Oxidation-Reduction (Redox) Reactions In oxidation-reduction (abbreviated as “redox”) reactions, electrons are transferred from one reactant to another. Oxidation I Lose electrons Reduction Gain electrons

Redox Reactions In the reaction between Na and Cl2: Na lost an electron, it has been oxidized electron (e-) Cl- Cl Cl gained an electron, it has been reduced 2 Na (s) + Cl2 (g)  2 NaCl (s)

Redox Reactions What about the reaction between Al and O2? O gained two electrons, it has been reduced O2- electrons (e-) Al Al lost 3 electrons, it has been oxidized Al3+ Al (s) + O2 (g)  Al2O3 (s) 4 Al (s) + 3 O2 (g)  2 Al2O3 (s)

Oxidation Numbers Oxidation Number (or Oxidation State): actual or hypothetical charge of an atom in a compound if it existed as a monatomic ion Common Oxidation Numbers: H+ = +1 Cl- = -1 O2- = -2 Al = 0 Na = 0 Na+ = +1 Oxidation numbers can also be assigned to atoms with in a more complex molecule.

Assigning Oxidation Numbers 1. The oxidation number of an element in its natural form is 0. Examples: 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. Examples: 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 neutral compound, the sum of all oxidation numbers must equal 0. In a polyatomic ion, the sum of all oxidation numbers must equal the charge.

Assigning Oxidation Numbers Examples: Determine the oxidation number for each element in the following:   CrO42–: Cr: ____, O: ____ H2SO4: H: ____, S: ____, O: ____ NO3-: N: ____, O: ____ CaCr2O7: Ca: ____, Cr: ____, O: ____ C2O42–: C: ____, O: ____ C3H8: C: ______________, H: ____

Redox Reactions 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!” (Though I prefer OIL RIG, it’s your choice). The element or reactant that is oxidized is the reducing agent. The element or reactant that is reduced is the oxidizing agent.

Examples Zn(s) + AgNO3(aq)  Zn(NO3)2(aq) + Ag(s) Al(s) + HCl(aq)  AlCl3(aq) + H2(g) C2H2(g) + O2(g)  CO2(g) + H2O(g)

Examples Ca(s) + H2O(l)  Ca(OH)2(aq) + H2(g) H2O2(aq) + Mn(OH)2(aq)  Mn(OH)3(aq)

Solution Concentration solution: homogeneous mixture of substances present as atoms, ions, and/or molecules   solute: component present in smaller amount solvent: component present in greater amount   Note: Unless otherwise stated, the solvent for most solutions considered in this class will almost always be water! Aqueous solutions are solutions in which water is the solvent.

How do we measure concentration? A concentrated solution has a large quantity of solute present for a given amount of solution. A dilute solution has a small quantity of solute present for a given amount of solution. SOLUTION CONCENTRATION = amount of solute amount of solvent The more solute in a given amount of solution  the more concentrated the solution Example: Explain the difference between the density of pure ethanol and the concentration of an ethanol solution.

How do we measure concentration? Concentration can be measured a number of ways: ppm (parts per million) – one part in a million parts ppb (parts per billion) – one part in a billion parts g/kg (grams per kilogram) – one gram solute per one kilogram of solvent The chemical standard most used is Molarity Molarity = moles of solute liters of solution units: M (molar = mol/L)

Solution Concentration Find the molarity of a solution prepared by dissolving 1.25 g of KOH in 150.0 mL of solution. Find the molarity of a solution prepared by dissolving 5.00 g of copper(II) sulfate in 250.0 mL of solution

Ion Concentrations When an ionic compound is dissolved in water, the concentration on the individual ions is based on their molecular formula… For example: 1 M NaCl solution contains 1 M Na+ and 1 M Cl- 2 M NaCl solution contains 2 M Na+ and 2 M Cl- 1 M CaCl2 solutions contains 1 M Ca2+ and 2 M Cl- 2 M CaCl2 solutions contains 2 M Ca2+ and 4 M Cl-

Solution Concentration Indicate the concentration of barium and chloride ions in a 1.00M barium chloride solution. Indicate the molarity of each ion in the solutions indicated below: In a 0.125M Na2SO4(aq) solution [Na+]=____________ and [SO42-]=____________. b. In a 0.500M Fe(NO3)3(aq) solution [Fe3+]=____________ and [NO3–]=___________. c. In a 1.250M Al2(SO4)3(aq) solution [Al+3]=____________ and [SO42-]=___________.

Solving Concentration Problems Keep in mind that if molarity and volume are both given, you can calculate # of moles since: volume  molarity = volume (in L)  moles of solute liters of solution   so volume units will cancel  # of moles! If you are given volume and molarity for a solution, multiply them together to get # of moles!

Solving Concentration Problems Calculate the mass of NaCl needed to make 1.00 L of a 1.00 M solution.

Preparing Solutions

Examples Calculate the mass of barium hydroxide required to make 250.0 mL of a 0.500M barium hydroxide solution. What volume (in mL) of a 0.125M silver nitrate solution contains 5.00 g of silver nitrate?

Examples Calculate the molarity of hydroxide ion in a solution prepared by diluting 50.0 mL of 1.50M potassium hydroxide with 100.0 mL of 0.500M calcium hydroxide.