POWER POINT PRESENTATION ON OXIDATION NUMBERS Redox Chemistry

POWER POINT PRESENTATION ON OXIDATION NUMBERS

Redox Chemistry Redox means reduction-oxidation reactions. Reduction - when an atom gains electrons (thereby “reducing” the oxidation number of that element). Oxidation - when an atom loses electrons (thereby increasing the oxidation number of that element).

OIL RIG O - Oxidation I - Is L - Losing electrons R - Reduction I - Is G - Gaining electrons

What are Oxidation Numbers? Oxidation Numbers are the charges that an atom or ion has or “appears” to have. Example: H 2 O Each H has a +1 and the O has a -2 Total = 2 (+1) + -2 = zero charge on the molecule

RULES FOR OXIDATION 1. The oxidation # of a free element is zero. Example: Na 0, C 0, He 0, O 20, Cl 20 Note that some elements are naturally diatomic.

RULES FOR OXIDATION 2. The oxidation # of a simple ion is its charge. Cl-1 is -1 O-2 is -2 Na+1 is +1

RULES FOR OXIDATION 3. Group 1 metals found in compounds have an oxidation # of +1. Example: in Na. Cl the Na is +1 (Na+1 Cl -1) Group 2 metals found in compounds have an oxidation # of +2. Example: in Mg. O the Mg is +2 (Mg+2 O-2)

RULES FOR OXIDATION 4. Hydrogen, found in compounds, will have an oxidation # of +1, except when combined with metals to form metal hydrides; then their oxidation # is -1. Examples: H 2 O the H is +1 (H+12 O -2) HF the H is +1 (H+1 F -1) Na. H the H is -1 (Na+1 H -1)

RULES FOR OXIDATION 5. Oxygen, found in compounds, will have an oxidation # of -2, except in peroxides like H 2 O 2 , where it is -1 or when combined with fluorine where it is +2. Examples: H 2 O the O is -2 (H+12 O-2) H 2 O 2 the O is -1 (H+12 O-12) OF 2 the O is +2 (O+2 F-12) (Side Rule-NOTE: ) F is ALWAYS -1

RULES FOR OXIDATION 6. In molecular or ionic compounds, the sum of the oxidation #s must equal zero. Examples: H 2 O => 2 H+1 and 1 O -2 = 0. (H+12 O-2) Na. F => 1 Na+1 and 1 F -1 = 0. (Na+1 F-1)

RULES FOR OXIDATION Example: Sulfuric Acid => H 2 SO 4 2 H+1 = +2 (see rule #4) 4 O -2 = -8 (see rule #5) therefore S must be +6 so that the sum of this molecule’s oxidation #s = zero. (H+12 S +6 O-24) 2 ( +1) + ( +6) + 4 ( -2) = 0

RULES FOR OXIDATION 7. In polyatomic ions, the sum of the oxidation #s equals the charge on the ion. Example: Ammonium ion => NH 4+1 the H must be +1 (see rule #4) 4 H+1 = +4 The total charge of this ion is +1. Therefore N must have a charge of -3. 4 ( +1) + ( -3) = +1

RULES FOR OXIDATION Example: Phosphate ion => PO 4 -3 the O must be -2 (see rule #5) 4 O-2 = -8. The total charge of this ion is -3. therefore P must have a charge of +5. ( +5) + 4 ( -2) = -3

OXIDATION NUMBERS Now you try some… Find the oxidation #s of all the elements in the following substances: H 2 CO 3 Na 2 CO 3 Mg(OH)2 PO 4 -3 NO 2 -1

OXIDATION NUMBERS Answers. . . H+1 2 C+4 O-2 3 Na +1 2 C +4 O -2 3 Mg +2(O -2 H +1 ) 2 (P +5 O -2 4)-3 (N +3 O -2 2)-1 Note (+2) + 2(-1) = 0 Note (+5) + 4(-2) = -3 Note (+3) + 2(-2) = -1

OXIDATION NUMBERS That’s all folks….