Section 13.3 Oxidation Numbers p. 583 - 595

Oxidation Numbers An oxidation number is a "fictitious" number assigned to an atom or ion to achieve "electron book - keeping".

Rules: 1) The oxidation number ( O.N. ) of elemental atoms is zero ( 0 ). Examples: Cl2(g) has two atoms of Cl in one molecule of Cl2(g). Each atom is assigned an oxidation number of zero (0). Fe(s) = 0 S8(s) = 0

2) The oxidation number ( O. N 2) The oxidation number ( O.N. ) of simple ions equals the charge of the ion. Examples: Cr 3+(aq) = + 3 S 2–(aq) = – 2 Fe 2+(aq) = + 2

3) The oxidation number ( O. N 3) The oxidation number ( O.N. ) of oxygen in compounds is – 2 , except in peroxides, where it is – 1. Examples: SO2(g): oxygen's O.N. = – 2 H2O(l): oxygen's O.N. = – 2 Na2O(s): oxygen's O.N. = – 2 *H2O2(l): oxygen's O.N. = – 1

4) The oxidation number ( O. N 4) The oxidation number ( O.N. ) of hydrogen in compounds is + 1 , except in metallic hydrides, where it is – 1. Examples: H2O(l): hydrogen's O.N. = + 1 NH3(g): hydrogen's O.N. = + 1 C4H10(g): hydrogen's O.N. = + 1 *NaH(s): hydrogen's O.N. = – 1 *CaH2(s): hydrogen's O.N. = – 1

5) The total of the ( O.N. ) of all the atoms in a molecule is zero ( 0 ).   Examples: N2O5(g) Note: The sum of O.N. from 2 nitrogens plus 5 oxygens = total charge of zero. 2 (x) + 5 (–2) = 0 2x – 10 = 0 2x = + 10 x = + 5   Therefore the O.N. of each atom of nitrogen is + 5.

P4O10(g): 4x + 10(–2) = 0 x = + 5 Note: The O.N. of halogens in compounds is usually –1.   S2Cl6(s): 2x + 6(–1) = 0  x = + 3 NH3(g): x + 3(+1) = 0  x = – 3

6) The total of the ( O.N. ) of all the atoms in a complex ion is equal to that complex ion's charge. Examples:   MnO4–(aq): Cr2O72–(aq): B4O72–(aq): C3H7Br(s): x + 4(–2) = –1  x = +7 2x + 7(–2) = –2  x = +6 4x + 7(–2) = –2  x = +3 3x + 7(+1) + (–1) = 0  x = – 2

Another example: Find the oxidation number of chlorine in NaClO3 (s) Ions are Na+ and ClO3- x + 3(-2) = -1 x = +5

Oxidation Numbers and Redox Reactions Oxidation number of an atom changes during a chemical reaction. Based on oxidation numbers: Oxidation – increase in oxidation number Reduction – decrease in oxidation number Important: In a redox reaction, both oxidation and reduction occur, therefore both, the increase and the decrease of the oxidation numbers have to be present.

Example 1. (Sample problem 13. 9, p Example 1. (Sample problem 13.9, p. 587) Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)

Example 2. (Sample problem 13. 10, p Example 2. (Sample problem 13.10, p. 587) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Example 3. (Communication example 2, p. 588) 2Cr2O72-(aq) + 16H+(aq) + 3C2H5OH(aq) → 4Cr3+(aq) + 11H2O(l) + 3CH3COOH(aq)

Activity: Questions #1 – 5, p.585 Practice Questions # 6 – 9, p. 589

Balancing Redox Equations Using Oxidation Numbers Steps in Balancing Redox Equations: Assign oxidation numbers and identify the atoms/ions whose oxidation numbers change. Using the change in oxidation numbers, write the number of electrons transferred per atom. Using the chemical formulas, determine the number of electrons transferred per reactant. (Use the formula subscripts to do this.) Calculate the simplest whole number coefficients for the reactants that will balance the total number the total number of electrons transferred. Balance the reactants and products. Balance the O atoms using H2O(l), and then balance the H atoms using H+(aq).

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Activity: Practice Questions #11 – 15, p Activity: Practice Questions #11 – 15, p. 593 Section Review # 1 – 12, p. 595