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Download 4.6 Oxidation-Reduction (Redox) Reactions Oxidation Reduction
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4.6 Oxidation-Reduction (Redox) Reactions Oxidation Reduction Reactions Oxidation-Reduction Reactions are commonly called Redox reactions and are those involving the oxidation of one chemical species with simultaneous reduction of another species. Oxidation and reduction must occur together. They cannot exist alone Is a reaction in aqueous solution involving an exchange of electrons between two species oxidation - the loss of one or more electrons by a substance reduction - the gain of one or more electrons by a substance. e.g., Zn (s) + 2H+ (aq) Zn (s) + 2H+ (aq) Zn2+ (aq) + H2 (g) Zn2+ (aq) + H2 (g) can be broken down into two half reactions: oxidation (loss of e-): Zn (s) Zn 2+ (aq) + 2e- reduction (gain of e-): 2H+ (aq) + 2e- H2 (g) 1 Oxidation Number can help identify a redox reaction by looking for change in oxidation number of an element in a compound oxidation number provides a measure of whether the atom is neutral, electron-rich or electron-poor comparing the oxidation number of an atom before and after reaction allows us to determine whether the atom has gained or lost electrons convenient way to keep track of electrons in a redox reaction Assigning Oxidation Numbers: All atoms have an oxidation number regardless of whether it carries an ionic charge. 1. An atom in its elemental state has an oxidation number of zero. 2. An atom in a monatomic ion has an oxidation number identical to its charge. 2 --Assigning Oxidation Number 3. Hydrogen, oxygen and halogens in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. Hydrogen has an oxidation number of +1 BUT in hydrides , hydrogen is 1 (i.e, NaH) Oxygen usually has an oxidation number of 2. BUT in peroxides, oxygen is 1 (i.e., H2O2) Halogens usually have an oxidation number of 1. BUT when bonded to oxygen, they have positive oxidation numbers (i.e, ClO2,,, BrO3- ) --assigning oxidation number 4. The algebraic sum of the oxidation numbers in a neutral compound must be zero. for H2SO4 2(+1) + (?) + 4( 2) = 0 (net charge) oxid. # of S = ? = 0 5. 2(+1) 4( 2) = +6 in a polyatomic ion, the sum must be equal to the ion charge for ClO4 (?) + 4( 2) = 1 (net charge) Oxid. # of Cl = ? = 1 4( 2) = +7 PROBLEM: assign an oxidation # for each atom in the following (a) Cr2O72- (b) Fe SO3 (c) AlH3 3 Other examples ? ? ? ? Identifying Redox Reactions First determine oxidation numbers of each species in the reaction and then identify the oxidation and reduction processes A. Oxidation and reduction occur together. Whenever an atom loses electrons (is oxidized) another atom must gain those electrons (be reduced). B. Reducing Agent- the substance that causes reduction to occur. loses one or more electrons and undergoes oxidation oxidation # of atom increases metals act as good reducing agents [ Na (s) , Fe (s) , Ca (s)] - Group 1A metals give up one electron to become monopositive ion (Na+) - Group 2A metals give up two electrons to become a dipositive ion (Ca2+) - Group 3A give up three electrons (Al3+) ; - Transition metals give up a variable number of electrons (hard to predict. C. Oxidizing Agent- the substance that causes oxidation to occur gains one or more electrons and undergoes reduction oxidation # of atom decreases reactive non-metals act as oxidizing agents [ Cl2 (g), O2 (g) ] 4 Take, for example, the following reaction: eg.4 Fe (s) + 4 Fe (s) 3 O2 (g) + 2 Fe2O3 (s) 3 O2 (g) 2 Fe (s) Identify what is oxidized, what is reduced. EXAMPLE: Identify the species oxidized, the species reduced, and the oxidizing and reducing agents in this net ionic equation. 16 H+ (aq) + 5 Sn2+ (aq) + 2 MnO4- (aq) Oxid. # +1 +2 2 Mn2+ (aq) + 5 Sn4+ (aq) + 8 H2O (l) +7 -2 +2 +4 +1 -2 reduction oxidation Here, Sn is being oxidized . Sn2+ is the reducing agent Mn is being reduced. MnO4- is the oxidizing agent PROBLEM: Identify the species oxidized and reduced and the oxidizing and reducing agent in the following net ionic equation: 3 SO42- (aq) + 12 H+ (aq) + 2 Al (s) Answer: 2 Al3+ (aq) + 3 H2SO3 (aq) + 3 H2O (l) S is being reduced; SO42- is the oxidizing agent Al is being oxidized; Al (s) is the reducing agent 5 This document was created with Win2PDF available at http://www.daneprairie.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only.