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Chapter 21 - Oxidation-Reduction Reactions 2: 38.40/49 = 78.37% 3: 40.36/49 = 82.37%

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Presentation on theme: "Chapter 21 - Oxidation-Reduction Reactions 2: 38.40/49 = 78.37% 3: 40.36/49 = 82.37%"— Presentation transcript:

1 Chapter 21 - Oxidation-Reduction Reactions 2: 38.40/49 = 78.37% 3: 40.36/49 = 82.37%

2 Overview Oxidation reactions are the principal source of energy on earth Most explosive reactions are oxidation reactions All oxidation reactions are accompanied by reduction reactions. The chemical changes that occur when electrons are transferred between reactants are known as oxidation-reduction reactions. Oxidation-reduction reactions are also known as redox reactions.

3 21.1 - Oxygen in Redox Reactions Oxidation originally meant the combination of an element with oxygen to give an oxide Iron rusting was considered oxidation Fe + O 2  Fe 2 O 3 We can also oxidize a compound. CH 4 + O 2  CO 2 + H 2 O Over time, reduction has meant the loss of oxygen from a compound Fe 2 O 3 + C  Fe + CO 2 The iron oxide loses oxygen, it is reduced to metallic iron. This came about because when a metal oxide is reduced, it loses a lot of volume. Oxidation and reduction occur simultaneously. As Fe 2 O 3 is reduced to iron by losing oxygen, the C is oxidized to CO 2 by gaining oxygen.

4 21.2 - Electron Transfer in Redox Reactions Now, chemists extend idea of oxidation and reduction to include all transfers/shifts of electrons Gives a much wider application of the term So oxidation is the complete or partial loss of electrons or the gain of oxygen Reduction is the complete or partial gain of electrons or the loss of oxygen

5 Mnemonic LEO the lion goes GER LEO = Losing Electrons is Oxidation GER = Gaining Electrons is Reduction

6 Example Mg(s) + S(s)  MgS(s) In a metal/non-metal reaction, electrons are always transferred from the metal to the non-metal This makes an ionic compound. Mg loses two electrons, sulfur gains two electrons Therefore, Mg is oxidized, while S was reduced General Rule In the formation of a metal/non-metal compound (i.e. forming ionic compound), the metal is oxidized, the non- metal is reduced

7 Agents The substance in a redox reaction that donates electrons is a reducing agent. In other words, whatever is oxidized, is the reducing agent Likewise, the substance in a redox reaction that receives electrons is the oxidizing agent In other words, whatever is reduced, is the oxidizing agent.

8 Another Example 2AgNO 3 + Cu  Cu(NO) 3 + 2Ag We want to know what is oxidized and what is reduced? First, re-write this to show ions 2Ag + + 2NO 3 - + Cu  Cu 2+ + 2NO 3 - + 2Ag The copper (Cu) has gone from a Cu to a Cu 2+, therefore it lost electrons. So it was oxidized. The silver ions (Ag + ) has gone from Ag + to Ag, so it gained electrons. It was reduced. The NO 3 - had no change

9 Ionic reactions are easy Look at all of the ionic compounds as their ions, and gain/loss of electrons is apparent Covalent compounds are harder 2H 2 + O 2  2H 2 O Initially, H-H bond, electrons are equally shared Initially, O-O bond, electrons are equally shared But H-O-H, oxygen pulls electrons towards it So the oxygen partially gains electrons, making it reduced And the hydrogens partially lose electrons, making them oxidized

10 Summary Processes Leading to Oxidation and Reduction OxidationReduction Complete loss of electrons (ionic reactions) Complete gain of electrons (ionic reactions) Shift of electrons away from an atom in a covalent bond Shift of electrons towards an atom in a covalent bond Gain of oxygenLoss of oxygen Loss of hydrogen by a covalent compound Gain of hydrogen by a covalent compound An increase in oxidation number A decrease in oxidation number

11 21.3 - Assigning Oxidation Numbers Oxidation numbers are a bookkeeping concept. An oxidation number is a positive or negative number assigned to an atom according to a set of arbitrary rules. We can actually balance complex redox reactions by using oxidation numbers

12 Rules The rules vary depending on the chemical we’re dealing with. 6 rules total 1. The oxidation number of a monatomic ion is equal to its ionic charge. If we have the compound CaBr 2, the Ca 2+ ion has an oxidation number of +2, while Br - has an oxidation number of -1

13 2. The oxidation number of hydrogen is a compound is always +1, unless it’s bonded to a metal (like NaH), then it is -1. So H 2 O, the hydrogens each have an oxidation number of +1 3. The oxidation umber of oxygen in a compound is always -2, except in peroxide (O 2 2- ), where it is -1 So in H 2 O, the oxygen has an oxidation number of 2- While in H 2 O 2 (hydrogen peroxide), each hydrogen has an oxidation number of +1, while each oxygen has an oxidation number of -1.

14 4. The oxidation number of an uncombined element is zero So the potassium atom in a lump of potassium metal, K, would be 0. The oxidation numbers of the oxygen in O 2 would be zero 5. For any compound, the oxidation numbers of each atom add up to the charge on the ion

15 Example What is the oxidation number of the atoms in the following compounds? S 2 O 3 O 2 Al 2 (SO 4 ) 3 Na 2 O 2

16 Example 2 What is the oxidation of phosphorus in each of the following substances? P 4 O 8 PO 4 3- P 2 O 5 P 4 O 6 H 2 PO 4 - PO 3 3-


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