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Chapter 20 Electrochemistry © 2012 Pearson Education, Inc. Electrochemical Reactions (Red-Ox) In electrochemical reactions, electrons are transferred from one species to another. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. (a) Bubbling is caused by water vapor escaping and (b) the exothermic reaction in water produces steam. (a) Bubbling is caused by water vapor escaping and (b) the endothermic reaction in water produces steam. (a) Bubbling is caused by hydrogen gas forming and (b) the endothermic reaction in water produces steam. (a) Bubbling is caused by hydrogen gas forming and (b) the exothermic reaction in water produces steam. Electrochemistry Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers. Electrochemistry © 2012 Pearson Education, Inc. Oxidation and Reduction • • A species is oxidized when it loses electrons. (the oxidation number increases) – Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion. A species is reduced when it gains electrons. (the oxidation number decreases) – Here, each of the H+ gains an electron, and they combine to form H2. Electrochemistry © 2012 Pearson Education, Inc. Oxidation and Reduction • The species reduced is the oxidizing agent. – H+ oxidizes Zn by taking electrons from it. • The species oxidized is the reducing agent. – Zn reduces H+ by giving it electrons. Electrochemistry © 2012 Pearson Education, Inc. Assigning Oxidation Numbers 1. 2. 3. 4. 5. Elements in their elemental form have an oxidation number of 0. The oxidation number of a monatomic ion is the same as its charge. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. – Oxygen has an oxidation number of 2, except in the peroxide ion, which has an oxidation number of 1. – Hydrogen is 1 when bonded to a metal, and +1 when bonded to a nonmetal. – Fluorine always has an oxidation number of 1. – The other halogens have an oxidation number of 1 when they are negative. They can have positive oxidation numbers, however; most notably in oxyanions. The sum of the oxidation numbers in a neutral compound is 0. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. N, +1; N, +2; N, +3; N, +4; O, –1 O, –2 O, –2 O, –2 Electrochemistry Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity: Cd(s) + NiO2(s) + 2 H2O(l) Cd(OH)2(s) + Ni(OH)2(s) Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent. Practice Exercise Identify the oxidizing and reducing agents in the reaction 2 H2O(l) + Al(s) + MnO4(aq) Al(OH)4(aq) + MnO2(s) Electrochemistry Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method. This method involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half-reactions, and then combining them to attain the balanced equation for the overall reaction. Electrochemistry © 2012 Pearson Education, Inc. The Half-Reaction Method 1. 2. 3. Assign oxidation numbers to determine what is oxidized and what is reduced. Write the oxidation and reduction half-reactions. Balance each half-reaction. a. Balance elements other than H and O. b. Balance O by adding H2O. c. Balance H by adding H+. d. Balance charge by adding electrons. Electrochemistry © 2012 Pearson Education, Inc. The Half-Reaction Method 4. 5. 6. 7. Multiply the half-reactions by integers so that the electrons gained and lost are the same. Add the half-reactions, subtracting things that appear on both sides. Make sure the equation is balanced according to mass. Make sure the equation is balanced according to charge. Electrochemistry © 2012 Pearson Education, Inc. The Half-Reaction Method Consider the reaction between MnO4 and C2O42: MnO4(aq) + C2O42(aq) Mn2+(aq) + CO2(aq) Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. MnO4- is reduced and the reducing agent is H+. MnO4- is reduced and the reducing agent is C2O42-. H+ is reduced and the reducing agent is C2O42-. H+ is reduced and the reducing agent is MnO4-. Electrochemistry The Half-Reaction Method First, we assign oxidation numbers: +7 +3 +2 +4 MnO4 + C2O42 Mn2+ + CO2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized. Electrochemistry © 2012 Pearson Education, Inc. Oxidation Half-Reaction C2O42 CO2 To balance the carbon, we add a coefficient of 2: C2O42 2CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side: C2O42 2CO2 + 2e Electrochemistry © 2012 Pearson Education, Inc. Reduction Half-Reaction MnO4 Mn2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side: MnO4 Mn2+ + 4H2O To balance the hydrogen, we add 8H+ to the left side: 8H+ + MnO4 Mn2+ + 4H2O Electrochemistry © 2012 Pearson Education, Inc. Reduction Half-Reaction 8H+ + MnO4 Mn2+ + 4H2O To balance the charge, we add 5e to the left side: 5e + 8H+ + MnO4 Mn2+ + 4H2O Electrochemistry © 2012 Pearson Education, Inc. Combining the Half-Reactions Now we evaluate the two half-reactions together: C2O42 2CO2 + 2e 5e + 8H+ + MnO4 Mn2+ + 4H2O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2: Electrochemistry © 2012 Pearson Education, Inc. Combining the Half-Reactions 5C2O42 10CO2 + 10e 10e + 16H+ + 2MnO4 2Mn2+ + 8H2O When we add these together, we get: 10e + 16H+ + 2MnO4 + 5C2O42 2Mn2+ + 8H2O + 10CO2 +10e Electrochemistry © 2012 Pearson Education, Inc. Combining the Half-Reactions 10e + 16H+ + 2MnO4 + 5C2O42 2Mn2+ + 8H2O + 10CO2 +10e The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16H+ + 2MnO4 + 5C2O42 2Mn2+ + 8H2O + 10CO2 Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. Yes, in each half-reaction Yes, on both sides of the balanced equation for a redox reaction No, the electrons do not show in half-reactions and therefore they do not appear in a balanced equation for a redox reaction. No, the electrons cancel in adding half-reactions to form a balanced equation for a redox reaction. Electrochemistry Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents Complete and balance this equation by the method of half-reactions: Cr2O72(aq) + Cl(aq) Cr3+(aq) + Cl2(g) (acidic solution) Practice Exercise Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution. (a) Cu(s) + NO3(aq) Cu2+(aq) + NO2(g) (b) Mn2+(aq) + NaBiO3(s) Bi3+(aq) + MnO4(aq) Electrochemistry Balancing in Basic Solution • If a reaction occurs in a basic solution, one can balance it as if it occurred in acid. • Once the equation is balanced, add OH to each side to “neutralize” the H+ in the equation and create water in its place. • If this produces water on both sides, you might have to subtract water from each side. Electrochemistry © 2012 Pearson Education, Inc. Sample Exercise 20.3 Balancing Redox Equations in Basic Solution Complete and balance this equation for a redox reaction that takes place in basic solution: CN(aq) + MnO4(aq) CNO(aq) + MnO2(s) (basic solution) Practice Exercise Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution: (a) NO2(aq) + Al(s) NH3(aq) + Al(OH)4(aq) (b) Cr(OH)3(s) + ClO(aq) CrO42(aq) + Cl2(g) Electrochemistry Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. The blue color of Cu2+(aq) lessens as it is reduced to Cu(s). The blue color of Cu(s) lessens as it is reduced to Cu2+(aq). The blue color of Zn2+(aq) lessens as it is reduced to Zn(s). The blue color of Zn(s) lessens as it is reduced to Zn(s). Electrochemistry Voltaic Cells • We can use that energy to do work if we make the electrons flow through an external device. • We call such a setup a voltaic cell. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. Cu, because it loses electrons in the chemical reaction. Cu, because it gains electrons in the chemical reaction. Zn, because it loses electrons in the chemical reaction. Zn, because it gains electrons in the chemical reaction. Electrochemistry Voltaic Cells • A typical cell looks like this. • The oxidation occurs at the anode. • The reduction occurs at the cathode. • Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop. Electrochemistry © 2012 Pearson Education, Inc. Voltaic Cells • Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. – Cations move toward the cathode. – Anions move toward the anode. • In the cell, then, electrons leave the anode and flow through the wire to the cathode. • As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment. © 2012 Pearson Education, Inc. Electrochemistry Voltaic Cells • As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. • The electrons are taken by the cation, and the neutral metal is deposited on the cathode. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. Zn2+(aq) ions formed in the reaction completely migrate through the salt bridge to the right compartment to maintain electrical charge balance. Anions from the right compartment migrate through the salt bridge to the left compartment to maintain electrical charge balance. NO3-(aq) ions in the salt bridge migrate into the left compartment and Zn 2+(aq) ions migrate into the salt bridge to maintain electrical charge balance. Electrochemistry + 2+ Na (aq) ions in the salt bridge migrate into the left compartment and Zn (aq) ions migrate into the salt bridge to maintain electrical charge balance. Sample Exercise 20.4 Describing a Voltaic Cell The oxidation-reduction reaction Cr2O72(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l) is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes. Practice Exercise The two half-reactions in a voltaic cell are Zn(s) Zn2+(aq) + 2 e ClO3(aq) + 6 H+(aq) + 6 e Cl(aq) + 3 H2O(l) (a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode isElectrochemistry consumed in the cell reaction? (c) Which electrode is positive? Electromotive Force (emf) • Water only spontaneously flows one way in a waterfall. • Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy. • The potential difference between the anode and cathode in a cell is called the electromotive force (emf). • It is also called the cell potential and is designated Ecell. Electrochemistry © 2012 Pearson Education, Inc. Cell Potential Cell potential is measured in volts (V). J 1V=1 C Electrochemistry © 2012 Pearson Education, Inc. Standard Cell Potential The cell potential under standard conditions is denoted E cell Zn(s) + Cu2+(aq, 1M) Zn2+(aq,1M) + Cu(s) E cell = +1.10 v Electrochemistry © 2012 Pearson Education, Inc. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated. Electrochemistry © 2012 Pearson Education, Inc. A. Yes B. No Electrochemistry Standard Cell Potentials The cell potential at standard conditions can be found through this equation: = Ered (cathode) Ered (anode) Ecell Because cell potential is based on the potential energy per unit of charge, it is an intensive property. Electrochemistry © 2012 Pearson Education, Inc. Standard Hydrogen Electrode • Their values are referenced to a standard hydrogen electrode (SHE). • By definition, the reduction potential for hydrogen is 0 V: 2 H+(aq, 1M) + 2e H2(g, 1 atm) Electrochemistry © 2012 Pearson Education, Inc. Cell Potentials • For the oxidation in this cell, = 0.76 V Ered • For the reduction, = +0.34 V Ered Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the increase in negative charge of nitrate ions formed in the reaction. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the decreasing positive charge of hydrogen ions consumed in the reaction. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the decreasing positive charge of zinc ions consumed in the reaction. Na+ migrates into the cathode half-cell to balance electrical charge by balancing the increasing positive charge of hydrogen ions formed in the reaction. Electrochemistry Cell Potentials = Ered (cathode) Ered (anode) Ecell = +0.34 V (0.76 V) = +1.10 V Electrochemistry © 2012 Pearson Education, Inc. A. 1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq). – B. 1 M solution for Cl2(g) and for Cl (aq). C. 1 atm pressure for Cl2(g) and for Cl–(aq). D. 1 torr pressure for Cl2(g) and 1 M solution for Cl–(aq). Electrochemistry Sample Exercise 20.5 Calculating from For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have Given that the standard reduction potential of Zn2+ to Zn(s) is 0.76 V, calculate the for the reduction of Cu2+ to Cu: Cu2+(aq, 1 M) + 2 e Cu(s) Practice Exercise The standard cell potential is 1.46 V for a voltaic cell based on the following half-reactions: In+(aq) In3+(aq) + 2 e Br2(l) + 2 e 2Br(aq) Using Table 20.1, calculate for the reduction of 3+ + In to In . Electrochemistry A. The electrode with the more positive Eredo value is associated with the cathode. B. The electrode with the less positive (more negative) Eredo value is associated with the cathode. Electrochemistry Sample Exercise 20.6 Calculating from Use Table 20.1 to calculate for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction Cr2O72(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l) Practice Exercise Using data in Table 20.1, calculate the standard emf for a cell that employs the overall cell reaction 2 Al(s) + 3 I2(s) 2 Al3+(aq) + 6 I(aq). Electrochemistry Sample Exercise 20.7 Determining Half-Reactions at Electrodes and Calculating Cell Potentials A voltaic cell is based on the two standard half-reactions Cd2+(aq) + 2 e Cd(s) Sn2+(aq) + 2 e Sn(s) Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential. Practice Exercise A voltaic cell is based on a Co2+/Co half-cell and an AgCl/Ag half-cell. (a) What half-reaction occurs at the anode? (b) What is the standard cell potential? Electrochemistry Oxidizing and Reducing Agents • The strongest oxidizers have the most positive reduction potentials. • The strongest reducers have the most negative reduction potentials. Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. A strong oxidizing agent is one that readily gains electrons and thus has a large positive value of Eredo for its half-reaction. The larger the positive value of Eredo the greater is the tendency for the reduction half-reaction to occur. A strong oxidizing agent is one that readily gains electrons and thus has a large negative value of Eredo for its half-reaction. The larger the negative value of Eredo the greater is the tendency for the reduction half-reaction to occur. A strong oxidizing agent is one that readily loses electrons and thus has a large positive value of Eredo for its half-reaction. The larger the positive value Eredo the greater is the tendency for the reduction halfreaction to occur. A strong oxidizing agent is one that readily loses electrons and thus has a large positive value of Eredo for its half-reaction. The larger the value Eredo the greater is the tendency for the reduction half-reaction to occur. Electrochemistry Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell. Electrochemistry © 2012 Pearson Education, Inc. Sample Exercise 20.8 Determining Relative Strengths of Oxidizing Agents Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3 (aq), Ag+ (aq), Cr2O72 (aq). Practice Exercise Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I (aq), Fe(s), Al(s). Electrochemistry Sample Exercise 20.9 Determining Spontaneity Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions. (a) Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g) (b) Cl2(g) + 2 I(aq) 2 Cl(aq) + I2(s) Practice Exercise Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions: (a) I2(s) + 5 Cu2+(aq) + 6 H2O(l) 2 IO3(aq) + 5 Cu(s) + 12 H+(aq) (b) Hg2+(aq) + 2 I(aq) Hg(l) + I2(s) (c) H2SO3(aq) + 2 Mn(s) + 4 H+(aq) S(s) + 2 Mn2+(aq) + 3 H2O(l) Electrochemistry A. B. C. D. Hg(l) is the stronger reducing agent because Hg(l) has the more negative Eredo. Pb(s) is the stronger reducing agent because Pb(s) has the more negative Eredo. Hg(l) is the stronger reducing agent because Hg(l) is a liquid metal and this changes Eredo. Pb(s) is the stronger reducing agent because Pb(s) is further right in the periodic table than Hg(l). Electrochemistry Free Energy G for a redox reaction can be found by using the equation G = nFE where n is the number of moles of electrons transferred, and F is a constant, the Faraday: 1 F = 96,485 C/mol = 96,485 J/V-mol Electrochemistry © 2012 Pearson Education, Inc. Free Energy Under standard conditions, G = nFE Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. The number of moles of the oxidizing agent in the balanced chemical equation. The number of moles of the reducing agent in the balanced chemical equation. The net number of moles of gas reacting in the balanced chemical equation. The number of moles of electrons transferred in the balanced chemical equation. Electrochemistry Sample Exercise 20.10 Using Standard Reduction Potentials to Calculate G and K (a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change, G,and the equilibrium constant, K, at 298 K for the reaction (b) Suppose the reaction in part (a) is written What are the values of E, G, and K when the reaction is written in this way? Practice Exercise For the reaction 3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH(aq) 3 Ni(s) + 2 CrO42(aq) + 8 H2O(l) (a) What is the value of n? (b) Use the data in Appendix E to calculate G. (c) Calculate K at T = 298 K. Electrochemistry Nernst Equation • Remember that G = G + RT ln Q • This means nFE = nFE + RT ln Q • Dividing both sides by nF, we get the Nernst equation: RT ln Q E = E nF • or, using base-10 logarithms, E = E 2.303RT nF log Q Electrochemistry © 2012 Pearson Education, Inc. Nernst Equation At room temperature (298 K), 2.303RT = 0.0592 V F Thus, the equation becomes 0.0592 log Q E = E n Electrochemistry © 2012 Pearson Education, Inc. Sample Exercise 20.11 Cell Potential under Nonstandard Conditions Calculate the emf at 298 K generated by a voltaic cell in which the reaction is Cr2O72(aq) + 14 H+(aq) + 6 I(aq) 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l) when [Cr2O72] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0 105 M. Practice Exercise Calculate the emf generated by the cell described in the practice exercise accompanying Sample Electrochemistry Exercise 20.6 when [Al3+] = 4.0 103 M and [I] = 0.0100 M. Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell If the potential of a Zn-H+ cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the H+ concentration? Practice Exercise What is the pH of the solution in the cathode half-cell in Figure 20.9 when PH2 = 1.0 atm, [Zn2+] in the anode half-cell is 0.10 M, and the cell emf is 0.542 V? Electrochemistry Concentration Cells • Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. would be 0, but Q would not. • For such a cell, Ecell • Therefore, as long as the concentrations Electrochemistry are different, E will not be 0. © 2012 Pearson Education, Inc. A. B. C. D. The Ni2+(aq) ions and the anions in the salt bridge migrate toward the anode. The NO3-(aq) ions and the cations in the salt bridge migrate toward the cathode. The Ni2+(aq) ions and the cations in the salt bridge migrate toward the anode. The NO3-(aq) ions and the anions in the salt bridge migrate toward the cathode. The Ni2+(aq) ions and the cations in the salt bridge migrate toward the cathode. The NO3-(aq) ions and the anions in the salt bridge migrate toward the anode. The Ni2+(aq) ions and the anions in the salt bridge migrate toward the cathode. The NO3-(aq) ions and the cations in the salt bridge migrate toward the anode. Electrochemistry Sample Exercise 20.13 Determining pH Using a Concentration Cell A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 atm, [H+] = 1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate for the solution at electrode 1.What is the pH of the solution? Practice Exercise A concentration cell is constructed with two Zn(s)Zn2+(aq) half-cells. In one half-cell [Zn2+] = 1.35 M, and in the other [Zn2+] = 3.75 104 M. (a) Which half-cell is the anode? (b) What is the emf of the cell? Electrochemistry Applications of Oxidation-Reduction Reactions Electrochemistry © 2012 Pearson Education, Inc. Batteries Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. +1 +2 +4 +6 Electrochemistry Alkaline Batteries Electrochemistry © 2012 Pearson Education, Inc. Hydrogen Fuel Cells Electrochemistry © 2012 Pearson Education, Inc. Corrosion and… Electrochemistry © 2012 Pearson Education, Inc. …Corrosion Prevention Electrochemistry © 2012 Pearson Education, Inc. A. B. C. D. O2 Fe Fe2+ H+ Electrochemistry A. B. C. D. Al and Cu Cu and Ni Al and Zn Ni and Zn Electrochemistry Electrolysis • Voltaic cells are spontaneous. Electrolytic cells are non-spontaneous, and work by using an applied electric current. • A battery or other source acts as an electron pump. Electrochemistry © 2012 Pearson Education, Inc. Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A. Practice Exercise (a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 103 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A? Electrochemistry