1. Reactions in Aqueous Solutions  a.k.a. Net Ionic Equations  Molecular Equations : shows complete formulas for reactants and products –Does not show what happens on the molecular level  Total (or Complete) Ionic Equations : All substances that are strong electrolytes (are soluble and dissociate) are written as their ions. –Some ions participate in the reaction –Some ions do NOT participate in the reaction-called spectator ions.  Net Ionic Equations : show only the ions that participate in the reaction

2. Writing Total Ionic Equations Once you write the molecular equation (synthesis, decomposition, etc.), you should check for reactants and products that are soluble or insoluble. Once you write the molecular equation (synthesis, decomposition, etc.), you should check for reactants and products that are soluble or insoluble. We usually assume the reaction is in water We usually assume the reaction is in water We can use a solubility table to tell us what compounds dissolve in water. We can use a solubility table to tell us what compounds dissolve in water. If the compound is soluble (does dissolve in water), then splits the compound into its component ions If the compound is soluble (does dissolve in water), then splits the compound into its component ions If the compound is insoluble (does NOT dissolve in water), then it remains as a compound If the compound is insoluble (does NOT dissolve in water), then it remains as a compound

3. Writing Total Ionic Equations Molecular Equation: K 2 CrO 4 + Pb(NO 3 ) 2  PbCrO 4 + 2 KNO 3 SolubleSolubleInsoluble Soluble Total Ionic Equation: 2 K + + CrO 4 -2 + Pb +2 + 2 NO 3 -  PbCrO 4 (s) + 2 K + + 2 NO 3 -

4. Net Ionic Equations These are the same as total ionic equations, but you should cancel out ions that appear on BOTH sides of the equation These are the same as total ionic equations, but you should cancel out ions that appear on BOTH sides of the equation Total Ionic Equation: 2 K + + CrO 4 -2 + Pb +2 + 2 NO 3 -  PbCrO 4 (s) + 2 K + + 2 NO 3 - (Spectator ions) Net Ionic Equation: CrO 4 -2 + Pb +2  PbCrO 4 (s)

5. Net Ionic Equations Try this one! Write the molecular, total ionic, and net ionic equations for this reaction: Silver nitrate reacts with Lead (II) Chloride in hot water. Try this one! Write the molecular, total ionic, and net ionic equations for this reaction: Silver nitrate reacts with Lead (II) Chloride in hot water. AgNO 3 + PbCl 2  Molecular: 2 AgNO 3 + PbCl 2  2 AgCl + Pb(NO 3 ) 2 Total Ionic: 2 Ag + + 2 NO 3 - + Pb +2 + 2 Cl -  2 AgCl (s) + Pb +2 + 2 NO 3 - Net Ionic: Ag + + Cl -  AgCl (s) Ag + + Cl -  AgCl (s)

6. Acid-Base Reactions  Acid: –produces hydrogen ions (H + ) in solution (Arrhenius) –proton donor (Lewis)  Base: –produces hydroxide ions (OH - ) in solution (Arrhenius) –proton acceptor (Lewis)  The reaction ALWAYS forms water and an ionic compound (sometimes aqueous).

7.  Example HNO 3 (aq) + KOH (aq)  Molecular: Total Ionic: H + (aq) + NO 3 - (aq) + K + (aq) + OH - (aq)  H 2 O ( l ) + K + (aq) + NO 3 - (aq) Net Ionic: H + (aq) + OH - (aq)  H 2 O ( l ) H + (aq) + OH - (aq)  H 2 O ( l ) Acid-Base Reactions H 2 O ( l ) + KNO 3 (aq)

8. Oxidation-Reduction Reactions  a.k.a. Redox Equations  Between a metal and a nonmetal forming an ionic compound Electron transfer occurs Electron transfer occurs Oxidation numbers: assigning an excess or deficiency in electrons for each element (the charge based on the compound). Oxidation numbers: assigning an excess or deficiency in electrons for each element (the charge based on the compound).

9. Rules for Oxidation Number (ox. #) Determination 1. The sum of the oxidation numbers add up to the charge a. all elements have an ox. # of 0 a. all elements have an ox. # of 0 b. ions of elements, ox. # is the charge ( Cl - ) b. ions of elements, ox. # is the charge ( Cl - ) c. the sum of the ox. # of a complex ion equals the charge ( CO 3 -2 ) c. the sum of the ox. # of a complex ion equals the charge ( CO 3 -2 ) 2. H is 1+ when combined with a nonmetal and 1- with a metal H 3 PO 4 CaH 2 H= 1+ PO 4 = 3- Ca = 2+ H= 1- H 3 PO 4 CaH 2 H= 1+ PO 4 = 3- Ca = 2+ H= 1-

10. 1. F is always 1-; Cl, Br, I are 1- except when combined with each other or O 2. O is 2- except when combined with F ( F 2 O ) 3. Group I is 1+ and Group II is 2+ in their compounds Rules for Oxidation Numbers (cont.)

11. Recognizing Redox Rxns.  2 HCl (aq) + Mg (s)  MgCl 2 (aq) + H 2 (q)  Net: 2 H + (aq) + Mg 2+ (aq)  Mg 2+ (aq) + H 2 0 (q)  Loss of electron = oxidation  Gain of electron = reduction “LEO the lion goes GER" “LEO the lion goes GER" reduction oxidation 1+ 0 2+ 0

12. Half Reactions  Separate the individual oxidation and reduction reactions.  Look at electron movement  Half rxn.: Mg 0 (aq)  Mg 2+ + 2 e- 2e- + 2 H +  H 2 0  Net: 2 H + (aq) + Mg 0 (aq)  Mg 2+ (aq) + H 2 0 (q)  Oxidizing agent: the one reduced (H+)  Reducing agent: the one oxidized (Mg 0 )

13. Recognition of Redox rxns.  Oxidation # changes  Reactions with oxygen  Reaction of any element (forms a new compound) Balancing  Balance by mass  Balance by charge  Balance net ionic equation

14. Fe (s) + Cl 2 (aq)  Fe 3+ (aq) + Cl - (aq) 1. Balance by mass Fe (s) + Cl 2 (aq)  Fe 3+ (aq) + Cl - (aq) Fe (s) + Cl 2 (aq)  Fe 3+ (aq) + Cl - (aq) 2. Write half reaction Fe 0 (s)  Fe 3+ + 3 e- 2e- + Cl 2  2 Cl - Example Problem : 2

15. 3. Balance by charge (want # of e- to cancel) (Fe 0 (s)  Fe 3+ + 3e-) (Fe 0 (s)  Fe 3+ + 3e-) (2e- + Cl 2  2 Cl - ) 2 Fe 0 (s) + 6e- + 3Cl 2  2 Fe 3+ + 6e- + 6 Cl - Final eqn.: 2 Fe (s) + 3Cl 2  2 Fe 3+ + 6 Cl - + 2 3