Chapter 21 Electrochemistry Chemical Change and Electrical Work

Chapter 21 Electrochemistry: Chemical Change and Electrical Work 21 -1

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 non-spontaneous reaction 21 -2

Key points about redox reactions p Oxidation (electron loss) always accompanies reduction (electron gain). p The oxidizing agent is reduced and the reducing agent is oxidized. p The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent. 21 -3

A summary of redox terminology OXIDATION One reactant loses electrons. Reducing agent is oxidized. Oxidation number increases. REDUCTION Other reactant gains electrons. Oxidizing agent is reduced. Oxidation number decreases. Figure 21. 1 21 -4 Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) Zn loses electrons. Zn is the reducing agent and becomes oxidized. The oxidation number of Zn increases from 0 to +2 (each Zn loses 2 electrons). Hydrogen ion gains electrons. Hydrogen ion is the oxidizing agent and becomes reduced. The oxidation number of H decreases from +1 to 0 (each H+ ion gains one electron).

Half-reaction method for balancing redox reactions This method divides the overall redox reaction into oxidation and reduction half-reactions. p Each reaction is balanced for mass (atoms) and charge. p One or both are multiplied by an integer to make the number of electrons gained and lost equal. p The half-reactions are recombined to give the balanced redox equation. Advantages The separation of half-reactions reflects the actual physical separations in electrochemical cells. p The half-reactions are easier to balance, especially if they involve acid or base. p It is usually unnecessary to assign oxidation numbers to those species not undergoing change. p 21 -5

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 ONs 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 7221 -6 2 Cr 3+ + 7 H 2 O(l) net: +6 - to left. Add 6 e� 2 Cr 3+ + 7 H 2 O(l)

6 e- + 14 H+(aq) + Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O(l) 2 I - I 2 + 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+ + 7 H 2 O(l) 3 I 2 + 6 e 2 Cr 3+(aq) + 3 I 2(aq) + 7 H 2 O(l) Do a final check on atoms and charges. 21 -7

Balancing redox reactions in basic solution p Balance the reaction in acid and then add OH- to neutralize the H+ ions. 14 H+(aq) + Cr 2 O 72 -(aq) + 6 I-(aq) + 14 OH-(aq) 14 H 2 O + Cr 2 O 72 - + 6 Ip 2 Cr 3+(aq) + 3 I 2(s) + 7 H 2 O(l) + 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 -8

The redox reaction between dichromate and iodide anions Cr 2 O 72 - Figure 21. 2 21 -9 I- Cr 3+ + I 2

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 anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMn. O 4 and Na 2 C 2 O 4 in basic solution. Mn. O 4 -(aq) + C 2 O 42 -(aq) PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: +7 Mn. O 44 H+ + Mn. O 4+ 3 e- 21 -10 Mn. O 2(s) + CO 32 -(aq) Mn. O 4 - R Mn. O 2 C 2 O 4 2 - O +4 Mn. O 2 +3 C 2 O 4 2 - Mn. O 2 + 2 H 2 O C 2 O 42 - + 2 H 2 O CO 32 - +4 2 CO 322 CO 32 - + 4 H+ + 2 e-

Sample Problem 21. 1 4 H+ + Mn. O 4 - +3 e- (continued) Mn. O 2+ 2 H 2 O C 2 O 42 - + 2 H 2 O x 3 x 2 8 H+ + 2 Mn. O 4 - + 6 e- 2 Mn. O 2+ 4 H 2 O 3 C 2 O 42 - + 6 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 -11 2 CO 32 - + 4 H+ + 2 e- 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)

Types of Electrochemical Cells Voltaic (or galvanic) cell: uses a spontaneous reaction (∆G < 0) to generate electrical energy. Electrolytic cell: uses electrical energy to drive a non-spontaneous reaction (∆G > 0). Contain two electrodes (anode and cathode) dipped into an aqueous electrolyte solution. The oxidation half-reaction occurs at the anode; the reduction half-reaction occurs at the cathode. 21 -12

General characteristics of voltaic and electrolytic cells VOLTAIC CELL Energy does is released from system work on its spontaneous redox reaction surroundings ELECTROLYTIC CELL Energy is absorbed tosupply) drive a surroundings (power non-spontaneous redox reaction do work on the system (cell) oxidation half-reaction: X X + + e- 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) 21 -13 overall (cell) reaction: A- + B + A + B (DG > 0) Figure 21. 3

The spontaneous reaction between zinc and copper(II) ion Zn(s) + Cu 2+(aq) 21 -14 Zn 2+(aq) + Cu(s) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21. 4

A voltaic cell based on the zinc-copper reaction oxidation half-reaction: Zn(s) Zn 2+(aq) + 2 ereduction half-reaction: Cu 2+(aq) + 2 e. Cu(s) Figure 21. 5 21 -15 overall (cell) reaction: Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s)

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 -16

A voltaic cell using inactive electrodes oxidation half-reaction: 2 I-(aq) I 2(s) + 2 e- Figure 21. 6 21 -17 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) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

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: oxidation half-reaction: Cr(s) Cr 3+(aq) + 3 ereduction half-reaction: Ag+(aq) + e. Ag(s) overall (cell) reaction: Cr(s) + Ag+(aq) Cr 3+(aq) + Ag(s) 21 -18 voltmeter e- salt bridge Cr K+ Ag NO 3 Cr 3+ + Ag� Cr(s) | Cr 3+(aq) || Ag+(aq) | Ag(s)

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 -19

Table 21. 1 Voltages of Some Voltaic Cells voltaic cell 21 -20 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

Determining an unknown Eohalf-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 -21 reduction half-reaction: 2 H 3 O+(aq) + 2 e. H 2(g) + 2 H 2 O(l) Figure 21. 7 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Calculating an unknown Eohalf-cell from Eocell Sample Problem 21. 3 PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2(aq) + Zn(s) Zn 2+(aq) + 2 Br-(aq) Eocell = 1. 83 V Calculate Eobromine given Eozinc = -0. 76 V PLAN: The reaction is spontaneous as written since the Eocell is (+). Zinc is being oxidized and is the anode. Therefore the Eobromine can be found using Eocell = Eocathode - Eoanode. SOLUTION: anode: Zn(s) - E = +0. 76 Zn 2+(aq) + 2 e� Eo. Zn as Zn 2+(aq) + 2 e- Zn(s) is -0. 76 V Eocell = Eocathode - Eoanode = 1. 83 = Eobromine - (-0. 76) Eobromine = 1. 86 - 0. 76 = 1. 07 V 21 -22

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p 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. p The reduction half-cell potential and the oxidation half-cell potential are added to obtain the Eocell. p p 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 -24 + Cu 2+(aq) Zn 2+(aq) stronger 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 Eocell 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) Eo = 0. 96 V Eo = -0. 23 V Eo = 1. 23 V Put the equations together in varying combinations so as to produce (+) Eocell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of Eo. Balance the number of electrons gained and lost without changing the Eo. In ranking the strengths, compare the combinations in terms of Eocell. 21 -25

Sample Problem 21. 4 (continued) SOLUTION: (1) NO 3 -(aq) + 4 H+(aq) + 3 e- (a) Rev (2) N 2 H 5+(aq) N 2(g) + 5 H+(aq) + 4 e- (1) NO 3 -(aq) + 4 H+(aq) + 3 e(2) N 2 H 5+(aq) (A) (1) NO(g) + 2 H 2 O(l) (B) 2 NO(g) + 3 Mn. O 2(s) + 4 H+(aq) 21 -26 4 NO(g) + 3 N 2(g) + 8 H 2 O(l) Eo = -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- Eocell = 1. 19 V x 3 NO 3 -(aq) + 4 H+(aq) + 3 e- (3) Mn. O 2(s) + 4 H+(aq) + 2 e- Eo = 0. 96 V Eo = +0. 23 V NO(g) + 2 H 2 O(l) x 4 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) Eo = 1. 23 V Eocell = 0. 27 V x 2 Mn 2+(aq) + 2 H 2 O(l) x 3 2 NO 3 -(aq) + 3 Mn 2+(aq) + 2 H 2 O(l)

Sample Problem 21. 4 Rev (2) N 2 H 5+(aq) (continued) N 2(g) + 5 H+(aq) + 4 e- (3) Mn. O 2(s) + 4 H+(aq) + 2 e(2) N 2 H 5 +(aq) N 2(g) + Eo = +0. 23 V Mn 2+(aq) + 2 H 2 O(l) 5 H+(aq) (3) Mn. O 2(s) + 4 H+(aq) + 2 e(C) N 2 H 5+(aq) + 2 Mn. O 2(s) + 3 H+(aq) + 4 e. Mn 2+(aq) + 2 H 2 O(l) Eo = 1. 23 V Eocell = 1. 46 V 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 -27

Sample Problem 21. 4 (continued) A comparison of the relative strengths of oxidizing and reducing agents produces the following overall ranking. oxidizing agents: Mn. O 2 > NO 3 - > N 2 reducing agents: N 2 H 5+ > NO > Mn 2+ 21 -28

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 -29

The reaction of calcium in water oxidation half-reaction: Ca(s) Ca 2+(aq) + 2 e- Figure 21. 8 21 -30 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) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Free energy and electrical work DG a -Ecell = -wmax DG = wmax = charge x (-Ecell) DG = -n F Ecell charge In the standard state: charge = n F 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 -31 DGo = -n F Eocell DGo = - RT ln K Eocell = - (RT/n F) ln K

The interrelationship of DGo, Eo, and K o DG� DGo = -n. FEocell reaction at standard-state conditions K o E� cell <0 >1 >0 spontaneous 0 1 0 at equilibrium >0 <1 <0 non-spontaneous DGo = -RT ln K By substituting standard state values into Eocell, we get: o E� cell K Eocell = (0. 0592 V/n) log K (at 25 o. C) Eocell = -RT ln K n. F Figure 21. 9 21 -32 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Calculating K and DGo from Eocell Sample Problem 21. 5 PROBLEM: 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 DGo at 25 o. C for this reaction. PLAN: Break the reaction into half-reactions, find the Eo for each half-reaction and then the Eocell. SOLUTION: 2 x Eocell = log K = 21 -33 Pb 2+(aq) + 2 e. Ag+(aq) + e- Pb(s) Ag(s) Eo = -0. 13 V Eo = 0. 80 V Pb(s) Pb 2+(aq) + 2 e. Ag+(aq) + e. Ag(s) 0. 592 V log K n n x Eocell 0. 592 V = (2)(0. 93 V) 0. 592 V Eo = 0. 13 V Eo = 0. 80 V Eocell = 0. 93 V . DGo = -n. FEocell = -(2)(96. 5 k. J/mol V)(0. 93 V) K = 2. 6 x 1031 DGo = -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 [H+] = 2. 5 M Calculate Ecell at 25 o. C. PH = 0. 30 atm 2 PLAN: Find Eocell and Q in order to use the Nernst equation. SOLUTION: Determining Eocell : 2 H+(aq) + 2 e. Zn 2+(aq) + 2 e. Zn(s) Ecell = Eocell - Eo = 0. 00 V H 2(g) Zn(s) Eo = -0. 76 V Zn 2+(aq) + 2 e- Eo = +0. 76 V 0. 0592 V n log Q Ecell = 0. 76 - (0. 0592/2) log (4. 8 x 10 -4) = 0. 86 V 21 -34 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 = DGo + RT ln Q -n. F Ecell = -n. F Ecell + RT ln Q Ecell = Eo cell - RT n. F ln Q When Q < 1 and thus [reactant] > [product], ln Q < 0, so Ecell > Eocell When Q = 1 and thus [reactant] = [product], ln Q = 0, so Ecell = Eocell When Q >1 and thus [reactant] < [product], ln Q > 0, so Ecell < Eocell Ecell = 21 -35 Eo cell - 0. 0592 n log Q

The relation between Ecell and log Q for the zinc-copper cell Figure 21. 10 21 -36 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

A concentration cell based on the Cu/Cu 2+ half-reaction oxidation half-reaction: Cu(s) Cu 2+(aq, 0. 1 M) + 2 e- Figure 21. 11 21 -37 reduction half-reaction: 2+(aq, 1. 0 M) + 2 e. Cu� Cu(s) overall (cell) reaction: Cu 2+(aq, 1. 0 M) Cu 2+(aq, 0. 1 M) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

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? Eocell 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 = Eocell - 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 -38

The laboratory measurement of p. H Pt glass electrode reference (calomel) electrode Hg Ag. Cl on Ag on Pt 1 M HCl Figure 21. 12 21 -39 thin glass membrane paste of Hg 2 Cl 2 in Hg KCl solution porous ceramic plugs Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

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The corrosion of iron Figure 21. 13 21 -41 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Enhanced corrosion at sea Figure 21. 14 21 -42

The effect of metal-metal contact on the corrosion of iron faster corrosion cathodic protection Figure 21. 15 21 -43 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

The use of sacrificial anodes to prevent iron corrosion Figure 21. 16 21 -44

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 ereduction half-reaction: Cu 2+(aq) + 2 e. Cu(s) overall (cell) reaction Sn(s) + Cu 2+(aq) Sn 2+(aq) + Cu(s) 21 -45 electrolytic cell oxidation half-reaction: Cu(s) Cu 2+(aq) + 2 ereduction half-reaction: Sn 2+(aq) + 2 e. Sn(s) overall (cell) reaction: Sn(s) + Cu 2+(aq) Sn 2+(aq) + Cu(s) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

The processes occurring during the discharge and recharge of a lead-acid battery VOLTAIC (discharge) Figure 21. 18 21 -46 ELECTROLYTIC (recharge).

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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. PLAN: Consider the metal and nonmetal components of each compound and 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 -48 anode Mg(s) + Br 2(g)

The electrolysis of water oxidation half-reaction: 2 H 2 O(l) 4 H+(aq) + O 2(g) + 4 ereduction half-reaction: 2 H 2 O(l) + 4 e 2 H 2(g) + 2 OH-(aq) overall (cell) reaction: 2 H 2 O(l) H 2(g) + O 2(g) Figure 21. 19 21 -49 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Sample Problem 21. 9 PROBLEM: Predicting the electrolysis products of aqueous ionic solutions What products form during electrolysis of aqueous solutions 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) Eo = -2. 93 V Eo = -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- Eo = 1. 07 V Eo = 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 -50

Sample Problem 21. 9 (continued) (b) Ag+(aq) + e 2 H 2 O(l) + 2 e- Eo = -0. 80 V Ag(s) H 2(g) + 2 OH-(aq) Eo = -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 Eo = -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 -51

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) Figure 21. 20 21 -52 Faraday constant (C/mol e-) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 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 9. 65 x 104 C/mol echarge (C) divide by time current (A) 21 -53 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 x (min) 12. 5 min x (60 s) = 6. 4 C/s = 6. 4 A

dry cell 21 -54

alkaline battery 21 -55

mercury and silver (button) batteries 21 -56

lead-acid battery 21 -57

nickel-metal hydride (Ni-MH) battery 21 -58

lithium-ion battery 21 -59

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