CHEM 102 1st Edition Lecture 32Outline of Last Lecture I. ElectrochemistryII. Molar Solubility ExampleIII. Precipitation ExampleIV. DemoV. Single Displacement ReactionVI. Oxidation NumbersVII. Largest Oxidation Number ExampleVIII. Oxidation and ReductionOutline of Current Lecture I. Transfer of ElectronsII. Oxidation-Reduction ReactionsIII. Balancing Redox ReactionsCurrent LectureI. Transfer of Electrons- Loss of electrons is Oxidation- Gain of electrons is Reduction- In the example: 2Ag+ + Cu 2Ag + Cu2+ Ag+ is reduced; it is the oxidizing agent Cu is oxidized; it is the reducing agent- Oxidation and reduction occur simultaneously- Electrons are conserved in a redox reactionsII. Oxidation-Reduction Reactions- Cu in AgNO3 ½ reactions: Cu Cu2+ + 2e- Ag+ + 2e- Ag Overall: Cu + 2Ag+ 2 Ag + Cu2+ Cu loses e-; it is oxidized Ag+ gains e-; it is reduced- Zn + 2H+ Zn2+ + H2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Zn Loses electrons Is oxidized Is the reducing agent Oxidizing number increases H+ Gains electron Is reduced Is the oxidizing agent Oxidation number increasesIII. Balancing Redox Equations- Use the unbalanced reaction to write two half-reactions- Balance elements except O and H- Balance O by adding H2O- Balance H by adding H+- Balance charges by adding electrons- Multiply each ½ reaction by coefficients to get the same # of electrons in each- Add the two ½ reactions- IF BASIC: Add OH- to each side of the equation to transform all the H+ to H2O then cancel all H2Os that appear on both sides.- Example Zn + NO3- Zn2+ + N2 in acidic solution Zn Zn2+NO3- N2 Zn Zn2+2NO3- N2 Zn Zn2+2NO3- N2 + 6H2O Zn Zn2+12H+ + 2NO3- N2 + 6H2O Zn Zn2+ + 2e-10e- + 12H+ + 2NO3- N2 + 6H2O 5(Zn Zn2+ + 2e-)1(10e- + 12H+ + 2NO3- N2 + 6H2O) 5Zn 5Zn2+ + 10e-10e- = 12H+ + 2NO3- N2 + 6H2O Overall: 5Zn + 12H+ + 2NO3- 5Zn2+ + N2 +
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