Balancing_Redox_Reaction_Review_Sheet

Balancing_Redox_Reaction_Review_Sheet - Oxidation Numbers...

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Oxidation Numbers and Balancing Redox Reactions Review I. Oxidation Numbers (Oxidation States): Read page AP5 in Appendix D in your textbook. For any given compound or molecule you should be able to determine the oxidation number. 1. Elements by themselves have oxidation numbers (ON) of 0. Example: Cu (s) and Cl 2 (g) both have ON = 0 Note: Chlorine is present naturally as a diatomic gas, Cl 2 . 2. When you are given a compound follow the below rules for determining the oxidation number of each species in the compound. a. The oxidation number of hydrogen (H) is almost always +1. b. The oxidation number of oxygen (O) is almost always -2. Example: What are the oxidation numbers of oxygen and hydrogen in OH - ? ON of O = -2 and ON of H = + 1. Notice that there is a net -1 charge on this molecule. This makes sense because (-2) + (+ 1) = -1. c. DO NOT confuse O and O 2 . Oxygen (O 2 ) is present as a diatomic gas, like Cl 2 so its oxidation number is 0. d. The oxidation number of alkali metals ( Group I : Li, Na, K, Rb, Cs, Fr) almost always have an oxidation number of +1. Example: In NaCl, sodium (Na) has an ON of + 1 e. The oxidation number of alkaline earth metals ( Group II : Be, Mg, Ca, Sr, Ba, Ra) are almost always +2 Example: In CaCl 2 , calcium (Ca) has an ON of +2 f. The oxidation number of halogens ( Group 7 : F, Cl, Br, I) are usually -1. Example: In NaF: ON of Na is +1 and ON of F is -1 Note: F 2 , Cl 2 , Br 2 , and I 2 are diatomic gases and all have ON = 0. If you use the Periodic Table, there is nothing to memorize here. 3. Oxidation Numbers of Polyatomic Species: The following are common polyatomic species that you should be familiar with. For the below species you can use the above rules to come up with the oxidation number of the ion.
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CN - SO 4 2- HSO 4 - CO 3 2- HCO 3 - OH - PO 4 3- HPO 4 2- H 2 PO 4 - 4. The sum of the oxidation numbers of each atom in a molecule must equal the charge of the molecule. Example: CaF 2 1 Ca = + 2 ON 2 F = ( 2 x -1) = -2 ON Net Charge = 0 If we look at CaF 2 we can see that there is no net charge on this molecule. Example: SO 4 2- This molecule has a net -2 charge so when I add the oxidation numbers I should have a net -2 charge. At this point we realize that we are not sure what the oxidation number of sulfur should be. This is not a problem because we know the ON of oxygen and we know it must have a net -2 charge so we can determine the ON of S. 1 S = ? Call this “+ X” 4 O = ( 4 x -2) = -8 Net Charge = - 2 (+ X) + (-8) = -2 so X must be + 6. The ON of S is + 6. NOTE:
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This note was uploaded on 11/03/2009 for the course CHEM 241 taught by Professor Tiani during the Spring '08 term at UNC.

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Balancing_Redox_Reaction_Review_Sheet - Oxidation Numbers...

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