2. Curriculum Standards National: Content Standard B: Physical Science: Chemical reactions – A large number of important reactions involve the transfer of either electrons (oxidation/reduction reactions) or hydrogen ions (acid/base reactions) between reacting ions, molecules, or atoms. (no related standard) Florida: SC.912.P.8.10 Describe oxidation-reduction reactions in living and non-living systems. Florida: SC.912.P.8.8 Characterize types of reactions, for example: redox, acid-base, synthesis, and single and double replacement reactions.

3. Curriculum Objectives for 7.06 regular Objectives (includes higher-order thinking): – Define oxidation and reduction – Apply rules for oxidation numbers – Assign oxidation numbers to elements in a reaction – Create algebra expressions to solve for an unknown oxidation number – Identify oxidation and reduction half reactions – Perform lab on your own

4. Vocabulary (Literacy Skill) What is an oxidation and reduction (redox) reaction? An exchange of ___________. Real life example: – Rusting: This is a redox reaction in which oxygen oxidizes (takes electrons from) iron to form iron (III) oxide, otherwise known as rust. 4Fe (s) + 3O 2 (g) → 2Fe 2 O 3 (s) – Iron is a very strong metal, whereas rust is a brittle ionic compound that flakes off of old cars and nails. What a difference a few electrons and bonds can make!

5. Vocabulary (Literacy Skill) Oxidation---Lose electrons Reduction---Gain electrons Way to remember: “LEO the lion says GER” Lose electrons is Oxidation (LEO) Gain electrons is Reduction (GER)

6. Vocabulary (Literacy Skill) For the “agents” think of the opposite result. Oxidizing Agent: A reactant that causes another substance to be oxidized (not being oxidized itself). The oxidizing agent is the reactant that is reduced (gain electrons). Reducing Agent: A reactant that causes another substance to be reduced (not being reduced itself). The reducing agent is the reactant that is oxidized (lose electrons).

7. Learn and apply the rules for Oxidation Numbers: Neutral elements in their standard state (alone) have a charge of zero (diatomic elements such as H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 ; solid Na, etc.) Oxygen has an oxidation number of -2 except in a peroxide which is -1 (H 2 O 2 exception) Ions alone (has a charge +/-) or in a compound will use their charge from 3.05). NaCl (Na +1 and Cl -1 ) For covalent compounds, pretend the compound is ionic with the more electronegative element (top right) forming the negative ion (anion). For example: Fluorine is always -1 in a compound, oxygen is almost always -2, and hydrogen is +1 in covalent compounds. The algebraic sum of all the oxidation numbers (multiplied by any subscripts) must add up to the total charge of the compound or ion. The element that gains electrons will be reduced and the element that loses electrons will be oxidized.

8. Review of ion charges from 3.05

9. Example to find which element is oxidized and which element is reduced What changes oxidation numbers? What gains electrons and what loses electrons? Oxygen stays the same oxidation number and hydrogen stays the same (from the rules). Therefore, Chromium and Nitrogen are going to be our focus (one is oxidized and one is reduced). Cr 2 O 7 2- (aq) + HNO 2 (aq) → Cr 3+ (aq) + NO 3 - (aq)

10. Split the reaction into half reactions and determine oxidation numbers Break up into half reactions (similar elements on each side). Cr 2 O 7 2- (aq) + HNO 2 (aq) → Cr 3+ (aq) + NO 3 - (aq) Cr 2 O 7 2- → Cr 3+ HNO 2 → NO 3 -

11. Split the reaction into half reactions and determine oxidation numbers Identify oxidation numbers (use the rules and create an algebra expression to solve for the unknown). Do the first half reaction. Cr 2 O 7 2- → Cr 3+ We know oxygen is -2 and the overall charge on the left is -2 so create an algebra expression to solve for the charge of Cr on the left. Cr is our variable (we need to find out the oxidation number of Cr) and we have a subscript of 2 so 2x is the variable on the left. 2x – 14 = -2 2x = 14 – 2 2x = 12 X = +6 (Cr is +6 on the left) On the right, Cr is already shown as a +3. Therefore, Cr goes from a +6 to a +3. Electrons are NEGATIVE. Did we GAIN or lose electrons? Is this first half reaction Oxidation or REDUCATION? What is the oxidizing agent (put the whole compound from the reactant side)?

12. Split the reaction into half reactions and determine oxidation numbers Now, do the second half reaction. Identify oxidation numbers (use the rules and create an algebra expression to solve for the unknown). HNO 2 → NO 3 – We know oxygen is -2 and the overall charge on the left is zero. On the right, the overall charge is -1. Hydrogen is +1. Nitrogen is the variable. Left calculation: Right calculation: +1 + x – 4 = 0x – 6 = -1 X = +3 on left x = +5 on right Therefore, nitrogen goes from +3 to +5. Electrons are NEGATIVE. The oxidation number became more positive. Did we gain or lose electrons? Is this second half reaction oxidation or reduction? What is the reducing agent?

13. Complete the lab in your kitchen You will use materials from your house or the grocery store. Please note your observations in the lab. Apply oxidation and reduction information to the lab at home.

14. Woohoo! Now it is your turn to complete the practice problems in the 7.06 regular lesson before submitting the 7.06 regular assessment. Feel free to watch this tutorial again and attend tutoring. Remember, you may resubmit 7.06 until you master this topic. Please review this tutorial for the module 7 exam and the final exam. Have a wonderful day!