Chemistry Lab Presenter: John R Kiser, MS Hickory Regional Director State Supervisor, Chemistry Lab

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1 Chemistry Lab 2010 Presenter: John R Kiser, MS Hickory Regional Director State Supervisor, Chemistry Lab

2 Introductions Topics for 2010: Solutions and Redox Regional vs State Topics Safety Requirements Long Sleeve Shirt! Must bring calculator! Need to know topics Formula Writing/Nomenclature Mole & Stoichiometry Calculations

3 Solution Terminology Solution: Homogeneous mixture Solvent: Component in greater/greatest amount Solute: Component(s) in lesser/least amount

4 Factors that influence solubility Polarity of Solute and Solvent Like Dissolves Like Polar solutes dissolve in polar solvents Nonpolar solutes dissolve in nonpolar solvents Nonpolar solutes do not dissolve well in polar solvents Temperature Solubility of most solids in water increases with temperature Solubility of gases in water decreases with temperature Gas Pressure As the pressure of a gas above a solution increases, the solubility of the gas in the solution increases. (Henry s Law) Sweet Tea and Soft Drinks

5 Amounts of Solute in Solution Saturated: The maximum amount of solute is dissolved in the solvent Unsaturated: Less than maximum amount of solute is dissolved in the solvent Supersaturated: More than the maximum amount of solute is dissolved in solvent To obtain a supersaturated solution, you heat solution until all solute dissolves. Carefully and slowly cooling the solution keeps all the solute dissolved in solvent. Solubility curves show maximum amount of solute that can be dissolved in 100 ml (sometimes 100 g) of water at a particular temperature. Above curve = supersaturated, Below curve = unsaturated

6 Saturation Pay attention to units of solubility on Y axis! Suppose a saturated solution of sodium nitrate is prepared at 60 o C in 200 ml of water. The solution is quickly cooled to 20 o C. What mass of sodium nitrates crystallizes during cooling?

7 Units of Concentration Moles of solute Molarity (M) = Liters of solution Mass Percent = Mass of component Total mass of solution x 100% Moles of solute Molality (m) = Mass of solvent (kg) Copyright 2010 Pearson Prentice Hall, Inc.

8 Units of Concentration Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is g/ml at 20 C, and the NaCl concentration is 3.50 mass % mass % = 3.50 grams of salt in grams of solution Assuming g of solution, calculate the volume: g solution 1 ml solution x g solution Convert the mass of NaCl to moles: x 1 L solution = L solution 1000 ml solution 3.50 g NaCl x 1 mole NaCl = moles NaCl 58.4 g NaCl Then, calculate the molarity: moles NaCl = M NaCl L solution

9 Units of Concentration In the previous example, what was the MOLALITY (m) of sodium chloride in seawater? Assume seawater contains only sodium chloride and water. Calculate the mass of water (solvent) in kg: g solution 3.50 grams NaCl (solute) = grams water (solvent) = kg Convert the mass of NaCl (solute) to moles: 3.50 g NaCl 1 mole NaCl x = moles NaCl 58.4 g Then, calculate the molality: moles NaCl = m NaCl kg Solvent

10 Concentrations - ppm Parts per million (ppm)= (Mass based) Mass of component Total mass of solution x ppm means a solution contains 50 grams of solute in 10 6 grams of solution (50 mg in 1 kg solution) In dilute aqueous solutions at 25 o C, ppm is also equivalent to mg solute in 1 L solution

11 Solving For Unknown Concentration: Titrations (Volumetric Analysis) In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Indicator substance that changes color at (or near) the equivalence point Equivalence point the point at which the reaction is complete Sometimes called stoichiometric point Endpoint The point at which the indicator changes color Slowly add reactants UNTIL the indicator changes color 4.7

12 Steps for Solving Titration Problems Be sure you have the correct balanced equation before beginning! STEP 1 Determine moles of starting compound STEP 2: Determine moles of desired compound STEP 3: Solve the problem Example Problem: Titration of Citric Acid in Fruit Juice State Level Question, 2010, commonly missed at many regionals! 3 NaOH (aq) + H 3 C 6 H 5 O 7 (aq) 3 H 2 O (l) + Na 3 C 6 H 5 O 7 (aq)

13 Solving for Unknown Concentration: Lambert-Beer Lambert-Beer Law: A = εbc A = Absorbance (unitless) b = path length (cm) c = concentration (M) ε = Molar absorbtivity (constant, units M -1 cm -1 ) Lambert-Beer or Beer s Law Plot: Provided all absorbance measurements are made on the same spectrophotometer with the same cell, a graph of absorbance vs. concentration will be linear. Lab activity at State in 2010! Commonly missed: Incorrect graphing, use of SpectroVis, and connecting the dots on the graph.

14 Solving for Unknown Concentration: Density Density of solution increases as solute concentration increases. The plot of density of solution versus concentration of solution should be linear. Can be used to solve for an unknown concentration.

15 Using Concentrations to Find Molar Mass: FP Depression Adding a solute to a solvent decreases the freezing point T f = k f *m T f = decrease in freezing point k f = Freezing point constant (1.86 o C m -1 for water) m = molality (mole solute per kg solvent) Assumes ideal behavior and that solute is NOT ionic. How to use to find molar mass of solute: Use T f and k f to find molality of solution Use mass of solvent to find moles solute present Mass solute dissolved divided by moles solute gives the molar mass of solute!

16 Freezing Point Depression Example When 2.50 grams of a covalent compound is dissolved in kg of water, the freezing point is determined to be ⁰C. What is the molar mass of the compound? (Assume Ideal Behavior) Molality = T f = ⁰C = m Kf 1.86 ⁰C m kg water * moles solute = moles solute kg water Molar mass= 2.50 grams solute = 62.0 grams per mole moles solute

17 Oxidation Reduction (Redox) Fundamental Concepts Oxidation is Loss of electrons, gain of O, loss of H Reduction is Gain of electrons, loss of O, gain of H Mnemonic Devices: LEO the lion goes GER! OIL RIG The species being reduced is the oxidizing agent The species being oxidized is the reducing agent

18 Activity Series of Elements Usually lists the best reducing agent (most easily oxidized) at top. Metal above reacts with ion below. Sample lab activity: The metal that reacts with everything else goes at the top; the ion that reacts with everything goes (as element, not ion) at bottom Aluminum Demonstration

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20 Determining if a species is oxidized or reduced: Oxidation Numbers Rules above take precedence over rules below! An atom in its elemental state has an oxidation number of 0. Oxidation number of H in H 2 = 0 An atom in a monatomic ion has an oxidation number identical to its charge. Oxidation number of Fe 3+ = +3 The total sum of oxidation numbers in a polyatomic ion is equal to the charge. Sum of oxidation numbers in a neutral molecule is 0. Pay attention to subscripts!

21 In a compound or polyatomic ion: 1A metal, ox #= +1; 2A metal, ox # = +2; 3A Metal, Ox # = +3 H, ox # = -1 if bonded to metal or Boron; H, ox # = +1 if bonded to another nonmetal Oxygen, ox# = -2 (EXCEPT IN PEROXIDES, -1) Fluorine, ox # = -1. Other halogens (if written to right in formula), ox # = -1 Other atoms not on this list can be deduced by following the rules above Finding oxidation number of Cl in ClO 4 Oxidation number of O = -2 Cl + 4*-2 = -1 Cl 8 = -1 Cl = +7 If oxidation number increases, oxidation is taking place If oxidation number decreases, reduction is taking place

22 Redox Half Reactions In many complex redox reactions, H + (or OH - ) and H 2 O are involved in the reaction and may not be obvious at first. Therefore, there is a systematic method to balancing complex redox reactions. First, separate into two half reactions, one for oxidation, one for reduction. Total Reaction: Cl - + Cr 2 O 7 2- Cl 2 + Cr 3+ Half Reactions: Cl - Cl 2 and Cr 2 O 7 2- Cr 3+

23 Balancing Half Reactions Using Coefficients, balance all atoms BUT H and O: Cr 2 O Cr 3+ Balance O by adding H 2 O Cr 2 O 7 2- Cr H 2 O Balance H by adding H + 14 H + + Cr 2 O Cr H 2 O Balance charge by adding e - 14 H + + Cr 2 O e - 2Cr H 2 O Number of electrons should correspond to change in oxidation number (keeping number of atoms in mind too!) No electrons in final answer, so multiply one (or both) half reaction(s) by a number so that electrons are equal. 2 Cl - Cl e - 2 Cl - Cl e - becomes 6 Cl - 3 Cl e -

24 Adding Half Reactions Together 6 Cl - 3 Cl e - 14 H + + Cr 2 O e - 2Cr H 2 O Be sure to cancel out electrons, water, and H + that appears on both sides 6 Cl H + + Cr 2 O Cl Cr H 2 O Be sure to double check charges and numbers of atoms! Electrons should not be left over! If you are balancing a half-reaction in basic solution, add OH - to both sides, convert H + to H 2 O, and cancel out!

25 Example Problem Balance the following half reaction in basic solution: Oxidation or reduction? OCl - Cl -

26 Galvanic Cells What happens when a redox reaction is spread out over 2 beakers? Oxidation at anode (-), reduction at cathode (+) Electrons flow from anode to cathode

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28 Standard Reduction Potential (E o red) E o cell =E o cathode E o anode Positive E o cell means the reaction is spontaneous More positive E o means a more favored reaction O Standard Conditions, 298 K, 1 atm, 1 M solutions All these values are measured with respect to 2H + /H 2 (These E o values are technically E o reduction. E o oxidation can also be written for reverse reactions)

29 Shorthand Notation for Galvanic Cells Anode half-reaction: Cathode half-reaction: Zn(s) Cu 2+ (aq) + 2e - Zn 2+ (aq) + 2e - Cu(s) Overall cell reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Anode half-cell Salt bridge Cathode half-cell Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Phase boundary Copyright 2010 Pearson Prentice Hall, Inc. Electron flow Phase boundary

30 Example Problem For the following Galvantic cell, determine the cathode, anode, and E o cell Ni (s) Ni 2+ (aq) Cu 2+ (aq) Cu(s)

31 The Effect of Concentration on Cell E What happens when conditions are not standard? Nernst Equation: E = E - RT nf ln Q or E = E - or 2.303RT nf log Q E = E V n log Q in volts, at 25 C Q = Reaction Quotient (from Equilibrium topics) Setup is just like an equilibrium constant, except the system is not at equilibrium. Products on top, reactants on bottom. Coefficients become exponents. Solutions in M, gases in atm, pure liquids and solids omitted.

32 Consider a galvanic cell that uses the reaction: Cu(s) + 2Fe 3+ (aq) Cu 2+ (aq) + 2Fe 2+ (aq) What is the potential of a cell at 25 C that has the following ion concentrations? [Fe 3+ ] = 1.0 x 10-4 M [Cu 2+ ] = 0.25 M [Fe 2+ ] = 0.20 M

33 Anode: Cathode: Batteries Lead Storage Battery Pb(s) + HSO 4- (aq) PbO 2 (s) + 3H + (aq) + HSO 4- (aq) + 2e - PbSO 4 (s) + H + (aq) + 2e - PbSO 4 (s) + 2H 2 O(l) Overall: Pb(s) + PbO 2 (s) + 2H + (aq) + 2HSO 4- (aq) 2PbSO 4 (s) + 2H 2 O(l) Copyright 2010 Pearson Prentice Hall, Chapter 17/33

34 Batteries Dry-Cell Batteries Leclanché cell Copyright 2010 Pearson Prentice Hall, Chapter 17/34

35 Fuel Cells Hydrogen-Oxygen Fuel Cell

36 Electrolysis and Electrolytic Cells Electrolysis: The process of using an electric current to bring about chemical change. Electrolysis of Molten Sodium Chloride Anode: 2Cl - (l) Cl 2 (g) + 2e - Cathode: 2Na + (l) + 2e - 2Na(l) Copyright 2010 Pearson Prentice Hall, Inc. Overall: 2Na + (l) + 2Cl - (l) 2Na(l) + Cl 2 (g)

37 Electrolysis and Electrolytic Cells Cathode has negative charge, connected to negative terminal on battery; anode has positive charge, connected to positive terminal. Reduction still at cathode, Oxidation still at anode Electrolysis of Molten Sodium Chloride

38 Electrolysis and Electrolytic Cells Electrolysis of Water Anode: Cathode: 2H 2 O(l) 4H 2 O(l) + 4e - O 2 (g) + 4H + (aq) + 4e - 2H 2 (g) + 4OH - (aq) Overall: 6H 2 O(l) 2H 2 (g) + O 2 (g) + 4H + + 4OH - (aq)

39 Electrolysis of Aqueous Solutions Electrolysis of NaI Consider reduction of water and reduction of cation Na + (aq) + e - Na (s) E o red = V 2 H 2 O (l) + 2e - H 2 (g) + 2 OH - (aq) E o red = V Less negative (or more positive) reduction is preferred! Consider oxidation of water and oxidation of anion 2 H 2 O (l) O 2 (g) + 4 H + (aq) + 4 e - E o ox = V 2 I - (aq) I 2 (s) + 2 e - E o ox = V Less negative (more positive) oxidation preferred! (Note: reversing from E o red table. If reading directly from table, without reversing oxidation, look for more negative E o red)

40 Overvoltage makes predictions difficult if both competing reactions have similar E o values. In electrolysis of aqueous NaCl, even though oxidation of water is less negative, Cl is oxidized to Cl 2 gas. Poorly understood, mainly due to kinetic factors and concentration effects. Prediction is difficult if values are close on table. Rules of Thumb for electrolysis of solutions Look on left side of SRP table. If cation is higher than water, cation will be reduced. Look to right side of SRP table. If anion is below water, anion will be reduced

41 Quantitative Aspects of Electrolysis Charge(C) = Current(A) x Time(s) Moles of e - = Charge(A) x 1 mol e - 96,500 C Faraday constant

42 Example Problem How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of A is passed through the cell for 1.5 hours? 19.8

43 Lab Time! Lab #1 Beer s Law Lab #2 Redox Titration Lab #3 Setup of a Galvanic cell and determining E O cell and E o for a half reaction

44 Thank you! Please me at For follow-up questions, concerns, etc..