1. Chapter 21 Electrochemistry: Chemical Change and Electrical Work

2. 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.

3. 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.

4. 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. The separation of half-reactions reflects actual physical separations in electrochemical cells.

5. 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 - +

6. 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.

7. 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.

8. The redox reaction between dichromate ion and iodide ion. Cr 2 O 7 2- I-I- Cr 3+ + I 2

9. Practice 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 )

10. Energy is absorbed to drive a nonspontaneous redox reaction 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

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

12. 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 ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13. 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

14. 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 )

15. Practice 2Diagramming Voltaic Cells PROBLEM: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.

16. 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. E cell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C)

17. 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

18. 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 ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

19. Practice 3Calculating an Unknown E 0 half-cell from E 0 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 0 cell = 1.83V Calculate E 0 bromine given E 0 zinc = 0.76V

20. 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

21. 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

22. Practice 4Writing 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: 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 )

23. 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

24. 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 25 0 C)

25. Practice 5 Calculating K and  G 0 from E 0 cell 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 25 0 C for this reaction. Pb( s ) + 2Ag + ( aq ) Pb 2+ ( aq ) + 2Ag( s )

26. Practice 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: [Zn 2+ ] = 0.010M [H + ] = 2.5M P = 0.30atm H2H2 Calculate E cell at 25 0 C.

27. 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

28. The relation between E cell and log Q for the zinc-copper cell Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

29. 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 ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

30. Practice 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 298K? Which electrode has a positive charge?

31. The corrosion of iron Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

32. Enhanced corrosion at sea

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

34. The use of sacrificial anodes to prevent iron corrosion

35. 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 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

36. The processes occurring during the discharge and recharge of a lead-acid battery VOLTAIC(discharge) ELECTROLYTIC(recharge) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

37. Comparison of Voltaic and Electrolytic Cells Cell Type GG E cell Electrode NameProcessSign Voltaic Electrolytic < 0 > 0 < 0 Anode Cathode Oxidation Reduction - - + +

38. Practice 8Predicting 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.

39. 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 ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

40. Practice 9Predicting 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 ?

41. 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)

42. Practice 10Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM:A technician is plating a faucet with 0.86g 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?