Chapter 18. Electrochemistry - PDF Free Download

Chapter 18 Electrochemistry

Section 17.1 Spontaneous Processes and Entropy

Section 17.1 http://www.bozemanscience.com/ap-chemistry/ Spontaneous Processes and Entropy

Section 17.1 Spontaneous Processes and Entropy

Section 18.1 Balancing Oxidation-Reduction Equations Review of Terms Electrochemistry the study of the interchange of chemical and electrical energy Oxidation reduction (redox) reaction involves a transfer of electrons from the reducing agent to the oxidizing agent Oxidation loss of electrons Reduction gain of electrons Reducing agent electron donor Oxidizing agent electron acceptor Copyright Cengage Learning. All rights reserved 5

Section 18.1 Balancing Oxidation-Reduction Equations Half Reactions The overall reaction is split into two half reactions, one involving oxidation and one involving reduction. 8H + + MnO 4 + 5Fe 2+ Mn 2+ + 5Fe 3+ + 4H 2 O Reduction: 8H + + MnO 4 + 5e Oxidation: 5Fe 2+ 5Fe 3+ + 5e Mn 2+ + 4H 2 O Copyright Cengage Learning. All rights reserved 6

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution 1. Write separate equations for the oxidation and reduction half reactions. 2. For each half reaction: A. Balance all the elements except H and O. B. Balance O using H 2 O. C. Balance H using H +. D. Balance the charge using electrons. Copyright Cengage Learning. All rights reserved 7

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution 3. If necessary, multiply one or both balanced half reactions by an integer to equalize the number of electrons transferred in the two half reactions. 4. Add the half reactions, and cancel identical species. 5. Check that the elements and charges are balanced. Copyright Cengage Learning. All rights reserved 8

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution Copyright Cengage Learning. All rights reserved 9

Section 18.1 Balancing Oxidation-Reduction Equations Cr 2 O 7 2- (aq) + SO 3 2- (aq) Cr 3+ (aq) + SO 4 2- (aq) How can we balance this equation? First Steps: Separate into half-reactions. Balance elements except H and O. Copyright Cengage Learning. All rights reserved 10

Section 18.1 Balancing Oxidation-Reduction Equations Method of Half Reactions Cr 2 O 7 2- (aq) 2Cr 3+ (aq) SO 3 2- (aq) SO 4 2- (aq) How many electrons are involved in each half reaction? Copyright Cengage Learning. All rights reserved 11

Section 18.1 Balancing Oxidation-Reduction Equations Method of Half Reactions (continued) 6e - + Cr 2 O 2-7 (aq) 2Cr 3+ (aq) SO 2-3 (aq) + SO 2-4 (aq) + 2e - How can we balance the oxygen atoms? Copyright Cengage Learning. All rights reserved 12

Section 18.1 Balancing Oxidation-Reduction Equations Method of Half Reactions (continued) 6e - + Cr 2 O 2-7 (aq) Cr 3+ (aq) + 7H 2 O H 2 O +SO 2-3 (aq) + SO 2-4 (aq) + 2e - How can we balance the hydrogen atoms? Copyright Cengage Learning. All rights reserved 13

Section 18.1 Balancing Oxidation-Reduction Equations Method of Half Reactions (continued) This reaction occurs in an acidic solution. 14H + + 6e - + Cr 2 O 2-7 2Cr 3+ + 7H 2 O H 2 O +SO 2-3 SO 2-4 + 2e - + 2H + How can we balance the electrons? Copyright Cengage Learning. All rights reserved 14

Section 18.1 Balancing Oxidation-Reduction Equations Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2-2Cr 3+ + 7H 2 O 3[H 2 O +SO 3 2- SO 4 2- + 2e - + 2H + ] Final Balanced Equation: Cr 2 O 7 2- + 3SO 3 2- + 8H + 2Cr 3+ + 3SO 4 2- + 4H 2 O Copyright Cengage Learning. All rights reserved 15

Section 18.1 Balancing Oxidation-Reduction Equations EXERCISE! Balance the following oxidation reduction reaction that occurs in acidic solution. Br (aq) + MnO 4 (aq) Br 2 (l)+ Mn 2+ (aq) 10Br (aq) + 16H + (aq) + 2MnO 4 (aq) 5Br 2 (l)+ 2Mn 2+ (aq) + 8H 2 O(l) Copyright Cengage Learning. All rights reserved 16

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution 1. Use the half reaction method as specified for acidic solutions to obtain the final balanced equation as if H + ions were present. 2. To both sides of the equation obtained above, add a number of OH ions that is equal to the number of H + ions. (We want to eliminate H + by forming H 2 O.) Copyright Cengage Learning. All rights reserved 17

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution 3. Form H 2 O on the side containing both H + and OH ions, and eliminate the number of H 2 O molecules that appear on both sides of the equation. 4. Check that elements and charges are balanced. Copyright Cengage Learning. All rights reserved 18

Section 18.1 Balancing Oxidation-Reduction Equations The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution Copyright Cengage Learning. All rights reserved 19

Section 18.2 Galvanic Cells Galvanic Cell Device in which chemical energy is changed to electrical energy. Uses a spontaneous redox reaction to produce a current that can be used to do work. Copyright Cengage Learning. All rights reserved 20

Section 18.2 Galvanic Cells Galvanic Cell Oxidation occurs at the anode. Reduction occurs at the cathode. Salt bridge or porous disk devices that allow ions to flow without extensive mixing of the solutions. Salt bridge contains a strong electrolyte held in a Jello like matrix. Porous disk contains tiny passages that allow hindered flow of ions. Copyright Cengage Learning. All rights reserved 22

Section 18.2 Galvanic Cells Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The pull, or driving force, on the electrons is called the cell potential ( E cell ), or the electromotive force (emf) of the cell. Unit of electrical potential is the volt (V). 1 joule of work per coulomb of charge transferred. Copyright Cengage Learning. All rights reserved 23

Section 18.2 Galvanic Cells Voltaic Cell: Cathode Reaction To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 24

Section 18.2 Galvanic Cells Voltaic Cell: Anode Reaction To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 25

Section 18.3 Standard Reduction Potentials Galvanic Cell All half-reactions are given as reduction processes in standard tables. Table 18.1 1 M, 1atm, 25 C When a half-reaction is reversed, the sign of E is reversed. When a half-reaction is multiplied by an integer, E remains the same. A galvanic cell runs spontaneously in the direction that gives a positive value for E cell. Copyright Cengage Learning. All rights reserved 26

Section 18.3 Standard Reduction Potentials Example: Fe 3+ (aq) + Cu(s) Cu 2+ (aq) + Fe 2+ (aq) Half-Reactions: Fe 3+ + e Fe 2+ E = 0.77 V Cu 2+ + 2e Cu E = 0.34 V To balance the cell reaction and calculate the cell potential, we must reverse reaction 2. Cu Cu 2+ + 2e E = 0.34 V Each Cu atom produces two electrons but each Fe 3+ ion accepts only one electron, therefore reaction 1 must be multiplied by 2. 2Fe 3+ + 2e 2Fe 2+ E = 0.77 V Copyright Cengage Learning. All rights reserved 27

Section 18.3 Standard Reduction Potentials Overall Balanced Cell Reaction 2Fe 3+ + 2e 2Fe 2+ E = 0.77 V (cathode) Cu Cu 2+ + 2e E = 0.34 V (anode) Balanced Cell Reaction: Cu + 2Fe 3+ Cu 2+ + 2Fe 2+ Cell Potential: E cell = E (cathode) E (anode) E cell = 0.77 V 0.34 V = 0.43 V Copyright Cengage Learning. All rights reserved 28

Section 18.3 Standard Reduction Potentials CONCEPT CHECK! Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why. Fe Na F - Na + Cl 2

Section 18.3 Standard Reduction Potentials Line Notation Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines which indicated salt bridge or porous disk. The concentration of aqueous solutions should be specified in the notation when known. Example: Mg(s) Mg 2+ (aq) Al 3+ (aq) Al(s) Mg Mg 2+ + 2e (anode) Al 3+ + 3e Al (cathode) Copyright Cengage Learning. All rights reserved 30

Section 18.3 Standard Reduction Potentials Description of a Galvanic Cell The cell potential (always positive for a galvanic cell where E cell = E (cathode) E (anode)) and the balanced cell reaction. The direction of electron flow, obtained by inspecting the half reactions and using the direction that gives a positive E cell. Copyright Cengage Learning. All rights reserved 31

Section 18.3 Standard Reduction Potentials Description of a Galvanic Cell Designation of the anode and cathode. The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half reaction is a conducting solid. Copyright Cengage Learning. All rights reserved 32

Section 18.3 Standard Reduction Potentials CONCEPT CHECK! Sketch a cell using the following solutions and electrodes. Include: The potential of the cell The direction of electron flow Labels on the anode and the cathode a) Ag electrode in 1.0 M Ag + (aq) and Cu electrode in 1.0 M Cu 2+ (aq) Copyright Cengage Learning. All rights reserved 33

Section 18.3 Standard Reduction Potentials CONCEPT CHECK! Sketch a cell using the following solutions and electrodes. Include: The potential of the cell The direction of electron flow Labels on the anode and the cathode b) Zn electrode in 1.0 M Zn 2+ (aq) and Cu electrode in 1.0 M Cu 2+ (aq) Copyright Cengage Learning. All rights reserved 34

Section 18.3 Standard Reduction Potentials CONCEPT CHECK! Consider the cell from part b. What would happen to the potential if you increase the [Cu 2+ ]? Explain. The cell potential should increase. Copyright Cengage Learning. All rights reserved 35

Section 18.4 Cell Potential, Electrical Work, and Free Energy Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted the actual work realized is always less than the calculated maximum. Copyright Cengage Learning. All rights reserved 36

Section 18.4 Cell Potential, Electrical Work, and Free Energy Maximum Cell Potential Directly related to the free energy difference between the reactants and the products in the cell. ΔG = nfe F = 96,485 C/mol e Copyright Cengage Learning. All rights reserved 37

Section 18.5 Dependence of Cell Potential on Concentration Nernst Equation The relationship between cell potential and concentrations of cell components At 25 C: 0.0591 E = E log Q n or E 0.0591 = log K n (at equilibrium) Copyright Cengage Learning. All rights reserved 39

Section 18.5 Dependence of Cell Potential on Concentration CONCEPT CHECK! Explain the difference between E and E. When is E equal to zero? When the cell is in equilibrium ("dead" battery). When is E equal to zero? E is equal to zero for a concentration cell. Copyright Cengage Learning. All rights reserved 40

Section 18.5 Dependence of Cell Potential on Concentration EXERCISE! A concentration cell is constructed using two nickel electrodes with Ni 2+ concentrations of 1.0 M and 1.00 10-4 M in the two half-cells. Calculate the potential of this cell at 25 C. 0.118 V Copyright Cengage Learning. All rights reserved 41

Section 18.5 Dependence of Cell Potential on Concentration CONCEPT CHECK! You make a galvanic cell at 25 C containing: A nickel electrode in 1.0 M Ni 2+ (aq) A silver electrode in 1.0 M Ag + (aq) Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential. 1.03 V Copyright Cengage Learning. All rights reserved 42

Section 18.6 Batteries A Common Dry Cell Battery

Section 18.6 Batteries Schematic of the Hydrogen-Oxygen Fuel Cell

Section 18.7 Corrosion Process of returning metals to their natural state the ores from which they were originally obtained. Involves oxidation of the metal. Copyright Cengage Learning. All rights reserved 47

Section 18.7 Corrosion Corrosion Prevention Application of a coating (like paint or metal plating) Galvanizing Alloying Cathodic Protection Protects steel in buried fuel tanks and pipelines. Copyright Cengage Learning. All rights reserved 49

Section 18.8 Electrolysis Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time moles of electrons grams of analyte quantity of charge moles of analyte Copyright Cengage Learning. All rights reserved 52

Section 18.8 Electrolysis Stoichiometry of Electrolysis current and time quantity of charge Coulombs of charge = amps (C/s) seconds (s) quantity of charge moles of electrons 1 mol e mol e = Coulombs of charge 96,485 C Copyright Cengage Learning. All rights reserved 53

Section 18.8 Electrolysis CONCEPT CHECK! An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO 3 ) 3. What is the metal? gold (Au) Copyright Cengage Learning. All rights reserved 54

Section 18.8 Electrolysis CONCEPT CHECK! Consider a solution containing 0.10 M of each of the following: Pb 2+, Cu 2+, Sn 2+, Ni 2+, and Zn 2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu 2+, Pb 2+, Sn 2+, Ni 2+, Zn 2+ Do the metals form on the cathode or the anode? Explain. Copyright Cengage Learning. All rights reserved 55

Section 18.9 Commercial Electrolytic Processes Production of aluminum Purification of metals Metal plating Electrolysis of sodium chloride Production of chlorine and sodium hydroxide Copyright Cengage Learning. All rights reserved 56