Chapter 12 Redox reactions and Electrochemistry 11 Balancing Redox Equations 12 Electrochemical Cells 13 Stoichiometry in Electrochemical Cells 14 (Skip) Metals and Metallurgy 15 (Skip) Electrometallurgy Note: See course website for text book error corrections Chapter 12 p. A-63 problem 29 OFB Chapter 12 1
This entire Chapter deals with Oxidation, Reduction, electrochemistry and electron transfer Balancing Redox Reactions Practice, Practice, Practice All reactions take place in either Acid (H + ) or Basic (OH - ) conditions Very important to remember the problem you are working. Is it acidic or basic conditions? We will be adding H 2 O and H + for acid reactions or H 2 O and OH- for basic reaction in order to properly balance the reactions Practice, Practice, Practice OFB Chapter 12 2
Term Oxidation Reduction Oxidizing Agent, does the oxidizing Reducing Agent, does the reducing Substance Oxidized Substance Reduced Oxidation Number Change Increase Decrease Decrease Increase Increase Decrease Electron Change Loss of Electrons Gain of Electrons Picks Up electrons Supplies Electrons Loses Electrons Gains Electrons OFB Chapter 12 3
Step 1 Write two unbalanced halfequations, one for the species that is oxidized and its product and one for the species that is reduced and its product Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. Step 3 Balance oxygen by adding H 2 O to the side deficient in O in each halfequation Step 4 Balance hydrogen. For half-reaction in acidic solution, add H + on to the side deficient in hydrogen. For a half-reaction in basic solution, add H 2 O to the side that is deficient in hydrogen and an equal amount of OH - to the other side OFB Chapter 12 4
Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication Step 7 Check for balance OFB Chapter 12 5
Balancing Redox Reactions Dithionate ion reacting with chlorous acid in aqueous acidic conditions S 2 O 6 (aq) +HClO 2 (aq) SO 4 (aq) + Cl 2 (g) Will have: Oxidation half reaction Reduction half reaction OFB Chapter 12 6
Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product S 2 O 6 (aq) +HClO 2 (aq) SO 4 (aq) + Cl 2 (g) S 2 O 6 SO 4 Dithionate (we will see eventually that this is the oxidation reaction) HClO 2 Cl 2 Chlorous acid (we will see eventually that this is the reduction reaction) OFB Chapter 12 7
Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. S 2 O 6 2SO 4 2HClO 2 Cl 2 OFB Chapter 12 8
Step 3 Balance oxygen by adding H 2 O to the side deficient in O in each half-equation 2H 2 O+ (Need 2 Os) S 2 O 6 2 SO 4 (6Os) (8Os) 2HClO 2 Cl 2 (4Os) + 4H 2 O (Need 4 Os) OFB Chapter 12 9
Step 4 Balance hydrogen. For half-reaction in acidic solution, add H + on to the side deficient in hydrogen. For a half-reaction in basic solution, add H 2 O to the side that is deficient in hydrogen and an equal amount of OH - to the other side Dithionate ion reacting with chlorous acid in aqueous acidic conditions 2H 2 O + S 2 O 6 2SO 4 + 4H + (4 Hs) Need 4 Hs 6H + + 2HClO 2 Cl 2 +4H 2 O Need 6 Hs (2 Hs) (8 Hs) OFB Chapter 12 10
Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. 2H 2 O + S 2 O 6-2 Oxidation reaction, electrons are lost on the product side 2SO 4 + 4H + + 2e - -4 +4 Need -2 6H + + 2HClO 2 + 6e - +6 Need -6 Cl 2 +4H 2 O Zero charge Reduction reaction, electrons are gained on the reactant side OFB Chapter 12 11
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication Oxidation Reaction has 2 electrons Reduction reaction has 6 electrons Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction 3 times 2H 2 O + S 2 O 6 2SO 4 + 4H + + 2e - OFB Chapter 12 12
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication 3 times 2H 2 O + S 2 O 6 2SO 4 + 4H + + 2e - equals 6H 2 O + 3S 2 O 6 6SO 4 + 12H + + 6e- Now add both Re-Ox reactions OFB Chapter 12 13
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication 6H 2 O + 3S 2 O 6 2 6SO 4 + 12H + + 6e- 6H + + 2HClO 2 + 6e- Cl 2 +4H 2 O 6H 2 O + 3S 2 O 6 + 6H+ + 2HClO 2 + 6e- 6SO 4 + 12H + + Cl 2 + 4H OFB Chapter 12 14 2 O + 6e- 6
Step 7 Check for balance 2H 2 O + 3S 2 O 6 + 2HClO 2 6SO 4 + 6H + + Cl 2 Left side Right side H 6 H 6 O 24 O 24 S 6 S 6 Cl 2 Cl 2 If it doesn t balance, look for simple mistakes first Omitted subscripts Omitted superscripts OFB Chapter 12 15
Typical Exam Question Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction Cr 2 O 7 + H + + I - Cr +3 + H 2 O + I 3 - OFB Chapter 12 16
Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product Cr 2 O 7 + H + + I - Cr +3 + H 2 O + I 3 - Cr 2 O 7 Cr +3 I - I 3 - OFB Chapter 12 17
Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. Cr 2 2 O 7 Cr +3 3 I - I 3 - OFB Chapter 12 18
Step 3 Balance oxygen by adding H 2 O to the side deficient in O in each half-equation Cr 2 O 7 2Cr +3 + 7H 2 O (7Os) (Need 7 Os) 3I - I 3 - (No Oxygen, okay) OFB Chapter 12 19
Step 4 Balance hydrogen. For half-reaction in acidic solution, add H + on to the side deficient in hydrogen. For a half-reaction in basic solution, add H 2 O to the side that is deficient in hydrogen and an equal amount of OH - to the other side Dichromate ion reacting with iodide ion in aqueous acidic conditions 14H + + Need 14 Hs Cr 2 O 7 2Cr +3 + 7H 2 O (14 Hs) 3I - I 3 - (No Hydrogen, okay) OFB Chapter 12 20
Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. 6e - + 14H + + Cr 2 O 7 Need -6 +14 +6-2 2Cr +3 + 7H 2 O Reduction reaction 3I - I - + 2e - 3-3 -1 Need -2 Oxidation reaction OFB Chapter 12 21
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication Reduction reaction has 6 electrons Oxidation Reaction has 2 electrons Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction 3 times 3I - I 3- +2e- OFB Chapter 12 22
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication 3 times 3I - I 3- +2e- equals 9I - 3 I 3- +6 e- Now add both Re-Ox OFB Chapter reactions 12 23
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication 6e- + 14H + + Cr 2 O 7 2Cr +3 + 7H 2 O equals 9I - 3 I 3- +6 e- 6e- + 14H + + Cr 2 O 7 + 9I - 2Cr +3 + 7H 2 O + 3 I 3- + 6 e- OFB Chapter 12 24
Step 7 Check for balance 14H + + Cr 2 O 7 + 9I - 2Cr +3 + 7H 2 O + 3 I 3 - Left side H 14 Cr 2 O 7 I 9 Right side H 14 Cr 2 O 7 I 9 OFB Chapter 12 25
Basic Solution Method 1 At Step 4 Instead of adding H + on the deficient side as did with the acid solution method Add H 2 O on the H deficient side and add an equal amount of OH - on the other side. OFB Chapter 12 26
Typical Exam Question Note: this is not Example 12, which is similar Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution Ag 2 S + Cr(OH) 3 Ag 0 + HS - + CrO 4 OFB Chapter 12 27
Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product Ag 2 S + Cr(OH) 3 Ag 0 + HS - + CrO 4 Ag 2 S Ag + HS - +1 0 Cr(OH) 3 CrO 4 +3 +6 Reduction reaction Oxidation reaction OFB Chapter 12 28
Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. Ag 2 S 2 Ag + HS - Cr(OH) 3 CrO 4 Already Balanced OFB Chapter 12 29
Step 3 Balance oxygen by adding H 2 O to the side deficient in O in each half-equation Ag 2 S 2Ag + HS - (Oxygen not needed) H 2 O + Cr(OH) 3 CrO 4 (need 1O) (3Os) (4Os) OFB Chapter 12 30
Step 4 Balance hydrogen. For half-reaction in acidic solution, add H + on to the side deficient in hydrogen. For a half-reaction in basic solution, add H 2 O to the side that is deficient in hydrogen and an equal amount of OH - to the other side This problem is in aqueous basic conditions H 2 O + Needs H Ag 2 S 2Ag + HS - (1 H) + OH - (need O) 5OH - + Add OH - H 2 O + Cr(OH) 3 (2 H) (3 H) CrO 4 + 5H 2 O Needs H OFB Chapter 12 31
Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. 2e - + H 2 O + Ag 2 S Need -2 2Ag + HS - + OH - Reduction reaction -1-1 5OH - + H 2 O + Cr(OH) 3 CrO 4 + 5H 2 O -5 Oxidation reaction -2 + 3e - Need -3 OFB Chapter 12 32
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication Reduction reaction has 2 electrons 2e x 3 = 6e Oxidation Reaction has 3 electrons 3e x 2 = 6e OFB Chapter 12 33
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication times Reduction Reaction = 6e - + 3H 2 O + 3Ag 2 S 6Ag + 3HS - + 3OH - Reduction reaction times Oxidation Reaction = 10OH - + 2H 2 O + 2Cr(OH) 3 2CrO 4 + 10H 2 O + 6e - Oxidation reaction OFB Chapter 12 34
.Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H + ion, OH - ion, or H 2 O appears on both sides of the final equation, cancel out the duplication Now add both Re-Ox reactions 6e - + 3H 2 O + 3Ag 2 S 6Ag + 3HS - + 3OH - 10OH - + 2H 2 O + 2Cr(OH) 3 2CrO 4 + 10H 2 O + 6e - OFB Chapter 12 35
Now add both Re-Ox reactions 6e - + 3H 2 O + 3Ag 2 S 6Ag + 3HS - + 3OH - 10OH - + 2H 2 O + 2Cr(OH) 3 2CrO 4 + 10H 2 O + 6e - equals 6e - + 3H 2 O + 3Ag 2 S + 10OH - + 2H 2 O + 2Cr(OH) 3 6Ag + 3HS - + 3OH - + 2CrO 4 + 10H 2 O + 6e - OFB Chapter 12 36
simplify 7 6e - + 3H 2 O + 3Ag 2 S + 10OH - + 2H 2 O + 2Cr(OH) 3 6Ag + 3HS - + 3OH - + 5 2CrO 4 + 10H 2 O + 6e - 3Ag 2 S + 7OH - + 2Cr(OH) 3 6Ag + 3HS - + 2CrO 4 + 5H 2 O OFB Chapter 12 37
Step 7 Check for balance 3Ag 2 S + 7OH - + 2Cr(OH) 3 6Ag + 3HS - + 2CrO 4 + 5H 2 O Left side Ag 6 S 3 O 13 H 13 Cr 2 Right side Ag 6 S 3 O 13 H 13 Cr 2 NOTE: This problem is the reverse reaction of Example 12 OFB Chapter 12 38
Chapter 12 Redox reactions and Electrochemistry Practice Practice Practice OFB Chapter 12 39
Balancing Disproportionation Reactions Same Steps For example Cl 2 ClO 3- + Cl - Step 1 Oxidation: Reduction: Cl 2 ClO 3 - Cl 2 Cl - OFB Chapter 12 40
Chapter 12 Redox reactions and Electrochemistry Review of Common Terms Term Oxidation State Change Electron Change 1 Oxidation increase loss of electrons e.g., Cu to Cu 2+ (0 to +2) 2 Reduction decrease gain of electrons e.g., Ni + to Ni (+1 to 0) 3 Oxidizing Agent decrease picks up electrons e.g., O 2 to oxide (0 to -2) 4 Reducing Agent increase supplies electrons e.g., H 2 to H 2 O (0 to +1) 5 Substance Oxidized increase loses electrons e.g., Mg to MgO (0 to +2) 6 Substance Reduced decrease gains electrons e.g., CuO to Cu (+2 to 0) OFB Chapter 12 41
Electrochemical Cells Cu(s) + 2Ag + (aq) Cu ++ (aq) + 2Ag(s) Silver plates spontaneously G f < 0 Called a galvanic or voltaic cell Oxidation Cu Cu ++ + 2e - Reduction Ag + + e - Ag OFB Chapter 12 42
Oxidation occurs here at the anode Reduction occurs here at the cathode Cu(s) + 2Ag + (aq) Cu ++ (aq) + 2Ag(s) Chemical energy converted to Electrical Energy. This potential difference can be measured. Current = charge/time or I=Q/t Units amperes = coulombs/sec. OFB Chapter 12 43
Cu(s) + 2Ag + (aq) Cu ++ (aq) + 2Ag(s) Convention for this reaction Cu(s) Cu +2 (aq) Ag+(aq) Ag(s) OFB Chapter 12 44
If an opposing electric force is applied the reverse reaction occurs, called an electrolytic cell Cu ++ (aq) + 2Ag(s) Cu(s) +2Ag + (aq) The electrodes are reversed Ag becomes the anode Cu becomes the cathode Note that the electron flow is still anode to cathode Ag + (aq) Ag(s) Cu(s) Cu +2 (aq) OFB Chapter 12 45
Stoichiometry in Electrochemical Cells Faraday s Law The quantities of substances produced or consumed at the electrodes are directly proportional to the amount of electric charge passed through the cell. When a given amount of electric charge passes through a cell, the quantity of a substance produced or consumed at an electrode is proportional to its molar mass divided by the number of moles of electrons required to produce or consume one mole of the substances. OFB Chapter 12 46
Faraday Constant (F ) = the electric charge carried by 1 mole of electrons. Charge 1 e = 1.602X10-19 Coulombs X Avogadro s number 1 Faraday = 96,485 Coulombs of charge per mole of electrons OFB Chapter 12 47
Problem 130 What number of coulombs of electricity is produced in each of the following oxidations? 2.00 mol of Zn(s) to Zn 2+ (aq) OFB Chapter 12 48
Problem 130 What number of coulombs of electricity is produced in each of the following oxidations? 1.00 mol of H 2 O 2 (aq) to O 2 (g) OFB Chapter 12 49
Problem 130 What number of coulombs of electricity is produced in each of the following oxidations? 2.50 mol of MnO 2 (s) to MnO 4- (aq) MnO 2 MnO 4 - OFB Chapter 12 50
Problem 130 What number of coulombs of electricity is produced in each of the following oxidations? 1.70 mol of H 2 S (aq) to 0.850 mol of H 2 S 2 O 3 (aq) H 2 S H 2 S 2 O 3 OFB Chapter 12 51
Stoichiometry in Electrochemical Cells M = Molar Mass M/n molar mass per # electrons e = Ag + + e - Ag M 107.87 = 1 n e Cu Cu ++ + 2e - M n = 63.54 2 OFB e Chapter 12 52
Example Chapter 12 Redox reactions and Electrochemistry How many grams of Cl 2 can be produced by the electrolysis of molten NaCl at a current of 10.0 amps for 5 minutes? 2Cl - Cl 2 + 2e - OFB Chapter 12 53
Reminder 1 mole electrons reduces 1 mol of Ag + 2 moles electrons reduces 1 mole of Mg 2+ 3 moles electrons reduces 1 mole of Al 3+ OFB Chapter 12 54
Example (not in book) Suppose a current of 100 amps is passed through a bath of molten alumina and cryolite ( a mineral Na 3 AlF 6 ) for 10 minutes. How many grams of elemental aluminum are produced? The molar mass of aluminum is 26.90 g/mol. What is the oxidation number of Al in Na 3 AlF 6? +3 Al +3 + 3e - Al OFB Chapter 12 56
Example How long would it take to plate 5.43 grams Nickel on to an electrode from a solution of NiCl 2 at a current of 12.34 amps? Molar mass of Nickel is 58.693 g mol -1. Ni +2 + 2e - Ni OFB Chapter 12 57
Chapter 12 Redox reactions and Electrochemistry Summary Balancing Redox reactions Understand electrochemical cells and the Faraday law calculations OFB Chapter 12 58
Chapter 12 Redox reactions and Electrochemistry Examples / Exercises 11, 12, 13, 14, 15 Problems 1, 3, 7, 9, 11, 13, 29, 31, 33 Note: see website for Problem 29 correct answers OFB Chapter 12 59