CHEMISTRY The Molecular Nature of Matter and Change

CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 21 Lecture Outlines* *See Power. Point Image Slides for all figures and tables pre-inserted into Power. Point without notes. 21 -1 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

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

Electrochemistry: Chemical Change and Electrical Work 21. 1 Half-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: A Case of Environmental Electrochemistry 21. 7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction 21 -3

Key Points About Redox Reactions • Oxidation (electron loss) always accompanies reduction (electron gain). • The oxidizing agent is reduced, and the reducing agent is oxidized. • The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent. 21 -4

Figure 21. 1 A summary of redox terminology Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) OXIDATION One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. Oxidation number increases. The oxidation number of Zn increases from x to +2. REDUCTION Other reactant gains electrons. Oxidizing agent is reduced. 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 0. 21 -5

Half-Reaction Method for Balancing Redox Reactions Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. • Each reaction is balanced for mass (atoms) and charge. • 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. 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. 21 -6

Balancing Redox Equations in Acidic Conditions 1. 2. 3. 4. 5. 6. 7. 8. Write the skeletons of the oxidation and reduction half-reactions. Balance all elements other than O and H. Balance the oxygen atoms by adding H 2 O 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 2 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 9. 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 H 2 O molecules. 12. Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1 -11. 21 -8

Balancing Redox Reactions in Acidic Solution Cr 2 O 72 -(aq) + I-(aq) Cr 3+(aq) + I 2(aq) 1. Divide the reaction into half-reactions Determine the O. N. s for the species undergoing redox. +6 -1 2 Cr 2 O 7 (aq) + I-(aq) Cr 2 O 72 I- Cr 3+ I 2 +3 0 3+ Cr (aq) + I 2(aq) Cr is going from +6 to +3 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction 14 H+(aq) + Cr 2 O 72 net: +12 6 e- + 14 H+(aq) + Cr 2 O 72 - 21 -9 2 Cr 3+ net: +6 2 Cr 3+ + 7 H 2 O(l) - to left. Add 6 e� + 7 H 2 O(l)

Balancing Redox Reactions in Acidic Solution 6 e- + 14 H+(aq) + Cr 2 O 722 I- 2 Cr 3+ I 2 continued + 7 H 2 O(l) + 2 e- Cr(+6) is the oxidizing agent and I(-1) is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary 2 I- I 2 + 2 e- X 3 4. Add the half-reactions together 6 e- + 14 H+ + Cr 2 O 726 I 14 H+(aq) + Cr 2 O 72 -(aq) + 6 I-(aq) 2 Cr 3+ 3 I 2 + 6 e 2 Cr 3+(aq) + 3 I 2(s) + 7 H 2 O(l) Do a final check on atoms and charges. 21 -10 + 7 H 2 O(l)

Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. 14 H+(aq) + Cr 2 O 72 -(aq) + 6 I-(aq) 2 Cr 3+(aq) + 3 I 2(s) + 7 H 2 O(l) + 14 OH-(aq) 14 H 2 O + Cr 2 O 72 - + 6 I- + 14 OH-(aq) 2 Cr 3+ + 3 I 2 + 7 H 2 O + 14 OH- Reconcile the number of water molecules. 7 H 2 O + Cr 2 O 72 - + 6 I- 2 Cr 3+ + 3 I 2 + 14 OH- Do a final check on atoms and charges. 21 -11

Figure 21. 2 The redox reaction between dichromate ion and iodide ion. Cr 2 O 72 - I- Cr 3+ + I 2 21 -12

Sample Problem 21. 1: PROBLEM: Balancing Redox Reactions by the Half-Reaction Method 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 Na. Mn. O 4 and Na 2 C 2 O 4 in basic solution: Mn. O 4 -(aq) + C 2 O 42 -(aq) PLAN: Mn. O 2(s) + CO 32 -(aq) Proceed in acidic solution and then neutralize with base. SOLUTION: Mn. O 4+7 Mn. O 44 H+ + Mn. O 4 - 21 -13 Mn. O 2 +4 Mn. O 2 C 2 O 42+3 C 2 O 42 - Mn. O 2 + 2 H 2 O C 2 O 42 - + 2 H 2 O +3 e- +2 e- CO 32+4 2 CO 322 CO 32 - + 4 H+

Sample Problem 21. 1: Balancing Redox Reactions by the Half-Reaction Method continued: 4 H+ + Mn. O 4 - +3 e- Mn. O 2+ 2 H 2 O C 2 O 42 - + 2 H 2 O 2 CO 32 - + 4 H+ + 2 e- 4 H+ + Mn. O 4 - +3 e- Mn. O 2+ 2 H 2 O C 2 O 42 - + 2 H 2 O 2 CO 32 - + 4 H+ + 2 e. X 3 X 2 8 H+ + 2 Mn. O 4 - +6 e- 2 Mn. O 2+ 4 H 2 O 8 H+ + 2 Mn. O 4 - +6 e 3 C 2 O 42 - + 6 H 2 O 2 Mn. O 2 -(aq) + 3 C 2 O 42 -(aq) + 2 H 2 O(l) + 4 OH 2 Mn. O 2 -(aq) + 3 C 2 O 42 -(aq) + 4 OH-(aq) 21 -14 3 C 2 O 42 - + 6 H 2 O 6 CO 32 - + 12 H+ + 6 e- 2 Mn. O 2+ 4 H 2 O 6 CO 32 - + 12 H+ + 6 e 2 Mn. O 2(s) + 6 CO 32 -(aq) + 4 H+(aq) + 4 OH 2 Mn. O 2(s) + 6 CO 32 -(aq) + 2 H 2 O(l)

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- 21 -15 ELECTROLYTIC CELL Surroundings(power Energy is absorbed tosupply) drive a nonspontaneous redox reaction do work on system(cell) A(s) B(s) Oxidation half-reaction AA + e- Reduction half-reaction Y+ + e - Y Reduction half-reaction B+ + e B Overall (cell) reaction X + Y+ X+ + Y DG < 0 Overall (cell) reaction A- + B + A + B DG > 0

Figure 21. 4 The spontaneous reaction between zinc and copper(II) ion Zn(s) + Cu 2+(aq) 21 -16 Zn 2+(aq) + Cu(s)

Figure 21. 5 A voltaic cell based on the zinc-copper reaction Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e- Reduction half-reaction Cu 2+(aq) + 2 e. Cu(s) Overall (cell) reaction Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) 21 -17

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 phase of lower oxidation state phase boundary between half-cells Zn(s) | Zn 2+(aq) || Cu 2+(aq) | Cu (s) Examples: Zn(s) Zn 2+(aq) + 2 e- Cu(s) graphite | I-(aq) | I 2(s) || H+(aq), Mn. O 4 -(aq) | Mn 2+(aq) | graphite inert electrode 21 -18

Figure 21. 6 A voltaic cell using inactive electrodes Oxidation half-reaction 2 I-(aq) I 2(s) + 2 e- Reduction half-reaction Mn. O 4 -(aq) + 8 H+(aq) + 5 e. Mn 2+(aq) + 4 H 2 O(l) Overall (cell) reaction 2 Mn. O 4 -(aq) + 16 H+(aq) + 10 I-(aq) 2 Mn 2+(aq) + 5 I 2(s) + 8 H 2 O(l) 21 -19

Sample Problem 21. 2: PROBLEM: PLAN: Diagramming Voltaic Cells 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 Ag. NO 3 solution, and a KNO 3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. 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). SOLUTION: Voltmeter e- Oxidation half-reaction Cr(s) Cr 3+(aq) + 3 e. Reduction half-reaction Ag+(aq) + e. Ag(s) salt bridge Cr K+ Ag NO 3 Cr 3+ + Ag� Overall (cell) reaction Cr(s) + Ag+(aq) Cr 3+(aq) + Ag(s) Cr(s) | Cr 3+(aq) || Ag+(aq) | Ag(s) 21 -20

Why Does a Voltaic Cell Work? 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. Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C) 21 -21

Table 21. 1 Voltages of Some Voltaic Cells Voltaic Cell 21 -22 Voltage (V) Common alkaline battery 1. 5 Lead-acid car battery (6 cells = 12 V) 2. 0 Calculator battery (mercury) 1. 3 Electric eel (~5000 cells in 6 -ft eel = 750 V) 0. 15 Nerve of giant squid (across cell membrane) 0. 070

Figure 21. 7 Determining an unknown E 0 half-cell with the standard reference (hydrogen) electrode Oxidation half-reaction Zn(s) Zn 2+(aq) + 2 e- Overall (cell) reaction Zn(s) + 2 H 3 O+(aq) Zn 2+(aq) + H 2(g) + 2 H 2 O(l) 21 -23 Reduction half-reaction 2 H 3 O+(aq) + 2 e. H 2(g) + 2 H 2 O(l)

Sample Problem 21. 3: PROBLEM: Calculating an Unknown E 0 half-cell from E 0 cell A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2(aq) + Zn(s) Zn 2+(aq) + 2 Br-(aq) E 0 cell = 1. 83 V Calculate E 0 bromine given E 0 zinc = -0. 76 V PLAN: The reaction is spontaneous as written since the E 0 cell is (+). Zinc is being oxidized and is the anode. Therefore the E 0 bromine can be found using E 0 cell = E 0 cathode - E 0 anode. SOLUTION: anode: Zn(s) Zn 2+(aq) + 2 e� E = +0. 76 E 0 Zn as Zn 2+(aq) + 2 e- Zn(s) is -0. 76 V E 0 cell = E 0 cathode - E 0 anode = 1. 83 = E 0 bromine - (-0. 76) E 0 bromine = 1. 86 - 0. 76 = 1. 07 V 21 -24

Table 21. 2 Selected Standard Electrode Potentials (298 K) Half-Reaction F 2(g) + 2 e 2 F-(aq) Cl 2(g) + 2 e 2 Cl-(aq) Mn. O 2(g) + 4 H+(aq) + 2 e. Mn 2+(aq) + 2 H 2 O(l) NO 3 -(aq) + 4 H+(aq) + 3 e. NO(g) + 2 H 2 O(l) Ag+(aq) + e. Ag(s) Fe 3+(g) + e. Fe 2+(aq) O 2(g) + 2 H 2 O(l) + 4 e 4 OH-(aq) Cu 2+(aq) + 2 e. Cu(s) 2 H+(aq) + 2 e. H 2(g) N 2(g) + 5 H+(aq) + 4 e. N 2 H 5+(aq) Fe 2+(aq) + 2 e. Fe(s) 2 H 2 O(l) + 2 e. H 2(g) + 2 OH-(aq) Na+(aq) + e. Na(s) Li+(aq) + e. Li(s) 21 -25 E 0(V) +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

• By convention, electrode potentials are written as reductions. • When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. • The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E 0 cell. • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) stronger reducing agent 21 -26 + Cu 2+(aq) stronger oxidizing agent Zn 2+(aq) weaker oxidizing agent + Cu(s) weaker reducing agent

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 E 0 cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents: (1) NO 3 -(aq) + 4 H+(aq) + 3 e(2) N 2(g) + 5 H+(aq) + 4 e(3) Mn. O 2(s) +4 H+(aq) + 2 e. PLAN: NO(g) + 2 H 2 O(l) N 2 H 5+(aq) Mn 2+(aq) + 2 H 2 O(l) E 0 = 0. 96 V E 0 = -0. 23 V E 0 = 1. 23 V Put the equations together in varying combinations so as to produce (+) E 0 cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E 0. Balance the number of electrons gained and lost without changing the E 0. In ranking the strengths, compare the combinations in terms of E 0 cell. 21 -27

Sample Problem 21. 4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (a) (1) NO 3 -(aq) + 4 H+(aq) + 3 e- Rev (2) N 2 H 5+(aq) (1) NO 3 -(aq) + 4 H+(aq) (2) N 2 H 5+(aq) (A) N 2(g) + 5 H+(aq) + 4 e+ 3 e- X 4 (B) 2 NO(g) + 3 Mn. O 2(s) + 4 H+(aq) E 0 cell = 1. 19 V X 3 4 NO(g) + 3 N 2(g) + 8 H 2 O(l) E 0 = -0. 96 V Mn 2+(aq) + 2 H 2 O(l) NO 3 -(aq) + 4 H+(aq) + 3 e- (3) Mn. O 2(s) +4 H+(aq) + 2 e- 21 -28 E 0 = +0. 23 V NO 3 -(aq) + 4 H+(aq) + 3 e- (3) Mn. O 2(s) +4 H+(aq) + 2 e(1) NO(g) + 2 H 2 O(l) N 2(g) + 5 H+(aq) + 4 e- 4 NO 3 -(aq) + 3 N 2 H 5+(aq) + H+(aq) Rev (1) NO(g) + 2 H 2 O(l) E 0 = 0. 96 V Mn 2+(aq) + 2 H 2 O(l) E 0 = 1. 23 V X 2 E 0 cell = 0. 27 V X 3 2 NO 3 -(aq) + 3 Mn 3+(aq) + 2 H 2 O(l)

Sample Problem 21. 4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) Rev (2) N 2 H 5+(aq) N 2(g) + 5 H+(aq) + 4 e- (3) Mn. O 2(s) +4 H+(aq) + 2 e(2) N 2 H 5+(aq) E 0 = +0. 23 V Mn 2+(aq) + 2 H 2 O(l) E 0 = 1. 23 V E 0 cell = 1. 46 V N 2(g) + 5 H+(aq) + 4 e- (3) Mn. O 2(s) +4 H+(aq) + 2 e(C) N 2 H 5+(aq) + 2 Mn. O 2(s) + 3 H+(aq) Mn 2+(aq) + 2 H 2 O(l) X 2 N 2(g) + 2 Mn 2+(aq) + 4 H 2 O(l) (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO 3 - > N 2 reducing agents: N 2 H 5+ > NO (B): oxidizing agents: Mn. O 2 > NO 3 - reducing agents: NO > Mn 2+ (C): oxidizing agents: Mn. O 2 > N 2 reducing agents: N 2 H 5+ > Mn 2+ 21 -29

Sample Problem 21. 4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: Mn. O 2 > NO 3 - > N 2 Reducing agents: N 2 H 5+ > NO > Mn 2+ 21 -30

Relative Reactivities (Activities) of Metals 1. Metals that can displace H 2 from acid 2. Metals that cannot displace H 2 from acid 3. Metals that can displace H 2 from water 4. Metals that can displace other metals from solution 21 -31

Figure 21. 8 The reaction of calcium in water Oxidation half-reaction Ca(s) Ca 2+(aq) + 2 e- Reduction half-reaction 2 H 2 O(l) + 2 e. H 2(g) + 2 OH-(aq) Overall (cell) reaction Ca(s) + 2 H 2 O(l) Ca 2+(aq) + H 2(g) + 2 OH-(aq) 21 -32

Free Energy and Electrical Work DG a -Ecell = DG = wmax = charge x (-Ecell) -wmax DG = -n F Ecell charge In the standard state charge = n F DG 0 = -n F E 0 cell n = # mols e. F = Faraday constant F = 96, 485 C/mol e- 1 V = 1 J/C F = 9. 65 x 104 J/V*mol e- 21 -33 DG 0 = - RT ln K E 0 cell = - (RT/n F) ln K

The interrelationship of DG 0, E 0, and K Figure 21. 9 DG 0 = -n. FEocell Reaction at standard-state conditions DG 0 K E 0 cell <0 >1 >0 0 1 0 at equilibrium >0 <1 <0 nonspontaneous DG 0 = -RT ln. K By substituting standard state values into E 0 cell, we get E 0 cell K E 0 cell = -RT ln. K n. F 21 -34 E 0 cell = 0. 0592(V/n) log K (at 250 C)

Sample Problem 21. 5: PROBLEM: Calculating K and DG 0 from E 0 cell Lead can displace silver from solution: Pb(s) + 2 Ag+(aq) Pb 2+(aq) + 2 Ag(s) As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DG 0 at 250 C for this reaction. PLAN: Break the reaction into half-reactions, find the E 0 for each half-reaction and then the E 0 cell. Substitute into the equations found on slide SOLUTION: 2 X E 0 cell = log K = 21 -35 Pb 2+(aq) + 2 e. Ag+(aq) + e- Pb(s) Ag(s) E 0 = -0. 13 V E 0 = 0. 80 V Pb(s) Pb 2+(aq) + 2 e. Ag+(aq) + e. Ag(s) 0. 592 V n n x E 0 cell 0. 592 V log K = (2)(0. 93 V) 0. 592 V E 0 = 0. 13 V E 0 = 0. 80 V E 0 cell = 0. 93 V DG 0 = -n. FE 0 cell = -(2)(96. 5 k. J/mol*V)(0. 93 V) K = 2. 6 x 1031 DG 0 = -1. 8 x 102 k. J

Sample Problem 21. 6: Using the Nernst Equation to Calculate Ecell 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+] = 0. 010 M Calculate Ecell at [H+] = 2. 5 M PH = 0. 30 atm 2 250 C. PLAN: Find E 0 cell and Q in order to use the Nernst equation. SOLUTION: Determining E 0 cell : 2 H+(aq) + 2 e- H 2(g) E 0 = 0. 00 V Zn 2+(aq) + 2 e- Zn(s) E 0 = -0. 76 V Zn 2+(aq) + 2 e- E 0 = +0. 76 V Zn(s) Ecell = E 0 cell - 0. 0592 V n log Q Ecell = 0. 76 - (0. 0592/2)log(4. 8 x 10 -4) = 0. 86 V 21 -36 Q= P x [Zn 2+] H 2 [H+]2 Q= (0. 30)(0. 010) (2. 5)2 Q = 4. 8 x 10 -4

The Effect of Concentration on Cell Potential DG = DG 0 + RT ln Q -n. F Ecell = -n. F Ecell + RT ln Q Ecell = E 0 cell - RT n. F ln Q • When Q < 1 and thus [reactant] > [product], ln. Q < 0, so Ecell > E 0 cell • When Q = 1 and thus [reactant] = [product], ln. Q = 0, so Ecell = E 0 cell • When Q >1 and thus [reactant] < [product], ln. Q > 0, so Ecell < E 0 cell Ecell = 21 -37 E 0 cell - 0. 0592 n log Q

The relation between Ecell and log Q for the zinc-copper cell Overall (cell) reaction Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) 21 -38 Figure 21. 10

Figure 21. 11 A concentration cell based on the Cu/Cu 2+ half-reaction Oxidation half-reaction Cu(s) Cu 2+(aq, 0. 1 M) + 2 e- Reduction half-reaction Cu 2+(aq, 1. 0 M) + 2 e. Cu(s) Overall (cell) reaction Cu 2+(aq, 1. 0 M) Cu 2+(aq, 0. 1 M) 21 -39

Sample Problem 21. 7: PROBLEM: PLAN: Calculating the Potential of a Concentration Cell A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0. 0100 M Ag. NO 3; in half-cell B, electrode B dips into 4. 0 x 10 -4 M Ag. NO 3. What is the cell potential at 298 K? Which electrode has a positive charge? E 0 cell 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. 010 M) half-cell A Ecell = E 0 cell - 0. 0592 V 1 log Ag+(aq, 4. 0 x 10 -4 M) half-cell B [Ag+]dilute [Ag+]concentrated Ecell = 0 V -0. 0592 log 4. 0 x 10 -2 = 0. 0828 V Half-cell A is the cathode and has the positive electrode. 21 -40

Figure 21. 12 The laboratory measurement of p. H Pt Glass electrode Reference (calomel) electrode Hg Ag. Cl on Ag on Pt 1 M HCl Thin glass membrane 21 -41 Paste of Hg 2 Cl 2 in Hg KCl solution Porous ceramic plugs

Table 21. 3 Some Ions Measured with Ion-Specific Electrodes 21 -42 Species Detected Typical Sample NH 3/NH 4+ Industrial wastewater, seawater CO 2/HCO 3 - Blood, groundwater F- Drinking water, urine, soil, industrial stack gases Br- Grain, plant tissue I- Milk, pharmaceuticals NO 3 - Soil, fertilizer, drinking water K+ Blood serum, soil, wine H+ Laboratory solutions, soil, natural waters

Figure 21. 13 The corrosion of iron 21 -43

Figure 21. 14 21 -44 Enhanced corrosion at sea

Figure 21. 15 The effect of metal-metal contact on the corrosion of iron faster corrosion 21 -45 cathodic protection

Figure 21. 16 21 -46 The use of sacrificial anodes to prevent iron corrosion

Figure 21. 17 The tin-copper reaction as the basis of a voltaic and an electrolytic cell voltaic cell Oxidation half-reaction Sn(s) Sn 2+(aq) + 2 e. Reduction half-reaction Cu 2+(aq) + 2 e. Cu(s) Overall (cell) reaction Sn(s) + Cu 2+(aq) Sn 2+(aq) + Cu(s) 21 -47 electrolytic cell Oxidation half-reaction Cu(s) Cu 2+(aq) + 2 e. Reduction half-reaction Sn 2+(aq) + 2 e. Sn(s) Overall (cell) reaction Sn(s) + Cu 2+(aq) Sn 2+(aq) + Cu(s)

Figure 21. 18 The processes occurring during the discharge and recharge of a lead-acid battery VOLTAIC(discharge) ELECTROLYTIC(recharge) 21 -48

Table 21. 4 Comparison of Voltaic and Electrolytic Cells Electrode Cell Type DG Ecell Name Process Sign Voltaic <0 >0 Anode Oxidation - Voltaic <0 >0 Cathode Reduction + Electrolytic >0 <0 Anode Oxidation + Electrolytic >0 <0 Cathode Reduction - 21 -49

Sample Problem 21. 8: PROBLEM: Predicting the Electrolysis Products of a Molten Salt Mixture A chemical engineer melts a naturally occurring mixture of Na. Br and Mg. Cl 2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced halfreactions and the overall cell reaction. Consider the metal and nonmetal components of each compound and PLAN: then determine which will recover electrons(be reduced; strength as an oxidizing agent) better. This is the converse to which of the elements will lose electrons more easily (lower ionization energy). SOLUTION: Possible oxidizing agents: Na+, Mg 2+ Possible reducing agents: Br-, Cl. Na, the element, is to the left of Mg in the periodic table, therefore the IE of Mg is higher than that of Na. So Mg 2+ will more easily gain electrons and is the stronger oxidizing agent. Br, as an element, has a lower IE than does Cl, and therefore will give up electrons as Br- more easily than will Cl-. Mg 2+(l) + 2 Br-(l) cathode 21 -50 anode Mg(s) + Br 2(g)

Figure 21. 19 The electrolysis of water Overall (cell) reaction 2 H 2 O(l) H 2(g) + O 2(g) Oxidation half-reaction 2 H 2 O(l) 4 H+(aq) + O 2(g) + 4 e- 21 -51 Reduction half-reaction 2 H 2 O(l) + 4 e 2 H 2(g) + 2 OH-(aq)

Sample Problem 21. 9: PROBLEM: Predicting the Electrolysis Products of Aqueous Ionic Solutions What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) Ag. NO 3; (c) Mg. SO 4? PLAN: Compare the potentials of the reacting ions with those of water, remembering to consider the 0. 4 to 0. 6 V overvoltage. The reduction half-reaction with the less negative potential, and the oxidation halfreaction with the less positive potential will occur at their respective electrodes. SOLUTION: (a) K+(aq) + e 2 H 2 O(l) + 2 e- K(s) H 2(g) + 2 OH-(aq) E 0 = -2. 93 V E 0 = -0. 42 V The overvoltage would make the water reduction -0. 82 to -1. 02 but the reduction of K+ is still a higher potential so H 2(g) is produced at the cathode. 2 Br-(aq) 2 H 2 O(l) Br 2(g) + 2 e. O 2(g) + 4 H+(aq) + 4 e- E 0 = 1. 07 V E 0 = 0. 82 V The overvoltage would give the water half-cell more potential than the Br-, so the Br- will be oxidized. Br 2(g) forms at the anode. 21 -52

Sample Problem 21. 9: continued (b) Ag+(aq) + e 2 H 2 O(l) + 2 e- Predicting the Electrolysis Products of Aqueous Ionic Solutions Ag(s) E 0 = -0. 80 V H 2(g) + 2 OH-(aq) E 0 = -0. 42 V Ag+ is the cation of an inactive metal and therefore will be reduced to Ag at the cathode. Ag+(aq) + e. Ag(s) The N in NO 3 - is already in its most oxidized form so water will have to be oxidized to produce O 2 at the anode. 2 H O(l) O (g) + 4 H+(aq) + 4 e 2 (c) Mg 2+(aq) + 2 e- Mg(s) 2 E 0 = -2. 37 V Mg is an active metal and its cation cannot be reduced in the presence of water. So as in (a) water is reduced and H 2(g) is produced at the cathode. The S in SO 42 - is in its highest oxidation state; therefore water must be oxidized and O 2(g) will be produced at the anode. 21 -53

Figure 21. 20 A summary diagram for the stoichiometry of electrolysis MASS (g) of substance oxidized or reduced M(g/mol) AMOUNT (MOL) of substance oxidized or reduced AMOUNT (MOL) of electrons transferred balanced half-reaction CHARGE (C) time(s) 21 -54 Faraday constant (C/mol e-) CURRENT (A)

Sample Problem 21. 10: PROBLEM: PLAN: Applying the Relationship Among Current, Time, and Amount of Substance A technician is plating a faucet with 0. 86 g of Cr from an electrolytic bath containing aqueous Cr 2(SO 4)3. If 12. 5 min is allowed for the plating, what current is needed? mass of Cr needed SOLUTION: Cr 3+(aq) + 3 e- Cr(s) divide by M mol of Cr needed 3 mol e-/mol Cr mol of e- transferred 0. 86 g (mol Cr) (3 mol e-) = 0. 050 mol e- (52. 00 g. Cr) (mol Cr) 0. 050 mol e- (9. 65 x 104 C/mol e-) = 4. 8 x 103 C 9. 65 x 104 C/mol echarge (C) divide by time current (A) 21 -55 4. 8 x 103 C (min) 12. 5 min (60 s) = 6. 4 C/s = 6. 4 A

Dry Cell 21 -56

Alkaline Battery 21 -57

Mercury and Silver (Button) Batteries 21 -58

Lead-Acid Battery 21 -59

Nickel-Metal Hydride (Ni-MH) Battery 21 -60

Lithium-Ion Battery 21 -61

21 -62