1. Do Now: What is current? Electric potential or voltage? An Electrode? REDOX? Which reaction below will take place spontaneously? Support your response. For the reaction that takes place spontaneously, how can we set up the reaction to generate an electric current? ZnSO 4 + Cu-> ??? CuSO 4 + Zn-> ??? Cu 2+ + 2e- -> Cu Zn -> Zn 2+ + 2e- Zn 2+ + 2e- -> Zn Cu -> Cu 2+ + 2e- WHICH???

2. By comparing a reduction half reaction to the standard hydrogen electrode (SHE) a species’ ability to gain electrons (be reduced) is determined. 2H + (aq) + 2e-  H 2(g) H+ will gain e- 2H + + 2e-  H 2 Zinc will lose e- Zn  Zn 2+ + 2e- Cu 2+ will gain e- Cu 2+ + 2e-  Cu H 2 will lose e- H 2  2H + + 2e-

3. How can we predict which substance will be oxidized and which substance will be reduced? Standard Reduction Potential compares a substance’s ability to be reduced = Activity Series All found by comparison to standard hydrogen electrode. Easily gains electrons (reduced) - Strongest oxidizing agent Easily loses electrons (oxidized) - Strongest reducing agent

4. What is an Electrochemical Cell? Devices that used REDOX reactions to either produce or use electricity Voltaic cell(Galvanic cell) – produce electricity as a result of a spontaneous REDOX reactions. (ΔG< 0) System does work on the surroundings. Electrolytic cell – electrical energy drives a nonspontaneous REDOX reaction to take place. (ΔG> 0) Surroundings do work on the system.

5. Zn (s) + CuSO 4(aq) -> ZnSO 4(aq) + Cu (s) What are the half-reactions? Oxidation Reduction

6. Voltaic Cell Zn Cu 2+ SO 4 2- Zn 2+ SO 4 2- Cu Zn -> Zn 2+ + 2e - Cu 2+ + 2e - -> Cu e-e- e-e- e-e- Oxidation Half Cell Reduction Half Cell wire Anode → electrode where oxidation takes place (-) ← Cathode electrode where reduction takes place (+) Na + NO 3 - Porous Plugs + + 1M + + + + - 1M - - - - - - ?V

7. Cell Potential, E The reading on the meter is the potential difference, or cell potential or voltage, between the two half cells. Spontaneous reactions will have a + cell potential as electrons move from higher to lower potential. The standard cell potential is designated E° cell where solution concentration is 1M, gases at a pressure of 1 atm, and temperature is 25 o C.

8. Zn (s) + CuSO 4(aq) -> ZnSO 4(aq) + Cu (s) From Standard Reduction Potential Table Zn -> Zn 2+ + 2e- Cu 2+ + 2e - -> Cu Oxidation Reduction

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

10. Prob 1: A voltaic cell is set up with an aluminum electrode in a solution of aluminum nitrate and a lead electrode in a solution of lead nitrate. Draw a labeled schematic and indicate in which direction the electrons flow? Write out the cell notation. Calculate the cell potential.

11. PROB 2: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 = -0.76 V PROb 3:(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° = -0.23 V (3) MnO 2 (s) +4H + (aq) + 2e - → Mn 2+ (aq) + 2H 2 O(l) E° = 1.23 V

12. Free Energy and Electrical Work For a spontaneous redox reaction,  G 0.  G = -nFE cell n = mol of e - transferred F is the Faraday constant = 9.65x10 4 J/V·mol e - Under standard conditions,  G° = -nFE° cell and

13. PROB 4: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  G° at 298.15 K for this reaction. What if cell is not at standard state?

14. Cell Potential and Concentration When Q [product], ln Q E° cell When Q = 1, [reactant] = [product], ln Q = 0, so E cell = E° cell When Q > 1, [reactant] 0, so E cell < E° cell Nernst Equation E cell = E° cell -ln Q RT nF

15. Sample Problem 21.6 Using the Nernst Equation to Calculate E cell PROB5: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 P = 0.30 atm H2H2 Calculate E cell at 298 K.

16. Figure 21.11AThe relation between E cell and log Q for the zinc- copper 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.

17. Figure 21.11BThe relation between E cell and log Q for the zinc- copper cell. A summary of the changes in E cell as any voltaic cell operates.

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

19. 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. Figure 21.12A concentration cell based on the Cu/Cu 2+ half-reaction. Overall (cell) reaction Cu 2+ (aq,1.0 M) → Cu 2+ (aq, 0.1 M) Oxidation half-reaction Cu(s) → Cu 2+ (aq, 0.1 M) + 2e - Reduction half-reaction Cu 2+ (aq, 1.0 M) + 2e - → Cu(s)

20. Figure 21.13Laboratory 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.

21. Electrochemical Processes in Batteries 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. A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added.

22. Alkaline battery.Figure 21.15 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

23. Fe(s) → Fe 2+ (aq) + 2e - [anodic region; oxidation] O 2 (g) + 4H + (aq) + 4e - → 2H 2 O(l) [cathodic region; reduction] The loss of iron: 2Fe(s) + O 2 (g) + 4H + (aq) → 2Fe 2+ (aq) + 2H 2 O(l)[overall] The rusting process: 2Fe 2+ (aq) + ½O 2 (g) + (2 + n)H 2 O(l) → Fe 2 O 3 ·nH 2 O(s) + 4H + (aq) Overall reaction: 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. Corrosion is the process where metals are oxidized

24. Figure 21.24The 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.

25. Figure 21.25The 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.

26. Electrolytic Cells An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction. Cu(s) → Cu 2+ (aq) + 2e - [anode; oxidation] Sn 2+ (aq) + 2e - → Sn(s) [cathode; reduction] Cu(s) + Sn 2+ (aq) → Cu 2+ (aq) + Sn(s) E° cell = -0.48 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.

27. Figure 21.27The processes occurring during the discharge and recharge of a lead-acid battery. VOLTAIC (discharge) ELECTROLYTIC (recharge) Switch

28. Figure 21.28The 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)

29. 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 PROB6: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?

30. Figure 21.29A summary diagram for the stoichiometry of electrolysis. MASS (g) of substance oxidized or reduced AMOUNT (mol) of substance oxidized or reduced CHARGE (C) CURRENT (A) AMOUNT (mol) of electrons transferred M (g/mol) balanced half-reaction Faraday constant (C/mol e - ) time (s)