Seattle Central CHEM 161 - Chapter04 Oxidation

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CHEM 161: Chapter 4 v0914 page 20 of 34 OXIDATION NUMBER: the actual or hypothetical charge of an atom in a compound if it (or OXIDATION STATE) existed as a monatomic ion – used to track changes in electron distribution in compounds and to determine electron transfer Guidelines for Assigning Oxidation Numbers 1. The oxidation number of an element in its natural form is 0. – e.g. the oxidation number is zero for each element in H2, O2, Cl2, P4, Na, etc. 2. The oxidation number of a monatomic ion is the charge on the ion. – e.g. In Na3N, the ions are Na+ and N3–, so oxidation #’s: Na = +1 and N = -3 In Al2O3, the ions are Al+3 and O2–, so oxidation #’s: Al = +3 and O = -2 3. In a compound or polyatomic ion, – Group I elements are always +1. – Group II elements are always +2. – Fluorine is always -1. – Oxygen is usually -2 (except in the peroxide ion, O22–, when O is -1) – Hydrogen is usually +1 (except when it is with a metal, like NaH or CaH2, then it is -1) 4. In a compound, the sum of all oxidation numbers must equal 0. In a polyatomic ion, the sum of all oxidation numbers must equal charge. Example: Determine the oxidation number for each element in the following: a. CrO42–: Cr: ____, O: ____ d. H2SO4: H: ____, S: ____, O: ____ b. NO3: N: ____, O: ____ e. CaCr2O7: Ca: ____, Cr: ____, O: ____ c. C2O42–: C: ____, O: ____ f. C3H8: C: ____, H: ____ Oxidation: process of losing electrons (oxidation number) Reduction: process of gaining electrons (oxidation number ) Electrons are negatively charged, so charge becomes confusing because more electrons results in a lower charge. Think of electrons the way you think of debt—more debt means lower net worth. – Thus, gaining electrons results in a lower oxidation number. In a redox reaction – One reactant Loses Electrons/is Oxidized (LEO) – Another reactant Gains Electrons/is Reduced (GER) An easy way to remember is “LEO the lion goes GER!” – The element or reactant that is oxidized is the reducing agent. – The element or reactant that is reduced is the oxidizing agent.CHEM 161: Chapter 4 Notes v0914 page 21 of 34 For each of the following reactions, 1. Balance the equation. 2. Identify the ion or element with its oxidation state in a compound that is oxidized and reduced. 3. Identify the oxidizing agent and the reducing agent. 4. Indicate the number of electrons transferred in the reaction. a. Zn(s) + AgNO3(aq)  Zn(NO3)2(aq) + Ag(s) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______, b. Al(s) + HCl(aq)  AlCl3(aq) + H2(g) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______, c. C2H2(g) + O2(g)  CO2(g) + H2O(g) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______,CHEM 161: Chapter 4 v0914 page 22 of 34 d. Ca(s) + H2O(l)  Ca(OH)2(aq) + H2(g) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______, e. H2O2(aq) + Mn(OH)2(aq)  Mn(OH)3(aq) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______, f. NO(g) + O3(g)  NO2(g) + O2(g) The reactant oxidized is ___________________, and the oxidizing agent is _______________. The reactant reduced is ___________________, and the reducing agent is _______________. The number of electrons transferred is ______,CHEM 161: Chapter 4 Notes v0914 page 23 of 34 Balancing Redox Reactions Using Half Reactions (in Acidic Solution) i. If given both the oxidation and reduction reactions, separate them into two half-reactions. ii. Balance atoms and use oxidation numbers to determine the numbers of electrons transferred. iii. Balance the oxygen atoms by adding H2O molecules to the other side of the equation. iv. Balance the number of hydrogen atoms by adding H+ ions. v. Confirm that the overall sum of charges on the left equals the overall sum on the right. Ex. 1: Use the steps above to balance each equation below. a. One of the half-reactions that occur in a lead-acid storage battery (e.g. a car battery) is shown in the following (unbalanced) half-reaction: _____ PbO2(s)  _____ Pb2+(aq) Use the steps above to balance the equation. Is the reaction above an oxidation or reduction half-reaction? oxidation reduction b. The other half-reaction is as follows: _____ Pb(s)  _____ Pb2+(aq) Balance the reaction using electrons. Is the reaction above an oxidation or reduction half-reaction? oxidation reduction c. Now, combine the two half-reactions, cancelling the electrons, and write the overall reaction below: Ex. 2. Consider the following unbalanced reaction: Cr2O72-(aq) + I(aq)  Cr3+(aq) + I2(aq) a. Indicate the oxidation number for each atom above, then separate it into half-reactions below, and finish balancing using steps i to iv above. b. Balance the electrons transferred, then write the overall reaction below.CHEM 161: Chapter 4 v0914 page 24 of 34 Balancing Redox Reactions Using Half Reactions (in Basic Solution) i. Indicate the electrons transferred using the oxidation numbers. (If given both the oxidation and reduction reactions, also balance the numbers of electrons transferred.) ii. Balance the oxygen atoms by adding H2O molecules to the other side of the equation. iii. Balance the number of


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