. 4. The oxidation number of hydrogen is +1 in its compounds while that of fluorine is –1. 5. Oxygen has an oxidation number of –2 in its compound. 28  General Rules for Calculating Oxidation Number 6. In their binary compounds with metals, the elements of group VIIA (KCl, NaBr) have oxidation number of –1; group 6A (CaO, MgO, ZnO, ZnS) -2 and group VA (AlN) -3. 7. The application of the rules are in order of 1, 2, 3……6, that is, rule 1 supercedes rule 2 and rule 2 surpercedes rule 3 and so on. NOTE: Some elements have more than one oxidation number in different compounds or reactions. E.g. Fe is +2 in FeCl2 but +3 in FeCl3. Nitrogen is -3 in NH3, +1 in N2O and +5 in N2O5 29 30 23Na (Group 1 element) Atomic Mass 23 Atomic number 11 (No. of proton) No. of electron 11 Electronic configuration 2,8,1 Na 1s2 2s2 2p6 3s1 Na will lose 1 electron Oxidation number +1 16O Group 6 element Atomic Mass 16 Atomic number 8 No. of electron 8 O Electronic configuration 2,6 1s2 2s2 2p4 O will need to gain 2 electrons Oxidation number -2 31  Example: Calculate the oxidation number of sulphur in the Following compounds a) S8 b) ZnS c) H2S d) Na2SO3 e) Na2SO4 f) Na2S2O3 Solution: a) S8 Oxidation number of S atom in S8 is equal to 0 (Rule 1, element in its uncombined state) 8 x (S) = 0. Therefore S = 0/8 = 0 b) ZnS Rule 6, S is in group 6 A. Hence in its binary compound with metals, Its oxidation number is –2. S = -2 c) H2S Rule 4, H = +1 Therefore 2(+1) + S = 0 32 2+ S=0 S = -2 d) Na2SO3 e) Na2S2O3 Rule 2, Na2SO3= 0 Rule 2, Na2SO3= 0 Rule 3, Na = +1; Rule 3, Na = +1; Rule 5, O = -2 Rule 5, O = -2 2(Na) + S + 3(O) = 0 2(Na) + 2(S) + 3(O) = 0 2(+1) + 2(S) + 3(-2) = 0 2(+1) + S + 3(-2) = 0 2 + 2S – 6 = 0 2+ S–6=0 2S – 4 = 0 S–4=0 2S = 4 S = +4 S = 4/2 S = +2 33  Example: What is the oxidation number of hydrogen in compounds (a) and (b), and that of oxygen in compound (c)? a) NaH b) H2O c) H2O2 a) NaH b) H2O c) H2O2 Rule 3, Na = +1 Rule 5, O = -2 Rule 3 & 4, H = +1 Na + H = 0 2(H) + O = 0 2(H) + 2O = 0 1+H=0 2H -2 = 0 2(1) + 2(O) = 0 2H = 2 2 + 2O = 0 H = -1 2O = -2 H = +1 O = -2/2 O = -1 34 In b&c) the determination of hydrogen’s oxidation number comes before that of oxygen. That is, Rule 4 should be applied before Rule 5. Hence H = +1 in both cases. b) H2O c) H2O2 Rule 4, H = +1 Rule 4, H = +1 Rule 5, O = -2 Answer: O = -1 H = +1 (Rule 4 is applied (but in Rule 5, O = -2) before Rule 5 Rule 5 does not apply here 35 Example: Calculate the oxidation number of oxygen in the following compounds a) H2O b) K2O c) KO2 a) H2O b) K2O c) KO2 Rule 4, H = +1 Rule 3, K = +1 Rule 3, K = +1 2H + O = 0 2K + O = 0 K + 2(O) = 0 2(+1) + O = 0 2(+1) + O = 0 1 + 2(O) = 0 2+O=0 2+O=0 2(O) = -1 O = -2 O = -2 O=-½ 36 Example: Calculate the oxidation number of the metal atom in each of the following ionic formular units a) MnO4- b) Al(H2O)63+ c) Cr2O72- a) MnO4- Rule 2, Mn + 4(O) = -1 Rule 5, O = -2 Mn + 4(-2) = -1 Mn – 8 = -1 c) Cr2O72- Mn = -1 + 8 Rule 2, 2(Cr) + 7(O) = -2 Mn = +7 Rule 5, O = -2 b) Al(H2O)63+ 2(Cr) + 7(-2) = -2 Rule 2, Al + 6(H2O) = +3 2(Cr) – 14 = -2 Rule 2, H2O = 0 2(Cr) = -2 + 14 Al + 6(0) = +3 2(Cr) = +12 Al + 0 = +3 Cr = +12/2 37 Al = +3 Cr = +6 Example: Calculate the oxidation number of the metal atom in each of the following ionic formular units a) Re2Cl82- b) Fe(CN)63- Solution: a) Re2Cl82- Rule 3, 2(Re) + 8(Cl) = -2 Rule 6, Cl = -1 2(Re) + 8(-1) = -2 c) Fe(CN)63- 2(Re) – 8 = -2 2(Re) = +6 Rule 3, Fe + 6(CN) = -3 Re = +3 CN is a formular unit on its own. Its charge is –1 (carbon is +4 and N = -5) Fe + 6(-1) = -3 Fe - 6 = -3 Fe = +3 38 Identifying Redox Reactions Redox reactions are identified by checking the oxidation numbers of the elements present in the reaction at both the reactant side and the product side, whether they change or not. If the oxidation number of some of the elements change, then it is a redox reaction. If there are no changes in oxidation number of the elements, then the reaction is not a redox reaction. 39 Example: State whether the following reactions are redox reactions or not. 1. 2Al(s) + 2NaOH(aq) + 6H2O → 2NaAl(OH)4(aq) + 3H2(g) Al changed from Al0 to Al3+ H changed from H+ to H0 Therefore it is a redox reaction. 2. Na2CO3(aq) + Ca(OH)2(s) → CaCO3(s) + 2NaOH(aq) Na is +1 in Na2CO3(aq) and NaOH Ca is +2 in Ca(OH)2(s) and CaCO3(s) H is +1 in Ca(OH)2(s) and NaOH O is -2 in Ca(OH)2(s), CaCO3(s) and NaOH There is no change in the oxidation states This is not a redox reaction 40 Example: State whether the following reactions are redox reactions or not. 3. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Fe has changed from Fe3+ to Fe (oxidation state 0) C has changed from C2+ to C4+ Therefore it is a redox reaction. 4. Ca3P2(s) + 6H2O(l) 3Ca(OH)2(aq) + 2PH3(g) No element in this reaction has a change in oxidation number. P is –3 in Ca3P2(s) and PH3(g) Ca is +2 in Ca3P2(s) and Ca(OH)2(aq). This is not a redox reaction 41 5. P4(s) + 5O2(g) → P4O10(s) P has changed from 0 in P4 to +5 in P4O10 O has changed from 0 in O2 to –2 in P4O10 This is therefore a redox reaction. 42  Balancing of Redox Equations Step 1:- Identify the species (atoms/ions) involved in the redox reaction (oxidation number changes) write out their half-reactions. Step 2:- Balance each of the half-reactions atomically in this order. i) atoms other than oxygen and hydrogen ii) oxygen atoms by adding H2O iii) hydrogen atoms by adding H+ Step 3:- Balance each half-reaction electrically by adding necessary number of electrons. 43  Balancing of Redox Equations Step 4:- Ensure the oxidation numbers of the two half-reactions are equal. Obtain the net redox reaction by combining the two half-reactions Step 5:- Simplify the net equation by bringing the same species together. Step 6:- Verify the final equation to be sure that it is balanced atomically and electrically. 44 For example: Balance the following redox reactions in acidic medium. i) MnO4- + Fe2+ Mn2+ + Fe3+ Step 1: Identify the oxidation and reduction half-reactions MnO4- → Mn2+ (reduction half-reaction) Fe2+ → Fe3+ (oxidation half-reaction) Step 2: Balancing each half-reaction atomically MnO4- → Mn2+ + 4H2O (4H2O added for 4 oxygen atoms) Since we now have 8 hydrogen atoms on the right, we need to add 8H+ to the left. MnO4- + 8H+ → Mn2+ + 4H2O Fe2+ → Fe3+ Step 3: Balancing each half-reaction electrically MnO4- + 8H+ → Mn2+ + 4H2O -1 + 8(+1) +2 + 0 (Oxidation number) +7 +2 45 The difference in oxidation numbers of the LHS and RHS of the equation is +5 (ie 7 – 2 = 5). Hence, 5 electrons are added to the LHS MnO4- + 8H+ + 5e- → Mn2+ + 4H2O ……………(1) For Fe2+ → Fe3+ Fe2+ → Fe3+ + e- …………………………………..………(2) (an electron is added to the right-hand side) Step 4: There are 5e- in equation (1) and one in equation (2). Therefore, equation (2) has to be multiplied by 5 so that it will have the same number of electrons as equation (1). MnO4- + 8H+ + 5e- → Mn2+ + 4H2O ………(1) 5Fe2+ → 5Fe3+ + 5e- ………………..…………….(3) MnO4- + 5Fe2+ + 8H+ + 5e- → Mn2+ + 4H2O + 5Fe3+ + 5e- The balanced redox reaction is therefore MnO4- + 5Fe2+ + 8H+ → Mn2+ + 4H2O + 5Fe3+ 46  ii) SO32- + MnO4- + H+ → SO42- + Mn2+ + H2O Step 1: SO32- → SO42- (oxidation) MnO4- → Mn2+ (reduction) Step 2: SO32- → SO42- (left-hand side is one oxygen less, hence H2O is added) SO32- + H2O → SO42- (add 2H+ to account for the H on the right-hand side, RHS) SO32- + H2O → SO42- + 2H+ MnO4- → Mn2+ (balance as in previous example to give) MnO4- + 8H+ → Mn2+ + 4H2O Step 3: Balancing the reaction electrically SO32- + H2O → SO42- + 2H+ -2 0 (2 electrons are to be added to the LHS) SO32- + H2O → SO42- + 2H+ + 2e- ……………(1) MnO4- + 8H+ → Mn2+ + 4H2O becomes MnO4- + 8H+ + 5e- → Mn2+ + 4H2O……………(2) 47  SO32- + H2O → SO42- + 2H+ + 2e- ……………..…(1) MnO4- + 8H+ + 5e- → Mn2+ + 4H2O……………(2) Step 4: Equation (1) has 2 electrons while equation (2) has 5 electrons, therefore equation (1) should be multiplied by 5 and equation (2) multiplied by 2. (1) x 5: 5SO32- + 5H2O → 5SO42- + 10H+ + 10e- (2) x 2: 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O 5SO32- + 5H2O + 2MnO4- + 16H+ + 10e- → 5SO42- + 10H+ + 2Mn2+ + 8H2O + 10e- Step 5: Simplifying the reaction, collecting like terms to the same side 5SO32- + 2MnO4- + 6H+ → 5SO42- + 2Mn2+ + 3H2O 48  Balancing Redox reaction in Basic medium It is easier to use a similar method as those in acidic medium, with extra step. The extra step involves addition of hydroxyl ion equivalent to the hydrogen ion, after the net reaction equation has been obtained. For example, balance the equation below in a basic medium. KMnO4 + Na2SO3 + H2O → MnO2 + Na2SO4 + KOH MnO4- + SO32- + H2O → MnO2 + SO42- + KOH Step 1: MnO4- → MnO2 SO32- → SO42- Step 2: Balancing each half-reaction atomically MnO4- → MnO2 + 2H2O (2H2O added to balance 2O atoms) MnO4- + 4H+ → MnO2 + 2H2O (4H+ added to balance the 4H from the 2H2O) SO32- + H2O → SO42- (H2O added to balance 1 oxygen atoms) SO32- + H2O → SO42- + 2H+ (2H+ added to balance the 2H from the 1H2O) 49 Step 3: Balancing each half reaction electrically MnO4- + 4H+ + 3e- → MnO2 + 2H2O ………………………….(1) (3e- added to the right-hand side to equate the charge to 0) For SO32- + H2O → SO42- + 2H+ + 2e- …………………………(2) (2e- added to the right-hand side to equate the charge to -2) Step 4: Combining the two half-equations (1) x 2: 2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O ……………..……..(3) (2) X 3: 3SO32- + 3H2O → 3SO42- + 6H+ + 6e- …………………………(4) 2MnO4- + 8H+ + 6e- + 3SO32- + 3H2O → 2MnO2 + 4H2O + 3SO42- + 6H+ + 6e- simplifying: 2MnO4- + 2H+ + 3SO32- → 2MnO2 + H2O + 3SO42- Step 5: To balance the reaction in a basic medium add 2OH- to both sides 2MnO4- + 2H+ + 3SO32- + 2OH- → 2MnO2 + H2O + 3SO42- + 2OH- Step 6: Simplying step 5 by collecting like terms, we have 2MnO4- + 3SO32- + 2H+ + 2OH- → 2MnO2 + H2O + 3SO42- + 2OH- 2MnO4- + 3SO32- + 2H2O → 2MnO2 + H2O + 3SO42- + 2OH- 2MnO4- + 3SO32- + H2O → 2MnO2 + 3SO42- + 2OH- 2KMnO4 + 3Na2SO3 + H2O → 2MnO2 + 3Na2SO42- + 2KOH 50 Balance in a basic medium: Cr3+ + IO3- → I - + CrO42- Step 1: Cr3+ → CrO42- IO3- → I- Step 2: Balancing each half-reaction atomically Cr3+ + 4H2O → CrO42- (4H2O added to balance 4 oxygen atoms) Cr3+ + 4H2O → CrO42- + 8H+ (8H+ added to balance for 8H introduced by 4H2O) IO3- → I- + 3H2O (3H2O added to balance 3 oxygen atoms) IO3- + 6H+ → I- + 3H2O (6H+ added to balance the 6H introduced by 3H2O) Step 3: Balancing each half reaction electrically Cr3+ + 4H2O → CrO42- + 8H+ +3 +6 this implies that 3 electrons should be added to the right-hand side Cr3+ + 4H2O → CrO42- + 8H+ + 3e-………………..(1)51  For IO3- + 6H+ → I- + 3H2O +5 -1 this implies that 6 electrons should be added to the left-hand side IO3- + 6H+ + 6e- → I- + 3H2O……………………….…(2) Step 4: Combining the two half-equations (1) x 2 2Cr3+ + 8H2O → 2CrO42- + 16H+ + 6e-……………..(3) (2) IO3- + 6H+ + 6e- → I- + 3H2O………………………….……(2) (3) + (2) 2Cr3+ + 8H2O + IO3- + 6H+ → 2CrO42- + 16H+ + I- + 3H2O simplifying we have 2Cr3+ + 5H2O + IO3- → 2CrO42- + 10H+ + I- Step 5: To balance the reaction in a basic medium add 10 OH- to both sides: 2Cr3+ + 5H2O + IO3- + 10 OH- → 2CrO42- + I- + 10H+ + 10 OH- 2Cr3+ + 5H2O + IO3- + 10 OH- → 2CrO42- + I- + 10H2O Step 6: Simplying step 5 by collecting like terms, we have 2Cr3+ + IO3- + 10 OH- → 2CrO42- + I- + 5H2O 52 Balancing Disproportional Reaction A disproportionation reaction is a reaction in which the same substance (on the reactant side) that is being oxidized is also being reduced. An example: P4(s) + OH-(aq) → H2PO2-(aq) + PH3(g) + H2O Step 1: P4 → H2PO2- P4 → PH3 Step 2: (a) balancing P P4 → 4H2PO2- P4 → 4PH3 (b) balancing O P4 + 8H2O → 4H2PO2- P4 → 4PH3 (c) balancing H P4 + 8H2O → 4H2PO2- + 8H+ P4 + 12H+ → 4PH3 Step 3: Balancing the half-equations electrically P4 + 8H2O → 4H2PO2- + 8H+ + 4e-……………….(1) → 4PH3………………………….(2) 53 P4 + 12H+ + 12e- Balancing a disproportionation reaction Step 4: Combining the half-reactions (1) x 3: 3P4 + 24H2O → 12H2PO2- + 24H+ + 12e-…….(3) (2): P4 + 12H+ + 12e- → 4PH3……………………..(2) (3) + (2): 4P4 + 24H2O → 12H2PO2- + 4PH3 + 12H+ Step 5: For the reaction to be in a basic medium, add 12 OH- to both sides 4P4 + 24H2O + 12 OH- → 12H2PO2- + 4PH3 + 12H+ + 12 OH- 4P4 + 24H2O + 12 OH- → 12H2PO2- + 4PH3 + 12H2O Step 6: Simplifying the net equation by collecting like terms we have 4P4 + 12H2O + 12 OH- → 12H2PO2- + 4PH3 (dividing through by 4 we have) P4 + 3H2O + 3 OH- → 3H2PO2- + PH3 54  Balancing a conproportionation reaction In this case, we may have the same element in two different compounds in the reactants coming together to form the only product or one of the products. In this process, one is oxidized to the product, while the other is reduced to the product. This example is known as conproportionation reaction. Such equations can be balanced in the same way as the disproportionation reaction. An example is HIO3 + HI + H+ → I2 + H2O There are a number of practical problems that involves redox reactions, in such problems the redox equations must be balanced first and then apply the mole ratio obtained from it 55 for calculations.  Exercise 1 Calculate the oxidation number of carbon in the following compounds a) CO b) CO2 c) COCl2 d) CaCO3 e) CCl4 f) CH4 g) H2CO h) HCO2H What is the oxidation number of manganese in each of the following compunds? a) KMnO4 b) Mn2O7 c) MgMnO3 d) MnO e) MnO2 Assign oxidation number to each of the elements in the following compounds a) OF2 b) HOF c) O2F2 d) PH3 e) CaH2 f) PH4+ g) ClF3 h) MnO42- 56  Oxidizing and Reducing agents The substance that is oxidized is the reducing agent because it is the one that made it possible for the other substance to be reduced. The substance that is reduced is therefore the oxidizing agent. A compound that is an oxidizing agent in a reaction can be a reducing agent in another reaction. This is essentially due to whether it is reduced or oxidized in the reaction. E.g, H2O2 can behave as both an oxidizing agent and a reducing agent depending on the substance it is reacting with as in the following reactions i) H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2H2O + 2Fe3+(aq) ii) 5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 8H2O + 2Mn2+(aq) + 5O2(g) iii) H2O2(aq) + Cl2(g) + 2 OH-(aq) → 2H2O + 2Cl-(aq) + O2(g) In (i) H2O2 oxidized Fe2+ to Fe3+, therefore it is an oxidizing agent in (i). (ii) H2O2 reduced MnO4- to Mn2+, it is therefore a reducing agent in (ii). (iii) H2O2 reduced Cl2 to Cl-, it is therefore a reducing agent in this (iii). 57 Some common strong oxidizing agents are Permanganate ion (MnO4-) Ozone (O3) Hypochloride ion (OCl-) Some common strong reducing agents are Carbon (C) Thiosulphate ion (S2O32-) Iodide ion (I-) 58  Exercise 2 State whether the following reactions are redox reactions or not. If they are, identify the oxidizing agents and the reducing agents. a) PH3(g) + HCl(g) PH4Cl(s) b) SiO2(aq) + 3C(s) SiC(s) + 2CO(g) c) Cu(s) + 2H2SO4(aq) CuSO4(aq) + SO2(g) + 2H2O(l) d) NaI(aq) + 3HOCl(aq) NaIO3(aq) + 3HCl(aq) e) I2(g) + 10HNO3(aq) 2HIO3(aq) + 10NO2(g) + 4H2O(l) f) 2HBr(aq) + H2SO4(aq) SO2(g) + Br2(g) + 2H2O(l) Write a balanced equation for each of these half-reactions in an acidic medium a) Cl2 ClO3- b) H4IO6- I c) Fe Fe(OH)2 d) BiO3- Bi3+ e) NO3- NH4+ 59  Exercise 3 Balance the following redox reactions, which occur in an acid medium. a) UO2+ + NO3- + H+ UO22+ + NO + H2O b) S2O32- + MnO4- SO42- + Mn2+ c) H2S + CrO42- S8 + Cr3+ d) Br2 + SO2 H2SO4 + HBr e) BrO3- + Cr3+ Br2 + HCrO4- Write a balanced equation for the following reactions, which occur in a basic medium a) As2S3 + H2O2 AsO43- + H2O + SO42- b) NO + MnO4- NO3- + MnO2 c) Fe(OH)2 + O2 Fe(OH)3 + H2O d) CN- + MnO4- MnO2 + CNO- + H2O e) NH3 + MnO4- N2 + MnO2 60 Some common strong oxidizing agents are Permanganate ion (MnO4-) Ozone (O3) Hypochloride ion (OCl-) Some common strong reducing agents are Carbon (C) Thiosulphate ion (S2O32-) Iodide ion (I-) 61 Example 9: oxidation H2S + Cl2 → S + 2HCl (R.A) (O.A) reduction H2S Cl2 is reduced to HCl (Cl2 is the O.A) 62  ELECTROCHEMISTRY Is the study of the relationship between chemical reactions and electrical energy Is the study of how chemical reactions can be used to produce electricity and how electricity can be used to produce chemical reactions. Chemical reaction Electrical Energy Electricity involves the flow of electrons. Applications of electrochemistry include: as electric power sources in batteries and fuel cells , manufacturing of chemicals, refining of metals, Electroplating, methods of controlling corrosion 1  ELECTROCHEMISTRY Electricity passes through a circuit under the influence of a potential or voltage, (the driving force of the movement of charge) The potential difference is the work done per unit charge to move a unit charge from one point to another in an electric field The tendency of a metal to loss or gain electron oxidation potential -tendency to lose electrons reduction potential - tendency to gain electrons Electrode- a conductor which allows the passage of electric current 2  ELECTRODE POTENTIAL Zn(s) in Cu(NO3)2(aq) Cu(s) in Zn(NO3)2(aq) Cu(s) in AgNO3(aq) Deposit of Cu(s) NO METAL DEPOSIT Deposit of Ag(s) The difference is due to their metal ion electrode potential. A metal cannot displace another metal with a higher negative electrode potential. Hence, Cu (E0 = +0.34V) cannot displace Zn (Eo = -0.76V) from Zn2+ solution, but can displace Ag (Eo = +0.80 V) from Ag+ solution 3  ELECTROCHEMISTRY When a strip of metal, M, is put in a solution containing metal ion Mn+, the metal, M, is called an electrode The combination of the metal, M, and the solution containing Mn+ is called a half-cell or a half-equation (electrode and solution). There are three types of possible interactions between the electrode and the solution (electrolyte). 4  POSSIBLE INTERACTIONS BETWEEN THE ELECTRODE AND ELECTROLYTE Mn+(aq) collide with M(s) no change Mn+ may gain ‘n’ electrons and be converted to M(s), (Mn+ is reduced). Mn+(aq) + ne- M(s) M(s) may lose ‘n’ electrons and enter into the solution. (M(s) is oxidized) M(s) Mn+(aq) + ne- These possibilities can be represented as: M(s)  Mn+(aq) + ne-. 5  POSSIBLE INTERACTIONS BETWEEN THE ELECTRODE AND ELECTROLYTE If we have two different metals, M (electrodes), electrons will flow from electrode of higher negative charge density (electrode potential) to electrode of lower negative potential. A small difference in electrode potentials is enough to set up electric current. This difference can only be measured when the two half-cells are connected. 6 This set-up is called electrochemical cell The electrodes are connected by a wire to permit the flow of electrons the solutions are connected through a salt bridge. It allows the passage of (anode, +ive electrode, (cathode, -ive electrode, electric current in oxidation reaction; reduction reaction, form of migration always on the LHS). always on the RHS). of ions. 7  The Electrochemical Cell Oxidation: Zn(s) Zn2+(aq) + 2e- Reduction: Cu2+(s) + 2e- Cu(s) Net Equation: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) The reading of potentiometer is 1.10V This is what is called the potential difference between the two half- cells. AKA cell potential or cell voltage or electromotive force (emf) because it is a driving force 8 Without a salt bridge… there will be excess Zn2+ (since Zn2+ is produced), i.e a net positive charge at the Zn half-cell A deficiency of Cu2+ (since Cu2+ is converted to Cu), , ie a net negative charge at With a salt bridge… the Cu half-cell Anions from the salt bridge (SO42-) migrates to the Zn half-cell electric current would Cations (Na+) migrates to the Cu not continue to flow. 9 half-cell. CURRENT FLOWS  POTENTIOMETER measures the potential difference in the cell precisely by generating enough voltage that opposes the current in the cell. It produces voltage of known values equal to the value of voltage from the cell. Special voltmeters that draw only a very little current can also be used to measure the voltage generated from the cell directly. 10  CELL NOTATION cell notation -to describe electrochemical cell. Boundaries (interface) between electrodes and solutions - a single vertical line (│). Boundaries between solutions - a double vertical line (║), for salt bridge and a series of vertical dots (  ) for porous plug or membrane. Zn(s) │ Zn2+(aq) ║ Cu2+(aq) │ Cu(s) Oxidation salt bridge Reduction (Anode) (Cathode) (LHS) (RHS) 11  TYPES OF ELECTROCHEMICAL CELLS Electrochemical Cells Voltaic/galvanic cells Electrolytic cells Spontaneous reactions. non-spontaneous reactions. The chemical changes Applied electricity produce electricity. produces chemical They have positive change voltage. They have negative voltage 12 Example: In the Daniel’s cell Zn(s) displaces copper(II) in aqueous solution. (a) write the half-cells and the net equation (b) write a cell notation for the spontaneous reaction. Solution: (a) Zn(s) displaces Cu(II) ion , Zn goes into solution to become Zn2+(aq) and Cu2+(aq) gets out of the solution to become Cu(s). Oxidation : Zn(s) → Zn2+(aq) + 2e- Reduction : Cu2+(aq) + 2e- → Cu(s) Net equation : Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). (b) The notation for reaction will be Zn│Zn2+(aq)║Cu2+(aq)│Cu(s) 13  STANDARD ELECTRODE POTENTIAL It would be very advantageous to be able to measure the potential of a single electrode For instance : Cu2+(aq) │ Cu(s). The potential would be a measure of the tendency of the reduction process to occur: Cu2+ + 2e- → Cu(s) A standard reference electrode is needed for this. This reference electrode is called standard hydrogen electrode (SHE). Its electrode potential is taken as zero. 14  STANDARD HYDROGEN ELECTRODE Pt SHE is made by e- passing H2(g) at 1 atm pressure H2 (g, 1 atm) Salt bridge over an inert Pt metal surface in a solution of H+ + whose activity H (a= 1 M) is 1 at 250C 15 The standard electrode (reduction) potential of Cu2+/Cu electrode in the diagram below is 0.337V. The emf obtained is the standard electrode potential or the standard reduction potential of the other cell (Eo). Oxidation occur in the H2 cell Reduction takes place in the other cell (Cu cell). 16 Pt,H2(g,1atm) │H+(1M) ║ Cu2+(1M)│Cu(s) (Eo =0.337 V) Oxidation: H2 (g, 1 atm) → 2H+(1M) + 2e- Reduction: Cu2+(aq)(1M) + 2e- → Cu(s) Net: H2(g,1atm) + Cu2+(1M) → 2H+(1M) +Cu(s) Eocell = +0.337 V. The positive value of EoCell = +0.337 V implies that Cu2+ is more easily reduced than H+(1M). In the case of Zn2+(1M)│Zn(s) Pt, H2 (g,1 atm) │H+(1M) ║ Zn2+(1M)│Zn(s) Eocell = -0.763V The negative value of EoCell = -0.763 V implies that Zn2+ is less easily reduced than H+(1M). (Oxidation, Current will flow in the opposite direction, from Zn 17  STANDARD ELECRODE (REDUCTION) POTENTIALS. Half-reactions in Acidic solution. Eo/V F2 + 2e- → 2F-(aq) 2.866 O3 + 2H+(aq) + 2e- → O2(g) + H2O 2.07 S2O82-(aq) + 2e- → 2SO42-(aq) 2.01 H2O2(aq) +2H+(aq) → 2H2O 1.77 Ce4+(aq) + e- → Ce3+(aq) 1.72 Au+(aq) + e- → Au(s) 1.692 MnO4-(aq) + 8H+ +5e- → Mn2+ + 4H2O 1.52 PbO2(s) + 4H+(aq) + 2e- → Pb2+ + 2H2O 1.455 Cl2 + 2e- → 2Cl-(aq) 1.360 Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O 1.332 O2(g) + 4H+(aq) + 4e- → 2H2O 1.229 18  STANDARD ELECRODE (REDUCTION) POTENTIALS. Half-reactions in Acidic solution. Eo/V MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O 1.224 2IO3-(aq) + 12H+(aq) + 10e- → I2 + 6H2O 1.195 Pt2+ + 2e- → Pt 1.18 Br2(l) + 2e- → 2Br-(aq) 1.065 NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O 0.96 Hg2+(aq) + 2e- → Hg(s) 0.851 Ag+(aq) + e- → Ag(s) 0.800 Fe3+(aq) + e- → Fe2+(aq) 0.771 O2(g) + 2H+(aq) + 2e- → H2O2(aq) 0.682 I2(s) + 2e- → 2I-(aq) 19 0.535  STANDARD ELECRODE (REDUCTION) POTENTIALS. Half-reactions in Acidic solution. Eo/V Cu+(aq) + e- → Cu(s) 0.52 H2SO3(aq) + 4H+(aq) + 4e- → S(s) +3H2O 0.45 Cu2+(aq) + 2e- → Cu(s) 0.337 Sn4+(aq) + 2e- → Sn2+(aq) 0.154 S(s) + 2H+(aq) +2e- → H2S(g) 0.141 2H+(aq) + 2e- → H2(g) 0.0000 Pb2+(aq) + 2e- → Pb(s) -0.126 Sn2+(aq) + 2e- → Sn(s) -0.136 Cr3+(aq) + e- → Cr2+(aq) -0.407 Fe2+(aq) + 2e- → Fe(s) 20 -0.440 STANDARD ELECRODE (REDUCTION) POTENTIALS. Half-reactions in Acidic solution. Eo/V Zn2+(aq) + 2e- → Zn(s) -0.763 Al3+(aq) + 3e- → Al(s) -0.1.66 Mg2+(aq) + 2e- → Mg(s) -2.375 Na+(aq) + e-→ Na(s) -2.714 Ca2+(aq) + 2e- → Ca(s) -2.76 K+(aq) + e- → K(s) -2.925 Li+(aq) + e- → Li(s) -3.045 21  STANDARD ELECRODE (REDUCTION) POTENTIALS. BASIC SOLUTION Eo/V O3(g) + H2O +2e- → O2(g) + 2OH- +1.24 OCl-(aq) + H2O + 2e- → Cl- + 2OH- +0.89 O2(g) + 2H2O + 4e- → 4OH-(aq) +0.401 CrO42-(aq) + 4H2O + 3e- → Cr(OH)3(s) + 5OH- -0.13 S(s) + 2e- → S2-(aq) -0.48 2H2O + 2e- → H2(g) + 2OH-(aq) -0.828 SO42-(aq) + H2O + 2e- → SO32-(aq) + 2OH-(aq)-0.93 22  CALCULATION OF STANDARD ELECTRODE (REDUCTION) POTENTIAL OF A CELL Eo is calculated by combining the standard electrode potential of the half-cell reactions. The steps are: Step 1: Write out the reduction equations for the half- cells and their standard reduction potential. Step 2: For oxidation half-cell reactions, reverse the reduction equation. Hence Eo will become -Eo. Step 3: Balance the number of electrons in the 2 half- equations. Note however that potential is an intensive property and therefore Eo values are not affected by the multiplying factors of the half-cell reactions. Step 4: Add the oxidation and reduction equations and do the same for their Eo. 23 Example: Calculate the Eo cell of a Daniel cell, the half-cells equation is Zn│Zn2+(aq) (1M)║Cu2+(aq) (1M)│Cu(s) Solution: The reduction half equations involved are Cu2+(aq) + 2e- → Cu(s) = 0.337 V (or EoCu2+/Cu = 0.337 V) Zn2+(aq) + 2e- → Zn(s) = -0.763 V (or EoZn2+/ Zn =-0.763V) The oxidation equation is Zn(s) → Zn2+(aq) + 2e- Eo = 0.763V (sign changed from negative to positive) The Eo cell for the cells will be EoZn/Zn2+ + EoCu2+/Cu = Eo cell. Oxidation: Zn(s) → Zn2+(aq) + 2e- Eo = 0.763V Reduction: Cu2+(aq) + 2e- → Cu(s) Eo = 0.337V Net: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Eo = 1.100V 24 Example: A new battery under test for commercial uses is the Zinc-Chlorine battery, its Eocell is 2.123 V. Given that the standard reduction potential of Zn is -0.763V, calculate the standard reduction potential of Chlorine. Solution: The net reaction of this cell is Zn(s) + Cl2(g) → ZnCl2(aq) Therefore, the half-equations are Oxidation: Zn(s) → Zn2+ + 2e- Eo = -(-0.763) Reduction: Cl2(g) + 2e- → 2Cl-(aq) Eo = x V Zn (s) + Cl2(g) → Zn2+ + 2Cl- (aq) Eocell =2.123V. EoZn/Zn2+ + EoCl2/2Cl- = Eo cell. 0.763V + Eo Cl2/2Cl- = 2.123V. Eo Cl2/2Cl- = 2.123 - 0.763V. 25 = 1.360V. It should be noted that, the oxidation and reduction standard potentials couldn’t just be added up or subtracted if there are electrons left in the net reaction. For example, calculating the Eo of Cu+ + e- → Cu From the following half-equations cannot just be treated like it was done in the previous example Cu2+ + e- → Cu+ Eo = 0.153 V Cu2+ + 2e- → Cu Eo = 0.337 V. giving Cu+ → Cu2+ + e- Eo = -0.153 V Cu2+ + 2e- → Cu Eo = 0.337 V. Cu+ + e- → Cu Eo  0.184 V. This is wrong! NOTE that once there are electrons in the net equation, Eo should be calculated through the Go procedure. 26 The correct solution in this case, where an electron is left in the net equation, has to be calculated through the Go of the two half reactions. Cu2+ + e- → Cu+ G1o = -nE1oF = -1 x 0.153 x 96485 J mol-1 Cu2+ + 2e- → Cu G2o = -nE2oF = - 2 x 0.337 x 96485 J mol-1. The Gxo for the reaction of Cu+ + e- → Cu is Gxo = G2o - G1o = {(- 2 x 0.337 x 96485) - (-1 x 0.153 x 96485)} J mol-1 = (0.153 –0.674) 96485 J mol-1 = -0.521 x 96485 J mol-1 Since Gxo for the reaction of Cu+ + e- → Cu = -nxExoF and nx = 1, F = 96485 C mol-1 -Exo x 96485 = -0.521 x 96485 Therefore Exo = 0.521 V For any of the previous examples, going through the Go will 27 still give the same answer as the direct additions For example, calculating the Eocell for the reaction Zn + Cu2+ → Zn2+ + Cu giving that Zn2+ + 2e- → Zn E1o = -0.763 V Cu2+ + 2e- → Cu E2o = 0.337 V. Gxo = G2o - G1o -nxExoF = (-n2E2oF) – (-n1E1oF) -2ExoF = {-2 x 0.337F} – {-2 x (-0.763)F} -2ExoF = -2F (0.337 + 763) Exo = 0.337 + 0.763 V = 1.100 V It is important to note that once there are electrons in the net equation, Eo should be calculated through the Go procedure. 28  FEASIBILITY OF REDOX REACTIONS AND GIBB’S ENERGY CHANGE IN A CELL. Feasibility of redox reactions simply implies the spontaneity of the reaction. If a reaction is spontaneous, it shows that the reaction occurs without any external input. The maximum useful work done by a system at constant pressure is equal to the change in Gibb’s free energy of the system. ∆G = Wmax(useful). Since work done by a system is negative, therefore the ∆G of a spontaneous reaction is negative, that is, ∆G < 0. For a non-spontaneous reaction ∆G > 0. Wmax(useful). = Welec (Welec is Welectrochemical cell) because it’s the only non pressure-volume work in this process. 29 FEASIBILITY OF REDOX REACTIONS AND GIBB’S ENERGY CHANGE IN A CELL. Welec = - nFEcell where ‘n’ is the number of moles of electrons F is Faraday’s constant, that is the electric charge per mole of electrons and is approximately 96485 Coulumbs per mole of