21-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 21 Electrochemistry: Chemical Change and Electrical.

21-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction

21-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.1 A summary of redox terminology. Zn( s ) + 2H + ( aq ) Zn 2+ ( aq ) + H 2 ( g ) OXIDATION Zn loses electrons. Zn is the reducing agent and becomes oxidized. The oxidation number of Zn increases from x to +2. REDUCTION One reactant loses electrons. Reducing agent is oxidized. Oxidation number increases. Hydrogen ion gains electrons. Hydrogen ion is the oxidizing agent and becomes reduced. The oxidation number of H decreases from +1 to 0. Other reactant gains electrons. Oxidizing agent is reduced. Oxidation number decreases.

21-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Balancing Redox Reactions in Acidic Solution Cr 2 O 7 2- ( 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. Cr 2 O 7 2- ( aq ) + I - ( aq ) Cr 3+ ( aq ) + I 2 ( aq ) +6+30 2. Balance atoms and charges in each half-reaction - I - I 2 2 + 7H 2 O( l ) Cr 2 O 7 2- Cr 3+ 14H + ( aq ) +Cr 2 O 7 2- Cr 3+ Cr is going from +6 to +3 I i s going from -1 to 0 net: +12 net: +6Add 6e - to left. 2Cr 2 O 7 2- Cr 3+ + 7H 2 O( l )14H + ( aq ) +6e - +

21-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Balancing Redox Reactions in Acidic Solution continued + 2e - I - I 2 2 Cr 2 O 7 2- Cr 3+ + 7H 2 O( l )14H + ( aq ) +6e - + 3. Multiply each half-reaction by an integer, if necessary - Cr(+6) is the oxidizing agent and I (-1) is the reducing agent. X 3 4. Add the half-reactions together - + 2e - I - I 2 2 3 6+ 6e - Cr 2 O 7 2- ( aq ) + 6 I - ( aq ) 2Cr 3+ ( aq ) + 3 I 2 ( s )+ 7H 2 O( l )14H + ( aq ) + 2 Cr 2 O 7 2- Cr 3+ + 7H 2 O( l )14H + +6e - +2 Do a final check on atoms and charges.

21-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH - so as to neutralize the H + ions. Cr 2 O 7 2- ( aq ) + 6 I - ( aq ) 2Cr 3+ ( aq ) + 3 I 2 ( s )+ 7H 2 O( l )14H + ( aq ) + + 14OH - ( aq ) Cr 2 O 7 2- + 6 I - 2Cr 3+ + 3 I 2 + 7H 2 O + 14OH - 14H 2 O + Reconcile the number of water molecules. + 14OH - Cr 2 O 7 2- + 6 I - 2Cr 3+ + 3 I 2 7H 2 O + Do a final check on atoms and charges.

21-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.1Balancing Redox Reactions by the Half-Reaction Method PROBLEM: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 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 ) PLAN:Proceed in acidic solution and then neutralize with base. SOLUTION: MnO 4 - MnO 2 C 2 O 4 2- CO 3 2- MnO 4 - MnO 2 C 2 O 4 2- CO 3 2- 2 MnO 4 - MnO 2 + 2H 2 O4H + + C 2 O 4 2- 2CO 3 2- + 2H 2 O+ 4H + +7+4 +3+4 +3e - +2e -

21-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.1Balancing Redox Reactions by the Half-Reaction Method continued: 4H + + MnO 4 - +3e - MnO 2 + 2H 2 OC 2 O 4 2- + 2H 2 O 2CO 3 2- + 4H + + 2e - 4H + + MnO 4 - +3e - MnO 2 + 2H 2 O X 2 C 2 O 4 2- + 2H 2 O 2CO 3 2- + 4H + + 2e - X 3 8H + + 2MnO 4 - +6e - 2MnO 2 + 4H 2 O3C 2 O 4 2- + 6H 2 O 6CO 3 2- + 12H + + 6e - 8H + + 2MnO 4 - +6e - 2MnO 2 + 4H 2 O 3C 2 O 4 2- + 6H 2 O 6CO 3 2- + 12H + + 6e - 2MnO 2 - ( aq ) + 3C 2 O 4 2- ( aq ) + 2H 2 O( l ) 2MnO 2 ( s ) + 6CO 3 2- ( aq ) + 4H + ( aq ) + 4OH - 2MnO 2 - ( aq ) + 3C 2 O 4 2- ( aq ) + 4OH - ( aq ) 2MnO 2 ( s ) + 6CO 3 2- ( aq ) + 2H 2 O( l )

21-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Energy is absorbed to drive a nonspontaneous redox reaction Figure 21.3 General characteristics of voltaic and electrolytic cells. VOLTAIC CELLELECTROLYTIC CELL Energy is released from spontaneous redox reaction Reduction half-reaction Y + + e - Y Oxidation half-reaction X X + + e - System does work on its surroundings Reduction half-reaction B + + e - B Oxidation half-reaction A - A + e - Surroundings(power supply) do work on system(cell) Overall (cell) reaction X + Y + X + + Y;  G < 0 Overall (cell) reaction A - + B + A + B;  G > 0

21-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.4 The spontaneous reaction between zinc and copper( II ) ion. Zn( s ) + Cu 2+ ( aq )Zn 2+ ( aq ) + Cu( s )

21-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.5 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 )

21-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Notation for a Voltaic Cell components of anode compartment (oxidation half-cell) components of cathode compartment (reduction half-cell) phase of lower oxidation state phase of higher oxidation state phase of lower oxidation state phase boundary between half-cells Examples:Zn( s ) | Zn 2+ ( aq ) || Cu 2+ ( aq ) | Cu ( s ) Zn( s ) Zn 2+ ( aq ) + 2e - Cu 2+ ( aq ) + 2e - Cu( s ) graphite | I - ( aq ) | I 2 ( s ) || H + ( aq ), MnO 4 - ( aq ) | Mn 2+ ( aq ) | graphite inert electrode

21-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.6 A voltaic cell using inactive electrodes. Reduction half-reaction MnO 4 - ( aq ) + 8H + ( aq ) + 5e - Mn 2+ ( aq ) + 4H 2 O( l ) Oxidation half-reaction 2 I - ( aq ) I 2 ( s ) + 2e - Overall (cell) reaction 2MnO 4 - ( aq ) + 16H + ( aq ) + 10 I - ( aq ) 2Mn 2+ ( aq ) + 5 I 2 ( s ) + 8H 2 O( l )

21-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.2Describing 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: SOLUTION: Identify the oxidation and reduction reactions and write each half- reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). Voltmeter Oxidation half-reaction Cr( s ) Cr 3+ ( aq ) + 3e - Reduction half-reaction Ag + ( aq ) + e - Ag( s ) Overall (cell) reaction Cr( s ) + Ag + ( aq ) Cr 3+ ( aq ) + Ag( s ) Cr Cr 3+ Ag Ag + K+K+ NO 3 - salt bridge e-e- Cr( s ) | Cr 3+ ( aq ) || Ag + ( aq ) | Ag( s )

21-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 21.1 Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) Common alkaline battery Lead-acid car battery (6 cells = 12V) Calculator battery (mercury) Electric eel (~5000 cells in 6-ft eel = 750V) Nerve of giant squid (across cell membrane) 2.0 1.5 1.3 0.15 0.070

21-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.8 Determining an unknown E 0 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 )

21-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.3Calculating an Unknown E 0 half-cell from E 0 cell PROBLEM:A voltaic cell houses the reaction between aqueous bromine and zinc metal: PLAN: SOLUTION: Br 2 ( aq ) + Zn( s ) Zn 2+ ( aq ) + 2Br - ( aq ) E 0 cell = 1.83V Calculate E 0 bromine given E 0 zinc = -0.76V 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. anode: Zn(s) Zn 2+ (aq) + 2e - E = +0.76 E 0 Zn as Zn 2+ (aq) + 2e - Zn(s) is -0.76V 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

21-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 21.2 Selected Standard Electrode Potentials (298K) Half-ReactionE 0 (V) 2H + ( aq ) + 2e - H 2 ( g ) 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 ) 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 ) Li + ( aq ) + e - Li( s ) +2.87 -3.05 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 -0.23 -0.44 -0.83 -2.71 strength of reducing agent strength of oxidizing agent

21-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 ) + Cu 2+ ( aq ) Zn 2+ ( aq ) + Cu( s ) stronger reducing agent weaker oxidizing agent stronger oxidizing agent weaker reducing agent Writing Spontaneous Redox Reactions

21-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4Writing Spontaneous Redox Reactions PROBLEM:(a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E 0 cell for each. PLAN: (b) Rank the relative strengths of the oxidizing and reducing agents: E 0 = 0.96V(1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) E 0 = -0.23V(2) N 2 ( g ) + 5H + ( aq ) + 4e - N 2 H 5 + ( aq ) E 0 = 1.23V(3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) 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.

21-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4Writing Spontaneous Redox Reactions continued (2 of 4) SOLUTION:(1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l )E 0 = 0.96V (1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) (2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - X4 X3 E 0 cell = 1.19V (a) E 0 = 1.23V(3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) (2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - E 0 = +0.23VRev E 0 = -0.96V(1) NO( g ) + 2H 2 O( l ) NO 3 - ( aq ) + 4H + ( aq ) + 3e - Rev (1) NO( g ) + 2H 2 O( l ) NO 3 - ( aq ) + 4H + ( aq ) + 3e - (3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) X2 X3 E 0 cell = 0.27V 4NO 3 - ( aq ) + 3N 2 H 5 + ( aq ) + H + ( aq ) 4NO( g ) + 3N 2 ( g ) + 8H 2 O( l )(A) 2NO( g ) + 3MnO 2 ( s ) + 4H + ( aq ) 2NO 3 - ( aq ) + 3Mn 3+ ( aq ) + 2H 2 O( l )(B)

21-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4Writing Spontaneous Redox Reactions continued (3 of 4) E 0 = 1.23V(3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) E 0 = +0.23V(2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - Rev (2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - (3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l )X2 E 0 cell = 1.46V (b) Ranking oxidizing and reducing agents within each equation: N 2 H 5 + ( aq ) + 2MnO 2 ( s ) + 3H + ( aq ) N 2 ( g ) + 2Mn 2+ ( aq ) + 4H 2 O( l )(C) (A): oxidizing agents: NO 3 - > N 2 reducing agents: N 2 H 5 + > NO (B): oxidizing agents: MnO 2 > NO 3 - reducing agents: NO > Mn 2+ (C): oxidizing agents: MnO 2 > N 2 reducing agents: N 2 H 5 + > Mn 2+

21-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4Writing Spontaneous Redox Reactions continued (4 of 4) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO 2 > NO 3 - > N 2 Reducing agents: N 2 H 5 + > NO > Mn 2+

21-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Relative Reactivities (Activities) of Metals 1. Metals that can displace H from acid 2. Metals that cannot displace H from acid 3. Metals that can displace H from water 4. Metals that can displace other metals from solution

21-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.89 The reaction of calcium in water. Overall (cell) reaction Ca( s ) + 2H 2 O( l ) Ca 2+ ( aq ) + H 2 ( g ) + 2OH - ( aq ) Oxidation half-reaction Ca( s ) Ca 2+ ( aq ) + 2e - Reduction half-reaction 2H 2 O( l ) + 2e - H 2 ( g ) + 2OH - ( aq )

21-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Free Energy and Electrical Work  G  -E cell -E cell = -w max charge charge = n F n = #mols e - F = Faraday constant F = 96,485 C/mol e - 1V = 1J/C F = 9.65x10 4 J/V*mol e -  G = w max = charge x (-E cell )  G = -n F E cell In the standard state -  G 0 = -n F E 0 cell  G 0 = - RT ln K E 0 cell = - (RT/n F) ln K

21-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.10 G0G0 E 0 cell K G0G0 K Reaction at standard-state conditions E 0 cell The interrelationship of  G 0, E 0, and K. < 0 spontaneous at equilibrium nonspontaneous 0 > 0 0 < 0 > 1 1 < 1  G 0 = -RT lnK  G 0 = -nFE o cell E 0 cell = -RT lnK nFnF By substituting standard state values into E 0 cell, we get E 0 cell = (0.0592V/n) log K (at 298 K)

21-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.5 Calculating K and  G 0 from E 0 cell PLAN: SOLUTION: PROBLEM:Lead can displace silver from solution: As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and  G 0 at 298 K for this reaction. Pb( s ) + 2Ag + ( aq ) Pb 2+ ( aq ) + 2Ag( s ) 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 E 0 = -0.13V E 0 = 0.80V 2X E 0 = 0.13V E 0 = 0.80V E 0 cell = 0.93V Ag + ( aq ) + e - Ag( s ) Pb 2+ ( aq ) + 2e - Pb( s ) Ag + ( aq ) + e - Ag( s ) Pb( s ) Pb 2+ ( aq ) + 2e - E 0 cell =log K 0.592V n log K =K = 2.6x10 31 n x E 0 cell 0.592V (2)(0.93V) 0.592V =  G 0 = -nFE 0 cell = -(2)(96.5kJ/mol*V)(0.93V)  G 0 = -1.8x10 2 kJ

21-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Effect of Concentration on Cell Potential  G =  G 0 + RT ln Q -nF E cell = -nF E cell + RT ln Q E cell = E 0 cell -ln Q RT nFnF When Q [product], lnQ E 0 cell When Q >1 and thus [reactant] 0, so E cell < E 0 cell When Q = 1 and thus [reactant] = [product], lnQ = 0, so E cell = E 0 cell E cell = E 0 cell - log Q 0.0592 n

21-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.6Using 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: PLAN: SOLUTION: [Zn 2+ ] = 0.010M [H + ] = 2.5M P = 0.30atm H2H2 Calculate E cell at 298 K. Find E 0 cell and Q in order to use the Nernst equation. Determining E 0 cell : E 0 = 0.00V2H + ( aq ) + 2e - H 2 ( g ) E 0 = -0.76VZn 2+ ( aq ) + 2e - Zn( s ) Zn( s ) Zn 2+ ( aq ) + 2e - E 0 = +0.76V Q = P x [Zn 2+ ] H2H2 [H + ] 2 Q = 4.8x10 -4 Q = (0.30)(0.010) (2.5) 2 E cell = E 0 cell - 0.0592V n log Q E cell = 0.76 - (0.0592/2)log(4.8x10 -4 ) =0.86V

21-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.12 A concentration cell based on the Cu/Cu 2+ half-reaction. Overall (cell) reaction Cu 2+ (aq,1.0M) Cu 2+ ( aq, 0.1M) Oxidation half-reaction Cu( s ) Cu 2+ ( aq, 0.1M) + 2e - Reduction half-reaction Cu 2+ ( aq, 1.0M) + 2e - Cu( s )

21-37 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.7Calculating the Potential of a Concentration Cell PROBLEM:A concentration cell consists of two Ag/Ag + half-cells. In half-cell A, electrode A dips into 0.0100M AgNO 3 ; in half-cell B, electrode B dips into 4.0x10 -4 M AgNO 3. What is the cell potential at 298 K? Which electrode has a positive charge? PLAN:E 0 cell will be zero since the half-cell potentials are equal. E cell 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, 0.010M) half-cell A Ag + (aq, 4.0x10 -4 M) half-cell B E cell = E 0 cell - 0.0592V 1 log [Ag + ] dilute [Ag + ] concentrated E cell = 0 V -0.0592 log 4.0x10 -2 = 0.0828V Half-cell A is the cathode and has the positive electrode.

21-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.13 The laboratory measurement of pH. Reference (calomel) electrode Glass electrode AgCl on Ag on Pt 1M HCl Thin glass membrane Porous ceramic plugs KCl solution Paste of Hg 2 Cl 2 in Hg Hg Pt

21-46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.21 The effect of metal-metal contact on the corrosion of iron. faster corrosion cathodic protection

21-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.23 The tin-copper reaction as the basis of a voltaic and an electrolytic cell. Oxidation half-reaction Sn( s ) Sn 2+ ( aq ) + 2e - Reduction half-reaction Cu 2+ ( aq ) + 2e - Cu( s ) Oxidation half-reaction Cu( s ) Cu 2+ ( aq ) + 2e - Reduction half-reaction Sn 2+ ( aq ) + 2e - Sn( s ) Overall (cell) reaction Sn( s ) + Cu 2+ ( aq ) Sn 2+ ( aq ) + Cu( s ) Overall (cell) reaction Sn( s ) + Cu 2+ ( aq ) Sn 2+ ( aq ) + Cu( s ) voltaic cellelectrolytic cell

21-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 21.3 Comparison of Voltaic and Electrolytic Cells Cell Type GG E cell Electrode NameProcessSign Voltaic Electrolytic < 0 > 0 < 0 Anode Cathode Oxidation Reduction - - + +

21-50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.25 The electrolysis of water. 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 ) Overall (cell) reaction 2H 2 O( l ) H 2 ( g ) + O 2 ( g )

21-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.8Predicting the Electrolysis Products of Aqueous Ionic Solutions PROBLEM:What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) AgNO 3 ; (c) MgSO 4 ? SOLUTION: PLAN:Compare the potentials of the reacting ions with those of water, remembering to consider the 0.4 to 0.6V overvoltage. The reduction half-reaction with the less negative potential, and the oxidation half- reaction with the less positive potential will occur at their respective electrodes. E 0 = -2.93V(a) K + ( aq ) + e - K( s ) E 0 = -0.42V2H 2 O( l ) + 2e - H 2 ( g ) + 2OH - ( aq ) 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. 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. E 0 = 1.07V2Br - ( aq ) Br 2 ( g ) + 2e - 2H 2 O( l ) O 2 ( g ) + 4H + ( aq ) + 4e - E 0 = 0.82V

21-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.8Predicting the Electrolysis Products of Aqueous Ionic Solutions continued E 0 = -0.42V2H 2 O( l ) + 2e - H 2 ( g ) + 2OH - ( aq ) E 0 = -0.80V(b) Ag + ( aq ) + e - Ag( s ) 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. 2H 2 O( l ) O 2 ( g ) + 4H + ( aq ) + 4e - 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 4 2- is in its highest oxidation state; therefore water must be oxidized and O 2 ( g ) will be produced at the anode. E 0 = -2.37V(c) Mg 2+ ( aq ) + 2e - Mg( s )

21-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.29 A summary diagram for the stoichiometry of electrolysis. MASS (g) of substance oxidized or reduced MASS (g) of substance oxidized or reduced AMOUNT (MOL) of electrons transferred AMOUNT (MOL) of electrons transferred AMOUNT (MOL) of substance oxidized or reduced AMOUNT (MOL) of substance oxidized or reduced CHARGE (C) CURRENT (A) balanced half-reaction Faraday constant (C/mol e - ) M(g/mol) time(s)

21-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.9Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM: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? PLAN: SOLUTION: mol of e - transferred divide by M 9.65x10 4 C/mol e - 3mol e - /mol Cr divide by time mass of Cr needed mol of Cr needed charge (C) current (A) Cr 3+ ( aq ) + 3e - Cr( s ) 0.86g (mol Cr) (3 mol e - ) (52.00 gCr) (mol Cr) = 0.050 mol e - 0.050 mol e - (9.65x10 4 C/mol e - ) = 4.8x10 3 C 4.8x10 3 C 12.5 min (min) (60s) = 6.4C/s = 6.4 A

21-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 21 Electrochemistry: Chemical Change and Electrical.

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21-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 21 Electrochemistry: Chemical Change and Electrical.

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Presentation on theme: "21-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 21 Electrochemistry: Chemical Change and Electrical."— Presentation transcript:

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About project

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