44510821-Midterm-Three-Study-Guide

44510821-Midterm-Three-Study-Guide - Chapter 21 and 23...

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Unformatted text preview: Chapter 21 and 23 Study Guide Midterm #3 Study Guide Chapter 21 Redox Reductions and Electrical Chemical Cells Process Oxidation y One reactant loses electrons y The reducing agent is oxidized y Oxidation numbers increase Reduction y Other reactant gains electrons y The oxidizing agent is reduced y Oxidation numbers decrease Zn(s) + 2H+ Zn+2(aq) + H2(g) Zinc loses its electrons and becomes the reducing agent and is oxidized. The oxidation number of zinc increased from 0 to +2. Hydrogen gains electrons and the hydrogen ion becomes the oxidizing agent and is reduced. The number of electrons for hydrogen decreases from +1 to 0. I. Review of Redox Equations A. Leo the lion says GER or OIL RIG 1. The loss of electrons is oxidation and the gaining of electrons is reduction, be sure to know and keep track of the changes during the balancing equations because one small error could cause a real problem in the balancing. 1. When balancing, the number of electrons gained and loss must always be equal so that the charges balance out and you can ultimately balance out the full equation. II. Balancing Redox Equations in Acidic Solutions A. Balance the redox reaction between the dichromate ion and the iodide ion to form chromium (III) ions and solid iodine. 1. Write the skeletal system first. Cr2O7-2(aq) + I- Cr+3(aq) + I2(s) [acidic solution] 2. Divide the reaction into half reactions in which contains both the oxidized and reduced forms of the species. Cr2O7-2 Cr+3 I- I2 3. Balance out the atoms and charges in each of the half reactions. We use water to balance out the O atoms, H+ to balance out the H atoms, and the e-s to balance out the positive charges. For the first half-reaction (Cr+3): Steps to Solving 1. Balance atoms other than O and H. We balance the two Cr on the left with a coefficient of two. Reaction Process Cr2O7-2 2 Cr+3 2. Balance the O atoms by adding water molecules. Each water has on O atom, so we added seven water on the right to balance the seven O in the Cr2O7-2. 3. Balance the H atoms by adding H+ ions. Water contains two hydrogen atoms, and since there are 14 hydrogen atoms on the right, we have to add 14 H on the left to balance those out. Cr2O7-2 2 Cr+3 + 7H2O 14H+ + Cr2O7-2 2 Cr+3 + 7H2O 4. Balance the charges by adding electrons. Each H ion has a +14 charge, and 14 H+ plus the Cr2O7-2 gives us +12 on the left side. The two Cr+3 give us a +6 charge on the right so we have to add six electrons (e-) on the left to balance it out. 6e- + 14H+ + Cr2O7-2 2 Cr+3 + 7H2O For the second half-reaction (I2): Steps to Solving 1. Balance atoms other than O and H. We balance the I on the right with a coefficient of two. Reaction Process 2 I- I2 2. Balance the O atoms by adding water molecules. Not needed, there aren¶t any O molecules. 2 I- I2 3. Balance the H atoms by adding H+ ions. Not needed, there aren¶t any H molecules 2 I- I2 4. Balance the charges by adding electrons. To balance out the -2 on the left, we add two electrons to the right to balance it out. 2 I- I2 + 2 e- 4. Multiply each of the half reactions by an integer (if necessary) so that the number of electrons is equal to the number of electrons gained. Since two electrons were loss in the oxidation and six were gained in the reduction, we multiply by 3 the iodine half reaction because it¶s the lowest common factor between 2 and 6. The same method is used to balance a redox equation in a basic solution. 3 (2 I- I2 + 2 e-) 6 I- 3 I2 + 6 eHowever, instead of H+ you will be using OH- instead. The rest is the same. 5. Add the half reactions together and cancel any substances that appear on both sides and include the states of matter (in this example, only the electrons cancelled). Also, be sure to check that the final equation is also balanced. 6e- + 14H+ + Cr2O7-2 2 Cr+3 + 7H2O 6 I- 3 I2 + 6e6I- (aq) + 14H+ (aq) + Cr2O7-2(aq) 3I2 (s) + 7H2O (l) + 2Cr+3 (aq) B. Things to Keep in Mind 1. Any redox reaction can be treated as the sum of a reaction and an oxidation reaction. Atoms and charges are conserved in each half-reaction. 2. Electrons lost in one half-reaction are gained in the other. 3. If a reaction has electrons on the left, then it¶s a reduction if it¶s written on the right side, then it¶s most likely an oxidation reaction. With the above sample, Iodine was the reduction and the Chromate was the oxidation. III. Electrical Cells A. Galvanic Cells (Voltaic Cells) 1. Uses a spontaneous reaction (¨G < 0) to generate electrical energy and the difference in chemical potential energy between higher energy reactions and lower energy products is converted into electrical energy. (i) The energy is converted to operate things (CD players, car starter motors, or other electrical voice). The system does the work on the surroundings and all batteries contain voltaic cells. (ii) Ecell becomes positive in these reactions. Be sure you are able to draw the relations of the reaction constant, with ¨G, and the voltaic cell charges. j ¨G0 = -RTln (K) j ¨G0 = (-n) (F) (E0cell) (iii)The anode is the oxidation side of the reaction and the cathode is the reduction side (an ox and red cat). +2 Zn, Zn The direction of the electrons will Cu, Cu move towards the cathode because it gains the electrons. Follow the red arrow!! B. Electrolytic Cell 1. Uses electrical energy to drive a non spontaneous reaction (¨G > 0). In this reaction, electrical energy from an external power supply converts lower energy reactants into higher energy products. (i) The surroundings do work on the system. Electroplating and the act of recovering metals involve the use of electrolytic cells. C. Properties 1. Voltaic cells occur when energy is released from spontaneous redox reactions and electrolytic cells are the ones that are absorbed to drive a non-spontaneous redox reaction. 2. Electrodes are defined as either anodes or cathodes and are dependent on half reaction conditions: (i) The oxidation half-reaction occurs at the anode. Electrons are lost by the substance being oxidized (the reduction agent) and they leave the cell at the anode. (ii) The reduction half-reaction occurs at the cathode. Electrons are gained by the substance being reduced (the oxidizing agent) and enter the cell at the cathode. Refer to Figure 21.2 on page 928 of the book. +2 One Example of a Voltaic Cell: 1. Cu+2, Cu and Zn, Zn+2 a. Find the half reactions of the pairs. Which represents the gaining of electrons and which is the loss of electrons? (i) Cu+2 (aq) + 2e- Cu (s) E0 = + 0.34V Reduction +2 0 (ii) Zn (aq) + 2e Zn (s) E = - 0.76V Reduction j Tip: Electrons on the left side signals a reduction reaction, if it¶s written on the right side, it¶s most likely an oxidation reaction. b. Which half reaction is more favored? (i) The copper is more favored because it¶s further to the right on the number line. The more positive it is, the more favored it is. 0.34 > -0.76. Therefore, the more favored reduction is the copper ions and zinc wants to be oxidized more. (See figure 1.1) (ii) Reverse the Zinc equation, since it wants to be oxidized instead. Zn (s) Zn+2 (aq) + 2eE0 = + 0.76V c. Find the cell potential! Be sure to keep track of which is reduction and oxidation so your calculations don¶t get off track. E0cell = E0cathode + E0annode E0cll = 0.34V + (0.76V) = 1.10 V d. Write the overall reaction by combining the two reactions. Cu+2 (aq) + Zn (s) Cu (s) + Zn+2 (aq) E0 = + 1.10 V *the electrons cancel out because you have 2 on the reactant and 2 on the products side. The fact that it came out a positive voltage, then the reaction is spontaneous. Had they been different, you would have to multiply both sides in order for them to cancel out by the lowest common factor (LCF). (i) IE: If copper had 3 electrons and zinc had 2, you would multiply the copper by 2 and the zinc by 3 to establish a LCF of six to cancel out the electrons. e. Figure out ¨G0, note that this is also the number of maximum amount of work you can find out of this voltaic cell because it is the measure of free energy. ¨G0 = (-2 mols) (96500 c/mol e-) (1.10 V) ¨G0 = -2.12 x 105 J/mol e- or -212 KJ/mol eTo find n, it is the moles of electrons that are being transferred. If you look at the number of electrons, copper uses two moles of electrons. So, n = 2. f. Find the equilibrium constant. ¨G0 = -RTln (K) -212 KJ = - (8.314/1000) (298K) ln (K) You have to divide the R constant by 1000 to convert it into kilojoules because the R constant is in joules, not KJ and ¨G0 is in KJ, so we must follow units. Remember, K does not have any units. K = e (-212KJ)/(-8.314/1000)(298K) K = 1.45 x 1037 Chapter 23 Coordination Compounds I. Review & Properties A. Lewis Acids & Bases Definition 1. An acid will donate a pair of electrons and bases will accept a pair of electrons in an acid and base reaction. Lewis bases are ligands because they are able to accept more than one pair of electrons and cause coordinate covalent binding. a. Coordinate covalent binding is when electrons bound are shared from the same entity or species. B. Properties of Transition Metals 1. All transition metals except for zinc, scandium, yttrium, cadmium, and lanthanum will give us colored complexes. Why? Because they have no electrons for promotions. Zinc and Cadmium will go to +2 with a D10 orbital that is completely full²this would be the same if it was Cu+1 instead of Cu+2. a. Electron Configuration (Chapter 8) 2. Transition elements have electrons available for promotions (the insides) and the outsides do not, that is why there is no color for the outside ones with the D shell full or completely empty. This causes the transition elements to have multiple oxidation states. a. When we bring ligands inside, we will have electron interactions (coming from the x, y, and z axes). This causes octahedral splitting when the ligands come in one the axes, when the ligands are brought in along the axes, there will be less interactions with D orbital that lie between the axes [D0 = Octahedral splitting energy]. (i) Octahedral is a d2sp3 hybridized orbital NOT sp3d2 because we deal with the lower, not the higher. And so the D orbital will come up first because it is the lowest available empty orbital. b. When ligands are brought in on the spaces in between the axes, this is called tetrahedral splitting [T0]. The three axes (ligands towards the spaces in between) between go high and the two ligands on the two axes go low. C. High & Low Spin Complexes 1. Dependent on the electron field of the ligand, strong field ligands will have a low spin complex (IE: The octahedral splitting is high). If we have a low field ligand, there will be a high spin complex. a. Water is a weak field ligand and always produces a high spin complex. Cyanide has a strong field ligand and will always have a low spin complex. b. Magnetic Susceptibility: Unpaired electrons will be paired in the electrical field and the paired will become unpaired. This allows us to see if it is a low spin or high spin complex. (i) These spin properties cause a series called spectrochemical series. j F-, Cl-, Br-, and I-, < C2O4-2 < H2O < NH3 = en They possess a small ¨0 and they are high spin. Large paramagnetic properties j < NH3 = en < phen < CN- < CO They possess a large ¨0 and have a low spin. Smaller paramagnetic properties II. Naming Coordination Complexes A. Common Prefixes 1. In this chapter, we won¶t be going over the deca but everything behind it is assumed to be remembered. B. When you list ligands you list them alphabetically. You change the ±ide to ±o. a. ClO4- = perchlorato b. Cl- = chloro C. More than one ligand 1. [Fe(C2H8N2)3]2f a. For multiple complex ligands we will use ±bis (2), --tris (3), -tetrakis (4), -pentakis (5), -hexakis (6) D. Be sure to know your nomenclature, here are some rules for chapter 23: 1. Fe(H2O)6Cl2 ± hexa aqua iron III chloride 2. K4[Fe(CN)6] ± potassium hexa cyanoate a. Instead of calling H2O water, you call it aquo. NH3 is an ammine and if it¶s complexed with a CO, then it¶s a carbonyl. Practice Names and Formulas 1. [Cu(NH3)4]SO4 tetraamminecopper(II) sulfate 2. [Co(en)2(H2O)Cl]Cl aquobisethylenediaminechlorocobalt(II) chloride *en = Ethylene Diamine , the non complexed Cl is free to disassociate, the other is not 3. [Pt(py)4][PtCl4] tetrakispyridineplatinum(II) tetracholoplatinate(II) *py = Pyridine , first determine the charge of platinum (Pt) 4. [Co(NH3)5Br]SO4 [Co(NH3)SO4]Br  This is a coordination isomer, they are composed of the same compounds but the structure is different. The first has a dark violet color and the second one has a red color. Write the Equation 1. Aquochlorobis(ethylenediamine)cobalt(III) chloride [Co(H2O)(en)2Cl]Cl2 For both of these chapters it¶s essential to know the naming/nomenclature of your elements because Prof. Webster expects students to know the nomenclature on the ³ion sheet´ he gave us which can be found on iLearn. Also, the two examples for the chapter 21 in this study guide can be found in the book if further clarification is needed and please continue reading in the book! This is just a guide on what you SHOULD want to cover, not what is going to be on the test. Good luck on the midterm. ...
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This note was uploaded on 02/07/2011 for the course CHEM 1C taught by Professor Webster during the Spring '11 term at UC Riverside.

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