1 / 19

4.4 Oxidation-Reduction Reactions

4.4 Oxidation-Reduction Reactions. An oxidation occurs when an atom or ion loses electrons. A reduction occurs when an atom or ion gains electrons. Oxidation-Reduction Reactions. One cannot occur without the other. Oxidation Numbers.

jessh
Download Presentation

4.4 Oxidation-Reduction Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 4.4 Oxidation-Reduction Reactions • An oxidation occurs when an atom or ion loses electrons. • A reduction occurs when an atom or ion gains electrons.

  2. Oxidation-Reduction Reactions One cannot occur without the other.

  3. Oxidation Numbers • To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity. • See rules for assigning oxidation numbers in text, page 139. You must remember these rules. • Oxidation numbers are written with the sign first, then the number. For example, O is -2, Ca is +2 • The oxidation numbers for a neutral compound must add up to zero. The oxidation numbers for an ionic compound must add up to the charge on the ion. • Remember OIL RIG for what is oxidized and reduced: Oxidation is loss of electrons Reduction is gain of electrons

  4. Assigning Oxidation Numbers

  5. Displacement Reactions • In displacement reactions (single displacement reactions), ions in solution are displaced through oxidation of another element. • The ions that caused the oxidation are then reduced. • In the case of copper reacting with silver ions in solution, the copper metal is placed in solution and silver metal is formed as a solid. See figure 4.14.

  6. Displacement Reactions In this reaction silver ions oxidize copper metal. The oxidation of copper is accompanied by the reduction of silver to Ag(s). Cu (s) + 2 Ag+ (aq) Cu2+(aq) + 2 Ag (s)

  7. Displacement Reactions The reverse reaction, however, does not occur. Cu2+ (aq) + 2 Ag (s)  Cu (s) + 2 Ag+ (aq) Why? x

  8. Reference the activity series table on the following slide or in the text. In it you will see that copper lies above silver in ease of oxidation. Since copper is already oxidized, as Cu+, silver cannot replace it since it is not as easily oxidized. • Metals in the activity series replace, or oxidize, metals below them in the series.

  9. Activity Series

  10. Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution. moles of solute Molarity (M) = volume of solution in liters 4.5 Concentration of Solutions

  11. Molarity Calculations

  12. Interconverting Molarity, Moles and Volume • The definition of molarity contains three quantities: molarity (M), moles of solute, and liters of solution. • To solve for moles, multiply liters of solution X molarity • Notice that the units for volume cancel out • To solve for volume of solution, divide moles of solute by the molarity • Notice that the units for moles cancel out

  13. Preparation of a Solution • See fig 4.16 in text • For example, you want to prepare 0.250 L of 1.00 M solution of CuSO4 • The molar mass of CuSO4 is 159.6 grams per mole • 159.6 grams in 1.00 L of solution = 1.00 M • 39.90 grams in 0.250 L of solution = 1.00 M • Weigh-out 39.90 grams of CuSO4 and place it into a 250-mL volumetric flask. • Add a small amount of water to the solute in the flask. Swirl to dissolve. • Add additional water to the calibration mark. • Shake the stoppered flask to ensure complete mixing.

  14. Dilution Mconc x Vconc = Mdilute x Vdilute How many mL of 1.00 M HCl must be diluted to make 350 mL of 0.250 M HCl solution? How many mL 2.50 M NaOH must be diluted to make 0.500 L of 0.750 M NaOH solution?

  15. 4.6 Solution Stoichiometry and Chemical Analysis

  16. In stoichiometric calculations, if you know the quantity of one reactant (or product) in a balanced chemical reaction, you can calculate the quantities of other reactants and products. • When working with solutions of known molarity, we use molarity and volume to determine the number of moles. Moles of solute = M x L soln • For example: how many grams of Ca(OH)2 are required to neutralize 25.0 mL of 0.100 M HN03? (See sample exercise 4.15. Also, try solving the practice exercises to see if you can get the posted answers.) Ca(OH)2 + 2 HNO3 2 H2O(l) + Ca(NO3)2 • Calculate number of moles of HNO3 Moles = Liters x M • Use moles of HNO3 and mole ratio to solve for grams of Ca(OH)2 grams Ca(OH)2 = moles HNO3 x mole ratio x molar mass of Ca(OH)2

  17. Titration • The analytical technique in which one can calculate the concentration of a solute in a solution. Titrations can be conducted for acid-base reactions, precipitation, or redox reactions. • An acid-base titration is shown below. The chemist is trying to determine the concentration of an acid solution.

  18. The sample of the solution of unknown concentration is combined with a reagent solution of known concentration, called a standard solution. • In the acid-base titration example shown, a standard NaOH solution is added to a specific volume of the acid solution of unknown concentration until the neutralization reaction between the HCl and the NaOH is complete. • The point at which stoichiometrically equivalent quantities are combined is known as the equivalence point of the titration. • How does one know the equivalence point has been reached? Acid-base titrations use dyes known as indicators. An indicator called phenolphthalein (phph) is colorless in an acidic solution but pink in a basic solution. • NaOH is added drop by drop to the acid solution containing phph. The drop of NaOH added after the equivalence point is reached turns the clear solution pink. The color change signals the end point of the titration which usually coincides very closely with the equivalence point.

  19. Knowing the total volume of NaOH added to the acid solution to bring about neutralization, a student/chemist can use stoichiometry calculations to determine the concentration of the acid. • See sample exercise 4.16 and practice exercises. • For example: A student is trying to determine the concentration of an NaOH solution. She neutralizes a 20.0-mL sample of the NaOH solution with with 45.7 mL of 0.500 M H2SO4. What is the molarity of the NaOH solution? • Write the balanced equation: H2SO4 + 2 NaOH Na2SO4 + 2 H2O • Calculate moles of H2SO4 used to neutralized the NaOH: • Use the mole ratio to calculate moles of NaOH in sample: • Calculate molarity of the NaOH:

More Related