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Electron-transfer reactions are called oxidation-reduction reactions or redox reactions. Oxidation – loss of electrons by one reactant. Reduction – gain.

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Presentation on theme: "Electron-transfer reactions are called oxidation-reduction reactions or redox reactions. Oxidation – loss of electrons by one reactant. Reduction – gain."— Presentation transcript:

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2 Electron-transfer reactions are called oxidation-reduction reactions or redox reactions. Oxidation – loss of electrons by one reactant. Reduction – gain of electrons by another reactant. Oxidation and reduction always occur together. Na → Na + + e - (oxidation) Cl 2 + 2e - → 2Cl - (reduction) The substance that is oxidized is the reducing agent. The substance that is reduced is the oxidizing agent.

3 Oxidation Is Losing e - Reduction Is Gaining e - =OIL RIG

4 Oxidation numbers provide a way to keep tabs on electron transfers. Oxidation numbers – the sign is written before the number. Electrical charges – the sign is written after the number. Na ion has a charge of 1+ and an oxidation number of +1. Since not all redox reactions produce ionic products, we can redefine redox reactions as chemical reactions in which changes in oxidation numbers occur.

5 Rules for Assigning Oxidation Numbers 1.The oxidation number of any free element (e.g., O 2, Ag, etc.) is zero, regardless of how complex its molecules may be. 2.The oxidation number for any simple, monatomic ion (e.g., Na + or Cl - ) is equal to the charge on the ion. 3.The sum of all the oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle. 4.In its compounds, fluorine has an oxidation number of -1. 5.In its compounds, hydrogen has an oxidation number of +1. 6.In its compounds, oxygen has an oxidation number of -2.

6 In binary ionic compounds with metals, the nonmetals have oxidation numbers equal to the charge on their anions. Example: Assigning Oxidation Numbers Molybdenum disulfide, MoS 2, has a structure that allows it to be used as a dry lubricant, much like graphite. What are the oxidation numbers of the atoms in MoS 2 ? Solution: Binary ionic compound: S(2 atoms)x(-2)=-4(Rule 6) Mo(1 atom)x(x)=x _________________________________ Sum=0(Rule 3) The value of x must be +4 for the sum to be zero. Therefore, Mo = +4andS = -2

7 Let’s redefine oxidation and reduction. Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number. To balance redox equations by the ion-electron method, we divide the oxidation and reduction processes into individual equations called half-reactions that are balanced separately.


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