CHEM 102 1nd Edition Lecture 31 Outline of Last Lecture Outline of Current Lecture I. ElectrochemistryII. Molar Solubility ExampleIII. Precipitation ExampleIV. DemoV. Single Displacement ReactionVI. Oxidation NumbersVII. Largest Oxidation Number ExampleVIII. Oxidation and ReductionCurrent LectureI. Electrochemistry- Deals with interconversion of electrical and chemical energy- Involves oxidation-reduction, or redox, reactions in electrochemical cellsII. Molar Solubility Example- Example 1 Find Ksp of (Ag)2CO3 if molar solubility is 2e-4 M(Ag)2CO32 Ag+CO32-- x 2x x Ksp = [Ag]2[CO32-] Ksp = (2x)2 Ksp = 4e-4 x 2e-4 Ksp = 3.2e-11- Example 2 Which has the largest molar solubility SrSO4; ksp = 2.8e-7 Zn(OH)2; ksp = 4.5e-17PbI2; ksp = 8.7e-9 MnS; ksp = 5e-15 The smaller the ksp, the more soluble a compound is However the compounds must be the same molecular structure inorder to compare them- 1st – x2- 2nd – 2x2x- 3rd – 2x2xThese 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.- 4th – x2 Solve for k where the ksp = the above- 1st = 5.29e-4- 2nd = 2.2e-6- 3rd = 1.63e-3- 4th = 7.1e-9 The largest x = 1.63e-3; largest molar solubilityIII. Precipitation Example- Example 3- If Na2C2O4 is added to the solution containing 0.1 M Ba2+, 0.001 M Ca2+, and 0.06 M Sr2+, the first salt to precipitate out will be: BaC2O4; ksp = 1.1e-7 CaC2O4; ksp = 2.3e-9SrC2O4; ksp = 5.6e-8- Since they all have different molarities; multiply their molecular structure in x’s by their molarity 1st – 0.1(x) = 1.1e-6 2nd – 0.001(x) = 2.3e-6 3rd – 0.06(x) = 9.3e-7- For the first to precipitate you want the smallest number aboveIV. Demo- The metal was placed in the metal ion solution to get the following Reactions or No Reactions (NR)Mg + Mg2+NRCu + Mg2+NRZn + Mg2+NRMg + Cu2+Mg2+ + CuCu + Cu2+NRZn + Cu2+Zn2+ + CuMg + Zn2+Mg2+ + ZnCu + Zn2+NRZn + Zn2+NRV. Single Displacement Reaction- Involves oxidation and reduction- Involves displacement of one element by another- More active metal displaces the lesser active oneVI. Oxidation Numbers- Mechanical aid in writing formulas and balancing equations- Indicates oxidation state of element in compound- Referred to as Oxidation Number- Represented as +n, -n while charges are +n and –n.- For free elements, O.N. = 0 Zn or H2- For monoatomic molecules O.N. = ion charge Na+1- For polyatomic ion, O.N. = sum of all atom’s charges (OH1-) For most compounds, H = +1 O = -2VII. Largest Oxidation Number Example- Which I has the largest O.N?- I2- IO-- IO2-- IO2-IO3- 1st = 0 2nd = -1 = O – I = -2- x = -1; x = 1 3rd = -1 = O – I = -4 – x = -1; x = 3 4th = 0 5th = -1 = O – I = -6 – x = -1; x = 5VIII. Oxidation and Reduction- LEO the lion goes GER LEO = Loses electrons = oxidation GER = gains electrons =
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