Chapter 21: Electrochemistry: Chemical Change and Electrical Work

Transcription

1 Chapter 21: Electrochemistry: Chemical Change and Electrical Work CHEM 1B: GENERAL CHEMISTRY Instructor: Dr. Orlando E. Raola Santa Rosa Junior College 21-1

2 Chapter 21 Electrochemistry: Chemical Change and Electrical Work 21-2

3 Electrochemistry: Chemical Change and Electrical Work 21.1 Redox Reactions and Electrochemical Cells 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries 21.6 Corrosion: An Environmental Voltaic Cell 21.7 Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions 21-3

4 Overview of Redox Reactions Oxidation is the loss of electrons and reduction is the gain of electrons. These processes occur simultaneously. Oxidation results in an increase in O.N. while reduction results in a decrease in O.N. The oxidizing agent takes electrons from the substance being oxidized. The oxidizing agent is therefore reduced. The reducing agent takes electrons from the substance being oxidized. The reducing agent is therefore oxidized. 21-4

5 Figure 21.1 A summary of redox terminology, as applied to the reaction of zinc with hydrogen ion Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 21-5

6 Half-Reaction Method for Balancing Redox Reactions The half-reaction method divides a redox reaction into its oxidation and reduction half-reactions. - This reflects their physical separation in electrochemical cells. This method does not require assigning O.N.s. The half-reaction method is easier to apply to reactions in acidic or basic solutions. 21-6

7 21-7 Steps in the Half-Reaction Method Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species. Balance the atoms and charges in each half-reaction. First balance atoms other than O and H, then O, then H. Charge is balanced by adding electrons (e - ) to the left side in the reduction half-reaction and to the right side in the oxidation half-reaction. If necessary, multiply one or both half-reactions by an integer so that number of e - gained in reduction = number of e - lost in oxidation Add the balanced half-reactions, and include states of matter.

8 Balancing Redox Reactions in Acidic Solution Cr 2 O 2-7 (aq) + I - (aq) Cr 3+ (aq) + I 2 (s) Step 1: Divide the reaction into half-reactions. Cr 2 O 2-7 Cr 3+ I - I 2 Step 2: Balance the atoms and charges in each half-reaction. For the Cr 2 O 2-7 /Cr 3+ half-reaction: Balance atoms other than O and H: Cr 2 O 2-7 2Cr 3+ Balance O atoms by adding H 2 O molecules: Cr 2 O 2-7 2Cr H 2 O 21-8

9 21-9 Balance H atoms by adding H + ions: 14H + + Cr 2 O 7 2-2Cr H 2 O Balance charges by adding electrons: 6e H + + Cr 2 O 7 2-2Cr H 2 O This is the reduction half-reaction. Cr 2 O 7 2- is reduced, and is the oxidizing agent. The O.N. of Cr decreases from +6 to +3. For the I - /I 2 half-reaction: Balance atoms other than O and H: 2I - I 2 There are no O or H atoms, so we balance charges by adding electrons: 2I - I 2 + 2e - This is the oxidation half-reaction. I - is oxidized, and is the reducing agent. The O.N. of I increases from -1 to 0.

10 Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e - lost in the oxidation equals the number of e - gained in the reduction. The reduction half-reaction shows that 6e - are gained; the oxidation half-reaction shows only 2e - being lost and must be multiplied by 3: 3(2I - I 2 + 2e - ) 6I - 3I 2 + 6e - Step 4: Add the half-reactions, canceling substances that appear on both sides, and include states of matter. Electrons must always cancel. 6e H + + Cr 2 O 2-7 2Cr H 2 O 6I - 3I 2 + 6e - 6I - (aq) + 14H + (aq) + Cr 2 O 7 2- (aq) 3I 2 (s) + 7H 2 O(l) + 2Cr 3+ (aq) 21-10

11 Balancing Redox Reactions in Basic Solution An acidic solution contains H + ions and H 2 O. We use H + ions to balance H atoms. A basic solution contains OH - ions and H 2 O. To balance H atoms, we proceed as if in acidic solution, and then add one OH - ion to both sides of the equation. For every OH - ion and H + ion that appear on the same side of the equation we form an H 2 O molecule. Excess H 2 O molecules are canceled in the final step, when we cancel electrons and other common species

12 Sample Problem 21.1 Balancing a Redox Reaction in Basic Solution PROBLEM: Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO 4 and Na 2 C 2 O 4 in basic solution: MnO 4- (aq) + C 2 O 4 2- (aq) MnO 2 (s) + CO 3 2- (aq) [basic solution] PLAN: We follow the numbered steps as described in the text, and proceed through step 4 as if this reaction occurs in acidic solution. Then we add the appropriate number of OH - ions and cancel excess H 2 O molecules. SOLUTION: Step 1: Divide the reaction into half-reactions. MnO 4 - MnO 2 C 2 O 4 2- CO

13 Sample Problem 21.1 Step 2: Balance the atoms and charges in each half-reaction. Balance atoms other than O and H: MnO - 4 MnO 2 C 2 O 2-4 2CO 2-3 Balance O atoms by adding H 2 O molecules: MnO 4 - MnO 2 + 2H 2 O 2H 2 O + C 2 O 4 2-2CO 3 2- Balance H atoms by adding H + ions: 4H + + MnO 4 - MnO 2 + 2H 2 O 2H 2 O + C 2 O 4 2-2CO H + Balance charges by adding electrons: 3e - + 4H + + MnO 4 - MnO 2 + 2H 2 O [reduction] 2H 2 O + C 2 O 4 2-2CO H + + 2e - [oxidation] 21-13

14 Sample Problem 21.1 Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e - lost in the oxidation equals the number of e - gained in the reduction. x 2 x 3 6e - + 8H + + 2MnO 4-2MnO 2 + 4H 2 O 6H 2 O + 3C 2 O 4 2-6CO H + + 6e - Step 4: Add the half-reactions, canceling substances that appear on both sides. 6e - + 8H + + 2MnO 4-2MnO 2 + 4H 2 O 2 6H 2 O + 3C 2 O 2-4 6CO H + + 6e - 2MnO H 2 O + 3C 2 O 4 2-2MnO 2 + 6CO H

15 Sample Problem 21.1 Basic. Add OH - to both sides of the equation to neutralize H +, and cancel H 2 O. 2MnO H 2 O + 3C 2 O OH - 2MnO 2 + 6CO [4H + + 4OH - ] 2MnO H 2 O + 3C 2 O OH - 2MnO 2 + 6CO Including states of matter gives the final balanced equation: 2 4H 2 O 2MnO 4- (aq) + 3C 2 O 4 2- (aq) + 4OH - (aq) 2MnO 2 (s) + 6CO 3 2- (aq) + 2H 2 O(l) 21-15

16 Electrochemical Cells A voltaic cell uses a spontaneous redox reaction (DG < 0) to generate electrical energy. - The system does work on the surroundings. A electrolytic cell uses electrical energy to drive a nonspontaneous reaction (DG > 0). - The surroundings do work on the system. Both types of cell are constructed using two electrodes placed in an electrolyte solution. The anode is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs

17 21-17 Figure 21.2 General characteristics of (A) voltaic and (B) electrolytic cells.

18 Spontaneous Redox Reactions A strip of zinc metal in a solution of Cu 2+ ions will react spontaneously: Cu 2+ (aq) + 2e - Cu(s) Zn(s) Zn 2+ (aq) + 2e - [reduction] [oxidation] Cu 2+ (aq) + Zn(s) Zn 2+ (aq) + Cu(s) Zn is oxidized, and loses electrons to Cu 2+. Although e - are being transferred, electrical energy is not generated because the reacting substances are in the same container

19 21-19 Figure 21.3 The spontaneous reaction between zinc and copper(ii) ion.

20 Construction of a Voltaic Cell Each half-reaction takes place in its own half-cell, so that the reactions are physically separate. Each half-cell consists of an electrode in an electrolyte solution. The half-cells are connected by the external circuit. A salt bridge completes the electrical circuit

21 Figure 21.4A A voltaic cell based on the zinc-copper reaction. Oxidation half-reaction Zn(s) Zn 2+ (aq) + 2e - Reduction half-reaction Cu 2+ (aq) + 2e - Cu(s) Overall (cell) reaction Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)

22 Operation of the Voltaic Cell Oxidation (loss of e - ) occurs at the anode, which is therefore the source of e -. Zn(s) Zn 2+ (aq) + 2e - Over time, the Zn(s) anode decreases in mass and the [Zn 2+ ] in the electrolyte solution increases. Reduction (gain of e - ) occurs at the cathode, where the e - are used up. Cu 2+ (aq) + 2e - Cu(s) Over time, the [Cu 2+ ] in this half-cell decreases and the mass of the Cu(s) cathode increases

23 Figure 21.4B A voltaic cell based on the zinc-copper reaction. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Oxidation half-reaction Zn(s) Zn 2+ (aq) + 2e - After several hours, the Zn anode weighs less as Zn is oxidized to Zn 2+. Reduction half-reaction Cu 2+ (aq) + 2e - Cu(s) The Cu cathode gains mass over time as Cu 2+ ions are reduced to Cu

24 Charges of the Electrodes Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) The anode produces e - by the oxidation of Zn(s). The anode is the negative electrode in a voltaic cell. Electrons flow through the external wire from the anode to the cathode, where they are used to reduce Cu 2+ ions. The cathode is the positive electrode in a voltaic cell

25 The Salt Bridge The salt bridge completes the electrical circuit and allows ions to flow through both half-cells. As Zn is oxidized at the anode, Zn 2+ ions are formed and enter the solution. Cu 2+ ions leave solution to be reduced at the cathode. The salt bridge maintains electrical neutrality by allowing excess Zn 2+ ions to enter from the anode, and excess negative ions to enter from the cathode. A salt bridge contains nonreacting cations and anions, often K + and NO 3-, dissolved in a gel

26 Flow of Charge in a Voltaic Cell Electrons flow through the wire from anode to cathode. Zn(s) Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu(s) Zn 2+ Cations move through the salt bridge from the anode solution to the cathode solution. SO 4 2- Anions move through the salt bridge from the cathode solution to the anode solution. By convention, a voltaic cell is shown with the anode on the left and the cathode on the right

27 Active and Inactive Electrodes An active electrode is an active component in its halfcell and is a reactant or product in the overall reaction. An inactive electrode provides a surface for the reaction and completes the circuit. It does not participate actively in the overall reaction. - Inactive electrodes are necessary when none of the reaction components can be used as an electrode. Inactive electrodes are usually unreactive substances such as graphite or platinum

28 Figure 21.5 A voltaic cell using inactive electrodes. Oxidation half-reaction 2I - (aq) I 2 (s) + 2e - Reduction half-reaction MnO 4- (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) Overall (cell) reaction 2MnO 4- (aq) + 16H + (aq) + 10I - (aq) 2Mn 2+ (aq) + 5I 2 (s) + 8H 2 O(l)

29 Notation for a Voltaic Cell The components of each half-cell are written in the same order as in their half-reactions. The anode components are written on the left. The cathode components are written on the right. Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) The single line shows a phase boundary between the components of a half-cell. The double line shows that the halfcells are physically separated. If needed, concentrations of dissolved components are given in parentheses. (If not stated, it is assumed that they are 1 M.) 21-29

30 Notation for a Voltaic Cell graphite I - (aq) I 2 (s) MnO 4- (aq), H + (aq), Mn 2+ (aq) graphite The inert electrode is specified. A comma is used to show components that are in the same phase

31 Sample Problem 21.2 Describing a Voltaic Cell with Diagram and Notation PROBLEM: Draw a 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(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3 solution, and a KNO 3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. PLAN: From the given contents of the half-cells, we write the halfreactions. To determine which is the anode compartment (oxidation) and which is the cathode (reduction), we note the relative electrode charges. Electrons are released into the anode during oxidation, so it has a negative charge. Since Cr is negative, it must be the anode, and Ag is the cathode

32 Sample Problem 21.2 SOLUTION: The half-reactions are: Ag + (aq) + e - Ag(s) [reduction; cathode] Cr(s) Cr 3+ (aq) + 3e - [oxidation; anode] The balanced overall equation is: 3Ag + + Cr(s) 3Ag(s) + Cr 3+ (aq) The cell notation is given by: Cr(s) Cr 3+ (aq) Ag + (aq) Ag(s) The cell diagram shows the anode on the left and the cathode on the right

33 Electrical Potential and the Voltaic Cell When the switch is closed and no reaction is occurring, each half-cell is in an equilibrium state: Zn(s) Zn 2+ (aq) + 2e - (in Zn metal) Cu(s) Cu 2+ (aq) + 2e - (in Cu metal) Zn is a stronger reducing agent than Cu, so the position of the Zn equilibrium lies farther to the right. Zn has a higher electrical potential than Cu. When the switch is closed, e - flow from Zn to Cu to equalize the difference in electrical potential The spontaneous reaction occurs as a result of the different abilities of these metals to give up their electrons

34 Cell Potential A voltaic cell converts the DG of a spontaneous redox reaction into the kinetic energy of electrons. The cell potential (E cell ) of a voltaic cell depends on the difference in electrical potential between the two electrodes. Cell potential is also called the voltage of the cell or the electromotive forces (emf). E cell > 0 for a spontaneous process

35 Table 21.1 Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) Common alkaline flashlight battery 1.5 Lead-acid car battery (6 cells 12 V) 2.1 Calculator battery (mercury) 1.3 Lithium-ion laptop battery 3.7 Electric eel (~5000 cells in 6-ft eel = 750 V) Nerve of giant squid (across cell membrane)

36 Figure 21.6 Measuring the standard cell potential of a zinccopper cell. The standard cell potential is designated E cell and is measured at a specified temperature with no current flowing and all components in their standard states

37 Standard Electrode Potentials The standard electrode potential (E half-cell ) is the potential of a given half-reaction when all components are in their standard states. By convention, all standard electrode potentials refer to the half-reaction written as a reduction. The standard cell potential depends on the difference between the abilities of the two electrodes to act as reducing agents. E cell = E cathode (reduction) - E anode (oxidation) 21-37

38 The Standard Hydrogen Electrode Half-cell potentials are measured relative to a standard reference half-cell. The standard hydrogen electrode has a standard electrode potential defined as zero (E reference = 0.00 V). This standard electrode consists of a Pt electrode with H 2 gas at 1 atm bubbling through it. The Pt electrode is immersed in 1 M strong acid. 2H + (aq; 1 M) + 2e - H 2 (g; 1 atm) E ref = 0.00V 21-38

39 Figure 21.7 Determining an unknown E half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) Zn 2+ (aq) + 2e Reduction half-reaction 2H 3 O + (aq) + 2e - H 2 (g) + 2H 2 O(l) Overall (cell) reaction Zn(s) + 2H 3 O + (aq) Zn 2+ (aq) + H 2 (g) + 2H 2 O(l)

40 Sample Problem 21.3 Calculating an Unknown E half-cell from E cell PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2 (aq) + Zn(s) Zn 2+ (aq) + 2Br - (aq) E cell = 1.83 V. Calculate E bromine, given that E zinc = V PLAN: E cell is positive, so the reaction is spontaneous as written. By dividing the reaction into half-reactions, we see that Br 2 is reduced and Zn is oxidized; thus, the zinc half-cell contains the anode. We can use the equation for E cell to calculate E bromine. SOLUTION: Br 2 (aq) + 2e - 2Br - (aq) [reduction; cathode] Zn(s) Zn 2+ (aq) + 2e - [oxidation; anode] E zinc = V 21-40

41 Sample Problem 21.3 E cell = E cathode E anode 1.83 = E bromine (-0.76) = E bromine E bromine = 1.07 V 21-41

42 Comparing E half-cell values Standard electrode potentials refer to the half-reaction as a reduction. E values therefore reflect the ability of the reactant to act as an oxidizing agent. The more positive the E value, the more readily the reactant will act as an oxidizing agent. The more negative the E value, the more readily the product will act as a reducing agent

43 strength of oxidizing agent Table 21.2 Selected Standard Electrode Potentials (298 K) Half-Reaction E (V) F 2 (g) + 2e 2F (aq) Cl 2 (g) + 2e 2Cl (aq) MnO 2 (g) + 4H + (aq) + 2e Mn 2+ (aq) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e NO(g) + 2H 2 O(l) Ag + (aq) + e Ag(s) Fe 3+ (g) + e Fe 2+ (aq) O 2 (g) + 2H 2 O(l) + 4e 4OH (aq) Cu 2+ (aq) + 2e Cu(s) 2H + (aq) + 2e H 2 (g) N 2 (g) + 5H + (aq) + 4e N 2 H 5+ (aq) Fe 2+ (aq) + 2e Fe(s) 2H 2 O(l) + 2e H 2 (g) + 2OH (aq) Na + (aq) + e Na(s) strength of reducing agent Li + (aq) + e Li(s)

44 Writing Spontaneous Redox Reactions Each half-reaction contains both a reducing agent and an oxidizing agent. The stronger oxidizing and reducing agents react spontaneously to form the weaker ones. A spontaneous redox reaction (E cell > 0) will occur between an oxidizing agent and any reducing agent that lies below it in the emf series (i.e., one that has a less positive value for E ). The oxidizing agent is the reactant from the half-reaction with the more positive E half-cell

45 Using half-reactions to write a spontaneous redox reaction: Sn 2+ (aq) + 2e - Sn(s) E tin = V Ag + (aq) + e - Ag(s) E silver = 0.80 V Step 1: Reverse one of the half-reactions into an oxidation step so that the difference between the E values will be positive. Here the Ag + /Ag half-reaction has the more positive E value, so it must be the reduction. This half-reaction remains as written. We reverse the Sn 2+ /Sn half-reaction, but we do not reverse the sign: Sn(s) Sn 2+ (aq) + 2e - E tin = V 21-45

46 Step 2: Multiply the half-reactions if necessary so that the number of e - lost is equal to the number or e - gained. 2Ag + (aq) + 2e - 2Ag(s) E silver = 0.80 V Note that we multiply the equation but not the value for E. Step 3: Add the reactions together, cancelling common species. Calculate E cell = E cathode - E anode. Sn(s) Sn 2+ (aq) + 2e - 2Ag + (aq) + 2e - 2Ag(s) Sn(s) + 2Ag + (aq) 2Ag(s) + Sn 2+ (aq) E tin = V E silver = 0.80 V E cell = 0.94 V E cell = E silver E tin = 0.80 (-0.14) = 0.94 V 21-46

47 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 balanced equations for spontaneous reactions (A, B, and C), and calculate E cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents. (1) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) E = 0.96 V (2) N 2 (g) + 5H + (aq) + 4e - N 2 H 5+ (aq) E = V (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E = 1.23 V PLAN: To write the redox equations, we combine the possible pairs of half-reactions. In each case the half-reaction with the less positive value for E will be reversed. We make e - lost equal to e - gained, add the half-reactions and calculate E cell. We can then rank the relative strengths of the oxidizing and reducing agents by comparing E values

48 Sample Problem 21.4 SOLUTION: (a) For (1) and (2), equation (2) has the smaller, less positive E value: (1) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) E = 0.96 V (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - E = V We multiply equation (1) by 4 and equation (2) by 3: (1) 4NO 3- (aq) + 16H + (aq) + 12e - 4NO(g) + 8H 2 O(l) E = 0.96 V (2) 3N 2 H 5+ (aq) 3N 2 (g) + 15H + (aq) + 12e - E = V (A) 4NO 3- (aq) + 3N 2 H 5+ (aq) + H + (aq) 3N 2 (g) + 4NO(g) + 8H 2 O(l) E cell = 0.96 V (-0.23 V) = 1.19 V 21-48

49 Sample Problem 21.4 For (1) and (3), equation (1) has the smaller, less positive E value: (1) NO(g) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - E = 0.96 V (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E = 1.23 V We multiply equation (1) by 2 and equation (3) by 3: (1) 2NO(g) + 4H 2 O(l) 2NO 3- (aq) + 8H + (aq) + 6e - E = 0.96 V (3) 3MnO 2 (s) +12H + (aq) + 6e - 3Mn 2+ (aq) + 6H 2 O(l) E = 1.23 V (B) 3MnO 2 (s) + 4H + (aq) + 2NO(g) 2Mn 2+ (aq) + 2NO 3- (aq) + 2H 2 O(l) E cell = 1.23 V (0.96 V) = 0.27 V 21-49

50 Sample Problem 21.4 For (2) and (3), equation (2) has the smaller, less positive E value: (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - E = V (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E = 1.23 V We multiply equation (3) by 2: (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - E = V (3) 2MnO 2 (s) +8H + (aq) + 4e - 2Mn 2+ (aq) + 4H 2 O(l) E = 1.23 V (C) N 2 H 5+ (aq) + 2MnO 2 (s) + 3H + (aq) N 2 (g) + 2Mn 2+ (aq) + 4H 2 O(l) E cell = 1.23 V (-0.23 V) = 1.46 V 21-50

51 Sample Problem 21.4: (b) We first rank the oxidizing and reducing agents within each equation, then we can compare E cell values. Equation (A) Oxidizing agents: NO 3 - > N 2 Reducing Agents: N 2 H 5 + > NO Equation (B) Oxidizing agents: MnO 2 > NO 3 - Reducing Agents: N 2 H 5 + > NO Equation (C) Oxidizing agents: MnO 2 > N 2 Reducing Agents: N 2 H 5 + > Mn 2+ Comparing the relative strengths from the E cell values: Oxidizing agents: MnO 2 > NO - 3 > N 2 Reducing agents: N 2 H + 5 > NO > Mn

52 The Activity Series of the Metals Metals that can displace H 2 from acid are metals that are stronger reducing agents than H 2. 2H + (aq) + 2e - H 2 (g) E = 0.00V Fe(s) Fe e - E = V Fe(s) + 2H + (aq) Fe 2+ (aq) + H 2 (g) E cell = 0.44 V The lower the metal is in the list of half-cell potentials, the more positive its E half-cell and the stronger it is as a reducing agent. The larger (more positive) the E half-cell of a metal, the more active a metal it is

53 The Activity Series of the Metals Metals that cannot displace H 2 from acid are metals that are weaker reducing agents than H 2. 2H + (aq) + 2e - H 2 (g) E = 0.00V 2Ag(s) 2Ag + + 2e - E = 0.80 V 2Ag(s) + 2H + (aq) 2Ag + (aq) + H 2 (g) E cell = V The higher the metal is in the list of half-cell potentials, the smaller its E half-cell and the weaker it is as a reducing agent. The smaller (more negative) the E half-cell of a metal, the less active a metal it is

54 The Activity Series of the Metals Metals that can displace H 2 from water are metals whose half-reactions lie below that of H 2 O: 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E = V 2Na(s) 2Na + (aq) + 2e - E = V 2Na(s) + 2H 2 O(l) 2Na + (aq) + H 2 (g) + 2OH - (aq) E cell = 2.29 V 21-54

55 The Activity Series of the Metals We can also predict whether one metal can displace another from solution. Any metal that is lower in the list of electrode potentials (i.e., has a larger E value) will reduce the ion of a metal higher up the list. Zn(s) Zn 2+ (aq) + 2e - Fe 2+ (aq) + 2e - Fe(s) E = -0.76V E = -0.44V Zn (s) + Fe 2+ (aq) Zn 2+ (aq) + Fe(s) E cell = 0.32 V 21-55

56 Figure 21.8 The reaction of calcium in water. Oxidation half-reaction Ca(s) Ca 2+ (aq) + 2e - Reduction half-reaction 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) Overall (cell) reaction Ca(s) + 2H 2 O(l) Ca 2+ (aq) + H 2 (g) + 2OH - (aq) 21-56

57 Figure 21.9 A dental voltaic cell. Biting down with a filled tooth on a scrap of aluminum foil will cause pain. The foil acts as an active anode (E aluminum = V), saliva as the electrolyte, and the filling as an inactive cathode as O 2 is reduced to H 2 O

58 Free Energy and Electrical Work For a spontaneous redox reaction, DG < 0 and E cell > 0. DG = -nfe cell n = mol of e - transferred F is the Faraday constant = 9.65x10 4 J/V mol e - Under standard conditions, DG = -nfe cell and RT E cell = ln K nf or V E cell = log K n for T = K 21-58

59 Figure The interrelationship of DG, E cell, and K. DG Reaction Parameters at the Standard State DG K E cell state conditions Reaction at standard- < 0 > 1 > 0 spontaneous at equilibrium > 0 < 1 < 0 nonspontaneous E cell K RT E cell = ln K nf 21-59

60 Sample Problem 21.5 Calculating K and DG from E cell PROBLEM: Lead can displace silver from solution, and silver occurs in trace amounts in some ores of lead. Pb(s) + 2Ag + (aq) Pb 2+ (aq) + 2Ag(s) As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ore. Calculate K and DG at K for this reaction. PLAN: We divide the spontaneous redox reaction into the half-reactions and use values from Appendix D to calculate E cell. From this we can find K and DG. SOLUTION: Writing the half-reactions with their E values: (1) Ag + (aq) + e - Ag(s) E = 0.80 V (2) Pb 2+ (aq) + 2e - Pb(s) E = V 21-60

61 Sample Problem 21.5 We need to reverse equation (2) and multiply equation (1) by 2: (1) 2Ag + (aq) + 2e - 2Ag(s) E = 0.80 V (2) Pb(s) Pb 2+ (aq) + 2e - E = V 2Ag + (aq) + Pb(s) 2Ag(s) + Pb 2+ (aq) E cell = 0.80 (-0.13) = 0.93 V RT E cell = ln K nf = V 2 log K = 0.93 V 0.93 V x 2 log K = = K = 2.6x V DG = -nfe cell = - 2 mol e- mol rxn x 96.5 kj V mol e - x 0.93 V = -1.8x10 2 kj/mol rxn 21-61

62 Cell Potential and Concentration Nernst Equation E cell = E cell - ln Q RT nf When Q < 1, [reactant] > [product], ln Q < 0, so E cell > E cell When Q = 1, [reactant] = [product], ln Q = 0, so E cell = E cell When Q > 1, [reactant] < [product], ln Q > 0, so E cell < E cell We can simplify the equation as before for T = K: E cell = E cell V n log Q 21-62

63 Sample Problem 21.6 Using the Nernst Equation to Calculate E cell PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn 2+ half-cell and an H 2 /H + half-cell under the following conditions: [Zn 2+ ] = M [H + ] = 2.5 M P = 0.30 atm H 2 Calculate E cell at 298 K. PLAN: To apply the Nernst equation and determine E cell, we must know E cell and Q. We write the equation for the spontaneous reaction and calculate E cell from standard electrode potentials. We must convert the given pressure to molarity in order to have consistent units. SOLUTION: (1) 2H + (aq) + 2e - H 2 (g) E = 0.00 V (2) Zn(s) Zn 2+ (aq) + 2e - E = V H + (aq) + Zn(s) H 2 (g) + Zn 2+ (aq) E cell = 0.00 (-0.76) = 0.76 V

64 Sample Problem 21.6 Converting pressure to molarity: n V = P RT = 0.30 atm atm L x K mol K = 1.2x10-2 M Q = [H 2 ][Zn 2+ ] [H + ] x = = 1.9x10-5 (2.5) 2 Solving for E cell at 25 C ( K), with n = 2: E cell = E cell V n = 1.10 V V 2 log Q log(1.9x10-5 ) = 0.76 (-0.14 V) = 0.90 V 21-64

65 Figure 21.11A The relation between E cell and log Q for the zinccopper cell If the reaction starts with [Zn 2+ ] < [Cu 2+ ] (Q < 1), E cell is higher than the standard cell potential. As the reaction proceeds, [Zn 2+ ] decreases and [Cu 2+ ] increases, so E cell drops. Eventually the system reaches equilibrium and the cell can no longer do work.

66 Figure 21.11B The relation between E cell and log Q for the zinccopper cell. A summary of the changes in E cell as any voltaic cell operates

67 Concentration Cells A concentration cell exploits the effect of concentration changes on cell potential. The cell has the same half-reaction in both cell compartments, but with different concentrations of electrolyte: Cu(s) Cu 2+ (aq; 0.10 M) + 2e - Cu 2+ (aq; 1.0 M) Cu(s) Cu 2+ (aq; 1.0 M) Cu 2+ (aq; 0.10 M) [anode; oxidation] [cathode; reduction] As long as the concentrations of the solutions are different, the cell potential is > 0 and the cell can do work

68 Figure A concentration cell based on the Cu/Cu 2+ half-reaction. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Oxidation half-reaction Cu(s) Cu 2+ (aq, 0.1 M) + 2e - Reduction half-reaction Cu 2+ (aq, 1.0 M) + 2e - Cu(s) Overall (cell) reaction Cu 2+ (aq,1.0 M) Cu 2+ (aq, 0.1 M) E cell > 0 as long as the half-cell concentrations are different. The cell is no longer able to do work once the concentrations are equal

69 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, the electrolyte is M AgNO 3 ; in half-cell B, it is 4.0x10-4 M AgNO 3. What is the cell potential at K? PLAN: The standard half-cell potentials are identical, so E cell is zero, and we find E cell from the Nernst equation. Half-cell A has a higher [Ag + ], so Ag + ions are reduced and plate out on electrode A, which is therefore the cathode. In half-cell B, Ag atoms of the electrode are oxidized and Ag + ions enter the solution. Electrode B is thus the anode. As for all voltaic cells, the cathode is positive and the anode is negative

70 Sample Problem 21.7 SOLUTION: The [Ag + ] decreases in half-cell A and increases in halfcell B, so the spontaneous reaction is: Ag + (aq; M) [half-cell A] Ag + (aq; 4.0x10-4 M) [half-cell B] E cell = E cell V 1 log [Ag + ] dil [Ag + ] conc = 0.0 V log 4.0x = V 21-70

71 Figure Laboratory measurement of ph. The operation of a ph meter illustrates an important application of concentration cells. The glass electrode monitors the [H + ] of the solution relative to its own fixed internal [H + ]. An older style of ph meter includes two electrodes. Modern ph meters use a combination electrode

72 Table 21.3 Some Ions Measured with Ion-Specific Electrodes Species Detected NH 3 /NH + 4 CO 2 /HCO - 3 F - Br - I - NO - 3 K + H + Typical Sample Industrial wastewater, seawater Blood, groundwater Drinking water, urine, soil, industrial stack gases Grain, plant tissue Milk, pharmaceuticals Soil, fertilizer, drinking water Blood serum, soil, wine Laboratory solutions, soil, natural waters 21-72

73 Figure Minimocroanalysis A microelectrode records electrical impulses of a single neuron in a monkey s visual cortex. The electrical potential of a nerve cell is due to the difference in concentration of [Na + ] and [K + ] ions inside and outside the cell.

74 Electrochemical Processes in Batteries A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added. A primary battery cannot be recharged. The battery is dead when the cell reaction has reached equilibrium. A secondary battery is rechargeable. Once it has run down, electrical energy is supplied to reverse the cell reaction and form more reactant

75 Figure Alkaline battery. Anode (oxidation): Zn(s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e - Cathode (reduction): MnO 2 (s) + 2H 2 O(l) + 2e - Mn(OH) 2 (s) + 2OH - (aq) Overall (cell) reaction: Zn(s) + MnO 2 (s) + H 2 O(l) ZnO(s) + Mn(OH) 2 (s) E cell = 1.5 V 21-75

76 Figure Silver button battery. Anode (oxidation): Zn(s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e - Cathode (reduction): Ag 2 O(s) + H 2 O(l) + 2e - 2Ag(s) + 2OH - (aq) Overall (cell) reaction: Zn(s) + Ag 2 O(s) ZnO(s) + 2Ag(s) E cell = 1.6 V The mercury battery uses HgO as the oxidizing agent instead of Ag 2 O and has cell potential of 1.3 V

77 Figure Lithium battery. The primary lithium battery is widely used in watches, implanted medical devices, and remote-control devices. Anode (oxidation): 3.5Li(s) 3.5Li e - Cathode (reduction): AgV 2 O Li e - Li 3.5 V 2 O 5.5 Overall (cell) reaction: AgV 2 O Li(s) Li 3.5 V 2 O

78 Figure Lead-acid battery. The lead-acid car battery is a secondary battery and is rechargeable

79 The reactions in a lead-acid battery: The cell generates electrical energy when it discharges as a voltaic cell. Anode (oxidation): Pb(s) + HSO 4- (aq) PbSO 4 (s) + H + (aq) + 2e - Cathode (reduction): PbO 2 (s) + 3H + (aq) + HSO 4- (aq) + 2e - PbSO 4 (s) + 2H 2 O(l) Overall (cell) reaction (discharge): PbO 2 (s) + Pb(s) + H 2 SO 4 (aq) 2PbSO 4 (s) + 2H 2 O(l) E cell = 2.1 V Overall (cell) reaction (recharge): 2PbSO 4 (s) + 2H 2 O(l) PbO 2 (s) + Pb(s) + H 2 SO 4 (aq) 21-79

80 Figure Nickel-metal hydride battery Anode (oxidation): MH(s) + OH - (aq) M(s) + H 2 O(l) + e - Cathode (reduction): NiO(OH)(s) + H 2 O(l) + e - Ni(OH) 2 (s) + OH - (aq) Overall (cell) reaction: MH(s) + NiO(OH)(s) M(s) + Ni(OH) 2 (s) E cell = 1.4 V 21-80

81 Figure Lithium-ion battery. Anode (oxidation): Li x C 6 (s) xli + + xe - + C 6 (s) Cathode (reduction): Li 1-x Mn 2 O 4 (s) + xli + + xe - LiMn 2 O 4 (s) Overall (cell) reaction: Li x C 6 (s) + Li 1-x Mn 2 O 4 (s) LiMn 2 O 4 (s) E cell = 3.7 V The secondary (rechargeable) lithium-ion battery is used to power laptop computers, cell phones, and camcorders

82 Fuel Cells In a fuel cell, also called a flow cell, reactants enter the cell and products leave, generating electricity through controlled combustion. Reaction rates are lower in fuel cells than in other batteries, so an electrocatalyst is used to decrease the activation energy

83 Figure Hydrogen fuel cell. Anode (oxidation): 2H 2 (g) 4H + (aq) + 4e - Cathode (reduction): O 2 (g) + 4H + (aq) + 4e - 2H 2 O(g) Overall (cell) reaction: 2H 2 (g) + O 2 (g) 2H 2 O(g) E cell = 1.2 V 21-83

84 Corrosion: an Environmental Voltaic Cell Corrosion is the process whereby metals are oxidized to their oxides and sulfides. The rusting of iron is a common form of corrosion. - Rust is not a direct product of the reaction between Fe and O 2, but arises through a complex electrochemical process. - Rusting requires moisture, and occurs more quickly at low ph, in ionic solutions, and when the iron is in contact with a less active metal

85 The Rusting of Iron The loss of iron: Fe(s) Fe 2+ (aq) + 2e - [anodic region; oxidation] O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) [cathodic region; reduction] 2Fe(s) + O 2 (g) + 4H + (aq) 2Fe 2+ (aq) + 2H 2 O(l) [overall] The rusting process: Overall reaction: 2Fe 2+ (aq) + ½O 2 (g) + (2 + n)h 2 O(l) Fe 2 O 3 nh 2 O(s) + 4H + (aq) H + ions are consumed in the first step, so lowering the ph increases the overall rate of the process. H + ions act as a catalyst, since they are regenerated in the second part of the process

86 Figure The corrosion of iron

87 Figure Enhanced corrosion at sea. The high ion concentration of seawater enhances the corrosion of iron in hulls and anchors

88 Figure The effect of metal-metal contact on the corrosion of iron. Fe in contact with Cu corrodes faster. Fe in contact with Zn does not corrode. The process is known as cathodic protection

89 Figure The use of sacrificial anodes to prevent iron corrosion. In cathodic protection, an active metal, such as zinc, magnesium, or aluminum, acts as the anode and is sacrificed instead of the iron

90 Electrolytic Cells An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction. Cu(s) Cu 2+ (aq) + 2e - Sn 2+ (aq) + 2e - Sn(s) Cu(s) + Sn 2+ (aq) Cu 2+ (aq) + Sn(s) [anode; oxidation] [cathode; reduction] E cell = V and ΔG = 93 kj As with a voltaic cell, oxidation occurs at the anode and reduction takes place at the cathode. An external source supplies the cathode with electrons, which is negative, and removes then from the anode, which is positive. Electrons flow from cathode to anode

91 Figure The tin-copper reaction as the basis of a voltaic and an electrolytic cell. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sn(s) Sn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu(s) Cu 2+ (aq) + Sn(s) Cu(s) + Sn 2+ (aq) voltaic cell Cu(s) Cu 2+ (aq) + 2e - Sn 2+ (aq) + 2e - Sn(s) Sn 2+ (aq) + Cu(s) Sn(s) + Cu 2+ (aq) electrolytic cell 21-91

92 Figure The processes occurring during the discharge and recharge of a lead-acid battery. VOLTAIC (discharge) Switch ELECTROLYTIC (recharge) 21-92

93 Table 21.4 Comparison of Voltaic and Electrolytic Cells Cell Type DG E cell Electrode Name Process Sign Voltaic < 0 > 0 Anode Oxidation - Voltaic < 0 > 0 Cathode Reduction + Electrolytic > 0 < 0 Anode Oxidation + Electrolytic > 0 < 0 Cathode Reduction

94 Products of Electrolysis Electrolysis is the splitting (lysing) of a substance by the input of electrical energy. During electrolysis of a pure, molten salt, the cation will be reduced and the anion will be oxidized. During electrolysis of a mixture of molten salts - the more easily oxidized species (stronger reducing agent) reacts at the anode, and - the more easily reduced species (stronger oxidizing agent) reacts at the cathode

95 Sample Problem 21.8 Predicting the Electrolysis Products of a Molten Salt Mixture PROBLEM: A chemical engineer melts a naturally occurring mixture of NaBr and MgCl 2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction. PLAN: We need to determine which metal and nonmetal will form more easily at the electrodes. We list the ions as oxidizing or reducing agents. If a metal holds its electrons more tightly than another, it has a higher ionization energy (IE). Its cation will gain electrons more easily, and it will be the stronger oxidizing agent. If a nonmetal holds its electrons less tightly than another, it has a lower electronegativity (EN). Its anion will lose electrons more easily, and it will be the reducing agent

96 Sample Problem 21.8 SOLUTION: Possible oxidizing agents: Na +, Mg 2+ Possible reducing agents: Br -, Cl - Mg is to the right of Na in Period 3. IE increases from left to right across the period, so Mg has the higher IE and gives up its electrons less easily. The Mg 2+ ion has a greater attraction for e - than the Na + ion. Mg 2+ (l) + 2e - Mg(l) [cathode; reduction] Br is below Cl in Group 7A. EN decreases down the group, so Br accepts e - less readily than Cl. The Br - ion will lose its e - more easily, so it is more easily oxidized. 2Br - (l) Br 2 (g) + 2e - [anode; oxidation] The overall cell reaction is: Mg 2+ (l) + 2Br - (l) Mg(l) + Br 2 (g) 21-96

97 Figure The electrolysis of water. Overall (cell) reaction 2H 2 O(l) H 2 (g) + O 2 (g) Oxidation half-reaction 2H 2 O(l) 4H + (aq) + O 2 (g) + 4e - Reduction half-reaction 2H 2 O(l) + 4e - 2H 2 (g) + 2OH - (aq) 21-97

98 Electrolysis of Aqueous Salt Solutions When an aqueous salt solution is electrolyzed - The strongest oxidizing agent (most positive electrode potential) is reduced, and - The strongest reducing agent (most negative electrode potential) is oxidized. Overvoltage is the additional voltage needed (above that predicted by E values) to produce gases at metal electrodes. Overvoltage needs to be taken into account when predicting the products of electrolysis for aqueous solutions. Overvoltage is V for H 2 (g) or O 2 (g)

99 Summary of the Electrolysis of Aqueous Salt Solutions Cations of less active metals (Au, Ag, Cu, Cr, Pt, Cd) are reduced to the metal. Cations of more active metals are not reduced. H 2 O is reduced instead. Anions that are oxidized, because of overvoltage from O 2 formation, include the halides, except for F -. Anions that are not oxidized include F - and common oxoanions. H 2 O is oxidized instead

100 Sample Problem 21.9 Predicting the Electrolysis Products of Aqueous Salt Solutions 21- PROBLEM: What products form at which electrode during electrolysis of aqueous solution of the following salts? (a) KBr (b) AgNO 3 (c) MgSO 4 PLAN: We identify the reacting ions and compare their electrode potentials with those of water, taking the V overvoltage into account. The reduction half-reaction with the less negative E occurs at the cathode, while the oxidation half-reaction with the less positive E occurs at the anode. SOLUTION: (a) KBr K + (aq) + e - K(s) E = H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E = -0.42V Despite the overvoltage, which makes E for the reduction of water between -0.8 and -1.0 V, H 2 O is still easier to reduce than K +, so H 2 (g) forms at the cathode.

101 Sample Problem Br - (aq) Br 2 (l) + 2e - 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - E = 1.07 V E = 0.82 V The overvoltage makes E for the oxidation of water between 1.2 and 1.4 V. Br - is therefore easier to oxidize than water, so Br 2 (g) forms at the anode. (b) AgNO 3 Ag + (aq) + e - Ag(s) E = 0.80 V 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E = -0.42V As the cation of an inactive metal, Ag + is a better oxidizing agent than H 2 O, so Ag(s) forms at the cathode. NO 3 - cannot be oxidized, because N is already in its highest (+5) oxidation state. Thus O 2 (g) forms at the anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - 21-

102 Sample Problem 21.9 (c) MgSO 4 Mg 2+ (aq) + 2e - Mg(s) E = V 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E = -0.42V Mg 2+ is a much weaker oxidizing agent than H 2 O, so H 2 (g) forms at the cathode. SO 4 2- cannot be oxidized, because S is already in its highest (+6) oxidation state. Thus O 2 (g) forms at the anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - 21-

103 Stoichiometry of Electrolysis Faraday s law of electrolysis states that the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell. The current flowing through the cell is the amount of charge per unit time. Current is measured in amperes. Current x time = charge 21-

104 Figure A summary diagram for the stoichiometry of electrolysis. MASS (g) of substance oxidized or reduced CURRENT (A) M (g/mol) time (s) AMOUNT (mol) of substance oxidized or reduced balanced half-reaction AMOUNT (mol) of electrons transferred Faraday constant (C/mol e - ) CHARGE (C) 21-

105 Sample Problem Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by electrolysis of aqueous Cr 2 (SO 4 ) 3. If 12.5 min is allowed for the plating, what current is needed? PLAN: To find the current, we divide charge by time, so we need to find the charge. We write the half-reaction for Cr 3+ reduction to get the amount (mol) of e - transferred per mole of Cr. We convert mass of Cr needed to amount (mol) of Cr. We can then use the Faraday constant to find charge and current. mass (g) of Cr needed divide by M mol of Cr 3 mol e - = 1 mol Cr mol e - transferred Charge (C) current (A) 21-1 mol e - = 9.65x10 4 C divide by time in s

106 Sample Problem SOLUTION: Cr 3+ (aq) + 3e - Cr(s) 0.86 g Cr x 1 mol Cr g Cr x 3 mol e- 1 mol Cr = mol e - Charge (C) = mol e - x 9.65x104 C 1 mol e - = 4.8x10 3 C Current (A) = charge (C) time (s) = 4.8x103 C 12.5 min x 1 min 60 s = 6.4 C/s = 6.4 A 21-

107 Chemical Connections Figure B21.1 The mitochondrion Mitochondria are subcellular particles outside the cell nucleus that control the electron-transport chain, an essential part of energy production in living organisms. 21-

108 Chemical Connections Figure B21.2 The main energy-yielding steps in the electrontransport chain (ETC). 21-

109 Chemical Connections Figure B21.3 Coupling electron transport to proton transport to ATP synthesis. 21-