Chemistry 110, Week 9
Chemistry 110, Week 9 CHEM 110
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This 2 page Class Notes was uploaded by Mikaela Notetaker on Sunday April 17, 2016. The Class Notes belongs to CHEM 110 at West Virginia University taught by Melissa G. Ely in Spring 2016. Since its upload, it has received 14 views. For similar materials see Introduction to Chemistry in Chemistry at West Virginia University.
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Date Created: 04/17/16
Quiz 9 Redox Reaction: A reaction where both oxidation and reduction occur at the same time. Electrons (e-) are transferred from one reactant to another. Oxidation: Loss of electrons (charge increase i.e. +2 -> +4) species is oxidized. Reduction: Gain of electrons (charge decreases i.e. +2 -> 0) species reduced. Oxidation OILRIG Is Loss of electons Reduction Is Gain of electrons Note: electrons always transferred from reactant being oxidized to that being reduced. -5 0 Gain e- Lose e- 5 In a balanced redox reaction the number of electrons lost = the number of electrons gained. Redox Reactions Must be Balanced: 1. Material balance 2. Charge balance Oxidation Number: Like a theoretical charge used to keep track of electrons during redox reaction. Rules for Oxidation Numbers: 1. Elements in molecules consisting of just that element are assigned an ox. Number of zero. a. Na (0), Fe (0), C2 (0) 2. Monatomic ions have an ox. Number equal to the ion charge. a. Cl (-1), S (-2) 3. In compound group IA metals assigned an ox. Number = +1 a. GIIA “ “ + “ “ = +2 4. In it compound Fluorine is assigned an ox. number = -1 5. In its compound Hydrogen is = +1 6. In its compound oxygen = -2 7. Apply the ox. Number rules in the given order. If an equation has two separate rules implemented in it than the rule with the lower number should be the one you use. 8. Sum of ox. Numbers is equal to the overall charge on the species. Note: for binary ionic compounds the charge of the nonmetal is the oxidation number. Ex. Fe Cl 31 A. OCl 2Cl does not have a rule, but O does) a. O(-2)Cl 2Because of rule 6 oxygens ox. Number is -2) b. -2+(2 Cl)2x = 0 (It is equal to 0 since the compound does not have a charge) c. 2x = 2 d. X = 1 e. O(-2)Cl (+1) 2 B. S 8Follows rule 1) a. S 80) -1 C. ClO (O has a rule and is equal to -2) a. X-2=-1 (It is -1 since the compound has an overall charge of -1) b. X = 1 -1 c. Cl(+1)O(-2) D. CO 2O = -2) a. X – 4 (-2 * the 2 O) = 0 b. X = 4 c. C(+4)O (22) Oxidation Number Method: 1. Assign ox. Numbers and determine species oxidized/reduced 2. Balance atoms oxidized/reduced 3. Determine the total number of electrons lost by the reactant oxidized (Reducing agent) and the total number of electrons gained by the reactant reduced (oxidizing agent). 4. Adjust coefficients of reactants and products so that the number of electrons lost = the number of electrons gained. +4 +3 +2 A. Ce + Zn -> Ce + Zn a. Ox numbers: Ce +4 (+4) + Zn (0) -> Ce +3(+3) + Zn (+2) b. Temporary Balance: All the reactants and products are balanced right now. c. Gain/Lose: i. Ce (+4) - > Ce (+3) so gain 1 e-, reduced, oxidizing agent ii. Zn (0) -> Zn (+2) so lose 2 e-, oxidized, reducing agent d. Common Multiple of Electrons: The common multiple of 1 and 2 is 2. So (1 x 2) and (2 x 1) e. Complete Equation: 2 Ce +4 (+4) + 1 Zn (0) -> 2 Ce +3 (+3) + 1 +2 Zn (+2) B. Zn + Br -2 Zn +2 + Br -1 a. Zn (0) + Br (0) -> Zn +2(+2)+ Br (-1) 2 +2 -1 b. Zn (0) + Br 20) -> Zn (+2)+ 2 Br (-1) c. Gain/Lose i. Zn (0) -> Zn (+2) so lose 2 e-, oxidized, reducing agent ii. Br (0) -> Br (-2) so gain 2 e-, reduced, oxidizing agent d. The common multiple of 2 is 2 so 2 x 1 +2 -1 e. 1 Zn (0) + 1 Br (2) -> 1 Zn (+2)+ 2 Br (-1)
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