64. V. ELECTROCHEMISTRY / A.Redox Reactions

59. V. ELECTROCHEMISTRY / A.Redox Reactions _V. ELECTROCHEMISTRYA. Redox Reactions1. Oxidation and Reduction a) Oxidation: the loss of electrons. b) Reduction: the gain of electrons. c) Oxidizing agent: the substance that causes something to become oxidized; the electron taker. d) Reducing agent: the substance that causes something to become reduced: the electron donor.Example(1): a) Write the following equations in net ionic form. b) With an arrow show the transfer of the electron(s). c) Identify which substance is oxidized, which is reduced, the oxidizing agent and the reducing agent.Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)Example(2): Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)Example(3): 2Fe(s) + 3SnCl4(aq)  2FeCl3(aq) + 3SnCl2(aq) 2. Half-reactions 60. V. ELECTROCHEMISTRY / A.Redox Reactions _Example(4): Write the half-reactions, including the electrons, for: Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)Example(5): Write the half-reactions, including the electrons, for: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)Example(6): Write the half-reactions, including the electrons, for: 2Fe(s) + 3SnCl4(aq)  2FeCl3(aq) + 3SnCl2(aq)3. Properties of elements as oxidizing and reducing agents (Compounds can also be oxidizing and reducing agents, as you will soon see.) a) The higher the EN of an element, the better the oxidizing agent (There are other factors, but this is the primary correlation.)Example(7): F2 Cl2 Br2 I2 Example(8): H2 O2 F2 b) The lower the IE, the better the reducing agent (There are other factors, but this is the primary correlation.) 61. V. ELECTROCHEMISTRY / A.Redox Reactions _Example(9): Na Mg Al4. The Ion-Electron Method a) Simple reactionsExample(10): Apply the ion-electron method to the equation below:Al + HCl  AlCl3 + H2Step 1: Find and write the half-reactions. (This can be done by checking for changes in oxidation states or by writing the equation in net ionic form.)Al  Al+3+1 H  H2Step 2: Balance the atoms in the half-reactions.Al  Al+3+1 2H  H2Step 3: Balance the charges in the half-reactions by adding electrons.Al  Al+3 + 3eˉ+1 2 eˉ + 2H  H2 Step 4: Make the number of electrons lost = the number gained by multiplying one or both half- reactions by a stoichiometric factor. 62. V. ELECTROCHEMISTRY / A.Redox Reactions _2(Al  Al+3 + 3eˉ)+1 3(2 eˉ + 2H  H2)Step 5: Add the half-reaction to arrive at the balanced net ionic equation.2Al  2Al+3 + 6eˉ+1 6 eˉ + 6H  3H2 ______+1 +3 2Al + 6H  2Al + 3H2 Step 6: Add the spectator ion back in, and associate the ion to arrive at the balanced molecular equation.+1 +3 2Al + 6H  2Al + 3H2 6Clˉ1 6Clˉ12Al + 6HCl  2AlCl3 + 3H2Example(11): Apply the ion-electron method to the equation below:F2 + Na3N → NaF + N2 b) Complex reactions, when the oxidizing agent contains oxygen – rules for an acidic solution: 63. V. ELECTROCHEMISTRY / A.Redox Reactions _Step 1: Write the half-reactions.Step 2: Balance the atoms: a) Balance the non-oxygen atoms first. b) Balance the oxygen by adding water. c) Balance the hydrogen by adding H+ to the opposite side of the water.Step 3: Balance the charges.Step 4: Make the number of electrons lost = the number gained.Step 5: Add the half-reaction. Step 6: Add the spectator ion back in (if ask for).Example(12): Apply the ion-electron method to the net ionic equation below taking place in an acid solution:+2 Cu + NO3ˉ → Cu + NO2 Example(13): Apply the ion-electron method to the net ionic equation below taking place in an acid solution: 64. V. ELECTROCHEMISTRY / A.Redox Reactions _+2 Clˉ + MnO4ˉ → Cl2 + Mn Example(14): Apply the ion-electron method to the net ionic equation below taking place in an acid solution:+2 Brˉ + PbO2 → Br2 + Pb5. Some common oxidizing reagents:+2 MnO4ˉ (acid) → Mn (base) → MnO22 +3 Cr2O7ˉ (acid) → Cr (base) → Cr2O32 +3 CrO4ˉ (acid) → Cr (base) → Cr2O32 2 2 SO4ˉ (acid) → SO3ˉ , SO2, S, Sˉ NO3ˉ (acid) → NO2ˉ, NO2, NO, N2 65. V. ELECTROCHEMISTRY / B. The Electrochemical Series _B. Electrochemical Series, ECS The Relative Driving Force of a Half-Reaction1. Reduction potentialsThe tendency for a half-reaction to occur can be measured directly in volts. The greater the voltage associated with the half-reaction, the stronger is the tendency for the reaction to occur.– Example(1): The half reaction F2(g) + 2e → 2Fˉ (aq) has a very strong tendency to occur. Write the series of Hess’s Law steps that add up to give the overall reaction. Indicate if the ΔH of each step is positive or negative. What can you say about the ΔS of the reaction?Example(2): All of the ΔH’s and the ΔS can be taken into account by measuring one thing. The voltage associated with the reaction: – F2(g) + 2e → 2Fˉ (aq) Eº = +2.87 voltsExample(3): a)Predict the relative voltage associated with the following half-reactions: b) Using the reduction table from your textbook assign each reaction its standard voltage.Na+(aq) + 2e– → Na(s)Mg+2(aq) + 2e– → Mg(s)Zn+2(aq) + 2e– → Zn(s) + – 2H (aq) + 2e → H2(g) – I2(s) + 2e → 2I ˉ(aq) – Br2(aq) + 2e → 2Brˉ(aq) – Cl2(g) + 2e → 2Clˉ(aq) – F2(g) + 2e → 2Fˉ(aq) 66. V. ELECTROCHEMISTRY / B. The Electrochemical Series _2. Predicting if a redox reaction will occur under standard conditions a) Using reduction potentialsExample(4): Will the following reaction take place as written? Cu+2(aq) + Zn(s) → Zn+2(aq) + Cu(s)Example(5): Will the following reaction take place as written? Cl2(g) + 2Brˉ(aq) → 2Clˉ(aq) + Br2(l)Example(6): Will the following reaction take place as written? + +2 10Clˉ(aq) + 16H (aq) + 2MnO4ˉ(aq) → 5Cl2(g) + 2Mn (aq) + 8H2O(l) b) Predicting by relative position on the tableExample(7): Will the following reaction take place as written? Mg+2(aq) + Zn(s) → Mg(s) + Zn+2(aq)Example(8): Will the following reaction take place as written? Ag+(aq) + Fe+3(aq) → Ag(s) + Fe+2(aq) 67. V. ELECTROCHEMISTRY / C. Voltaic Cells _Note that there are two types of electrochemical cells. A cell that produces a voltage, by a spontaneous chemical reaction, is called a Voltaic Cell. A cell that forces a nonspontaneous reaction to take place, by applying an external voltage, is called an Electrolysis Cell.C. Voltaic Cells 1. Principle of operationA voltaic cell produces a voltage by an oxidation-reduction reaction. In a voltaic cell, the half- reactions are kept separate from one another. Thus, the electrons transferred must pass through an external circuit, from the oxidation reaction to the reduction reaction.2. Cell construction and operationExample(1): Show the construction of a cell made from Zn, Zn+2, Cu, and Cu+2. a) As the cell operates what happens to the [Zn+2] and [Cu+2]? b) In which direction do the electrons move? c) Which electrode is negative, which is positive? d) In which direction do the ions move? e) Which electrode is the anode, which is the cathode? f) What is the voltage of the cell under standard conditions? 68. V. ELECTROCHEMISTRY / C. Voltaic Cells _Example(2): Show the construction of a cell made from Mg, Mg+2, Al, and Al+3. a) As the cell operates what happens to the [Mg+2] and [Al+3]? b) In which direction do the electrons move? c) Which electrode is negative, which is positive? d) In which direction do the ions move? e) Which electrode is the anode, which is the cathode? f) What is the voltage of the cell under standard conditions?3. Standard cell notation.Example(3): Write the standard notation for the cells in example(1) and (2). 69. V. ELECTROCHEMISTRY / C. Voltaic Cells _4. The Standard Hydrogen ElectrodeA single half-reaction cannot produce a voltage, since no oxidation or reduction can take place. The voltage of a half-reaction is measured relative to the standard hydrogen electrode. The Eº is arbitrarily taken to be zero. a) Electrode constructionExample(4): When a half-cell made from Mg(s) in contact with 1M Mg+2(aq) is connected with the standard hydrogen electrode, the voltage produced is +2.38 volts. a) What is the voltage of the Mg half reaction? b) Is the Mg half-reaction oxidation or reduction? 5. Determining the voltage of a cell given the standard notation.Example(5): What is the Eº of the cell: Zn| Zn+2 (1M) || Ni+2 (1M) | Ni 70. V. ELECTROCHEMISTRY / C. Voltaic Cells _+2 – Example(6): What is the Eº of the cell: Cu| Cu (1M) || Cl (1M) | Cl2|Pt6. Determining the voltage of a cell without the standard notation.Example(7): What is the Eº of the cell made from Cr, Al, Cr+3, and Al+3?Example(8): What is the Eº of the cell made from Cl2, Br2, Clˉ, and Brˉ? 71. V. ELECTROCHEMISTRY / D. Effect of Concentration on Cell Potential _D. Effect of Concentration on Cell Potential1. LeChatelier’s PrincipleExample(1): The standard voltage for the cell Zn| Zn+2 (1M) || Cu+2 (1M) |Cu is +1.10 volts. If the concentration of the Cu+2 is increased to greater than 1M, what happens to the voltage of the cell?Example(2): The standard voltage for the cell Zn| Zn+2 (1M) || Cu+2 (1M) |Cu is +1.10 volts. If the concentration of the Zn+2 is increased to greater than 1M, what happens to the voltage of the cell?Example(3): The standard voltage for the cell Zn| Zn+2 (1M) || Cu+2 (1M) |Cu is +1.10 volts. a) As the cell operates, what happens to [Zn+2] and [Cu+2]. b) What happens to the voltage of the cell as it operates? 72. V. ELECTROCHEMISTRY / D. Effect of Concentration on Cell Potential _2. Nernst Equation a) ΔG = -nFEWhere n is the total number of moles of electrons transferred in the balanced redox reaction. F (Faraday) is the charge on one mole of electrons. One F is 96,5000 coulombs of charge. E is the voltage of the cell. b) ΔG = ΔGº + RT lnQ -nFE = -nFEº + RT lnQE = Eº –RT lnQ nFE = Eº –0.059 logQ nExample(4): Using the Nernst Equation calculate the voltage of the following cell:Zn| Zn+2 (0.1M) || Cu+2 (0.001M) |Cu 73. V. ELECTROCHEMISTRY / D. Effect of Concentration on Cell Potential _Example(5): Using the Nernst Equation calculate the voltage of the following cell:Zn| Zn+2 (10M) || Ag+ (0.1M) |Ag 74. V. ELECTROCHEMISTRY / E. Electrolysis _E. Electrolysis1. Cell construction and principle of operationExample(1): Draw an electrolysis cell for molten NaClExample(2): a) For the above cell, write the two half reactions. b) What is the minimum voltage required to do the electrolysis? c) Label the anode and the cathode.2. Electrolysis in an aqueous solutionWhen water is present, the water maybe electrolyzed instead of the intended ion(s).Example(3): What are the products in the electrolysis of aqueous 1M LiBr? 75. V. ELECTROCHEMISTRY / E. Electrolysis _Example(4): What are the products in the electrolysis of aqueous 1M SnF2?3. Faraday’s Law a) Faraday’s Law: The mass of a substance produced at an electrolysis electrode is directly proportional to the amount of electricity passed through the cell. b) Recall that 1 Faraday is one mole of electrons. One F carries 96,500 coulombs of charge.Example(5): If molten NaCl is electrolyzed with 1 F, what mass of Na and Cl2 will be produced?Example(6): If ZnBr2 is electrolyzed with 1 F, what mass of Zn and Br2 will be produced? 76. V. ELECTROCHEMISTRY / E. Electrolysis _ c) Charge = current x timeCurrent is measured in amperes (amps); 1 amp = 1 coulomb/secExample(7): a) How many coulombs are produced by a current of 4.00 amps for exactly 2 hours? b) How many Faradays are produced?Example(8): How many grams of Cu can be plated out of a CuCl2 solution using 10.0 amps for 3.5 hours?Example(9): How many hours are required to plate 1.5 grams of Cr from a CrCl3 solution using 2.5 amps?

64. V. ELECTROCHEMISTRY / A.Redox Reactions

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