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Chapter 22 Oxidation – Reduction Rxns

Chapter 22 Oxidation – Reduction Rxns. Section 22.1 The Meaning of Oxidation and Reduction. OBJECTIVES: Define the terms “oxidation” and “reduction” in terms of electron loss or gain. State the characteristics of a RedOx Rxn, and recognize oxidizing and reducing agents.

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Chapter 22 Oxidation – Reduction Rxns

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  1. Chapter 22Oxidation – Reduction Rxns

  2. Section 22.1The Meaning of Oxidation and Reduction • OBJECTIVES: • Define the terms “oxidation” and “reduction” in terms of electron loss or gain. • State the characteristics of a RedOx Rxn, and recognize oxidizing and reducing agents.

  3. Oxygen in RedOx Reactions • “Oxidation” has traditionally meant the combination of elements (or compounds) with oxygen to form oxides. • Oxidation is most closely associated with combustion (burning). • “Reduction” traditionally meant the loss of oxygen from a compound. • When ores (especially iron) have their oxygen removed they are “reduced” back to their original elements • Reduction and oxidation occur simultaneously and the process is termed a “RedOx Reaction”

  4. Electron Transfer in RedOx Reactions • There are 2 mnemonics for helping you memorize the definitions for reduction and oxidation: OIL RIG = Oxidation Is Loss of Electrons Reduction Is Gain of Electrons LEO goes GER = Loss of Electrons is Oxidation Gain of Electrons is Reduction

  5. Example of a RedOx Reaction In this example elemental magnesium and sulfur combine to produce the compound magnesium sulfide (a synthesis reaction). The magnesium atom becomes a magnesium cation by losing two electrons. The two electrons are needed on the product side to balance the charge of this OXIDATION HALF REACTION.

  6. The sulfur atom gains the two electrons lost by the magnesium atom to become a sulfide anion. This is the REDUCTION HALF REACTION. The over all balanced reaction is the sum of the two half reactions. NO electrons can show, and it must balance by both mass and charge!

  7. More Good Stuff • The oxidizing reagent is itself reduced (and thereby causes oxidation). • The reducing agent is itself oxidized (and thereby causes reduction). • In the preceding example, Mg was oxidized, so it is the reducing agent. The S was reduced, so it is the oxidizing agent.

  8. Table 22.1 Processes Leading to Oxidation and Reduction • Oxidation can occur by: Loss of electrons Shift of electrons away from a covalent bond Gain of Oxygen Increase in oxidation number (charge) Loss of hydrogen (biology definition)

  9. Table 22.1 Processes Leading to Oxidation and Reduction • Reduction can occur by: Gain of electrons Shift of electrons toward an atom in a covalent compound Loss of Oxygen Decrease in Oxidation Number (charge) Gain of hydrogen (biology definition)

  10. Some Important Facts • On the NYS Regents Exam the answer to any RedOx question will be one of the reactants, never the products! • If you are asked to identify a RedOx reaction from a list of reactions, the answer can simply be found by identifying which reaction has an element on one side the arrow, and then a compound containing that element on the other side of the arrow! Simple!! • Synthesis and Single Replacement reactions are always RedOx reactions. • Decomposition reactions can be RedOx, but Double Replacement are rarely RedOx.

  11. Corrosion Facts • Iron reacts with atmospheric oxygen gas to form iron (III) oxide under dry conditions. • Under moist conditions, especially in the presence of salts, iron forms iron hydroxide(s). • Iron corrodes all the way through and forms pits or holes as it rusts. • Not all metals corrode - Gold and platinum (Noble Metals) are examples. • Some metals, such as silver or aluminum, form thin layers of oxide (rust) that protect them from further damage.

  12. How Can We Prevent Corrosion? • Rust free metal surfaces can be cleaned and varnished or painted to keep molecular oxygen from reacting with the metal. • A “sacrifice metal” can be placed in contact with the metal we are trying to protect. In this case, the sacrifice metal rusts faster than its partner, and supplies electrons to it such that it will not be easily oxidized. Typically Mg or Zn are used. If Zn is the sacrifice metal, we say the metal we are trying to protect has been “galvanized”.

  13. One More Method • If the metal object we are trying to protect is really valuable, we can use a D.C. current source (a battery) to supply a feed of electrons to the metal we are trying to protect. • This process is usually done on expensive automobiles, underground pipelines, storage tanks, etc.

  14. Section 22.2Oxidation Numbers • OBJECTIVES: • Determine the oxidation number of an atom in pure substances. • Define RedOx in terms of changes in oxidation number

  15. Rules for Assigning Oxidation Numbers • The oxidation number of a monatomic ion is equal in magnitude and sign to its ionic charge. Ex: The oxidation number of bromide is -1. • The oxidation number of hydrogen is +1 except in hydrides where it is -1. Ex: In NaH hydrogen is -1. • The oxidation number of oxygen is normally -2, except in peroxides where it is -1. Ex: In H2O2, the oxygen is -1. • The oxidation number of any uncombined element (including diatomic elements) is 0. Ex: The charges on S, Na, and N2 are all 0.

  16. Rules for Assigning Oxidation Numbers • The sum of all the oxidation numbers in an electrically neutral compound is 0. Ex: In SO2, S = +4 and O = -2 (S) + 2(O) = 0 or (+4) + 2(-2) = 0 • The sum of all the oxidation numbers in a polyatomic ion is equal to the charge on that ion. Ex: In CO32-, C = +4 and O = -2 (C) + 3(O) = -2 or (+4) + 3(-2) = -2

  17. Try These! • What are the charges (oxidation numbers) of all the elements in these examples? S2O3 S = +3, O = -2 2(S) + 3(O) = 0 or 2(+3) + 3(-2) = 0 KMnO4 K = +1, Mn = +7, O = -2 K + Mn + 4(O) = 0 or (+1) + (+7) + 4(-2) = 0 N2 N = 0 (remember, diatomic elements are all zero!) Cr2O72- Cr = +6, O = -2 2(Cr) + 7(O) = -2 or 2(+6) + 7(-2) = -2

  18. Write the Net Ionic Rxn and Identify what is Reduced and Oxidized in this Equation: 2AgNO3(aq) + Cu(s) 2Ag(s) + Cu(NO3)2(aq) For this single replacement rxn, the balanced complete ionic rxn is: 2Ag1+(aq) + Cu0(s) + 2NO31-(aq) 2Ag0(s) + Cu2+(aq) + 2NO31-(aq) The spectator ion must be 2NO31-(aq), so that leaves the net ionic rxn as: 2Ag1+(aq) + Cu0(s) 2Ag0(s) + Cu2+(aq) (Note that this rxn is balanced by both mass and charge!)

  19. 2Ag1+(aq) + Cu0(s) 2Ag0(s) + Cu2+(aq) The half rxns are: 2Ag1+(aq) + 2e1- 2Ag0(s)Reduction and Cu0(s) Cu2+(aq) + 2e1- Oxidation Thus, 2Ag1+(aq) is reduced (making it the oxidizing agent), and Cu0(s) is oxidized (making it the reducing agent). Notice that all the answers to a potential NYS Regents question on this rxn are on the reactant (left) side of the reaction!

  20. Section 22.3Balancing RedOx Rxns • OBJECTIVES: • Balance RedOx rxns by either “inspection” or the “half-rxn” method

  21. Balancing by “Inspection” • Most chemical rxns can be balanced by simply examining the unbalanced reaction, picking out one particular species that is unbalanced, and iteratively changing the coefficients until balance is achieved • This is the method you have used all along, and it will be more than sufficient to balance any RedOx reaction you may find on the NYS Regents exam

  22. Balancing by the “Half-Reaction” Method • This method is far more useful than what is stated on pg. 663 – 665 of your text book • Not only can you balance simple Regents level rxns, you can begin to think about far more involved IB course rxns • The basic idea is to: (1) break up the rxn into its Red and Ox parts, (2) multiple each half by what ever factor is needed to make the number of electrons in both half-rxns equal, and the (3) adding the two half-rxns back together while canceling any species that is found to be in common

  23. Example of Half-Rxn Method Li(s) + Al(NO3)3(aq) LiNO3(aq) + Al(s) Write the complete ionic reaction: Li(s) + Al3+ (aq) + 3(NO3)1-(aq) Li1+ (aq) + NO31-(aq) + Al(s) Eliminate the spectator ions. Write the net ionic reaction: Li(s) + Al3+ (aq) Li1+ (aq) + Al(s) Now write the two half-rxns: 3e1- + Al3+ (aq) Al(s) Reduction Li(s) Li1+ (aq) + 1e1- Oxidation

  24. As you can see, the number of electrons is not the same, so we will multiple the oxidation half by 3: 3e1- + Al3+ (aq) Al(s) Reduction 3Li(s) 3Li1+ (aq) + 3e1- Oxidation Now add the two half-reactions back together to get: 3Li(s) + Al3+ (aq) 3Li1+ (aq) + Al(s) Put the spectator ions back in and you get: 3Li(s) + Al(NO3)3(aq) 3LiNO3(aq) + Al(s) Of course you could have done this by simple “inspection”, but the technique we just used can be applied to solve far more difficult problems!

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