Study Guide for Module 17 Oxidation-Reduction Reactions and Electrochemistry

Transcription

1 Chemistry 1020, Module 17 Name Study Guide for Module 17 Oxidation-Reduction Reactions and Electrochemistry Reading Assignment: Chapter 17 in Chemistry, 6th Edition by Zumdahl. Guide for Your Lecturer: 1. Review of Oxidation-Reduction 2. Review of Oxidation Numbers 3. Review of Balancing Oxidation-Reduction Equations 4. An Overview of Galvanic Cells 5. Calculation of the EMF of Electrochemical Cell, Predicting Spontaneity 6. Oxidizing and Reducing Agents 7. Overview of Electrolytic Cells & Electrolysis 8. Stoichiometry in Electrochemical Cells 9. The Relationship of Standard Potentials to Thermodynamics and Equilibrium 10. The Effect of Concentration Changes on Voltage Produced by a Voltaic Cell Homework Note: indicates problems to be stressed on drill quizzes and hour exams. Review of Oxidation-Reduction 1a) What is generally happening when an oxidation-reduction reaction occurs? (Electrons are being transferred.) b) Define oxidation. (p ) c) Define reduction. (p ) d) Define oxidizing agent. (p ) e) Define reducing agent. (p ) Review of Oxidation Numbers 2a) Define oxidation number (p ) Xavier University of Louisiana 263

2 Chemistry 1020, Module 17 Review of Oxidation Numbers [continued] ********************************************************************* Rules for assigning oxidation numbers 1: The sum of the oxidation numbers must equal the overall charge. Corollaries: a) The oxidation number of a pure element is zero. b) The oxidation numbers of a compound must sum to zero. c) The oxidation numbers of an ion must sum to the charge of the ion. 2: The last remaining unassigned element is assigned an oxidation number consistent with rule #1. 3: Oxidation numbers for specific atoms in molecules and ions must be assigned using the following set of rules, in order of priority until one unassigned atom remains(see rule #2). a) The oxidation number of Group IA metals is always +1 in a compound. b) The oxidation number of Group IIA metals is always +2 in a compound. c) The oxidation number of F is always 1 in a compound. d) The oxidation number of Al is always +3 in a compound. e) The oxidation number of H is usually +1 in a compound. (Exception: 1 in binary compounds with metals.) f) The oxidation number of O is usually 2 in a compound. (Exception: 1 in peroxides such as H2O2.) g) In cases where two unassigned atoms remain and all previous rules have been applied, the element which is closer to the right and top of the periodic table has an oxidation number which matches its charge as an ion. ********************************************************************* 2b) Use the rules above to determine the oxidation numbers of each element in the following compounds and be prepared to do so for similar ones. (pp ) Chemical Oxidation numbers of the elements in the chemical species. species S. ClO3 - (x)(-2) <- oxidation number of O is -2 and that of Cl is unknown. ClO3 - so 1(x) + 3(-2) = -1 ; x -6 = -1 x = +5, Therefore oxidations numbers are Cl = +5 & O = -2 A. NO2 - B. ClO2 - C. BrO3 - D. H2SO4 E. HClO F. MnO4-264 Xavier University of Louisiana

3 Chemistry 1020, Module 17 Name Review of Oxidation Numbers (continued) 2c) Do the following for each of the following chemical reactions: Determine the oxidation numbers of each species. Draw arrows from reactants to products connecting elements whose oxidation numbers changed. Above the arrows show the electron transfer which would have accounted for change in oxidation number. Label the oxidation and reduction above the arrows. Identify the species being oxidized and the species being reduced. Identify the oxidizing agent and the reducing agent in the reaction. S. 2 K(s) + 2 NH3(l) 2 KNH2(s) + H2(g) 1 e - lost per atom K (x)(+1) (+1)(x)(+1) (0) (-3)(+1) (+1)(-3)(+1) (0) NH3 KNH2 2 K(s) + 2 NH3(l) 2 KNH2(s) + H2(g) x + 3(1) = x + 2(1) = 0 x = -3 x = -3 1 e - gained per atom H Substance oxidized: K Substance reduced: H in NH3 Oxidizing agent: NH3 Reducing agent: K A. 2 CaS(s) + 3 O2(g) 2 CaO(s) + 2 SO2(g) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: B. Fe2O3(s) + 6 Na(s) 3 Na2O(s) + 2 Fe(s) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: C. 2 Fe2O3(s) + 3 C(s) 3 CO2(g) + 4 Fe(s) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: Xavier University of Louisiana 265

4 Chemistry 1020, Module 17 Review of Oxidation Numbers (continued) 2c) Do the following for each of the following chemical reactions: Determine the oxidation numbers of each species. Draw arrows from reactants to products connecting elements whose oxidation numbers changed. Above the arrows show the electron transfer which would have accounted for change in oxidation number. Label the oxidation and reduction above the arrows. Identify the species being oxidized and the species being reduced. Identify the oxidizing agent and the reducing agent in the reaction. D. CO2(g) + 2 H2(g) C(s) + 2 H2O(g) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: E. 2 C2H6(g) + 7 O2(g) 6 H2O(l) + 4 CO2(g) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: F. Mg(s) + 2 HNO3(aq) Mg(NO3)2 + H2(g) Substance oxidized: Substance reduced: Oxidizing agent: Reducing agent: 266 Xavier University of Louisiana

5 Chemistry 1020, Module 17 Name Review Balancing Oxidation-Reduction Equations 3a) Balance each of the following oxidation-reduction equations by the half-reaction method used in class. (pp ) ****************************************************************************** Algorithm for Balancing Oxidation-Reduction Equations Using Half-Reaction Method ❶ Separate overall equation into two parts using the chemical symbols. ❷ For each half-reaction: a) Balance all elements except O and H. b) Balance O using H2O. c) Balance H using H +. d) Balance charge using electrons. (i.e. Determine overall charge on both sides and then add electrons to the more positive side.) e) If the solution is acidic, skip to Step #3. If the solution is basic, add OH - to get rid of H + anywhere (by combining to produce water) and then add an equal number of OH - to the other side of the equation to keep it balanced. If the solution is neutral, add OH - to get rid of H + on the reactants side (by combining to produce water) and then add an equal number of OH - to the product side of the equation to keep it balanced. ❸ Multiply each half-reaction by coefficients so that electrons gained = electrons lost. ❹ Add half-reactions back together. ❺ Combine like terms and check to make certain the coefficients are still the lowest set possible. ****************************************************************************** S. Cr + MnO2 Mn 2+ + Cr 3+ (acidic) Separate, balance each half-reaction, & multiply by coefficients so e - s lost = e - s gained (2e - + 4H + + 1MnO2 1Mn H2O)3 <----Balanced half-reaction for reduction (1Cr 1Cr e - )2 <----Balanced half-reaction for oxidation Next add the two together and combine like terms 6e H + + 3MnO2 + 2Cr 3Mn H2O + 2Cr e - Since reaction is acidic the balanced equation is 12H + + 3MnO2 + 2Cr 3Mn H2O + 2Cr 3+ Oxidation half-reaction: 1Cr 1Cr e - Reduction half-reaction: 2e - + 4H + + 1MnO2 1Mn H2O Oxidizing agent: MnO2 Reducing agent: Cr A. Cl - + MnO4 - Mn 2+ + Cl2 (acidic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: Xavier University of Louisiana 267

6 Chemistry 1020, Module 17 Review of Balancing Oxidation-Reduction Equations (continued) 3a) Balance each of the following oxidation-reduction equations by the half-reaction method used in class. (pp ) B. H2O2 + MnO4 - O2 + Mn 2+ + H2O (acidic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: C. As + NO3 - H3AsO3 + NO (acidic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: 268 Xavier University of Louisiana

7 Chemistry 1020, Module 17 Name Review of Balancing Oxidation-Reduction Equations (continued) 3a) Balance each of the following oxidation-reduction equations by the half-reaction method used in class. (pp ) D. H2O2 + NO NO3 - + H2O (acidic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: E. Cr 3+ + O2 Cr2O H2O (acidic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: Xavier University of Louisiana 269

8 Chemistry 1020, Module 17 Review of Balancing Oxidation-Reduction Equations (continued) 3b) Balance each of the following oxidation-reduction equations by the half-reaction method. (pp ) S. NO + MnO4 - MnO2(s) + NO2 - Separate, balance 1st half-reaction (basic solution) 3e - + 4H + + 1MnO4-1MnO2(s) + 2H2O <----1st half-rxn balanced if in acid. 3e - + 4OH - + 4H + + 1MnO4-1MnO2(s) + 2H2O + 4OH - <----Add OH - to both sides 2 to cancel H + s since basic. 3e - + 1MnO H2O 1MnO2(s) + 2H2O + 4OH - <----Combine H + s and OH - s to make H2O & combine like terms. 3e - + 1MnO H2O 1MnO2(s) + 4OH - <----1st half-rxn totally balanced. Next, balance 2nd half-reaction 1H2O + 1NO 1NO H + + 1e - <----2nd half-rxn balanced if in acid. 2OH - + 1H2O + 1NO 1NO H + + 1e - + 2OH - <----Add OH - to both sides 1 to cancel H + s since basic. 1NO + 1H2O + 2OH - 1NO e - + 2H2O <----Combine H + s and OH - s to make H2O & combine like terms. 1NO + 2OH - 1NO e - + 1H2O <----2nd half-rxn totally balanced. Then, multiply by coefficients so e - s lost = e - s gained (3e - + 1MnO H2O 1MnO2(s) + 4OH - )1 (1NO + 2OH - 1NO e - + 1H2O )3 Next, add the two together and combine like terms 2 1 3e + 1MnO H2O + 3NO + 6OH - 1MnO2(s) + 4OH - + 3NO e - + 3H2O 1MnO NO + 2OH - 1MnO2(s) + 3NO H2O <----Balanced equation. Oxidation half-reaction: 1NO + 2OH - 1NO e - + 1H2O Reduction half-reaction: 3e - + 1MnO H2O 1MnO2(s) + 4OH - Oxidizing agent: MnO4 - Reducing agent: NO A. SO MnO4 - MnO2(s) + SO4 2- (basic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: 270 Xavier University of Louisiana

9 Chemistry 1020, Module 17 Name Review of Balancing Oxidation-Reduction Equations (cont d) 3b) Balance each of the following oxidation-reduction equations by the half-reaction method. (pp ) B. Zn + BrO4 - Zn 2+ + Br - (basic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: C. MnO4 - + S 2- MnO2(s) + S (basic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: Xavier University of Louisiana 271

10 Chemistry 1020, Module 17 Review of Balancing Oxidation-Reduction Equations (cont d) 3b) Balance each of the following oxidation-reduction equations by the half-reaction method. (pp ) D. ClO4 - + NO2 - ClO2 - + NO3 - (basic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: E. ClO - + CrO2 - Cl - + CrO4 2- (basic) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: 272 Xavier University of Louisiana

11 Chemistry 1020, Module 17 Name An Overview of Galvanic Cells 4a) Define spontaneous reaction. (A reaction which can occur without the addition of energy (i.e. without heat or work from the surroundings into the system.) b) Can a reaction be spontaneous in both the forward and the reverse directions? (NO) c) Which spontaneous chemical reactions can be used as the basis of a battery? (In theory, any spontaneous oxidation reduction reaction could be used to make a battery. In practice, however, we usually make the ones which are cheapest and easiest to make.) d) What is meant by the phrase metal X displaces metal Y from solution? (Metal X reacts with cations of Y to produce cations of X and metallic Y.) e) There are two basic types of chemical cells. List the two and state what is happening to a) electrons and b) energy in each. (1. Galvanic cell: a) A spontaneous chemical reaction causes electrons to flow. b) Chemical energy is changed to electrical energy as a spontaneous chemical reaction occurs. 2. Electrolytic cell: a) The flow of electrons causes a nonspontaneous chemical reaction to occur. b) Electrical energy is used to cause a nonspontaneous chemical reaction to occur.) Type of Cell What happens to electrons What happens to energy f) Define electrode. (Wire or piece of metal at which reaction occurs in an electrochemical cell.) g) The two electrodes (and the corresponding half-reactions which occur at them) must be separated in a galvanic cell. Why? I.E. What would happen if they were not separated? (If the two half-reactions are not separated, the chemical reaction would give off heat instead of causing electrons to flow.) h) What are the two common ways of separating half-reactions in galvanic cells? (Using either a) a porous disk or b) a salt bridge.) i) The item used to separate half-reactions in a galvanic cell must have two properties. What are they? (They must a) allow ions to flow so the circuit is complete, while at the same time, b) preventing the reactants from mixing directly and giving up heat.) Xavier University of Louisiana 273

12 Chemistry 1020, Module 17 An Overview of Galvanic Cells (continued) 4j) What is a salt bridge? (A salt bridge consists of a U-shaped tube that contains a strong electrolyte solution, such as NaNO3(aq), whose ions will not react with other ions in the cell or with the electrode materials. The U-tube may be loosely plugged with glass wool, or the electrolyte may be incorporated into a gel so that the electrolyte solution does not run out when the U-tube is inverted and placed so as to provide electrical contact between the two half-reactions.) k) List two additional names for galvanic cells. (Battery, voltaic cell) l) Distinguish between anode and cathode. (Anode = electrode at which oxidation occurs. Cathode = electrode at which reduction occurs. ) m) What term might a chemist use to a describe a battery which is dead? (It is in equilibrium.) n) How many electrons flow in the solution in an electrochemical cell? (NONE, only ions flow in solution. Electrons flow in the wire connecting the two electrodes and in the electrodes.) 274 Xavier University of Louisiana

13 Chemistry 1020, Module 17 Name An Overview of Galvanic Cells (continued) 4o) Spontaneous reactions can be used to produce galvanic cells (batteries). On the diagram provided, show how the given spontaneous reaction can be used to produce a battery by doing each of the following (pp ): Break the overall reaction into half-reactions and place them at the appropriate electrodes (by tradition, the oxidation is on the left and the reduction is on the right of such a diagram). Label the anode and cathode and indicate the metal from each is made. Indicate the ions present in each beaker (the solutions are made from metal nitrates). Indicate the ions present in the salt bridge. Assume it was made by dissolving sodium nitrate in water. Indicate (with arrows) the direction each ion in the beakers and the salt bridge flows. Indicate (with arrows attached to ions) whether the electrode gains or loses mass as the battery operates. Indicate (with arrows) where and in which direction the electrons flow as the battery operates. Write the line notation for the cell. (Rules: Anode on left, reactants before products, for salt bridge, and for change in state. S. Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) Line notation: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Anode > < Cathode Oxidation half-reaction: Zn(s) Zn e - Reduction half-reaction: Cu e - Cu(s) What happens to mass of each of the two electrodes as the battery operates? The mass of Zn decreases as it dissolves, the mass of Cu increases as Cu 2+ is plated from the solution. Where and in which direction does each of the cations move? Cations in the solutions or salt bridge (Cu 2+, Zn 2+, and Na + ) move toward the Cu electrode (i.e. move toward the cathode). Where and in which direction does each of the anions move? NO3 - ions in the two solutions and in the salt bridge move toward the Zn electrode (i.e. move toward the anode). Where and in which direction do electrons flow? Electrons flow in the wire from the anode (where they are given up by Zn(s) to form Zn 2+ ) to the cathode (where they are picked up by Cu 2+ to form Cu metal.) A. 2 Al(s) + 3 Cd 2+ (aq) 2 Al 3+ (aq) + 3 Cd(s) Line notation: wire salt bridge Oxidation half-reaction: Reduction half-reaction: What happens to mass of each of the two electrodes as the battery operates? Where and in which direction does each of the cations move? Where and in which direction does each of the anions move? Where and in which direction do electrons flow? Xavier University of Louisiana 275

14 Chemistry 1020, Module 17 An Overview of Galvanic Cells (continued) 4o) Complete each of the following cells providing the information itemized on the previous page. B. Pb(s)+ Cu 2+ (aq) Cu(s) + Pb 2+ (aq) Line notation: wire salt bridge Oxidation half-reaction: Reduction half-reaction: What happens to mass of each of the two electrodes as the battery operates? Where and in which direction does each of the cations move? Where and in which direction does each of the anions move? Where and in which direction do electrons flow? C. Zn(s) + 2 Ag + (aq) Zn 2+ (aq) + 2 Ag(s) Line notation: wire salt bridge Oxidation half-reaction: Reduction half-reaction: What happens to mass of each of the two electrodes as the battery operates? Where and in which direction does each of the cations move? Where and in which direction does each of the anions move? Where and in which direction do electrons flow? 276 Xavier University of Louisiana

15 Chemistry 1020, Module 17 Name An Overview of Galvanic Cells (continued) 4o) Complete each of the following cells providing the information itemized on the previous page. D. Co(s) + Sn 2+ (aq) Co 2+ (aq) + Sn(s) Line notation: wire salt bridge Oxidation half-reaction: Reduction half-reaction: What happens to mass of each of the two electrodes as the battery operates? Where and in which direction does each of the cations move? Where and in which direction does each of the anions move? Where and in which direction do electrons flow? E. Zn(s) + Co 2+ (aq) Zn 2+ (aq) + Co(s) Line notation: wire salt bridge Oxidation half-reaction: Reduction half-reaction: What happens to mass of each of the two electrodes as the battery operates? Where and in which direction does each of the cations move? Where and in which direction does each of the anions move? Where and in which direction do electrons flow? Xavier University of Louisiana 277

16 Chemistry 1020, Module 17 Calculation of the EMF of an Electrochemical Cell, Predicting Spontaneity 5a) What are standard-state conditions for electrochemical measurements. (1 M concentrations, 1 atm pressure for all gases in the reaction, and 25 o C) b) Define standard reduction potential. (p. 833) c) Write the chemical reaction which is the standard on which standard reduction potentials are based. (p. 833) d) Draw a diagram of the cell which could be used to determine the standard reduction potential for zinc.(p. 833) 278 Xavier University of Louisiana

17 Chemistry 1020, Module 17 Name Calculation of the EMF of an Electrochemical Cell, Predicting Spontaneity (cont.) 5e) Define electromotive force. (A measure of the tendency for a reaction to occur. ) What are its units? (volts) What is the symbol for electromotive force? (E) f) How is the voltage produced by a cell related to the spontaneity of the reaction occurring in the cell? (If the voltage of the cell is >0, the reaction is spontaneous. If the voltage is <0, the reaction is not spontaneous as written.) g) How is the voltage associated with a cell related to work? (If E>0, it represents the maximum work which could be obtained if the reaction occurs under ideal conditions. If E<0, it represents the minimum work which would have to be done to cause the reaction to occur under ideal conditions.) h) How can you calculate the voltage associated with a cell from reduction half-potentials. (Add the voltage associated with the anode to that associated with the cathode.) i) Calculate the maximum EMF generated by a voltaic cell or the minimum EMF required to cause an electrolytic cell reaction to proceed. Assume the solutions were prepared from the nitrate salts and to be at standard conditions. (pp ) S. 3 Cd Cr(s) 2 Cr Cd(s) First, put together half-reactions from the table of SRP s to get given equation 3(Cd e - Cd), E o = volts 2(Cr Cr e - ), E o = 0.74 volts (Note: This is reverse of equation in table so sign is changed) 3 Cd Cr 3 Cd + 2 Cr 3+, E o = 0.34 volts (Note: The standard reduction potential is an intensive property (does not depend on how many times the reaction occurs), so the potential is not multiplied by the integer required to balanced the cell reaction. In this example, the reduction half-reaction was multiplied by 3 and the oxidation half-reaction was multiplied by 2 so that electrons lost equal the electrons gained. However, the E o s were not multiplied by these integers.) -Is the reaction spontaneous? Explain how you know and what this means. The reaction is spontaneous because E o is > 0. This means that a battery based on this reaction could produce a maximum of 0.34 volts. (Note: If E o had been <0, this would have been the minimum voltage required to cause the reaction to occur.) A. Cu 2+ + Pb(s) Cu(s) + Pb 2+ -Is the reaction spontaneous? Explain how you know and what this means. Xavier University of Louisiana 279

18 Chemistry 1020, Module 17 Calculation of the EMF of Electrochemical Cell, Predicting Spontaneity (continued) 5i) Calculate the maximum EMF generated by a voltaic cell or the minimum EMF required to cause an electrolytic cell reaction to proceed. Assume the solutions were prepared from the nitrate salts and to be at standard conditions. (pp ) B. Fe Hg(liq) Fe(s) + Hg2 2+ -Is the reaction spontaneous? Explain how you know and what this means. C. 2 Ag(s) + Sn 2+ 2 Ag + + Sn(s) -Is the reaction spontaneous? Explain how you know and what this means. D. Ni(s) + Co 2+ Co(s) + Ni 2+ -Is the reaction spontaneous? Explain how you know and what this means. E. Zn(s) + Co 2+ Co(s) + Zn 2+ -Is the reaction spontaneous? Explain how you know and what this means. F. Cd(s) + Ni 2+ Ni(s) + Cd 2+ -Is the reaction spontaneous? Explain how you know and what this means. 280 Xavier University of Louisiana

19 Chemistry 1020, Module 17 Name Oxidizing and Reducing Agents 6a) Which of the following is the better oxidizing agent under standard conditions? 1st pair Better oxidizing agent and justification S. O2 or F2 E o F2 = 2.87 volts (from table of Standard Reduction Potentials) E o O2 = 1.23 volts, Since E o F2 > E o O2, F2 is the better oxidizing agent A. H2O2 or I2 B. ClO3 - or Cr2O7 2- C. H2O2 or O2 D. Cu 2+ or I2 E. NO3 - or ClO3 - F. Cl2 or MnO4 - b) Which of the following is the better reducing agent? 1st pair Better reducing agent and justification S. Zn or H2 E o Zn = 0.76 volts (by reversing value in table of SRP s) E o H2 = 0 volts, Since E o Zn > E o H2, Zn is the better reducing agent A. Mg or Zn B. Fe or Co C. Cr or Al D. Co or Pb E. Zn or Ni F. Fe or Cd Xavier University of Louisiana 281

20 Chemistry 1020, Module 17 Oxidizing and Reducing Agents (continued) 6c) List six common oxidizing agents and three common reducing agents and the balanced half-reactions for each by completing the table. (Oxidizing agents: MnO4 -, Cl2, O2, Cr2O7 2-, NO - 3, H 2O2 ; Reducing agents: Zn, H2, charcoal.) Oxidizing agents Balanced reaction when the oxidizing agent reacts 1) 2) 3) 4) 5) 6) Reducing agents Balanced reaction when the reducing agent reacts 1) 2) d) In general, which elements are good reducing agents? ( Most metals are good reducing agents.) 282 Xavier University of Louisiana

21 Chemistry 1020, Module 17 Name Overview of Electrolytic Cells, Electrolysis 7a) What happens to energy in an electrolytic cell? (The flow of electrons causes a nonspontaneous chemical equation to occur. That is, electrical energy is transformed into chemical energy.) b) Define electrolysis. (855) c) Diagram an electrolytic cell which might be used to make the following nonspontaneous reaction occur showing ions flowing, metal of which electrodes is made, reactions at each electrode, name of each electrode, etc. as for the galvanic cells in #4. Then calculate the minimum voltage needed to cause the reaction to occur. Reaction, Diagram: S. Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + Zn(s) electrons battery electrons Calculation of minimum voltage needed/meaning Calculation of voltage: Zn 2+ (aq) + 2e - Zn(s), Ec o = volt Cu(s) Cu 2+ (aq) + 2e -, Ea o = volt Cu porous disk Zn Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + Zn(s), Zn 2+ Ecell o = v v = volts NO 3 - Anode Cu 2+ NO - 3 Cathode Cathode half-rxn: Zn 2+ (aq) + 2e - Zn(s) Anode half-rxn: Cu(s) Cu 2+ (aq) + 2e - A. Pb(s) + Sn 2+ (aq) Pb 2+ (aq) + Sn(s) battery Meaning: The reaction is nonspontaneous, the battery must produce at least 1.10 volts to cause it to occur. What happens to the mass of each of the two electrodes as the reaction proceeds? The mass of the Cu electrode decreases since it is dissolving while that of the Zn increases since Zn 2+ is being plated out. Calculation of voltage: porous disk Meaning: Cathode half-rxn: Anode half-rxn: What happens to the mass of each of the two electrodes as the reaction proceeds? Xavier University of Louisiana 283

22 Chemistry 1020, Module 17 Overview of Electrolytic Cells, Electrolysis (continued) 7c) Diagram an electrolytic cell which might be used to make the following nonspontaneous reaction occur showing ions flowing, metal of which electrodes is made, reactions at each electrode, name of each electrode, etc. as for the galvanic cells in #4. Then calculate the minimum voltage needed to cause the reaction to occur. B. Ni(s) + Fe 2+ (aq) Ni 2+ (aq) + Fe(s) Calculation of voltage: battery porous disk Meaning: Cathode half-rxn: Anode half-rxn: What happens to the mass of each of the two electrodes as the reaction proceeds? Reaction, Diagram: C. Cu(s) + Ni 2+ (aq) Cu 2+ (aq) + Ni(s) battery Calculation of minimum voltage needed/meaning: Calculation of voltage: porous disk Meaning: Cathode half-rxn: Anode half-rxn: What happens to the mass of each of the two electrodes as the reaction proceeds? D. Co(s) + Zn 2+ (aq) Co 2+ (aq) + Zn(s) battery Calculation of voltage: porous disk Meaning: Cathode half-rxn: Anode half-rxn: What happens to the mass of each of the two electrodes as the reaction proceeds? 284 Xavier University of Louisiana

23 Chemistry 1020, Module 17 Name Stoichiometry in Electrochemical Cells 8a) Scientists often compare the flow of electrons in a wire to the flow of water in a pipe. Define each of the following terms used in discussing the flow of electrons in a wire and state the analogous term for flow of water in a pipe. Electrical flow term: coulomb (p. 865) Analogous term from flow of water: Definition of coulomb: Electrical flow term: ampere (p. 865) Analogous term from flow of water: Definition of ampere: Electrical flow term: voltage (p. 865) Analogous term from flow of water: Definition of voltage: b) State Faraday s Law of Electrolysis. (The amount of a substance which reacts or is produced is proportional to the amount of electricity which flows through the cell.) c) In Chemistry 1010 (Module 3), you learned to work gram-gram stoichiometry problems. Faraday s Law now makes it possible to perform charge gram problems. However, in order to do so, you need a way to convert from coulombs to moles of electrons. What is conversion factor between the two? What is it called? (1 mole electrons = 96,500 coulombs. This quantity is called a Faraday. i.e. 1 Faraday = 1 mole e - = 96,500 coulombs.) Xavier University of Louisiana 285

24 Chemistry 1020, Module 17 Stoichiometry in Electrochemical Cells (continued) 8d) In Module 3 you learned to solve gram-gram stoichiometry problems in three steps: (pp ): ❶Write the balanced half-reaction. ❷Convert grams of known to moles of known using the definition of molecular weight. ❸Convert moles of known to moles of unknown using the balanced equation. ❹Convert moles of unknown to grams of unknown using the definition of molecular weight. Because of Faraday s Law of Electrolysis (from #7b), you can also work charge-gram problems in a similar manner: ❶Write the balanced half-reaction. ❷Convert ampere and seconds to coulombs using the 1 amp = 1 coul/sec (from #7a. Then convert coulombs to moles of electrons using 1 mole e - = 96,500 coul (from #7c). ❸Convert moles of known to moles of unknown using the balanced equation. ❹Convert moles of unknown to whatever is desired using one of the two types of conversions in step ❶. Perform the indicated charge-gram stoichiometry calculation using the method above. (pp ): S. How many grams of aluminum can be produced by the electrolysis of aluminum oxide if a current of 3.2 amps runs through the molten oxide for 36 minutes? ❶Balanced equation for deposition of Al is, Al e - Al ❷mol e - 60 sec coul/sec = (3.2 amps)(36 min)( 1 min )(1 1 amp )( 1 mol e coul ) = mol e- ❸mol Al = mol e - 1 mol Al ( 3mol e- ) = mol Al 27.0 g Al ❹g Al = mol Al ( 1 mol Al ) = g Al which rounds to 0.64 grams Al A. A 0.23 M solution of AgNO3 is electrolyzed for 10.0 minutes with a current of 5.2 amps. What is the maximum number of grams of silver which could be plated out during the process? B. If Cu is produced by electrolyzing molten copper(ii) oxide, how many grams of copper will be deposited by a current of 0.25 amps in 36 minutes? C. If lead metal is produced by the electrolysis of lead(iv) oxide, how many grams of lead will be deposited by 4.1 amps during 15 minutes? 286 Xavier University of Louisiana

25 Chemistry 1020, Module 17 Name Stoichiometry in Electrochemical Cells (continued) 8d) D. If a 0.52 M solution of Cr(NO3)3 is electrolyzed for one hour with a current of 1.2 amps, how many grams of chromium will be deposited during the process? E. If a 0.52 M solution of Cu(NO3)2 is electrolyzed for 1.5 hours with a current of 0.22 amps, how many grams of copper will be deposited during the process? e) Work the following problems which are related to the charge-gram stoichiometry problems above. (pp ) S. If lead metal is produced by the electrolysis of lead(iv) oxide, how long will it take to produce 2.6 grams of lead with a current of 0.52 amps? ❶Balanced equation for deposition of Pb is, Pb e - Pb 1 mol Pb ❷mol Pb= 2.6 g Pb( 207 g Pb )= mol Pb ❸mol e - 4 mol e- = mol Pb( 1 mol Pb ) = mol e- ➍coul = mol e coul ( 1 mol e- ) = 4.86 x 10 3 couls ❺We know amp = coul sec so sec = coul amp = 4.86 x 103 couls 0.52 amp = 9.35 x 10 3 seconds time = 9.35 x 10 3 sec ( 1 min 60 sec ) = 156 min which rounds to 1.6 x 102 min A. What amperage is needed to produce 1.5 grams of silver by deposition from a 0.23 M solution of AgNO3 by electrolysis in 12 minutes? B. If Cu is produced by electrolyzing molten copper(ii) oxide, what must be the amperage through the cell to deposit 5.1 grams of copper in 21 minutes? Xavier University of Louisiana 287

26 Chemistry 1020, Module 17 Stoichiometry in Electrochemical Cells (continued) 8e) C. If Aluminum is produced by electrolyzing molten aluminum oxide, how long will it take for 1.5 grams of Al to plate out if 6.0 amps are run through the cell? D. How long will be required to cause 6.4 grams of chromium to deposit from a 0.52 M solution of Cr(NO3)3 if a current of 0.76 amps is run through the solution? E. How long will be required to cause 2.1 grams of nickel to deposit from a 0.22 M solution of Ni(NO3)2 if a current of 0.25 amps is run through the solution? The Relationship of Standard Potentials to Thermodynamics and Equilibrium 9a) What is the relationship between the Joule (a energy unit), the volt (a unit of electromotive force), and the coulomb (a measure of the amount of charge)? (1 volt = 1 Joule/coulomb) b) The voltage produced by a cell under standard conditions is relatively easy to measure. Further, it can rather easily be shown that E o is related to the equilibrium constant for the reaction on which the cell is based. Thus, scientists often measure cell voltages to obtain a value for K. What is the equation which relates E o and K... in exponential form? in log form? in ln form? (K = 10 nfe /2.303RT ) (log K = nfe o /2.303RT) (ln K = nfe o /RT) Define all terms in the equations. 288 Xavier University of Louisiana

27 Chemistry 1020, Module 17 Name The Relationship of Standard Potentials to Thermodynamics and Equilibrium (continued) 9c) Perform the indicated conversion of equilibrium constant to standard voltage (or vice-versa) at 25 o C. (p. 840) S. Calculate the equilibrium constant for the reaction, Co(s) + Zn 2+ Zn(s) + Co 2+ First, put together half-reactions from the table of SRP s to get given equation Zn e - Zn, E o = volts Co Co e -, E o = 0.28 volts (note this is reverse of equation in table so sign changed) Zn 2+ + Co Zn + Co 2+, E o = volts Now calculate K from K = 10 nfe /2.303RT =10 (2 mol electrons)(96,500 coul/1 mol electrons)(-0.48 v) ( 1J/coul 1 volt ) /(2.303)(8.314 J/mol K)(298 K) = where n = # electrons transferred in overall balanced equation K = 5.5*10-17 A. Calculate the equilibrium constant for the reaction, Ni(s) + 2 H + Ni 2+ + H2(g) B. Calculate the equilibrium constant for the reaction 10 Cl MnO H + 2 Mn H2O + 5 Cl2 C. Calculate the equilibrium constant for the reaction, Ni(s) + Zn 2+ Ni 2+ + Zn(s). D. Calculate the equilibrium constant for the reaction Cu(s) + Mg 2+ Cu 2+ + Mg(s). Xavier University of Louisiana 289

28 Chemistry 1020, Module 17 The Relationship of Standard Potentials to Thermodynamics and Equilibrium (continued) 9c) Perform the indicated calculation at 25 o C. (p. 840) E. Calculate the equilibrium constant for the reaction Cu(s) + Ni 2+ Cu 2+ + Ni(s). d) Perform the indicated calculation at 25 o C. FYI: 1 volt = 1 Joule/coulomb. (p. 846) S. What would be the standard voltage associated with a electrochemical cell based on the reaction A + 2B + = A B if the equilibrium constant for the reaction is 1.2*10-12 at 25 o C? Solve for E o in K = 10 nfeo /2.303RT log K = nfe o /2.303RT 2.303RTlogK = nfe o E o = 2.303RTlog K nf E o = (2.303)(8.314 J/mol K)( K)(log 1.2*10-12 ) (2 moles of electrons)(96,500 coul/mole of electrons) E o = volts which rounds to E o = volts A. What would be the standard voltage associated with a electrochemical cell based on the reaction 2A + 3B 2+ = 2A B if the equilibrium constant for the reaction is 6.7*10 23 at 25 o C? B. What would be the standard voltage associated with a electrochemical cell based on the reaction 3A + B 3+ = 3A + + B if the equilibrium constant for the reaction is 7.8*10-37 at 25 o C? C. What would be the standard voltage associated with a electrochemical cell based on the reaction A + 2B + = A B if the equilibrium constant for the reaction is 2.3*10-45 at 25 o C? 290 Xavier University of Louisiana

29 Chemistry 1020, Module 17 Name The Relationship of Standard Potentials to Thermodynamics and Equilibrium (continued) 9d) D. What would be the standard voltage associated with a electrochemical cell based on the reaction 3A + B 3+ = 3A + + B if the equilibrium constant for the reaction is 9.2*10 27 at 25 o C? E. What would be the standard voltage associated with a electrochemical cell based on the reaction 2A + B 2+ = 2A + + B if the equilibrium constant for the reaction is 2.7*10-26 at 25 o C? Xavier University of Louisiana 291

30 Chemistry 1020, Module 17 The Effect of Concentration Changes on Voltage Produced by a Voltaic Cell 10a) What is the major factor which determines the voltage produced by a battery? (The nature of the substances which are reacting.) b) List three factors which can affect the voltage produced by a battery to a lesser extent than the factor in a? (1. Temperature, 2. Concentration of the ions in solution, 3. Pressure of any gases.) c) What happens to the voltage of a cell as concentration of a reactant or product increases or decreases? (Voltage increases as concentration of a product decreases and increases as concentration of a reactant increases.) d) For the reaction indicated use the diagram of the voltaic cell to determine the effect of the specified change BEING SURE TO EXPLAIN YOUR REASONING. Assume the solutions were prepared from the nitrates of the metal ions. (pp ) Zn(s) + Cu 2+ Zn 2+ + Cu(s) Zn(s) Zn e - Cu e - Cu(s) S. Water is added to the half-cell with Zn in it. If water is added to the Zn half-cell, the concentration of Zn 2+ will decrease. Since Zn 2+ is a product, the voltage will increase since voltage always increases when concentration of a product decreases. A. Water is added to the half-cell with Cu in it. B. Zn (s) is added to the half-cell with Zn in it. C. Zn(NO3)2 (s) is added to the Zn half-reaction. 292 Xavier University of Louisiana

31 Chemistry 1020, Module 17 Name The Effect of Concentration Changes on Voltage Produced by a Voltaic Cell (continued) 10d) For the reaction indicated use the diagram of the voltaic cell to determine the effect of the specified change BEING SURE TO EXPLAIN YOUR REASONING. Assume the solutions were prepared from the nitrates of the metal ions. (pp ) D. Cu(NO3)2 (s) is added to the Cu half-reaction. E. NaNO3(s) is added to the Cu half-reaction. e) What is a concentration cell? (A galvanic cell whose driving force is the difference in concentration of two solutions of the same ion.) f) Write an equation which might occur in a concentration cell based on the indicated half-cell. (p ) S. Fe/Fe 2+ Fe(s) + Fe 2+ (1.0 M) Fe 2+ (0.5 M) + Fe(s) This is one example. Any other in which the concentration of Fe 2+ in the reactants is greater than concentration of Fe 2+ in the products would produce voltage. A. Zn/Zn 2+ B. Ni/Ni 2+ C. Cu/Cu 2+ D. Ag/Ag + E. Cd/Cd 2+ Xavier University of Louisiana 293

32 Chemistry 1020, Module 17 Bonding/Model Activity to Improve Ability to Visualize in 3-D A. a) Write the balanced equation if hydrocyanic acid is placed in water. b) Draw the Lewis structure of all of the species in the above reaction. B. a) Write the balanced equation if perchloric acid is placed in water. b) Draw the Lewis structure of all of the species in the above reaction. C. a) Write the balanced equation if acetic acid acid is placed in water. b) Draw the Lewis structure of all of the species in the above reaction. 294 Xavier University of Louisiana

33 Chemistry 1020, Module 17 Name Bonding/Model Activity to Improve Ability to Visualize in 3-D (continued) D. a) Write the balanced equation if ammonia is placed in water. b) Draw the Lewis structure of all of the species in the above reaction. A-D c) When you go to drill be prepared to assemble models of each of the reactants in the equations above and then use those models to demonstrate to your instructor what happens as the reactions occur. Challenge Questions A. Three metals, labeled A, B, & C, react with each other and with acids in the following way: C displaces A & B from solution. B displaces hydrogen from 1 M HCl but A does not. Which is the best reducing agent? Show your reasoning using appropriate equations and comparisons. B. Which of the ions, A +, B +, or C + in #A is the best oxidizing agent? Show your reasoning. C. Arrange the three metals in #A in an activity series together with hydrogen. That is, set up a table of standard reduction potentials for the three metals and hydrogen. D. Four metals labeled A, B, C & D, react with each other and with acids in the following manner: a) A + B + = N.R. b) B + C + = B + + C c) B + D + = B + + D d) C + D + = N.R. e) A + C + = N.R. On the basis of this information, which is the best reducing agent? Present your rationale. E. On the basis of the information in #D, which is the best oxidizing agent: A +, B +, C +, or D +? Show your rationale. F. Although all of the batteries discussed previously in this module were based on oxidation-reduction reactions, in theory, any spontaneous reaction or process could be used as the basis of a battery. What would be the standard voltage associated with a galvanic cell based on the spontaneous precipitation of CuS? G. Although all of the batteries discussed previously in this module were based on oxidation-reduction reactions, in theory, any spontaneous reaction or process could be used as the basis of a battery. What would be the standard voltage associated with a galvanic cell based on the spontaneous precipitation of calcium fluoride? Xavier University of Louisiana 295

34 Chemistry 1020, Module 17 Revised by Ann Privett, Spring, 1998; Ann Privett, Spring, 2000; Ann Privett, Spring, 2001; RI Spring Xavier University of Louisiana