Chemistry 141 Friday, October 6, 2017 Lecture 14

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Presentation transcript:

Chemistry 141 Friday, October 6, 2017 Lecture 14 Solution Chemistry, Part 4: Redox Reactions

Objectives for today Begin to gain chemical intuition about reactions in solution Assign oxidation states to elements in compounds Identify the elements that are oxidized and reduced in redox reactions Write half reactions and balance redox reactions Perform stoichiometric calculations with solutions

Rules for determining oxidation states The oxidation state of an atom in its elemental form is 0. The oxidation state of a monatomic (free) ion is equal to its charge. The sum of the oxidation numbers of the atoms in any uncharged compound is 0. The sum of the oxidation numbers of the atoms in a charged species (such as a polyatomic ion) is equal to the charge of the species. Within compounds, the following rules apply in order : Alkali metals have oxidation number +1 (e.g., NaCl). Alkaline earth metals have oxidation number +2 (e.g., BaCl2). Hydrogen (H) has oxidation number +1, except in compounds with alkali metals or alkaline earth metals. Fluorine (F) has oxidation number –1. Oxygen (O) has oxidation number –2, except in compounds with fluorine. The other halogens have oxidation number –1, except in compounds with fluorine or oxygen.

Oxidation – Reduction (Redox) Reactions Oxidation – Reduction reactions involve a change in oxidation states Two elements must change oxidation states electrons are transferred total number of electrons must be conserved If one element is oxidized, another must be reduced Oxidation – loses electron(s), oxidation state increases Reduction – gains electron(s), oxidation state decreases The oxidizing agent is reduced, and reducing agent is oxidized in the reaction

Displacement Reactions 2 AgNO3 (aq) + Cu (s)  Cu(NO3)2 (aq) + 2 Ag (s)

Activity Series Elements higher on the activity series are more reactive. They are more likely to exist as ions.

Balancing Redox Equations Make two half-reactions (oxidation and reduction). Balance atoms other than O and H. Then, balance O and H using H2O/H+. Add electrons to balance charges. Multiply by common factor to make electrons in half-reactions equal. Add the half-reactions. Simplify by dividing by common factor or converting H+ to OH– if basic. Double-check atoms and charges balance!

Stoichiometry with solutions How many mL of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)2 solution? Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O