Unit #8, Chapter 10 Outline Electrochemistry and Redox Reactions Lesson Topics Covered Homework Questions and Assignments 1 Introduction to Electrochemistry definitions 1. Read pages 462 467 2. On page 467, do questions 1 4 full ionic and net ionic equations 3. On page 468, do questions 5 7 half reactions 4. On page 472, do questions 1 3, 6 Oxidation and Reduction definitions oxidizing and reducing agents redox reactions Activity Series for Metals using the activity series to predict single displacement reactions 2 Oxidation Numbers the Rules using oxidation numbers to identify oxidizing and reducing agents in chemical reactions 3 Balancing Redox Reactions in Acidic Conditions the half-reaction method 4 Balancing Redox Reactions in Basic Conditions the half-reaction method 1. Read pages 476 477 2. On page 478, do questions 9 11 3. Read pages 478 480 4. On page 480, do questions 13 15 5. Complete Handout: Oxidation Numbers 1. Read pages 482 484, 486 490 (don t worry about balancing under basic conditions for now) For the questions below, there are various instructions. Please just balance the redox reactions as you have been taught using the half reaction method: 2. On page 491, do questions 27a,b,c, 28c,d 3. On page 494, do questions 2, 3, 6 4. On page 497, do questions 30 32 5. On page 498, do questions 1 3 Balance the net ionic equations below using the half reaction method under acidic and basic conditions: 1. MnO 2 + Cl Mn 2+ + Cl 2 2. NO + Sn NH 2 OH + Sn 2+ 3. Cd 2+ + V 2+ Cd + VO 3 4. S 2 O 3 2-5. Mn 2+ + BiO 3 + NiO 2 Ni(OH) 2 + SO 3 2- MnO 4 6. I 2 O 5 + CO I 2 + CO 2 7. NO 2 NO 2 + NO 3 8. P 4 H 2 PO 2 + PH 3 + Bi 3+
Unit #8, Chapter 10 Outline Electrochemistry and Redox Reactions 5 Introduction to Electrochemical Cells Galvanic cells Electrolytic cells Electrodes Electrolytes Direction of electron motion Galvanic cell notation 6 Quiz on Balancing Redox Reactions Calculating Standard Cell Potentials standard reduction potentials, Eº calculating standard cell potentials interpretting values of Eº 1. Read pages 504 508. 2. On page 509, do questions 1a, 2, 3 3. On page 515, do question 8 4. For the redox reactions in Q 5 8 on page 521 of your text: write the two half reactions taking place, calculate the Eº for the reaction, identify the anode and cathode, and identify the oxidizing agent and the reducing agent. 5. Make a labeled sketch of a simple electrochemical cell using gold and aluminum electrodes. a) Write the equations for the oxidation and the reduction half-reactions for this cell. b) On your sketch, label the anode and the cathode and indicate the direction of electron flow. c) Write the equation for the overall cell reaction and calculate the Eº for this cell. 1. On page 521, do questions 5 7 (just calculate Eº the way we did in class) 2. On page 523, do questions 1 and 2a Electrolytic Cells
Unit 8, Lesson 01: Activity Series for Metals Based on experimental evidence, metals can be arranged in an activity series, from most reactive (lose electrons most easily) to least reactive (gain electrons most easily). A neutral metal atom will be oxidized (lose electrons) by any metal ion below it on the activity series. If a neutral metal atom is below the metal ion on the activity series, no reaction will occur. Use the Activity Series to predict if a reaction will occur between the following metal atoms and ions. If a reaction will occur, complete the equations. If no reaction will occur, write NR (no reaction): 1. Calcium metal with nickel (II) nitrate solution a) Balanced chemical equation: b) Full ionic equation: c) Net ionic equation: d) Oxidation half reaction: e) Reduction half reaction: f) Oxidizing agent: g) Reducing agent: 2. Tin with zinc sulfate solution: a) Balanced chemical equation: b) Full ionic equation: c) Net ionic equation: d) Oxidation half reaction: e) Reduction half reaction: f) Oxidizing agent: g) Reducing agent: 3. Silver nitrate solution with lead: Activity Series lithium potassium barium calcium sodium magnesium aluminum zinc chromium iron cadmium cobalt nickel tin lead copper mercury silver platinum gold a) Balanced chemical equation: b) Full ionic equation: c) Net ionic equation: d) Oxidation half reaction: e) Reduction half reaction: f) Oxidizing agent: g) Reducing agent:
Unit 8, Lesson 02: Oxidation Numbers Oxidation numbers are. They are used as a tracking system to follow during chemical reactions. In a bond, the atom with the is assigned a oxidation number, because the electrons will spend more time to this atom, giving it a charge. The less electronegative atom is assigned a oxidation number. During redox reactions, the oxidation numbers of the atoms. If the oxidation numbers don t change, then the reaction is. If the oxidation number of an atom increases, the atom has been. If the oxidation number of an atom decreases ( ), then the atom has been. The following rules are used to assign oxidation numbers: 1. The oxidation numbers of all pure elements are eg. 2. For monoatomic (single atom) ions, the oxidation number is equal to the eg. oxidation number of Cl is oxidation number of Ca 2+ is oxidation number of Pb 4+ is oxidation number of P 3- is 3. The oxidation number of fluorine in a compound is ALWAYS 4. The oxidation number of hydrogen is except when it is in a metal hydride such as. In metal hydrides, because hydrogen has the it is assigned an oxidation number of. 5. The oxidation number of oxygen is except: a) in peroxides such as, when its oxidation number is b) in OF 2. Because fluorine s oxidation number in compounds is always, oxygen must be 6. The sum of the oxidation numbers of all of the atoms in a compound must equal the overall charge for the compound a) if the compound is neutral overall (uncharged) then the sum of the oxidation number is b) if the formula is for an ion, then the sum of the oxidation numbers is equal to the Examples: Find the oxidation numbers of the elements in the following compounds: a) NiCl 2 c) CrO 4 2- b) K 2 Cr 2 O 7 d) MnO 4
Identify the oxidizing and reducing agents in the following unbalanced reactions: 1. Mg (s) + N 2 (g) Mg 3 N 2 (s) 2. KClO 3 (s) KCl (s) + O 2 (g) 3. C 2 H 4 (g) + O 2 (g) CO 2 (g) + H 2 O (g) 4. Cu(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) NaNO 3 (aq) + CuCO 3 (aq) 5. H 2 O 2 (l) O 2 (g) + H 2 O (l)
Unit 8, Lesson 02: Practice with Oxidation Numbers 1. Find the oxidation numbers of the elements in bold print. a) HClO d) PbSO 4 g) Na 2 O 2 b) KClO 3 e) NaIO 4 h) K 2 SO 4 c) MnO 2 f) ClO 4 i) NH 4 1+ 2. State whether the change is an oxidation or a reduction. a) MnO 4 1 becomes MnO 4 2 d) P 4 O 6 becomes P 4 O 10 b) N 2 becomes NH 3 e) NH 3 becomes N 2 O c) O 2- becomes O 2 f) SO 4 2 becomes S 2 O 3 2 3. Identify the oxidizing and reducing agents in the following unbalanced reactions: a) I 2 + H 2 S HI + S b) Zn + HNO 3 Zn(NO 3 ) 2 + NO 2 + H 2 O c) Ag 2 O + NH 3 Ag + H 2 O + N 2 d) H 2 O + ClO 3 1 + SO 2 SO 4 2 + HCl e) K 2 Cr 2 O 7 + HBr KBr + CrBr 3 + H 2 O + Br 2 f) SnCl 2 + PbCl 4 SnCl 4 + PbCl 2 g) Sb + Cl 2 SbCl 3 h) NaI + H 2 SO 4 H 2 S + I 2 + Na 2 SO 4 + H 2 O
Unit 8, Lesson 04: Balancing Redox Reactions in Acidic Conditions Redox reactions can be extremely complex and difficult to balance, for example: NaCN + KMnO 4 + H 2 O NaCNO + MnO 2 + KOH There are many systematic approaches to balancing these reactions. We will use the half-reaction method because this seems to be what is used at most universities. we will usually work from a net ionic equation- the spectator ions do not participate so we ignore them water and its ions (H + and OH - ) are often involved, so they will be added as needed. eg. Balance this redox reaction: HSO 3 + IO 3 I 2 + SO 4 2- Step 1: Assign oxidation numbers to all atoms to determine which species are being oxidized and reduced (do this on the reaction above). Step 2: Divide the net ionic reaction into oxidation and reduction half reactions. Step 3: In the half reactions, balance all atoms EXCEPT oxygen and hydrogen (see above). Step 4: In each half reaction, balance oxygen by adding H 2 O to whatever side needs more oxygen. Step 5: In each half reaction, balance hydrogen by adding H+ to whatever side needs more hydrogen (see above). Step 6: In each half reaction, balance the charges by adding electrons to whatever side needs them. Step 7: Multiply the half reactions by the lowest common multiple (LCM) to make the number of electrons lost in the oxidation half reaction equal to the number of electrons gained in the reduction half reaction (see above). Step 8: Multiply the LCM multiple(s) through both half reactions then add both half reactions together. Step 9: Simplify the reaction by canceling out species that appear on both sides of the equation. Step 10: Double check that all atoms and charges are correctly balanced.
Summary of Steps: 1. Assign oxidation numbers 2. Write two half reactions 3. Balance all atoms except oxygen and hydrogen using species from the original reaction (if necessary) 4. Balance oxygen by adding water where needed (if necessary) 5. Balance hydrogen by adding H+ where needed (if necessary) 6. Balance charge by adding electrons where needed 7. Multiply the half reactions by a lowest common factor so the number of electrons in each is equal 8. Add the half reactions together 9. Simplify 10. Double check that all atoms and charges are balanced eg. Balance this redox reaction: Cr 2 O 7 2- + C 2 H 4 O HC 2 H 3 O 2 + Cr 3+ eg. Balance the net ionic reaction from the front of the page CN + MnO 4 CNO + MnO 2 eg. NH 3 + Br 2 NH 4 Br + N 2
Unit 8, Lesson 05: Electrochemical (Galvanic) Cells During a redox reaction, electrons are transferred from the substance being (the agent) to the substance being (the agent). For example, when a piece of zinc is placed in a solution containing Cu 2+ ions, the following reactions take place: Overall: Zn (s) Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e - Cu (s) The oxidation and the reduction occur where the zinc atoms and Cu 2+ ions are in contact with one another on the surface of the piece of zinc. The electrons from the zinc atoms to the Cu 2+ ions. In an electrochemical cell, the same reactions take place but the half-reactions are. The oxidation and reduction reactions take place at different surfaces called. Electron transfer occurs through a conducting wire that connects these electrodes, and this creates an. The salt bridge contains an solution such as potassium nitrate. This permits the flow of from one half-cell to the other and is necessary to maintain the electrical of the two solutions. The construction of a simple electrochemical cell is illustrated below. voltmeter zinc strip salt bridge copper strip Zn(NO 3 ) 2 solution Cu(NO 3 ) 2 solution Using the activity series on page 470 of your text, we can predict the direction that the electrons will flow. Zinc is reactive than copper, that is, zinc its electrons more easily. Zinc will donate electrons to the copper and electrons will flow from the half-cell toward the halfcell. will be oxidized and will be reduced. The electrode at which oxidation occurs is called the while the electrode at which reduction occurs is called the. Because electrons flow from the anode, the anode is the electrode. Because electrons flow toward the cathode, the cathode is the electrode. We can represent this cell using Galvanic Cell Notation. The anode is always written on the :
Unit 8, Lesson 06: Using Reduction Potentials to Calculate the Voltage of an Electrochemical Cell An activity series is used to compare how easily an element electrons relative to other elements. The most reactive metals lose their electrons most easily and are at the of the metal activity series. Alternatively, we can measure and compare the level of that an element has for new electrons. This is called the element s. In this series, the elements with the attraction for electrons (the highest reduction potentials) are at the of the list. Standard reduction potentials ( ) are measured for elements in their and at standard conditions, ie. in M solution, at ºC and atm pressure. They are measured against the hydrogen electrode,, which has been assigned a reduction potential of exactly V. Standard reduction potentials (Eº) are reported on page : eg. Na + (aq) + e - Na (s) Eº= V (very attraction for electrons) Cl 2 (g) + 2e - 2 Cl- (aq) Eº= V (very attraction for electrons) Fe 3+ (aq) + 3e - Fe (s) Eº= V ( attraction for electrons) If you are given two half reactions, you can calculate the voltage that will be produced when these reactions are combined in an electrochemical cell: 1. Determine which half reaction will be the anode ( electrons). The half reaction with the (more ) Eº will be the anode. 2. Write this half reaction as an and the sign on the Eº value. 3. Write the other half reaction as a, keeping the Eº as reported. 4. Balance the number of electrons so that the number of electrons at the anode equals the number of electrons at the cathode. *** Do multiply the Eº values (the strength of attraction for electrons does not increase as the number of moles goes up). 5. Add the Eº values for the half reactions. The and more the Eº for an electrochemical cell, the the voltage it will produce. eg. Calculate the voltage produced by an electrochemical cell made of Br 2 Br and Fe Fe 2+ eg. Calculate the voltage produced by an electrochemical cell made of Ni Ni 2+ and Al Al 3+