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Electrochemistry : Oxidation and Reduction

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1 Electrochemistry : Oxidation and Reduction
Electrochemical Reaction - Chemical reaction that involves the flow of electrons. Redox Reaction (oxidation-reduction reaction) - A reaction in which at least one atom changes in oxidation state. Reduction - Any process in which the oxidation number of an atom decreases (becomes more negative). Oxidation - Any process in which the oxidation number of an atom increases (becomes more positive). Oxidation Number - The charge that an atom would have if the compound in which it were found were ionic. (Next page is a refresher on “How to”.) To help remember oxidation and reduction, remember the following: OILRIG: Oxidation Is Loss Reduction Is Gain Types of Redox Reactions Corrosion - A type of redox reaction in which a metal is destroyed. 4 Fe(s) + 3 O2(g) 2 Fe2O3 • 3 H2O Metathesis Reaction - A reaction in which atoms are interchanged and there is no change in oxidation number. Disproportionation Reaction - A reaction in which a single reactant undergoes both oxidation and reduction Disproportionation Don't confuse this with the following definitions: Oxidizing Agent - Causes oxidation; undergoes reduction (gains electrons). Reducing Agent - Causes reduction; undergoes oxidation (loses electrons).  **In spontaneous redox reactions, stronger oxidizing and reducing agents are converted into weaker oxidizing and reducing agents. Good Oxidizing Agents Atoms, ions, and molecules with large electron affinities. e.g. F2, Cl2 Compounds with large oxidation states. WHY? - The electronegativity increases as oxidation state increases Electronegativity - The tendency of an atom to draw electrons toward itself. e.g. MnO4-, CrO42- Good Reducing Agents Active metals e.g. Na, Mg, Al, Zn Metal hydrides e.g. NaH, CaH2   H2 can act as either: Oxidizing agent when it combines with metals. Reducing agent when it combines with nonmetals.

2 Assigning Oxidation Numbers
Before we balance a Redox equation lets first refresh our memory on how to calculate oxidation numbers. Oxidation Number - The charge that an atom would have if the compound in which it were found were ionic. The rules: ) The sum of the oxidation numbers of the atoms in a molecule must be equal to the overall charge on the molecule ) To assign a number to a transition metal ion (not listed in the table below) start with the overall charge, add the total number of negative charges for oxygen (if there were four as in the case of MnO4- then you would add 8 for a total of +7 for Mn), continue until all other species listed in the table below are considered (subtract if it is a positive value.) The result is the oxidation number of the transition metal ion. 3)The most electronegative element will have a negative oxidation number. Assigning Oxidation Numbers Category Oxidation # Example 1) Neutral substances containing only a single element N2, He 2) Monatomic ions same as the charge Na+ = +1 3) Hydrogen combined with a nonmetal HBr, CH4, OH- 4) Hydrogen combined with a metal NaH, CaH2 5) Metals in Group IA Li3N, Na2S 6) Metals in Group IIA Mg3N2 7) Oxygen H2O, NO (Exceptions: H2O2, O22-) -1 8) Halogens AlF3, HCl

3 Al(s) + OH-(aq) Al(OH)-4(aq) + H2(g)
Balancing Redox equations using the Oxidation number method (Basic solution is demonstrated) Al(s) + OH-(aq) Al(OH)-4(aq) + H2(g) H2O 1. What are the reduction and oxidation pairs? a) Al(s) and Al(OH)-4(aq) (oxidized) b) ? and H2(g) (reduced) Hint: 1. The reaction is taking place in a basic, aqueous media. 2. Look for a reduction potential for H2(g) in a table. 2. Calculate the Oxidation numbers and transfer to the redox partner 3 e Al(s) + OH-(aq) Al(OH)-4(aq) + H2O 2H2O + H2(g) 2 2 x x -1e x 0 3. Mass Balance 4. Charge Balance H2O Al(s) + OH-(aq) Al(OH)-4(aq) + H2(g) H2O 2 3 6 2

4 Pb(OH)-3(aq) + OCl-(aq) PbO2(s) + Cl-(aq)
Balancing Redox equations using the Half Reaction method (Basic solution is demonstrated) Pb(OH)-3(aq) + OCl-(aq) PbO2(s) + Cl-(aq) OH- 1. What are the reduction and oxidation pairs? 2. Mass Balance both equations (add H2O to balance extra oxygens then add extra H+ to balance extra hydrogens from the added H2O) 3. Charge balance both equations (add extra e-) 4. Cancel any common terms. 5. Is the reaction taking place in a basic solution? Are there are any H+ left? add OH- to both sides. H+ and OH- will make H2O on one side. 6. Add the two half reactions and cancel any extra water. Pb(OH)-3(aq) PbO2(s) OH- + H2O + H+ + 2e- OCl-(aq) Cl-(aq) OH- H2O + 2e- + OH- + 2e- + 2 H+ + + H2O + OH- OH- H2O + Pb(OH)-3(aq) + OCl-(aq) PbO2(s) + Cl-(aq) + 2H2O + OH-


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