Electrochemical Cells

Transcription

1 Electrochemical Cells There are two types: Galvanic and Electrolytic Galvanic Cell: a cell in which a is used to produce electrical energy, i.e., Chemical energy is transformed into Electrical energy. Electrolytic Cell: a cell in which electrical energy is used to bring about, i.e., electrical energy is transformed into chemical energy. will be dealt with later on.) (This type Refer to the cover of the CHEM*1050 lab manual which shows the Zinc Copper cell as well as Figures 19.2 and Consider the Zinc Copper Cell: If you place Zn metal in an aqueous CuSO 4 solution: 1. Cu 2+ ions plate out on the Zn surface; 2. At the same time the Zn metal dissolves to yield Zn 2+ ions; 3. The solution also becomes warmer. 4. The blue colour fades. (You will be studying this in Experiment 5: Galvanic Cells.) What is happening? Reaction? If we arrange things properly we can generate electrical energy. To "see" how this cell operates in detail, view the animated Zn/Cu galvanic cell: 72

2 Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-7 (Pg 871) What does this process tell us? 1. Zn(s) has a greater tendency to than Cu(s) 2. Cu 2+ (aq) has a greater tendency to than Zn 2+ (aq). 3. Zn(s) is a stronger than Cu(s). 4. Cu 2+ (aq) is a stronger than Zn 2+ (aq). NOTE, the cell is constructed such that have a Zn/Zn 2+ (aq) and a Cu/Cu 2+ (aq) metal electrodes are connected via a. links the solutions. Oxidation occurs at the (connected to negative terminal of voltmeter) Reduction occurs at the (connected to positive terminal of voltmeter) Note: If the anode is connected to the negative terminal of the voltmeter, the voltage reading is always positive. Thus, the anode and cathode can be determined experimentally by using a voltmeter (i.e., if hooked up backwards, the voltage reading will be negative). Why a Salt Bridge? Zn 2+ (aq) is produced in the compartment 73

3 Cu 2+ (aq) is consumed in the compartment To maintain electroneutrality, we need to get into the anode (to counter the build up of positive charge) and into the cathode (to counter the loss of positive charge). This is where the SALT BRIDGE comes in (e.g., KCl and gel in a glass tube). It completes the electrical circuit by allowing (anions to anode and cations to cathode); maintains electroneutrality. It also prevents between Cu 2+ (aq) and Zn(s). At the ANODE, The Zn(s) goes into solution and 2e go along the wire to the Cu plate. Anode compartment Cl (aq) ions pop out of the Salt Bridge to maintain. At the CATHODE, The 2 electrons entice a Cu 2+ ion to plate out of the solution Cathode compartment pop out of the Salt Bridge to maintain neutrality. Note: A porous separator can also be used to separate the two solutions (as shown above). OTHER POINTS: If we have 1.0 M Cu 2+ (aq) and 1.0 M Zn 2+ (aq), we get a voltage of exactly volts, no matter how (the anode is connected to the negative terminal of the voltmeter and the cathode is connected to the positive terminal). The "products", Zn 2+ (aq) and Cu(s), are ; they will be made as soon as the current flows. The concentrations of BOTH cations [Cu 2+ ] and [Zn 2+ ] DO affect the cell voltage more on this later. 74

4 Another Voltaic cell: Hydrogen - Copper Cell Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure (page 878). Anode: An inert platinum surface is placed in H + (aq) with H 2 (g) bubbling past it. (When we have 1.0 M solutions, this is a standard hydrogen electrode-see below.) H 2 (g) 2H + (aq) + 2e ( occurs at the anode) Cu 2+ (aq) + 2e Cu(s) ( occurs at the cathode) H 2 (g) + Cu 2+ (aq) 2H + (aq) + Cu(s) If we have 1.0 M solutions and H 2 (g) at 1 atm, we get a cell voltage of exactly volts. Other types of electrodes and half cells: 75

5 Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-9 (page 872) Reaction: H 2 (g) + 2Fe 2+ (aq) 2H + (aq) + 2Fe 3+ (aq) Oxidation Rxn: Reduction Rxn: Note: Within the two half cell reactions there are no metals. This creates a problem because we need an electrode in each half cell to allow for the flow of electrons from the anode compartment to the cathode compartment. (Another example: Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) The two half cells are made up of only ions in solution no metals present electrodes.) to act as ANSWER: INERT ELECTRODES Use a solid, inert conductor in the solution, e.g.,, and let the ions in solution transfer electrons to or from the surface of the inert electrode. The inert electrode does not participate in the redox reaction taking place within the halfcell, but it is included in the design of the galvanic cell since it completes the electrical circuit. Question : The galvanic cell below uses the half-cells Z + /Z and Y 2+ /Y, bridge containing MX(aq). The voltmeter gives a positive voltage reading. and a salt 76

6 Identify C. A. Z + B. Z C. M + D. X E. Y Cell Diagrams & Conventional Shorthand Notation (Refer to section 19.3 ) Instead of cumbersome cell drawings we usually use a shorthand notation specify the structure of the cell. to e.g. Zinc - Copper Cell: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Anode Salt Cathode Bridge The Anode is ALWAYS on the (and the Cathode on the ). " " - indicates a junction between different, e.g., sol & metal, sol n & gas or metal & compound as in Ag(s) & AgCl(s). " " indicates the that connects the two sol ns. States are indicated, but in the eq ns are not. Usual convention is for each half rxn. Examples: Zinc - Chlorine Cell: Zn(s) Zn 2+ (aq) Cl (aq) Cl 2 (g) Pt(s) Tin (II) - Iron (III) Cell: Au(s) Sn 4+ (aq), Sn 2+ (aq) Fe 3+ (aq), Fe 2+ (aq) C(s, graphite) Note: Tin(IV) and tin(ii), as well as iron (III) and iron (II), are separated by a commas since there is no phase change (i.e., both are in aqueous solution). 1. Write the cell notation for: 2Ag(s) + Hg 2 Cl 2 (s) 2AgCl(s) + 2Hg(l) 77

7 2. Write the cell notation for: a) b) Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure (pg. 884) Ref: Jones & Atkins, Chemistry: Molecules, Matter and Change, 4th ed., Fig (p. 803). a) b) Question #3: The following reaction occurs in basic solution, 2Ag + (aq) + Cu(s) 2Ag(s) + Cu 2+ (aq) The cell notation for this voltaic cell is A) Ag + (aq) Cu(s) Ag(s) Cu 2+ (aq). B) Cu(s) Cu 2+ (aq) Ag + (aq) Ag(s). C) Cu(s) Cu 2+ (aq) 2Ag + (aq) 2Ag(s). D) Ag(s) Ag + (aq) Cu 2+ (aq) Cu(s). E) Cu(s) Cu 2+ (aq) Ag(s) Ag + (aq). Question #4: Consider the following cell notation: Pt(s) H 2 (g) H + (aq) Cl 2 (g) Cl (aq) Pt(s). The balanced reaction for this voltaic cell is 78

8 A) Pt(s) + H 2 (g) + Cl 2 (g) H + (aq) + Cl (aq) + Pt(s). B) Pt(s) + H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl (aq) + Pt(s). C) H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl (aq). D) 2H + (aq) + 2Cl (aq) H 2 (g) + Cl 2 (g). E) Pt(s) + 2H + (aq) + 2Cl (aq) H 2 (g) + Cl 2 (g) + Pt(s). Standard Cell Potentials, E is a measure of the driving force, or spontaneity, of a chemical reaction. STANDARD conditions are the same as in thermodynamics, i.e., solutes, gases, pure. We cannot measure potentials of individual half cells, we can only measure the potential of two half cells coupled together. To solve this problem, we arbitrarily assign a voltage of 0 to the, then measure/report all other standard half cells relative to it. Standard Hydrogen Electrode : 2H + + 2e H 2 (g), E = V In a Galvanic Cell, the standard cell potential (E ) is the difference between the standard reduction potentials of the two electrodes: E (cell) = E (cathode) E (anode) where the E values at the cathode (where reduction takes place) and the anode (where oxidation takes place) are both written as. 79

9 EXAMPLE: Zinc-Hydrogen Cell: Zn(s) Zn 2+ (aq) H + (aq) H 2 (g) Pt(s) for which E (cell) = V For this cell, reduction takes place in the H + /H 2 half-cell and oxidation in the Zn 2+ /Zn half-cell. takes place Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) E = 0.76 (measured) This gives, E (cell) = E (H + /H 2 ) E (Zn 2+ /Zn) = 0.00 E (Zn 2+ /Zn) E (Zn 2+ /Zn) = 0.76 V. Thus, the standard reduction potential for Zn 2+ (aq) + 2e Zn(s) is 0.76 V. CONVENTION: Always report half cell potentials in the form: Ox + ne Red i.e., Standard Reduction Potentials. Now look back at the Standard Zn(s)/Zn 2+ (aq) and Cu(s)/Cu 2+ (aq) cell: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) for which E (cell) = 1.10 V. For this cell, reduction takes place in the Cu 2+ /Cu half-cell and oxidation in the Zn 2+ /Zn half-cell. takes place Ox: Red: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s); E = V E (Cu 2+ /Cu) = V. Therefore, Cu 2+ (aq) + 2e Cu(s) E = V By proceeding in this way we can build up a Table of Standard Electrode Potentials or Standard Reduction Potentials. These tables are VERY USEFUL! 80

10 Sample of a Standard Reduction Potential Table Ox + ne Red Potential in Volts F 2 (g) + 2e 2F (aq) Au 3+ (aq) + 3e Au(s) Cl 2 (g) + 2e 2Cl (aq) Ag + (aq) + e Ag(s) Fe 3+ (aq) + e Fe 2+ (aq) Cu 2+ (aq) + 2e Cu(s) Sn 4+ (aq) + 2e Sn 2+ (aq) H + (aq) + 2e H 2 (g) (by convention) Cd 2+ (aq) + 2e Cd(s) 0.40 Zn 2+ (aq) + 2e Zn(s) 0.76 Na + (aq) + e Na(s) 2.71 Li + (aq) + e Li(s) 3.05 Check out the table of standard reduction potentials in Appendix I. In this table the Standard Potentials are listed in electrochemical order very useful when comparing reactivity. 81

11 Reference: Appendix 2B, page A14 from the text Chemistry - Molecules, Matter and Change, P. Atkins & L. Jones, 4th ed., W.H. Freeman and Company, New York,

12 OBSERVATIONS from the table above: is the BEST oxidizing agent (and the most easily reduced) is the WORST oxidizing agent (and the most difficult to reduce) is the BEST reducing agent (and the most easily oxidized) is the WORST reducing agent (and the most difficult to oxidize) Consider the following: Which metals were first known and used by early civilizations? Why? Comment on their relative positions in the table. How do we obtain Standard Cell Potentials from Standard Reduction Potentials? What would the Standard Cell Potential be for a Cu Cu 2+ & Zn Zn 2+ cell? Step 1: Write out the two standard reduction potentials (SRP) equations: Step 2: The half reaction with the more positive SRP goes as written (is the reduction reaction) and thus the lower SRP (the oxidation reaction) is always subtracted from the higher SRP. E (cell) = E (cathode) E (anode) E (cell) = E ( / ) E ( / ) Step 3: Now, Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq), E (cell) = ( 0.76) = 1.10V Comments: Note that the E value is POSITIVE (this is true for all voltaic or galvanic cells), i.e., the reaction is spontaneous as we have written it (under Standard Conditions). 83

13 Now look at the Ni Ni 2+ and Ag Ag + cell. What is the spontaneous cell reaction, anode, cathode, E? Step 1: Get SRPs from Tables Ni 2+ (aq) + 2e Ni(s) E = Ag + (aq) + e Ag(s) E = Step 2: The half reaction with the more positive SRP goes as written and thus the lower SRP is always subtracted from the higher SRP. E (cell) = Step 3: Write the overall reaction equation and the cell potential Does anything strike you as odd about this?? Although we multiplied the coefficients in the silver half cell by 2, we did NOT double the half cell voltage. Compare: Why NOT?? 1) Ag + (aq) + e Ag(s) 2) 2Ag + (aq) + 2e 2Ag(s) The ENERGY release by equation (2) is certainly twice that of (1), but remember that Voltage = Joules/Coulombs (energy per coulomb of charge transferred) Equation (2) involves twice the energy, but it also involves twice the coulombs so the VOLTAGE does not change. Voltage is an Intensive property. 84

14 Question: Would the reaction: Al 3+ (aq) + Mg(s) Al(s) + Mg 2+ (aq) go as written under standard sonditions, or would the reverse reaction occur? Step 1: Get SRP equations Step 2: The half reaction with the more positive SRP goes as written and thus the lower SRP is always subtracted from the higher SRP. E (cell) = Step 3: Write the overall equation and cell potential: Answer: (Yes/No) the reaction (WOULD/WOULD NOT) occur as written. Question #5: Consider the following standard electrode potentials: Ag + (aq) + e Ag(s) E = 0.80 V Mn 2+ (aq) + 2e Mn(s) E = 1.18 V For the standard cell: Mn(s) Mn 2+ (aq) Ag + (aq) Ag(s) which of the following statements are true? A. The cell voltage will be 1.98 V. B. The half-cell reaction is Mn(s) Mn 2+ (aq) + 2e. C. The overall cell potential will decrease with time. Question #6: For a galvanic cell using Fe Fe 2+ (1.0 M) and Pb Pb 2+ (1.0 M) half-cells, which of the following statements is correct? Fe e Fe; E = 0.44 V Pb e Pb; E = 0.13 V A) The mass of the iron electrode increases during discharge. B) Electrons leave the Pb electrode to pass through the external circuit during discharge. C) The concentration of Pb 2+ decreases during discharge. D) The iron electrode is the cathode. E) When the cell has completely discharged (to 0 V), the concentration of Pb 2+ is zero. 85

15 Problem 1: With access to an SRP table, deduce what reaction (if any) will occur when: (a) Metallic Sn is added to 1 M HCl(aq) solution. (b) A solution of 1.0 M Sn 2+ (aq) is treated with (i) metallic Zn: (ii) 1.0 M FeCl 3 (aq) sol n (no solids are formed): (c) A solution of 1.0 M Sn 4+ (aq) is treated with metallic Zn. Reduction Potentials: Fe 3+ (aq) + e Fe 2+ (aq); E = V Fe 3+ (aq) + 3e Fe(s); E = 0.04 V Sn 4+ (aq) + 2e Sn 2+ (aq); E = V Sn 2+ (aq) + 2e Sn(s); E = 0.14 V 2H + (aq) + 2e H 2 (g); E = 0.0 V Zn 2+ (aq) + 2e Zn(s); E = 0.76 V Problem 2: You are told the following: 1. Solid iodine and solid sodium DO react. 4. Au(s) and Ag + (aq) do NOT react 2. Au + (aq) and iodide ion DO react. 5. Ag(s) and I 2 (s) do NOT react 3. Ag + (aq) and sodium DO react. Deduce from these data alone WITHOUT access to SRP tables the correct ORDER of the relevant SRPs for the processes involved. Problem 3: You are told the following: 1. Br 2 (l) and K(s) DO react. 4. Hg(l) and Co 2+ (aq) do NOT react 2. Hg 2+ (aq) and solid potassium DO react. 5. Hg(l) and Br 2 (l) DO react 3. Co(s) and K + (aq) do NOT react. Deduce from these data alone without access to SRP tables the correct ORDER of the relevant SRPs for the processes involved. 86

16 Problem #4: All the following statements are true EXCEPT A) most metal oxides are basic. B) most metals have positive reduction potentials. C) most metals are dense solids at 500 K. D) metals are good heat conductors. E) metals are good electrical conductors. Problem #5: The overall reaction for the lead storage battery is: Pb(s) + PbO 2 (s) + 4H + 2 (aq) + 2SO 4 (aq) 2PbSO 4 (s) + 2H 2 O(l) When the battery discharges, which of the following is (are) true? 1. PbSO 4 is formed at both electrodes. 2. PbO 2 is the oxidizing agent. 3. The ph increases. A) a only B) b only C) c only D) a & b only E) a, b, & c Problem #6: A piece of iron half-immersed in a sodium chloride solution will corrode more rapidly than a piece of iron half-immersed in pure water because A) the sodium ions oxidize the iron atoms. B) the chloride ions oxidize the iron atoms. C) the chloride ions form a precipitate with iron. D) the chloride ions increase the ph of the solution. E) the sodium ions and chloride ions carry a current through the solution. Answers to Problem #1: a) Metallic Sn is added to 1 M HCl(aq) solution: Sn(s) + 2H + (aq) Sn 2+ (aq) + H 2 (g), E (reaction) = E (H + /H 2 ) E (Sn 2+ /Sn) = V (b) A solution of 1.0 M Sn 2+ (aq) is treated with (i) metallic Zn: Sn 2+ (aq) + Zn(s) Sn(s) + Zn 2+ (aq), E = V (ii) FeCl 3 (aq) solution: Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq), E = V (c) A solution of 1.0 M Sn 4+ (aq) is treated with metallic Zn: Sn 4+ (aq) + Zn(s) Sn 2+ (aq) + Zn 2+ (aq), E = V Answer to Problem 2: Based on the information given, write chemical equations for the spontaneous reactions. (i) I 2 (s) + 2e 2I (aq), E (i) (iii) Au + (aq) + e Au(s), E (iii) (ii) Na + (aq) + e Na(s), E (ii) (iv) Ag + (aq) + e Ag(s), E (iv) From observation 1, E (i) is more positive than E (ii). From observation 2, E (iii) is more positive than E (i). From observation 3, E (iv) is more positive than E (ii). From observation 4, E (iii) is more positive than E (iv). From observation 5, E (iv) is more positive than E (i). Therefore, E (iii) > E (iv) > E (i) > E (ii). 87