Oxidation Numbers

C12-1-09 REACTIVITY AND OXIDATION NUMBERS OUTCOME QUESTION(S): C12-1-09 REACTIVITY AND OXIDATION NUMBERS Outline the development of our understanding of oxidation and reduction reactions. Develop an activity series and predict spontaneity of reactions. Determine the oxidation numbers for atoms in compounds and ions. Identify reactions as redox or non-redox and identify species involved. Vocabulary & Concepts Oxidizing agent Reducing agent 

Balancing REDOX reactions: Both atoms and electrons must be balanced in redox reactions: gains 1e- x 2 atoms = 2e- oxidation numbers: +1 -1 Cu + Cl2 → Cu+ + Cl- 2 2 2 loses 1e- x 2 = 2e- Balancing is done by assigning oxidation numbers to all elements

Oxidation number represents the charge the atom would have if every bond were ionic. Pretend all bonds are give/take like an ionic bond, consider the electronegativity for each element and write the charges -1 -1 +2 MgCl2 Cl is higher EN - Mg loses 2 electrons - Each Cl gains 1 electron Cl  Mg (ionic) -2 S O  2- -2 -2 +6 SO42- O is higher EN - Each O gains 2 electrons - S loses 6 electrons (covalent) -2

Mg (s) + Cl2 (g) → MgCl2 (s) Mg → Mg2+ + 2e- Cl + 1e- → Cl1- **There is a written difference between ion charge and oxidation number** Mg (s) + Cl2 (g) → MgCl2 (s) +2 -1 Mg → Mg2+ + 2e- Cl + 1e- → Cl1- Magnesium ion ion charge: 2+
 oxidation #: +2 Chlorine ion ion charge: 1-
 oxidation #: -1 Sometimes these numbers are the same (like above) sometimes they are very different – which is why we write them differently

Rules for assigning Oxidation Numbers 1. Assign known numbers first, calculate others. Any element in its “natural” state – zero Cu O2 All metal ions – number equals ion charge Fe3+ +3 CuCl2 + 2 *Use reverse criss-cross to find metals charge* Hydrogen – usually +1 Oxygen – usually -2 Fluorine – is -1 Except metal hydrides (NaH) Except peroxides (H2O2)

K2Cr O7 2 2. Neutral compounds: Sum of O# must equal zero Assign O#s for each atom in K2Cr2O7 Step 1: Start with atoms which are known. O: –2
 K: +1 Step 2: Solve for other atoms. +1 +6 -2 Multiple the O#s by the subscripts for each atom to get the sum K2Cr O7 2 +2 +12 ?? -14 = 0 K = +1 Cr = +6 O = –2.

Sum of O# must equal charge of compound 3. Charged compounds: Sum of O# must equal charge of compound Assign O#s for each atom in NO31- Step 1: Start with atoms which are known. O: –2
 Step 2: Solve for other atoms. +5 -2 NO31- +5 ?? -6 = -1 N = +5 O = –2.

Assign oxidation numbers and determine if the following reaction is a redox reaction. +4 -2 +1 -2 +1 +4 -2 SO2 + H2O → H2SO3 +4 -4 +2 -2 +2 +4 -6 Remember: Redox involves the transfer of electrons – no change in oxidation numbers during the reaction means not a redox reaction NOT a redox reaction

Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) Show that the following is a redox reaction and list the element oxidized and reduced. +1 +5 -2 +1 +5 -2 Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) +1 +5 -6 +1 +5 -6 Oxidation cannot occur without reduction Redox reaction • Cu – oxidized (loss of electrons)
 • Ag – reduced (gain of electrons)

Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) Oxidizing agent – compound causing the oxidation of another substance. Reducing agent - compound causing the reduction of another substance. Think opposites: if the compound is reduced (gains) it must have been the “oxidizing agent” of the reaction and vice versa +1 +1 Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) • Cu – oxidized Cu – reducing agent • Ag – reduced AgNO3 – oxidizing agent

2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) Identify the substance oxidized and reduced and the oxidizing and reducing agent. Substance - means element Agent – means compound +1 +5 -2 +1 -2 +2 -2 +1 -2 2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) +1 +5 -6 +2 -2 +2 -2 +2 -2 S – oxidized H2S – reducing agent N – reduced HNO3 – oxidizing agent

6 electrons transferred (Number lost and gained have to match!) How many electrons are transferred in this reaction? R: gains 3e- x (2 atoms) = 6e- gained +1 +5 -2 +1 -2 +2 -2 +1 -2 HNO3(aq) + H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) 2 3 O: loses 2e- x (3 atoms) = 6e- lost Stoichiometry (coefficients and subscripts) used to determined total electrons transferred 6 electrons transferred (Number lost and gained have to match!)

C12-1-09 REACTIVITY AND OXIDATION NUMBERS CAN YOU / HAVE YOU? C12-1-09 REACTIVITY AND OXIDATION NUMBERS Outline the development of our understanding of oxidation and reduction reactions. Develop an activity series and predict spontaneity of reactions. Determine the oxidation numbers for atoms in compounds and ions. Identify reactions as redox or non-redox and identify species involved. Vocabulary & Concepts Oxidizing agent Reducing agent