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 7.06H Objectives (includes higher-order thinking): – Review 7.06 regular concepts – 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 – Balance oxidation and reduction half reactions using eight steps

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. Scaffolding: Breaking directions/calculations into easier steps (Literacy Skill) This is a challenging eight-step calculation. However, with this tutorial, you will rock this calculation! I will break the calculation into easier steps and discuss the directions for each step. You will see examples for each step. Please take notes, pause, rewind, etc. You might want to watch this tutorial twice, complete several practice problems yourself in the lesson, and attend tutoring for this concept. Let’s get started.

8. 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.

9. Review of ion charges from 3.05

10. Balancing redox equations (8 total steps): 1. Identify the element that is oxidized and the element that is reduced (example from the lesson) 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)

11. 2. 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 -

12. 2. 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?

13. 2. 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?

14. 3. Balance elements except hydrogen and oxygen by using coefficients Cr 2 O 7 2− → 2Cr 3+ (balanced Cr) HNO 2 → NO 3 − (N is already balanced)

15. 4. Balance any oxygen atoms by adding H 2 O to the other side (you are allowed to do this because there is plenty of water available in the aqueous solution). Cr 2 O 7 2− → 2Cr 3+ + 7H 2 O This half reaction needs seven oxygen atoms on the right, so we add seven H 2 O molecules to the right. HNO 2 + H 2 O → NO 3 − You have three oxygen atoms on the right and two on the left so this half reaction needs one more oxygen atom on the left (add one H 2 O molecule to the left).

16. 5. Balance hydrogen by adding H+ to the other side (you are allowed to do this because the solution is acidic, so there are H + ions available). Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O This half reaction needs 14 hydrogen atoms on the left to balance the 14 hydrogen atoms in the seven H 2 O molecules, so we add 14 H+ ions to the left. HNO 2 + H 2 O → NO 3 − + 3H + This half reaction needs three hydrogen atoms on the right to balance the three hydrogen atoms on the left, so we add three H + ions to the right.

17. 6. Balance the total charges on each side by adding electrons to the side with the higher charge to make the charges equal. Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O -2 + 14 → +6 +12 → +6 Electrons are NEGATIVE. Need 6 electrons on the left (6e-). 6e− + Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O HNO 2 + H 2 O → NO 3 − + 3H + 0 = -1 + 3 0 = +2 Need 2 electrons on the right (2e-). HNO 2 + H 2 O → NO 3 − + 3H + + 2e−

18. 7. Balance the electrons Make the electrons equal in both half reactions. 6e− + Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O HNO 2 + H 2 O → NO 3 − + 3H + + 2e− When multiplying a half reaction by a number, that number gets multiplied by every coefficient in that half reaction. The first half reaction has 6 electrons and the second half reaction has 3 electrons so multiply EVERYTHING in the second half reaction by 3. 6e− + Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O 3(HNO 2 + H 2 O → NO 3 − + 3H + + 2e−) equals 3HNO 2 + 3H 2 O → 3NO 3 − + 9H + + 6e−

19. 8. Combine the two half reactions and simplify Put both of the left half reactions together on the left and both of the right half reactions on the right. 6e− + Cr 2 O 7 2− + 14H + → 2Cr 3+ + 7H 2 O 3HNO 2 + 3H 2 O → 3NO 3 − + 9H + + 6e− 6e− + Cr 2 O 7 2- + 3HNO 2 + 3H 2 O + 14H + → 2Cr 3+ + 3NO 3 - + 9H + + 7H 2 O + 6e- Now, time to simplify to get the final answer. The electrons simplify (here they cancel each other out). The three H 2 O on the left cancel three of the seven H 2 O to yield four H 2 O on the right of the final equation. The nine H + on the right cancel nine of the 14 H + on the left leaving five H + on the left of the final equation. FINAL ANSWER (SHOW THE CHARGES) Cr 2 O 7 2- + 3HNO 2 + 5H + → 2Cr 3+ + 3NO 3 - + 4H 2 O

20. Tips for the last step – If you have equal amounts of the same substance on both sides of the balanced equation (like you will for the electrons), they cancel out and do not get written in the final equation. – If the same substance is on both sides of the balanced equation with different coefficients, you can reduce them by subtracting the lower number from both (one side will cancel, the other will reduce). – If the same substance appears more than once on the same side of the equation, you can combine them by adding the coefficients together.

21. Summary and review of eight-step calculation to balance redox equations (Literacy Skill) 1. Identify the element that is oxidized and the element that is reduced (example from the lesson). 2. Split the reaction into half reactions and determine oxidation numbers 3. Balance elements except hydrogen and oxygen by using coefficients 4. Balance any oxygen atoms by adding H 2 O to the other side (you are allowed to do this because there is plenty of water available in the aqueous solution). 5. Balance hydrogen by adding H+ to the other side (you are allowed to do this because the solution is acidic, so there are H + ions available). 6. Balance the total charges on each side by adding electrons to the side with the higher charge to make the charges equal. 7. Balance the electrons 8. Combine the two half reactions and simplify You are done!

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