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CHM1046 Week 1 Lecture Notes

by: Jeneene Romero

CHM1046 Week 1 Lecture Notes

Marketplace > Florida International University > > CHM1046 Week 1 Lecture Notes
Jeneene Romero
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Class Notes
Chemistry, chm, 1046, CHM1046
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This 6 page Class Notes was uploaded by Jeneene Romero on Tuesday July 26, 2016. The Class Notes belongs to at Florida International University taught by in Summer 2016. Since its upload, it has received 9 views.


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Date Created: 07/26/16
Experiment 32: Galvanic Cells Purpose: The purpose of this experiment is to measure the relative reduction potentials for a number of redox couples, develop and understanding of the movement of electrons, anions, and cations in a galvanic cell, study factors affecting cell potentials, and to estimate the concentration of ions in solution using the Nernst equation. Lab Procedure A. Reduction Potentials of Several Redox Couples 1. Collect the electrodes, solutions, and equipment. Obtain four small beakers and fill them three-fourths full of the 0.1 M solutions as shown in Figure 32.3. Share these solutions with other chemists/groups of chemists in the laboratory. 2. Set up the Copper-Zinc cell. Place a Cu strip (electrode) in the CuSO4 solution and a Zn strip (electrode) in the Zn(NO3)2 solution. Roll and flatten a piece of filter paper, wet the filter paper with a 0.1 M KNO3 solution. Fikd and insert the ends of the filter paper into the solutions in the two beakers; this is the salt bridge. Set the voltmeter on the alligator clips; one positive and one negative. 3. Determine the copper-zinc potential. If the voltmeter reads a negative potential, make sure to switch the connections to the electrodes. Read and record the positive cell potential. Identify the metal strips that serve as the cathode (positive terminal) and anode (negative terminal). 4. Repeat for the remaining cells. Determine the cell potentials for all possible galvanic cells that can be constructed from the four redox couples. Refer to the Report Sheet for the various galvanic cells. Prepare a new salt bridge for each galvanic cell. 5. Determine the relative reduction potentials. Assuming the reduction potential of the Zn2+ (0.1 M)/Zn redox couple is -0.79 V, calculate the reduction potentials of all other redox couples. 6. Determine the reduction potential of the unknown redox couple. Place a 0.1M solution and electrode obtained from your instructor in a small beaker. Determine the reduction potential, relative to the Zn2+ (0.1 M)/Zn redox couple, for your unknown redox couple. B. Effect of Concentration Changes on Cell Potential 1. Effect of different molar concentrations. Set up the galvanic cell using 1 M CuSO4 and 0.001 M CuSO4 solutions. Immerse a polished copper electrode in each solution. Prepare a salt bridge to connect the two half-cells. Measure the cell potential. Determine the anode and the cathode. Write an equation for the reaction occurring at each electrode. 2. Effect of complex formation. Add 2-5 mL of 6M NH3 to the 0.001 M CuSO4 solution until any precipitate re-dissolves. Observe and record any changes in the half-cell and the cell potential. 3. Effects of precipitate formation. Add 2-5 mL of 0.2 M Na2S to the 0.001 M and what happens to the cell potential? Record your observations. C. The Nernst Equation and an Unknown Concentration 1. Prepare the diluted solutions. Prepare solutions 1 through 4 using a 1 mL pipet and 100 mL volumetric flasks. Be sure to rinse the pipet with the more concentrated solution before making the transfer. Use deionized water for dilution to the mark in the volumetric flasks. Calculate the molar concentration of the Cu2+ ion for each solution and record. 2. Measure and calculate the cell potential for solution 4. Set up the experiment using small 50 mL beakers. The Zn2+/Zn redox couple is the reference half-cell for this part of the experiment. Connect the two half-cells with a new salt bridge. Reset the voltmeter to the lowest range. Connect the electrodes to the voltmeter and record the potential difference. Calculate the theoretical cell potential Ecell table of standard reduction potentials and the Nernst equation. 3. Measure and calculate the cell potentials for solutions 3 and 2. Repeat Part C.2 with solutions 3 and 2, respectively. A freshly prepared salt bridges is required for each cell. See data from Part A.3 from potential solution 1. 4. Plot the data. Plot Ecell expt and Ecell calc (ordinate) versus pCu (abscissa) on appropriate software for the four concentrations of CuSO4. 5. Determine the concentration of the unknown. Obtain a CuSO4 solution with an unknown copper ion concentration from your instructor and set up a like galvanic per (II) ion concentration in the solution. Experiment 32 Pre-Lab Questions 1. In a galvanic cell, a. Oxidation occurs at the (name of electrode) – anode b. The cathode is the (sign) electrode – positive c. Cations flow in solution toward the (name of electrode) – anode d. Electrons flow from the (name of electrode) to (name of electrode) – anode to cathode 2. What is the purpose of the salt bridge? Explain. a. The salt bridge makes the half-cells electrically neutral within the internal circuit and prevents it from reaching equilibrium at a faster rate. b. The salt bridge is prepared by folding a piece of filter paper and wetting it with 0.1 M KNO3 solution. The ends of the filter paper are then inserted into the two beakers. 0.1M CuSO ×1mL 3. 4 =0.1M (molar concentration) 100mL 4. Refer to figure 32.2 and equations 32.8 and 32.9 a. Y = E ̊cell – 0.0592/2 log b. Y = E ̊cell – 0.0592/2 −6g Kc0.96/0.59 = log Kc = 32.54 10 mol c. cell potential=1× =0.0000001 L −0.0592 −6 E° cell log1.0×10 ( )(.04139 =).66V 2 5. Consider the two redox couples −¿→Cr (s) ¿ a. Cathode: 3+¿+3e C r ¿ −¿ ¿ b. Anode: +¿+e 3Ag→3Ag ¿ +3 c. 3Ag+C r →3Ag+Cr (s) d. E ̊cell = +0.80 – 0.74 = 0.06 e. n represents the moles of electrons exchanged according to the cell reaction. Experiment 32 Table of Results A. Reduction Potentials of Several Redox Couples Measur Equation for Equation for Galvan ed Anod Anode Catho Cathode ic Cell e de EcellV) Reactio2+ R2+ction Cu-Zn 0.930 Zn Zn → Zn + Cu Cu + 2e → 2e Cu Ni → Ni2+ + Cu 2+ + 2e → Cu-Ni 0.638 Ni Cu 2e 2+ 2+Cu Cu-Fe 0.644 Fe Fe → Fe + Cu Cu + 2e → 2e Cu Ni → Ni2+ + Zn 2+ + 2e → Zn-Ni 0.847 Ni Zn 2e2+ 2+Zn Fe-Ni 0.600 Ni Ni → Ni + Fe Fe + 2e → 2e Fe Zn → Zn 2++ Fe2++ 2e → Zn-Fe 0.261 Zn Fe 2e Fe 2+¿ aq +Cu (s) 1. Cu-Zn: 2+¿→Zn ¿ ¿ Zn(s+C u 2+¿+Cu (s) ¿ a. Cu-Ni: 2+¿→ N i¿ ¿(s)+C u 2+¿+Cu (s) ¿ b. Cu-Fe: 2+¿→F e Fe(s+Cu ¿ 2+¿+Zn s) ¿ c. Zn-Ni: 2+¿→N i ¿(s)+Zn ¿ 2+¿+Fe(s)¿ d. Fe-Ni: 2+¿→N i ¿( )+Fe¿ 2+¿+Fe (¿) e. Zn-Fe: 2+¿→Zn Zn(s+Fe ¿ 2. Zn is the oxidizing agent. 3. Cu-Zn + Zn-Ni: 0.930V + 0.847V = 1.78V a. Cu-Ni: 0.638V b. Values are not the same. The sum of Cu-Zn and Zn-Ni is higher than Cu-Ni 4. Zn-Fe + Zn-Ni: 0.261V + 0.847V = 1.10V a. Fe-Ni: 0.600V b. Values are not the same. The sum of Zn-Fe and Zn-Ni is higher than Fe-Ni. 5. Galvanic Measur For the Reduction Potential % Error Cell ed Redox Potential (theoretical) Ecell) Couple (experimental) Cu-Zn 0.930 2+¿/Cu 0.930+ −0.79 )=0.14V31 .31−.14 ¿ ×100=54.84 C u .31 Zn-Fe 0.638 2+¿/Fe 0.638+ −0.79 )=−0.15V47 −0.48−(−.15) F e¿ −0.48 ×100=68.75 Zn-Zn 0 2+¿/Zn -0.79V -0.79V 0.00 Zn ¿ Zn-Ni 0.847 2+¿/¿ 0.847+ −0.79 )=0.057V40 −2.4−.057 ×100=102.3 N i −2.4 Zn- 0.675 X2+/X -0.76 −0.76−(−.115) Unkno 0.675+ −0.79 )=−.115V ×100=84.87 −0.76 wn 6. Reduction potential of the unknown redox couple = -.115V B. Effect of Concentration Changes on Cell Potential 1. Cell potential of concentration cell: .0409V −¿ ¿ i. Anode half-reaction/ 1 M CuSO 4 : −¿→O +2H O+42 ¿ 4O H −¿→Cu(s) ii. Cathode half-reaction/ .001 M CuSO 4 : 2+¿+2e ¿ ¿ Cu iii. A potential is recorded because of the difference in the solutions concentration. 2. Cell potential from complex formation: 0.111V i. Observation: White precipitate ii. Potential changed with the addition of NH (aq3 because there are more Cu 2+ions flow from cathode and the 2+ concentration of Cu ions decrease. 3. Cell potential from precipitate formation: 0.900V i. Observation: Turned solution brown ii. Potential changed with the addition of Na S b2cause of the formation of CuS released electrons. As there are more electrons flowing from anode to cathode, thus potential increased. C. The Nernst Equation and an Unknown Concentration Soluti Conc. Of Ecell) 2+¿ Ecelllculated on # Cu(NO3)2 experiment ¿ al C u ¿ ¿ −log¿ 1 0.1 mol/L 0.930 1 1.1V 2 0.001 mol/L 0.809 -3 1.04V 3 0.00001 0.597 -5 0.9816V mol/L 4 0.0000001 .379 -7 0.9224V mol/L Sample calculation: E celllculated Solution 1 2 0.0592 [Zn ] log 2 E cell E ocell 2 [Cu ] 0.0592 [0.1 ] 2+ 2+ o 2+ o 2+ 2 log [0.1] E Cu /Cu– EZn /Zn= E Cu /Cu– E Zn /Zn- = 0.34 – (-0.76) = 1.1 V. 3. Ecellor the solution of unknown concentration: 0.675 a. pCu= -log(0.675) = 0.17


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