CHAPTER_18_SUPPLEMENTS_I

CHAPTER_18_SUPPLEMENTS_I - x =-1 So x = 6 Other results are...

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CHAPTER 18 SUPPLEMENTS I Periodic Table and oxidation state Composition of Ionic Compounds
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Oxidation State Rules Examples of the application of the rules given below. If 2 rules are in conflict, the rule with the smaller number wins or takes precedence. This means the O rule is the least important rule. 1. What is the oxidation state of F in F 2 ? Answer 0 (rule 1) 2. What is the oxidation state of O in H 2 O 2 Answer -1 (twice the charge on the H (+1) + twice the charge on O (x) = the charge on the compound (0) , that is, (2)1 + 2x = 0. So, x = -1.) 3. What is the oxidation state of S in HSO 4 1- ? Answer 6 (the charge on the H (+1) + 4 times the charge on O (-2) + the charge on the sulfur (x) = the charge on the compound (-1) , that is, 1 + 4(-2)
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Unformatted text preview: + x =-1. So, x = 6.) Other results are given with the next question Examples Using Oxidation State Rules Oxidation State by Analogy Oxidation and Reduction Balancing Half Reactions If a half-reaction is multiplied by a number, each part of the reaction is multiplied by this number. For example, if the reduction half reactions use up 10 electrons then the oxidation half reaction must produce 10 electrons. Here the oxidation of nickel produces 2 electrons Ni(s) → Ni 2+ (aq) +2e-So, it is multiplied by 5 to get 10 electrons: 5(Ni(s) → Ni 2+ (aq) +2e-) = 5Ni(s) → 5Ni 2+ (aq) +10e-Balancing Half Reaction Example Balancing Half Reaction in Basic Solution...
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