1. CHEM 152L -- Experiment 20 Oxidation-Reduction Reactions and Galvanic Cells Goals: To set up a series of Galvanic Cells and compare the voltage produced by different electrodes. To establish the standard electrode potentials for a variety of electrodes. To calculate equilibrium constants for oxidation- reduction reactions. To construct and examine a concentration cell.

2. ELECTROCHEMISTRY Electrochemical Cells are everywhere. www.electroprod2.com/batt_1203.html Electroplating is used throughout manufacturing www.philip.greeenspun.com/humor/ eecs-differences-explained www.electronetwork.org

3. Electrochemistry is the study of the interchange of chemical and electrical energy. Electrochemists focus on two processes that make up oxidation-reduction reactions. Oxidation Reactions generate electric current from a spontaneous chemical reaction. Reduction Reactions consume electric current to produce a chemical change. When these two types of reactions are combined, electrons are transferred from one chemical species to another and the system is referred to as a redox reaction (Reduction-Oxidation reaction). O.I.L. R.I.G. L.E.O. G.E.R.

4. Lets talk about these two types of electrochemical reactions by looking at some metal examples. Oxidation Reactions generate electric current from a spontaneous chemical reaction. Reduction Reactions consume electric current to produce a chemical change. Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) The species loses electrons and is said to be OXIDIZED. Its OXIDATION NUMBER is increased (0  +2). The species gains electrons and is said to be REDUCED. Its OXIDATION NUMBER is lowered (+2  0).

5. As mentioned, redox reactions are the result of coupling an oxidation reaction with a reduction reaction. So for our zinc and copper example: Zn(s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu(s) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) The net number of electrons in the system does not change, but electrons are transferred from one species to the other. Meaning the final reaction must be charge balanced! For a SPONTANEOUS redox reaction, we can tap into the electron flow by constructing a GALVANIC or VOLTAIC cell.

6. Electrochemical Cells -- Galvanic Cells An electrochemical cell consists of two ELECTRODES connected by an external wire and placed in a conducting solution. Each electrode consists of a metal strip in contact with a solution containing ions of the metal. M1M1 M2M2 anodecathode M 1 + (aq)X - (aq)M 2 + (aq)X - (aq)

7. ZnCu anode (oxidation) cathode (reduction) The two electrodes are known as HALF-CELLS and the PROCESSES that take place are known as HALF REACTIONS. OXIDATION takes place at the ANODE Zn 0 (s)  Zn 2+ (aq) + 2e - REDUCTION takes place at the CATHODE. Cu 2+ (aq) + 2e -  Cu 0 (s) Zn 2+ (aq)NO 3 - (aq)Cu 2+ (aq)SO 4 - (aq)

8. ZnCu anode (oxidation) cathode (reduction) For a spontaneous net reaction, electrons will flow through the wire from ANODE to CATHODE. Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- Electrons flow through the wire from ANODE to CATHODE BUT without completing the circuit, only a small amount of electrons can flow from anode to cathode until a charge is built up each of the electrodes. This build up of charge prevents further flow of electrons.

9. ZnCu anode (oxidation) cathode (reduction) So to complete the circuit and allow continued flow of electrons, the two solutions must be in ionic contact. Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- Electrons flow through the wire from ANODE to CATHODE The circuit is completed by a NET migration of ANIONS from CATHODE to ANODE. NO 3 -

10. ZnCu anode (oxidation) cathode (reduction) So to complete the circuit and allow continued flow of electrons, the two solutions must be in ionic contact. Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- Electrons flow through the wire from ANODE to CATHODE The circuit is completed by a NET migration of ANIONS from CATHODE to ANODE. NO 3 -

11. Electrochemical Cells -- Galvanic Cells An electrochemical cell consists of two ELECTRODES connected by an external wire and placed in a conducting solution. We represent an electrochemical cell by the shorthand notation Zn(s)Zn 2+ (aq, 1M) Cu 2+ (aq,1M) Cu(s) ZnCu anode (oxidation) Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- Electrons flow through the wire from ANODE to CATHODE NO 3 - cathode (reduction)

12. ZnCu anode (oxidation) cathode (reduction) Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- Electrons flow through the wire from ANODE to CATHODE NO 3 - Anions flow across salt bridge from CATHODE to ANODE To compete the circuit we must have a net migration of ANIONS and ELECTRONS. However, the solutions can’t be in direct contact because then there would be no flow of electrons across the wire. The examples in Zumdahl use glassware with a “frit” or “salt bridge” to connect the two solutions. Salt Bridges allow the migration of small anions but not bulky metal ions. In our lab, we will use cellulose membranes in the same way -- to allow passage of ions but prevent gross mixing of electrode solutions. e-e- NO 3 - KNO 3

13. For the experiment, you will be provided with four half-cells from which six different combinations may be constructed. The half-cells are: Ag(s) Ag + (aq, 0.1M) Cu(s)Cu 2+ (aq, 0.1M) Pb(s)Pb 2+ (aq, 0.1M) Zn(s)Zn 2+ (aq, 0.1M)

14. In any galvanic cell, the species being oxidized is a stronger reducing agent than the species being reduced. By examining the behavior of many different half-cell combinations, we can arrange the various elements in order of their reducing capacity. This ordering is known as the Electrochemical Series. We construct a mini-series for the four metals in use. Which metal will be the anode or cathode? Oxidation (dissolution) occurs at the anode. Reduction (plating) occurs at the cathode. Part 1

15. Before constructing the various cells, we carry out some simple tests to tell us which half-cell will act as the anode and which the cathode. We take each electrode metal in turn and place it in contact with a small amount of solution of each of the other metals. If plating takes place on the metal surface, we know that the metal ion in solution is being reduced. The metal electrode is therefore a STRONGER REDUCING AGENT than the metal in solution. Ag + Pb 2+ Zn 2+ Cu 2+ e.g. Zn(s)  What will you see if plating occurs?

16. Zn 2+ Cu 2+ Zn Cu Anode: Cathode: Cu 2+ Zn 2+ 2e-2e- 2e-2e- e- Oxidation Dissolution Reduction Plating Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) Flash Animation

17. So….if metal A plates onto metal B, which is the stronger reducing agent? Metal B

18. In a cell constructed from half-cells containing these two metals, the stronger reducing agent will function as the ANODE, the other half-cell as the CATHODE. Part I of the DATA & REPORT SHEET: Record your observations and predictions of the relative strengths of the four metals acting as reducing agents. Part II of the DATA & REPORT SHEET: Now that you know the reduction strength, you now know which roles the metals will play when they are combined to form a galvanic cell. Write down the reactions.

19. HALF-CELL POTENTIALS Measurements of the voltages produced by various half-cell combinations indicates that the observed voltage is the difference between two individual contributions, each characteristic of one of the half-cells involved. ZnCu anode (oxidation) cathode (reduction) Zn 0 (s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu 0 (s) e-e- NO 3 - volt meter

20. At a given temperature, the voltage of the cell depends on the CHEMICAL NATURE & CONCENTRATIONS of the SALT SOLUTIONS and the PRESSURES of any GASES involved, but is independent of the molar amounts of the various chemical species. STANDARD TEMPERATURE is 25 °C. STANDARD CONCENTRATIONS are 1 M. STANDARD PRESSURE is 1 atm. STANDARD CELL POTENTIAL is denoted by E° and is measured in volts. 1 Volt = 1 Joule/Coulomb (1 V = 1 J/C) same electrodes same electrolytes same concentrations 0.5 V

21. Convention assigns a half-cell potential of ZERO VOLTS to the reduction potential taking place at the STANDARD HYDROGEN ELECTRODED (SHE) H 2 (g, 1atm)Pt/H + (aq, 1M) 2H + (g) + 2e -  H 2 (g) Under standard conditions, the measured overall cell voltage for any given half-cell combined with the standard hydrogen electrode is attributed to the reaction taking place in that half-cell. We have no way of measuring the absolute value of any one of these HALF-CELL POTENTIALS, but if we assign an arbitrary potential to any one of them, we can derive potentials for all of the others relative to it.

22. If a half-cell acts as an anode (compared to SHE), the observed potential is the STANDARD OXIDATION POTENTIAL (SOP) for the half-cell. If a half-cell acts as an cathode (compared to SHE), the observed potential is the STANDARD REDUCTION POTENTIAL (SRP) for the half-cell. E anode = SOP=-SRPE cathode = SRP Overall cell voltage is given by E ° = SRP cathode – SRP anode or E ° = SRP cathode + SOP anode

23. Conventionally, we list half-cell potentials as STANDARD REDUCTION POTENTIALS or SRP’s. In any cell, the electrode with the numerically higher SRP will function as the cathode. For barium, the positive potential indicates a spontaneous oxidation process relative to the hydrogen half-cell. The corresponding SRP would be equal in magnitude but opposite in sign to the observed SOP. SRP = -SOP Ba(s)  Ba 2+ (aq) + 2e - SOP = 2.90V Ba 2+ (aq) + 2e -  Ba(s) SRP = -2.90V

24. If you couple any half-cell with the SHE and the SHE acts as the cathode, then the half-cell must be the anode and will have a lower reduction potential. e.g.Ba 2+ (aq) + 2e -  Ba(s) E° = SRP = -2.90V If the SHE acts as a cathode when coupled with a half-cell, then by default the half-cell is the cathode and produce a higher reduction potential. e.g.Cu 2+ (aq) + 2e -  Cu(s) E ° = SRP = 0.34V

25. Overall cell voltage is given by E ° = SRP cathode – SRP anode or E ° = SRP cathode + SOP anode Ba 2+ (aq) + 2e -  Ba(s) E° = -2.90V Cu 2+ (aq) + 2e -  Cu(s)E° = 0.34V E ° = 0.34 – (-2.90) E° = 3.24 Anode Cathode Cu 2+ (aq) + Ba (s)  Cu (s) + Ba 2+ (aq) Part II of the DATA & REPORT SHEET: Now that you know the reduction strength, you now know which roles the metals will play when they are combined to form a galvanic cell. Write down the reactions.

26. Half-Cell Potentials So How do we figure out the half cell potentials of our metals? If we are given 1 metal’s SRP, then we can figure out the other 3 metals by taking the voltage of the Galvanic cell, and solving for the unknown SRP E cell = SRP cathode – SRP anode Part 3 and 4

27. Complete Cell Potentials How do we predict the cell potential (Voltage) of a galvanic cell If we know the SRP’s of the metals, then we can solve for voltage E cell = SRP cathode – SRP anode Part 5

28. The cell potential is a measure of the FREE ENERGY CHANGE for the overall reaction.  G = -nFE with  G measured in JOULES. The Faraday Constant, F, is the charge on 1 mol of electrons F = 9.65  10 4 C n is the moles of electrons transferred in the balanced chemical equation describing the reaction. For a spontaneous reaction, is  G positive or negative? For a spontaneous reaction, is E positive or negative? Part 6

29. The cell voltage at standard conditions measures the work that can be obtained from the chemical system at standard conditions as it proceeds to EQUILIBRIUM.  G = -RT ln K and  G = -nFE Thus E = (RT/nF) ln K or E = (0.0257/n) ln K E = (0.0591/n) log K In the experiment (Part 6), we will use the measured cell voltages for several reactions taking place under standard conditions to calculate the values of their equilibrium constants.

30. Under non-standard conditions, the voltage is given by the Nernst Equation: E cell = E° – (RT/nF) ln Q where Q is the reaction quotient. Note: that where the two electrode concentrations are the same, the log term goes to zero and E cell = E° R = gas constant = 8.314 J/(mol K) n = # of electrons transferred T = temperature (K) F = charge on an electron = 96485 C/mol

31. E =E  - (RT/nF) ln {[Cu 2+ (anode)] / [Cu 2+ (cathode)]} PART 7: Concentration and voltage E cell = E° – (RT/nF) ln Q Recall that E cell is dependant on chemical nature & concentrations of the salt solutions In Part 7 we will make two solutions of CuSO 4 by diluting the original 0.1M solution. The cell can then be set up with a 0.01M solution on one side and a 0.1M solution on the other. Use the Nernst equation to calculate the expected voltage: Cu (s) | Cu 2+ (aq, 0.01M) || Cu 2+ (aq, 0.10M) | Cu (s) cell. E = 0 -(0.0257/2) ln (0.01/0.10) =0.03 V

32. A few practical matters: 1.Sand electrodes before beginning to clean of any old plated metal 2.After sanding, rinse electrodes to remove any contaminants 3.Keep cellulose membranes in solution to prevent drying out 4.The electrolyte connecting the two cells is KNO 3 (not DI water)

33. Run Wavetek in DC Mode (measure V)

34. 1.Measure voltage immediately, then disconnect 2.In Part 7, make fresh Cu solutions 3.Red goes to cathode and black to anode 4.When recording shorthand cells anode cathode 5.Remember (aq) and (s) notations 6.Waste: DO NOT MIX solutions!!! 5 Waste Carboys (Pb, Zn, Cu, Ag, KNO 3 ) 7.Parts 2,4, and 6…….there is no data collection

35. DO NOT MIX WASTE!!! DO NOT SEND ANYTHING DOWN THE DRAIN!!! (improper disposal will result in fines) Work in pairs for experimental. Reports are due at the end of the lab period.