Bài giảng hóa học đại cương i

THUY LOI UNIVERSITY Lecture Notes GENERAL CHEMISTRY (The contents of the lectures may be changed without notice) HÀ NỘI, 01 - 2016 24-Feb-16 Chapter 16 Chemistry Kinetics: Rates and Mechanisms The Molecular Nature of Matter and Change 24-Feb-16 of Chemical Reactions 24-Feb-16 Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Factors That Influence Reaction Rate 16.2 Expressing the Reaction Rate 16.1 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Concentration Changes over Time Factors That Influence Reaction Rate 16.5 The Effect of Temperature on Reaction Rate 16.6 Explaining the Effects of Concentration and Temperature 16.7 Reaction Mechanisms: Steps in the Overall Reaction 16.8 Catalysis: Speeding Up a Chemical Reaction 24-Feb-16 24-Feb-16 24-Feb-16 Reactant Concentration: C ↑→ reaction rate ↑ Physical State: the greater the surface area per unit volume, the more contact it makes with other reactants, and the faster the reaction Temperature: T ↑→ reaction rate ↑ 24-Feb-16 24-Feb-16  Reaction: aA + bB → cC + dD v (mol L-1 s-1); C (mol/L); t (s) 16.3 where a, b, c, and d are the coefficients for the balanced equation The Rate Law & Reaction Orders  Example 1: H2 + I2 → 2HI 24-Feb-16 24-Feb-16 Example 2: For any general reaction aA + bB → cC + NO(g) + O3(g) → NO2(g) + O2(g) dD vrxn = k[NO][O3] The rate law: → the reaction order with respect to NO is vrxn = k[A]m.[B]n → the reaction order with respect to O3 is → the reaction is second order overall • The term k is the rate constant, which is specific for a - given reaction at a given temperature 2NO(g) + 2H2(g) • The exponents m and n are individual reaction orders vrxn = and are determined by experiment → N2(g) + 2H2O(g) k[NO]2[H2]1 → the reaction is second order with respect to NO • m, n are also called to be reaction orders with respect to → the reaction is first order with respect to H2 A, B •24-Feb-16(m+n) is called to be overall reaction order → the reaction is third order overall 24-Feb-16 10 How to determine Exercise Reaction Orders from experiments Determine the individual order with respect to each reactant and the overall reaction order from the given rate law To determine the values of m and n, (a)2NO(g) + O2(g) → 2NO2(g); • we run a series of experiments in which one rate = k[NO]2[O2] reactant concentration changes while the other (b)CH3CHO(g) → CH4(g) + CO(g); is kept constant, rate = k[CH3CHO]3/2 • and we measure the effect on the initial rate in (b)H2O2(aq) + 3I-(aq) + 2H+(aq) → I3-(aq) + 2H2O(l); rate = 24-Feb-16 k[H2O2][I-] each case 24-Feb-16 11 12 24-Feb-16 Example 3: Determine the individual order with respect to each Exercise reactant and the overall reaction order from the given table Determine the individual order with respect to each reactant below: and the overall reaction order from the given table below: aA + bB → products O2(g) + 2NO(g) → 2NO2(g) Experiment Initial Rate (mol/L·s) Initial [A] (mol/L) Initial [B] (mol/L) 1.75x10-3 2.50x10-2 3.00x10-2 3.50x10-3 5.00x10-2 3.00x10-2 Experiment Initial Rate (mol/L·s) [O2] [NO] 3.50x10-3 2.50x10-2 6.00x10-2 3.21x10-3 1.10x10-2 1.30x10-2 7.00x10-3 5.00x10-2 6.00x10-2 6.40x10-3 2.20x10-2 1.30x10-2 12.48x10-3 1.10x10-2 2.60x10-2 9.60x10-3 3.30x10-2 1.30x10-2 28.8x10-3 1.10x10-2 3.90x10-2 Solution: Initial Reactant Concentrations (mol/L) vrxn = k[A]m.[B]n From the data, we make mathematical equations, solve the 24-Feb-16 Example 4: equations → the value of k, m, n 13 24-Feb-16 aA + bB → products SOLUTION: we count the number of particles as concentrations (a) For reactant A (red): experiments and have the same number of particles of B, but the number of particles of A doubles The rate doubles Thus the order with respect to A is For reactant B (blue): experiments and show that when the number of particles of B doubles (while A remains constant), the rate quadruples The order with respect to B is The overall order is + = (b) Rate = k[A][B]2 (c) Between experiments and 4, the number of particles of A doubles while the number of particles of B does not change The rate should double, so rate = x 2.0x10-4 = 4.0x10-4mol/L·s (a) What is the reaction order with respect to A (red), B (blue)? The overall order? (b) Write the rate law for the reaction (c) Predict the initial rate of experiment Expt # Initial rate (mol/L·s) 14 0.50x10-4 1.0x10-4 2.0x10-4 ? 24-Feb-16 15 24-Feb-16 16 First-order rate equation: Second-order rate equation: 16.4 Integrated Rate Laws Zero-order rate equation: 24-Feb-16 17 24-Feb-16 18 24-Feb-16 Example 5: Units of k At 1000oC, cyclobutane (C4H8) decomposes in a first-order The value of k is easily determined from reaction, with the very high rate constant of 87 s-1, to two experimental rate data molecules of ethylene (C2H4) (a) If the initial C4H8 concentration is 2.00 M, what is the The units of k depend on the overall reaction order concentration after 0.010 s? Overall Reaction Order Units of k (t in seconds) mol/(L·s) 1/s L/(mol·s) L2/(mol2·s) (b)What fraction of C4H8 has decomposed in this time? SOLUTION: 24-Feb-16 19 24-Feb-16 Graphs of 0th, 1st, 2nd reaction order Reaction Integrated Rate Law Order Zero-order reaction Linear form Linear graph [A]t= -kt + [A]o y = ax + b [A]t & time ln[A]t= -kt + ln[A]o y = ax + b ln[A]t & time 1/[A]t= +kt + 1/[A]o y = ax + b 1/[A]t & time [A]t & time 24-Feb-16 21 20 First-order reaction ln[A]t & time Second-order reaction 1/[A]t & time 24-Feb-16 22 Example 6: Reaction Half-life Find the rate constant of the reaction, if substance A (green) • The half-life (t1/2) for a reaction is the time taken for the decomposes to two other substances, B (blue) and C (yellow), in a first- concentration of a reactant to drop to half its initial value • First-order reaction, t1/2 does not depend on the initial conc • Second-order reaction, t1/2 is inversely proportional to the order gaseous reaction The following scenes represent the reaction mixture at the start of the reaction (at 0.0 s) and after 30.0 s initial concentration • Zero-order reaction, t1/2 is directly proportional to the initial concentration Zero-order reaction First-order reaction Second-order reaction SOLUTION: 24-Feb-16 23 24-Feb-16 → find t1/2 →k 24 24-Feb-16 Exercise Draw the molecular scenes of the reaction mixture at t = 60.0 s and Overview about Reaction Rate 90.0 s, if substance A (green) decomposes to two other substances, B (blue) and C (yellow), in a first-order gaseous reaction The following Zero Order First Order Second Order Rate law rate = k rate = k[A] rate = k[A]2 Units for k mol/L·s 1/s L/mol·s Half-life [A]o/2k ln2/k 1/(k[A]o) Integrated rate law [A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0 1/[A]t = kt +1/[A]0 Plot for straight line [A]t vs t ln[A]t vs t 1/[A]t vs t Slope, y intercept -k, [A]0 -k, ln[A]0 k, 1/[A]0 scenes represent the reaction mixture at the start of the reaction (at 0.0 s) and after 30.0 s 24-Feb-16 24-Feb-16 25 26 • Temperature has a dramatic effect on reaction rate For many reactions, an increase of 10°C will double or triple the rate Experimental data shows that k increases exponentially as T increases 16.5 • The Arrhenius equation: Temperature and the Rate Constant k = rate constant A = frequency factor Ea = activation energy R = 8.314 J/mol.K Higher T larger k increased rate Smaller Ea 24-Feb-16 27 24-Feb-16 28 Activation Energy • In order to be effective, collisions between particles must exceed a certain energy threshold • When particles collide effectively, they reach an activated state The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction • The lower the activation energy, the faster the reaction 24-Feb-16 Ea 29 24-Feb-16 30 24-Feb-16 Exothermic reaction, ∆Hrxn < ∆Hrxn > ∆Hrxn < ∆Hrxn = Ea,for – Ea,rev 24-Feb-16 31 24-Feb-16 Endothermic reaction, ∆Hrxn > 32 Calculating Activation Energy Ea can be calculated from the Arrhenius equation: At two different temperatures k = rate constant Ea = activation energy R = 8.314 J/mol.K ∆Hrxn = Ea,for – Ea,rev 24-Feb-16 33 24-Feb-16 34 Graphs of Activation Energy Integrated Rate Laws Linear form Linear graph y = ax + b lnk & 1/T 16.6 Effects of Concentration & Temperature 24-Feb-16 24-Feb-16 35 36 24-Feb-16 Temperature and Collision Energy Collision Theory • The basic principle of collision theory is that particles • An increase in temperature causes an increase in the must collide in order to react kinetic energy of the particles This leads to more frequent • collisions and reaction rate increases An increase in the concentration of a reactant leads to a larger number of collisions, hence increasing reaction rate • • sufficient energy equal to or greater than Ea increases The number of collisions depends on the product of the numbers of reactant particles, not their sum • At a higher temperature, the fraction of collisions with Reaction rate therefore increases Concentrations are multiplied in the rate law, not added 24-Feb-16 37 24-Feb-16 38 Example 7: Transition State Theory Find Ea if the decomposition of hydrogen iodide, • 2HI(g) → H2(g) + I2(g) An effective collision between particles leads to the formation of a transition state or activated complex has rate constants of 9.51x10-9 L/mol·s at 500 K, and 1.10x10-5 • L/mol·s at 600 K partial bonds It is a transitional species partway between SOLUTION: reactants and products • The transition state is an unstable species that contains Transition states cannot be isolated, and exists at the point of maximum potential energy • The energy required to form the transition state is the activation energy 24-Feb-16 39 24-Feb-16 40 Transition State & Intermediate State • An intermediate is a short-lived unstable molecule in a reaction which is formed between the reaction when reactants change into products • Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones) • An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle 24-Feb-16 24-Feb-16 41 42 24-Feb-16 •Transition states are local energy maximums and have partial bonds This might be one of the reasons why they can’t be isolated as intermediates 16.7 Reaction Mechanisms 24-Feb-16 24-Feb-16 43 44 Reaction Mechanisms • Rate Laws for General Elementary Steps The mechanism of a reaction is the sequence of single reaction steps that make up the overall equation • Elementary Step The individual steps of the reaction mechanism are called Molecularity Rate Law elementary steps because each one describes a single A → product Unimolecular Rate = k[A] molecular event 2A → product Bimolecular Rate = k[A]2 • A + B → product Bimolecular Rate = k[A][B] 2A+B → product Termolecular Rate = k[A]2[B] Each elementary step is characterized by its molecularity, the number of particles involved in the reaction • The rate law for an elementary step can be deduced from the The molecularity of each step equals the total number of reactant particles reaction stoichiometry – reaction order equals molecularity for an elementary step only 24-Feb-16 45 24-Feb-16 46 Example 8: The following elementary steps are proposed for a reaction mechanism: The Rate-Determining Step of a Reaction (1) NO2Cl(g) → NO2(g) + Cl(g) (2) NO2Cl(g) + Cl(g) → NO2(g) + Cl2(g) • (a) Write the overall balanced equation The slowest step in a reaction is the rate-determining or rate- limiting step (b) Determine the molecularity of each step • (c) Write the rate law for each step SOLUTION: The rate law for the rate-determining step becomes the rate law for the overall reaction a) Make the sum of the elementary steps to find the overall equation Example 9: the reaction NO2(g) + CO(g) → NO(g) + CO2(g) has 2NO2Cl(g) → 2NO2(g) + Cl2(g) been proposed to occur by a two-step mechanism: b) The molecularity of each step equals the total number of reactant particles Step(1) is unimolecular Step(2) is bimolecular c) We write the rate law for each step using the molecularities as reaction orders (1) NO2(g) + NO2(g) → NO3(g) + NO(g) [slow; rate-determining] (2) NO2(g) + CO(g) → NO2(g) + CO2(g) [fast] Overall rate law = slow rate law → vrxn = k[NO2]2 ratestep1 = k1[NO2Cl] ratestep2 = k2[NO2Cl][Cl] 24-Feb-16 47 24-Feb-16 48 10-Apr-16 Sample Problem 20.6 Using Molecular Scenes to Determine the Signs of DH, DS, and DG PROBLEM: The following scenes represent a familiar phase change for water (blue spheres) Sample Problem 20.6 SOLUTION: (a) (a) What are the signs of DH and DS for this process? Explain The scene represents the condensation of water vapor, so the amount of gas decreases dramatically, and the separated molecules give up energy as they come closer together (b) Is the process spontaneous at all T, no T, low T, or high T? Explain PLAN: (a) From the scenes, we determine any change in the amount of gas, which indicates the sign of DS, and any change in the freedom of motion of the particles, which indicates whether heat is absorbed or released (b) To determine reaction spontaneity we need to consider the sign of DG at different temperatures 20-55 Sample Problem 20.7 DS < and DH > (b) Since DS is negative and DH is positive, the –TDS term is positive In order for DG to be < 0, the temperature must be low The process is spontaneous at low temperatures 20-56 Determining the Effect of Temperature on ΔG PROBLEM: A key step in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g): 2SO2(g) + O2(g) → 2SO3(g) Sample Problem 20.7 SOLUTION: (a) At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K (a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how DG will change with increasing T (b) Assuming DH and DS are constant with increasing T, is the reaction spontaneous at 900.° C? DG < at 209 K (= 25°C), so the reaction is spontaneous With DS < 0, the term -TDS > and this term will become more positive at higher T DG will become less negative, and the reaction less spontaneous, with increasing T (b) DG = DH - TDS Convert T to K: 900 + 273.15 = 1173 K Convert S to kJ/K: -187.9 J/K = 0.1879 kJ/K PLAN: We note the sign of DG to see if the reaction is spontaneous and the signs of DH and DS to see the effect of T We can then calculate whether or not the reaction is spontaneous at the higher temperature DG = -198.4 kJ – [1173 K)(-0.187.9 kJ/K) = 22.0 kJ DG > 0, so the reaction is nonspontaneous at 900.°C 20-57 20-58 Figure 20.15 The effect of temperature on reaction spontaneity The sign of DG switches at T = DH ΔS Sample Problem 20.8 Finding the Temperature at Which a Reaction Becomes Spontaneous PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is nonspontaneous DG = 8.9 kJ) Calculate the temperature at which the reaction becomes spontaneous PLAN: We need to calculate the temperature at which DG crosses over from a positive to a negative value We set DG equal to zero, and solve for T, using the values for DH and DS from the text SOLUTION: T= DH DS = 58.1 kJ = 352 K 0.165 kJ/K At any temperature above 352 K (= 79°C), the reaction becomes spontaneous 20-59 20-60 10-Apr-16 Chemical Connections Figure B20.1 The coupling of a nonspontaneous reaction to the hydrolysis of ATP Chemical Connections Figure B20.2 The cycling of metabolic free energy A spontaneous reaction can be coupled to a nonspontaneous reaction so that the spontaneous process provides the free energy required to drive the nonspontaneous process The coupled processes must be physically connected 20-61 20-62 Chemical Connections DG, Equilibrium, and Reaction Direction Figure B20.3 ATP is a high-energy molecule A reaction proceeds spontaneously to the right if Q < K; Q < so ln Q < and DG < K K A reaction proceeds spontaneously to the left if Q > K; Q > so ln Q > and DG > K K A reaction is at equilibrium if Q = K; Q = so ln Q = and DG = K K When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A) and the increase in resonance stabilization (B) causes a large amount of energy to be released 20-64 20-63 DG, Q, and K ΔG and the Equlibrium Constant DG = RT ln Q = RT lnQ – RT lnK K For standard state conditions, Q = and If Q and K are very different, DG has a very large value (positive or negative) The reaction releases or absorbs a large amount of free energy If Q and K are nearly the same, DG has a very small value (positive or negative) The reaction releases or absorbs very little free energy DG° = -RT lnK A small change in DG° causes a large change in K, due to their logarithmic relationship As DG° becomes more positive, K becomes smaller As DG° becomes more negative, K becomes larger To calculate DG for any conditions: DG = DG° + RT lnQ 20-65 20-66 10-Apr-16 Sample Problem 20.9 Using Molecular Scenes to Find DG for a Reaction at Nonstandard Conditions PROBLEM: These molecular scenes represent three mixtures in which A2 (black) and B2 (green) are forming AB Each molecule represents 0.10 atm The equation is Table 20.2 The Relationship Between DG° and K at 298 K DG° (kJ) K Significance 100 3x10-18 50 2x10-9 10 2x10-2 7x10-1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035 no forward reaction; {Essentially reverse reaction goes to completion and reverse reactions proceed {Forward to same extent A2(g) + B2(g) REVERSE REACTION 9x10-36 FORWARD REACTION 200 reaction goes to completion; {Forward essentially no reverse reaction 20-67 (a) If mixture is at equilibrium, calculate K (b) Which mixture has the most negative DG, and which has the most positive? (c) Is the reaction spontaneous at the standard state, that is PA2= PB2 = PAB = 1.0 atm? 20-68 Sample Problem 20.9 Sample Problem 20.9 PLAN: (a) Mixture is at equilibrium, so we first write the expression for Q and then find the partial pressure of each substance from the number of molecules and calculate K (b) To find DG, we apply equation 20.13 We can calculate the value of T using K from part (a), and substitute the partial pressure of each substance to get Q (c) Since 1.0 atm is the standard state for gases, DG = DG° SOLUTION: (a) K= 2AB(g) DGo = -3.4 kJ/mol A2(g) + B2(g) (0.40)2 (0.20)(0.20) 2AB(g) Q= PAB2 PA2 x PB2 = 4.0 -3.4 kJ mol -3.4 kJ 1000 J 8.314 J T ln 4.0 =mol·K mol T= - kJ 8.314 J = 295 K ln 4.0 mol·K Mixture is at equilibrium, so ΔG = 0; DG = DG° + RT lnQ = -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 4.0) = -3400 J + 3400 J = 0.0 J Mixture 2: Q= (0.20)2 (0.30)(0.30) = 0.44 DG = DG° + RT lnQ = -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 0.44) = -3400 J – 2010 J = -5.4x10-3 J 20-70 20-69 Sample Problem 20.9 Mixture 3: Q= (b) DG° = -RT lnK (0.60)2 = 36 (0.10)(0.10) Sample Problem 20.10 Calculating DG at Nonstandard Conditions PROBLEM: The oxidation of SO2(g) is too slow at 298 K to be useful in the manufacture of sulfuric acid, so the reaction is run at high T 2SO2(g) + O2(g) → 2SO3(g) DG = DG° + RT lnQ = -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 36) = -3400 J + 8800 J = +5.4x10-3 J (a) Calculate K at 298 K and at 973 K, (DG°298 = -141.6 kJ/mol of reaction as written; using DH° and DS°values at 973 K, DG°973 = -12.12 kJ/mol for reaction as written.) Mixture has the most negative DG, and mixture has the most positive DG (b) Two containers are filled with 0.500 atm of SO2, 0.0100 atm of O2, and 0.100 atm of SO3; one is kept at 25°C and the other at 700.°C In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Under standard conditions, DG = DG° = -3.4 kJ/mol Yes, the reaction is spontaneous when the components are in their standard states 20-71 (c) Calculate DG for the system in part (b) at each temperature 20-72 10-Apr-16 Sample Problem 20.10 Sample Problem 20.10 PLAN: (a) We know DG°, T, and R, so we can calculate the values of K at each temperature (b) To determine if a net reaction will occur, we find Q from the given partial pressures and compare Q to K from part (a) of each substance to get Q (c) These are not standard-state pressures, so we find DG at each temperature using the equation DG = DG° + RT lnQ At 973 K, -(DG°)/RT = K at 973 K = e1.50 = 4.5 (b) Calculating the value of Q Q= (a) DG° = -RT lnK so At 298 K, -(DG°)/RT = PSO x P PSO O SOLUTION: K=e -(DG°/RT) -141.6x103 J/mol -12.12x103 J/mol = 1.50 8.314 J/mol·K x 973 K = (0.100)2 = 4.00 (0.500)2(0.0100) Since Q < K at both temperatures, the reaction will proceed toward the products (to the right) in both cases until Q equals K = 57.2 8.314 J/mol·K x 298 K K at 298 K = e57.2 = 7x1024 20-73 At 298 K, the system is far from equilibrium and will proceed far to the right At 973 K the system is closer to equilibrium and will proceed only slightly toward the right 20-74 Sample Problem 20.10 Figure 20.16 Free energy and the extent of reaction (c) DG298 = DG° + RT ln Q = -141.6 kJ/mol + (8.314x10-3 kJ/mol·K x 298 K x ln 4.00) = -138.2 kJ/mol DG973 = DG° + RT ln Q = -12.12 kJ/mol + (8.314x10-3 kJ/mol·K x 973 K x ln 4.00) = -0.9 kJ/mol Each reaction proceeds spontaneously (green curved arrows) from reactants or products to the equilibrium mixture, at which point DG = After that, the reaction is nonspontaneous in either direction (red curved arrows) Free energy reaches a minimum at equilibrium 20-75 20-76 10-Apr-16 Lecture PowerPoint Chapter 21 Chemistry Electrochemistry: The Molecular Nature of Matter and Change Chemical Change and Electrical Work Martin S Silberberg 21-1 Copyright  The McGraw-Hill Companies, Inc Permission required for reproduction or display 21-2 Electrochemistry: Chemical Change and Electrical Work Key Points About Redox Reactions 21.1 Half-Reactions and Electrochemical Cells •Oxidation (electron loss) always accompanies reduction (electron gain) 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell •The oxidizing agent is reduced, and the reducing agent is oxidized 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries •The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent 21.6 Corrosion: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction 21-3 21-4 Half-Reaction Method for Balancing Redox Reactions Figure 21.1 A summary of redox terminology Zn(s) + 2H+(aq) Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions Zn2+(aq) + H2(g) •Each reaction is balanced for mass (atoms) and charge OXIDATION One reactant loses electrons Zn loses electrons Reducing agent is oxidized Zn is the reducing agent and becomes oxidized Oxidation number increases REDUCTION Other reactant gains electrons Oxidizing agent is reduced •One or both are multiplied by some integer to make the number of electrons gained and lost equal •The half-reactions are then recombined to give the balanced redox equation The oxidation number of Zn increases from x to +2 Advantages: •The separation of half-reactions reflects actual physical separations in electrochemical cells •The half-reactions are easier to balance especially if they involve acid or base •It is usually not necessary to assign oxidation numbers to those species not undergoing change Hydrogen ion gains electrons Hydrogen ion is the oxidizing agent and becomes reduced Oxidation number decreases The oxidation number of H decreases from +1 to 21-5 21-6 10-Apr-16 Balancing Redox Equations in Acidic Conditions Write the skeletons of the oxidation and reduction half-reactions Balance all elements other than O and H Balance the oxygen atoms by adding H2O molecules on the side of the arrow where O atoms are needed Balance the hydrogen atoms by adding H+ ions on the side of the arrow where H atoms are needed Balance the charge by adding electrons, e- If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction halfreaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number lost Add the half-reactions as if they were mathematical equations Check to make sure that the atoms and the charge are balanced Balancing Redox Equations in Basic Conditions Steps #1-8 Begin by balancing the equation as if it were in acid solution If you have H+ ions in your equation at the end of these steps, proceed to Step #9 Otherwise, skip to Steps 9-12 Add enough OH- ions to each side to cancel the H+ ions Be sure to add the OH- ions to both sides to keep the charge and atoms balanced • 10 Combine the H+ ions and OH- ions that are on the same side of the equation to form water 11 Cancel or combine the H2O molecules 12 Check to make sure that the atoms and the charge balance If they balance, you are done If they not balance, re-check your work in Steps 1-11 21-8 Balancing Redox Reactions in Acidic Solution Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) 6e- Divide the reaction into half-reactions - + 14H+(aq) Determine the O.N.s for the species undergoing redox +6 -1 Cr2O72-(aq) + I-(aq) Cr2O72I- 2 I- I2 Cr is going from +6 to +3 I- I2 + 2e- X3 Add the half-reactions together 6e- + 14H+ + Cr2O72- 14H+(aq) + Cr2O72- Cr3+ + + 7H2O(l) Cr2O72- Cr3+ Cr3+ I- + 7H2O(l) I2 + 6e- Add 6e - to left net: +6 net: +12 + + 7H2O(l) + 2e- I is going from -1 to I2 14H+(aq) continued Multiply each half-reaction by an integer, if necessary - Balance atoms and charges in each half-reaction - 6e- Cr3+ Cr(+6) is the oxidizing agent and I(-1) is the reducing agent +3 Cr3+(aq) + I2(aq) Cr3+ + Cr2O72- 14H+(aq) + Cr2O72-(aq) + I-(aq) + 7H2O(l) 2Cr3+(aq) + 3I2(s) + 7H2O(l) Do a final check on atoms and charges 21-9 21-10 Balancing Redox Reactions in Basic Solution Figure 21.2 The redox reaction between dichromate ion and iodide ion Balance the reaction in acid and then add OH- so as to neutralize the H+ ions 14H+(aq) + Cr2O72-(aq) + I-(aq) + 14OH-(aq) 14H2O + Cr2O72- + I- Cr2O72- 2Cr3+(aq) + 3I2(s) + 7H2O(l) + 14OH-(aq) 2Cr3+ + 3I2 + 7H2O + 14OHCr3+ + I2 Reconcile the number of water molecules 7H2O + Cr2O72- + I- 2Cr3+ + 3I2 + 14OH- Do a final check on atoms and charges 21-11 I- 21-12 10-Apr-16 Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4 PLAN: -(aq) + C 2O 2-(aq) MnO2(s) + CO3 Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method continued: 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 2CO32- + 4H+ + 2e- 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 2CO32- + 4H+ + 2e- 2-(aq) X3 X2 Proceed in acidic solution and then neutralize with base 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- SOLUTION: MnO4+7 MnO44H+ + MnO4- 8H+ + 2MnO4- +6e- MnO2 +4 MnO2 C2O42+3 C2O42- CO32+4 CO32- MnO2 + 2H2O C2O42- + 2H2O +3e- +2e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2CO32- + 4H+ 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH- + 4OH- 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 21-13 2MnO2(s) + 6CO32-(aq) + 2H2O(l) 21-14 Figure 21.4 The spontaneous reaction between zinc and copper(II) ion Figure 21.3 General characteristics of voltaic and electrolytic cells VOLTAIC CELL System Energydoes is released work on from its spontaneous surroundings redox reaction X(s) Y(s) Oxidation half-reaction X X+ + e- ELECTROLYTIC CELL Surroundings(power Energy is absorbed tosupply) drive a nonspontaneous redox reaction work on system(cell) A(s) B(s) Oxidation half-reaction AA + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ eB Overall (cell) reaction X + Y+ X+ + Y DG < Overall (cell) reaction A- + B+ A + B DG > 21-15 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-16 Figure 21.5 A voltaic cell based on the zinc-copper reaction Notation for a Voltaic Cell components of anode compartment components of cathode compartment (oxidation half-cell) (reduction half-cell) phase of lower phase of higher oxidation state oxidation state phase of higher oxidation state phase of lower oxidation state phase boundary between half-cells Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Examples: Zn(s) Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- graphite | Reduction half-reaction Cu2+(aq) + 2eCu(s) I-(aq) inert electrode Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-17 Zn2+(aq) + 2e- 21-18 | I2(s) || Cu2+(aq) + 2eH+(aq), MnO4 -(aq) Cu(s) | Mn2+(aq) | graphite 10-Apr-16 Figure 21.6 A voltaic cell using inactive electrodes Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l) Sample Problem 21.2: Diagramming Voltaic Cells PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge Measurement indicates that the Cr electrode is negative relative to the Ag electrode PLAN: Identify the oxidation and reduction reactions and write each halfreaction Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction) Voltmeter e- SOLUTION: salt bridge Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Cr Ag K+ NO3- Reduction half-reaction Ag+(aq) + eAg(s) Cr3+ Ag + Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l) Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) 21-20 21-19 Why Does a Voltaic Cell Work? Table 21.1 Voltages of Some Voltaic Cells The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit Voltaic Cell Voltage (V) Ecell > for a spontaneous reaction Volt (V) = Joule (J)/ Coulomb (C) 1.5 Lead-acid car battery (6 cells = 12V) 2.0 Calculator battery (mercury) 1.3 Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070 21-22 21-21 Figure 21.7 Common alkaline battery Determining an unknown E0half-cell with the standard reference (hydrogen) electrode Sample Problem 21.3: Calculating an Unknown E0half-cell from E0cell PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Calculate E0bromine given E0zinc = -0.76V PLAN: The reaction is spontaneous as written since the E0cell is (+) Zinc is being oxidized and is the anode Therefore the E0bromine can be found using E0cell = E0cathode - E0anode SOLUTION: anode: Zn(s) Zn2+(aq) + 2e E0Zn as Zn2+(aq) + 2eOverall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) 21-23 Reduction half-reaction 2H3O+(aq) + 2eH2(g) + 2H2O(l) - E = +0.76 Zn(s) is -0.76V E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76) E0bromine = 1.86 - 0.76 = 1.07V 21-24 10-Apr-16 Table 21.2 Selected Standard Electrode Potentials (298K) •By convention, electrode potentials are written as reductions E0(V) Half-Reaction F2(g) + 2e2F-(aq) Cl2(g) + 2e2Cl-(aq) MnO2(g) + 4H+(aq) + 2eMn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3eNO(g) + 2H2O(l) Ag+(aq) + eAg(s) Fe3+(g) + eFe2+(aq) O2(g) + 2H2O(l) + 4e4OH-(aq) Cu2+(aq) + 2eCu(s) 2H+(aq) + 2eH2(g) N2(g) + 5H+(aq) + 4eN2H5+(aq) Fe2+(aq) + 2eFe(s) 2H2O(l) + 2eH2(g) + 2OH-(aq) Na+(aq) + eNa(s) Li+(aq) + eLi(s) •When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell Reverse the sign of the potential +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 -0.23 -0.44 -0.83 -2.71 -3.05 •The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell •When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents Example: Cu2+(aq) + Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM: (a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B, and C), and calculate E0cell for each + weaker oxidizing agent Cu(s) weaker reducing agent Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (a) (b) Rank the relative strengths of the oxidizing and reducing agents: (1) NO3 -(aq) + 4H+(aq) + 3e- (2) N2(g) + 5H+(aq) + 4e- E0 NO(g) + 2H2O(l) N2H5+(aq) (3) MnO2(s) +4H+(aq) + 2e- = 0.96V Mn2+(aq) + 2H2O(l) Rev (2) N2H5 +(aq) (2) N2H5+(aq) E0 = 1.23V Put the equations together in varying combinations so as to produce (+) E0cell for the combination Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0 Balance the number of electrons gained and lost without changing the E0 (1) NO3-(aq) + 4H+(aq) + 3eN2(g) + (1) NO3-(aq) + 4H+(aq) + 3e- E0 = -0.23V (A) Rev (1) NO(g) + 2H2O(l) NO(g) + 2H2O(l) E0 = +0.23V X4 (3) MnO2(s) +4H+(aq) + 2e- E0cell = 1.19V X3 4NO(g) + 3N2(g) + 8H2O(l) E0 = -0.96V Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (B) 2NO(g) + 3MnO2(s) + 4H+(aq) 21-27 + 4e- NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) +4H+(aq) + 2e(1) NO(g) + 2H2O(l) NO(g) + 2H2O(l) E0 = 0.96V 5H+(aq) N2(g) + 5H+(aq) + 4e- 4NO3-(aq) + 3N2H5+(aq) + H+(aq) In ranking the strengths, compare the combinations in terms of E0cell Mn2+(aq) + 2H2O(l) E0 = 1.23V X2 E0cell = 0.27V X3 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) 21-28 Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e(2) N2H5+(aq) E0 = 1.23V (C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of E0cell = 1.46V N2(g) + 5H+(aq) + 4eMn2+(aq) + 2H2O(l) Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) E0 = +0.23V Mn2+(aq) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Oxidizing agents: MnO2 > NO3- > N2 X2 Reducing agents: N2H5+ > NO > Mn2+ N2(g) + 2Mn2+(aq) + 4H2O(l) (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO3- > N2 (B): oxidizing agents: MnO2 > NO3 (C): oxidizing agents: MnO2 > N2 21-29 Zn2+(aq) stronger oxidizing agent 21-26 21-25 PLAN: Zn(s) stronger reducing agent reducing agents: N2H5+ > NO - reducing agents: NO > Mn2+ reducing agents: N2H5+ > Mn2+ 21-30 10-Apr-16 Figure 21.8 The reaction of calcium in water Relative Reactivities (Activities) of Metals Oxidation half-reaction Ca(s) Ca2+(aq) + 2e- Reduction half-reaction 2H2O(l) + 2eH2(g) + 2OH-(aq) Metals that can displace H2 from acid Metals that cannot displace H2 from acid Metals that can displace H2 from water Metals that can displace other metals from solution Overall (cell) reaction Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq) 21-31 21-32 Free Energy and Electrical Work DG a -Ecell DG = wmax = charge x (-Ecell) K E0cell 1 >0 at equilibrium >0 E0cell •When Q = and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell •When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell 0.0592 Ecell = E0cell - log Q n Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-37 Figure 21.11 21-38 A concentration cell based on the Cu/Cu2+ half-reaction Sample Problem 21.7: Calculating the Potential of a Concentration Cell PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3 What is the cell potential at 298K? Which electrode has a positive charge? PLAN: E0cell will be zero since the half-cell potentials are equal Ecell is calculated from the Nernst equation with half-cell A (higher [Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+ SOLUTION: Ag+(aq, 0.010M) half-cell A 0.0592V Ecell = E0cell Oxidation half-reaction Cu(s) Cu2+(aq, 0.1M) + 2e- Reduction half-reaction Cu2+(aq, 1.0M) + 2eCu(s) log [Ag+]concentrated Ecell = V -0.0592 log 4.0x10-2 = 0.0828V Half-cell A is the cathode and has the positive electrode Overall (cell) reaction Cu2+(aq,1.0M) Cu2+(aq, 0.1M) 21-40 21-39 Figure 21.12 Ag+(aq, 4.0x10-4M) half-cell B [Ag+]dilute The laboratory measurement of pH Table 21.3 Some Ions Measured with Ion-Specific Electrodes Pt Glass electrode Reference (calomel) electrode Hg AgCl on Ag on Pt 1M HCl Thin glass membrane 21-41 Paste of Hg2Cl2 in Hg KCl solution Porous ceramic plugs 21-42 Species Detected Typical Sample NH3/NH4+ Industrial wastewater, seawater CO2/HCO3- Blood, groundwater F- Drinking water, urine, soil, industrial stack gases Br- Grain, plant tissue I- Milk, pharmaceuticals NO3- Soil, fertilizer, drinking water K+ Blood serum, soil, wine H+ Laboratory solutions, soil, natural waters 10-Apr-16 Figure 21.14 Figure 21.13 The corrosion of iron 21-43 Enhanced corrosion at sea 21-44 Figure 21.15 The effect of metal-metal contact on the corrosion of iron faster corrosion The use of sacrificial anodes to prevent iron corrosion cathodic protection 21-45 Figure 21.17 Figure 21.16 21-46 The tin-copper reaction as the basis of a voltaic and an electrolytic cell Figure 21.18 The processes occurring during the discharge and recharge of a lead-acid battery VOLTAIC(discharge) voltaic cell Oxidation half-reaction Sn(s) Sn2+(aq) + 2eReduction half-reaction Cu2+(aq) + 2eCu(s) Overall (cell) reaction 2+ 2+ Sn(s) + Cu (aq) Sn (aq) + Cu(s) 21-47 electrolytic cell Oxidation half-reaction Cu(s) Cu2+(aq) + 2eReduction half-reaction Sn2+(aq) + 2eSn(s) Overall (cell) reaction 2+ Sn(s) + Cu (aq) Sn2+(aq) + Cu(s) ELECTROLYTIC(recharge) 21-48 10-Apr-16 Sample Problem 21.8: Table 21.4 Comparison of Voltaic and Electrolytic Cells Electrode Cell Type DG Ecell Name Process Sign Voltaic 0 Anode Oxidation - Voltaic 0 Cathode Reduction + Electrolytic >0 0