Electrochemical Cells Post-Lab
8 Pages

Electrochemical Cells Post-Lab

Course Number: CHE 2C, Summer 2010

College/University: UC Davis

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Q# Question Text 5) Part I Analysis. Scoring Scheme: 3-2-1-1 Given the Nernst equation, Ecell = (E0cathode - E0anode) - [0.0257/n]lnQ, and the equation, E0cell = E0cathode E0anode, substitute E0cell into the Nernst equation and solve for E0cell . Which of the following would be the correct expression for E0cell ? Your Answer: c. E0cell = Ecell + [0.0257/n]lnQ You Scored 3 points out of 3 Possible 6) Part I...

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Text Q# Question 5) Part I Analysis. Scoring Scheme: 3-2-1-1 Given the Nernst equation, Ecell = (E0cathode - E0anode) - [0.0257/n]lnQ, and the equation, E0cell = E0cathode E0anode, substitute E0cell into the Nernst equation and solve for E0cell . Which of the following would be the correct expression for E0cell ? Your Register to View AnswerE0cell = Ecell + [0.0257/n]lnQ You Scored 3 points out of 3 Possible 6) Part I Analysis. Scoring Scheme: 3-2-1-1 For the cell, Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s), which of the following is the correct spontaneous overall cell reaction: Your Register to View AnswerZn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) You Scored 3 points out of 3 Possible 7) Part I Analysis. Data Entry - No Scoring Please enter the value of Ecell you measured in volts for the spontaneous cell reaction for Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s). Ecell = Your Answer: 0.7940 No Points Possible 8) Part I Analysis. Scoring Scheme: 3-2-1-1 For the cell, Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s), which of the following is the correct expression for Q, the reaction quotient in the Nernst equation, for the spontaneous overall cell reaction: Your Register to View AnswerQ = [Zn2+]/[Cu2+] You Scored 3 points out of 3 Possible 9) Part I Analysis. Scoring Scheme: 3-3-2-1 Using your cell concentrations and the reaction quotient expression, calculate and enter the value of Q for the Nernst equation for this cell, Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s) Q= Your Answer: 1 You Scored 3 points out of 3 Possible 10 Part I analysis. ) Scoring Scheme: 3-3-2-1 Please enter the value of lnQ that appears in the Nernst equation for this cell, Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s) . ln Q = Your Answer: 0 You Scored 3 points out of 3 Possible 11 Part I analysis. ) Scoring Scheme: 3-3-2-1 For the cell, Zn(s) | Zn(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s), put all the terms together that enter into the equation for E0cell, and enter the value you calculated for it. E0cell = Your Answer: 0.7940 You Scored 3 points out of 3 Possible 12 Part I analysis. ) Data Entry - No Scoring Please enter the accepted value from the table in your text for the standard electrode potential, E0, for the half-cell reaction for copper: Cu2+(aq) + 2e- -> Cu(s) in volts. You will use this value as a reference to calculate the standard electrode potentials for the other halfreactions involved in the cells you measured. E0(Cu2+/Cu) = Your Answer: 0.34 No Points Possible 13 Part I analysis. ) Scoring Scheme: 3-3-2-1 Using the equation E0cell = E0cathode - E0anode, and the reference value of E0 for Cu(II)/Cu(s) for the role that copper plays in this overall cell reaction, determine the value of the standard potential for the half-reaction Zn2+(aq) + 2e- -> Zn(s) and enter it here. Standard potential = Your Answer: -0.4540 You Scored 3 points out of 3 Possible 14 Part I analysis. ) Scoring Scheme: 3-2-1-1 For cell Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s), which of the following is the correct spontaneous overall cell reaction: Your Register to View AnswerPb(s) + Cu2+(aq) -> Pb2+(aq) + Cu(s) You Scored 3 points out of 3 Possible 15 Part I analysis. ) Data Entry - No Scoring Please enter the value of Ecell you measured in volts for the spontaneous cell reaction for cell Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s). Ecell = Your Answer: 0.4760 No Points Possible 16 Part I analysis. ) Scoring Scheme: 3-2-1-1 For the cell, Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s) which of the following is the correct expression for Q, the reaction quotient in the Nernst equation, for the spontaneous overall cell reaction: Your Register to View AnswerQ = [Pb2+]/[Cu2+] You Scored 3 points out of 3 Possible 17 Part I analysis. ) Scoring Scheme: 3-3-2-1 Using your cell concentrations and the reaction quotient expression, calculate and enter the value of Q for the Nernst equation for this cell: Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s). Q= Your Answer: 1 You Scored 3 points out of 3 Possible 18 Part I analysis. ) Scoring Scheme: 3-3-2-1 Please enter the value of lnQ that appears in the Nernst equation for this cell, Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s) . lnQ = Your Answer: 0 You Scored 3 points out of 3 Possible 19 Part I analysis. ) Scoring Scheme: 3-3-2-1 For the cell, Pb(s) | Pb(NO3)2(0.10M) || Cu(NO3)2(0.10M) | Cu(s), put all the terms together that enter into the equation for E0cell, and enter the value you calculated for it. E0cell = Your Answer: 0.4760 You Scored 3 points out of 3 Possible 20 Part I analysis. ) Scoring Scheme: 3-3-2-1 Using the equation E0cell = E0cathode - E0anode, and the reference value of E0 for Cu(II)/Cu(s) for the role that copper plays in this overall cell reaction, determine the value of the standard potential for the half-reaction Pb2+(aq) + 2e- -> Pb(s) and enter it here. Standard potential = Your Answer: -0.1360 You Scored 3 points out of 3 Possible 21 Part I Analysis. ) Scoring Scheme: 3-2-1-1 For cell C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s), which of the following is the correct spontaneous overall cell reaction: Your Register to View Answer2Fe3+(aq) + Cu(s) -> Cu2+(aq) + 2Fe2+(aq) You Scored 3 points out of 3 Possible 22 Part I analysis. ) Data Entry - No Scoring Please enter the value of Ecell you measured in volts for the spontaneous (i.e., a positive number) cell reaction for cell C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s). Ecell = Your Answer: 0.3710 No Points Possible 23 Part I analysis. ) Scoring Scheme: 3-2-1-1 For the cell, C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s), which of the following is the correct expression for Q, the reaction quotient in the Nernst equation, for the spontaneous overall cell reaction: Your Register to View AnswerQ = [Fe2+]2[Cu2+]/[Fe3+]2 You Scored 3 points out of 3 Possible 24 Part I analysis. ) Scoring Scheme: 3-3-2-1 Using your cell concentrations and the reaction quotient expression, calculate and enter the value of Q for the Nernst equation for this cell C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s). Q= Your Answer: 0.10 You Scored 3 points out of 3 Possible 25 Part I analysis. ) Scoring Scheme: 3-3-2-1 Please enter the value of lnQ that appears in the Nernst equation for this cell, C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s). lnQ = Your Answer: -2.30 You Scored 3 points out of 3 Possible 26 Part I analysis. ) Scoring Scheme: 3-3-2-1 For the cell, C(gr) | Fe(NO3)3(0.10M), FeSO4(0.10M) || Cu(NO3)2(0.10M) | Cu(s), put all the terms together that enter into the equation for E0cell, and enter the value you calculate for it. E0cell = Your Answer: 0.3710 You Scored 3 points out of 3 Possible 27 Part I analysis. ) Scoring Scheme: 3-3-2-1 Using the equation E0cell = E0cathode - E0anode, and the reference value of E0 for Cu(II)/Cu(s) for the role that copper plays in this overall cell reaction, determine the value of the standard potential for the half-reaction Fe3+(aq) + e- -> Fe2+(aq) and enter it here. Standard potential = Your Answer: 0.7110 You Scored 3 points out of 3 Possible 28 Part II analysis. ) Scoring Scheme: 3-2-1-1 Using facts the that a spontaneous cell reaction gives a positive measured voltage when the red lead of the voltmeter is connected to the cathode, and that reduction takes place at the cathode while oxidation takes place at the anode, which of the following is the spontaneous reaction for the copper concentration cell you measured in Part II of the electrochemistry experiment? Your Register to View AnswerCu2+(0.1M) + Cu(s) -> Cu2+(0.01M) + Cu(s) You Scored 3 points out of 3 Possible 29 Part II analysis. ) Scoring Scheme: 3-3-2-1 Because the standard potentials are the same for both the anode and the cathode reactions for a concentration cell, the Nernst equation for a concentration cell becomes simply, Ecell = -(0.0257/n)lnQ, where Q is to be calculated for the spontaneous cell reaction. What is the value of Q for the copper concentration cell? Q= Your Answer: 0.10 You Scored 3 points out of 3 Possible 30 Part II analysis. ) Scoring Scheme: 3-3-2-1 Ecell = -(0.0257/n)lnQ Now, using your value for Q calculate Ecell from the Nernst equation for the copper concentration cell, and enter it here. Ecell = Your Answer: 0.03 You Scored 3 points out of 3 Possible 31 Scoring Scheme: 3-2-1-1 ) For Part II of the electrochemistry experiment, you measured the potential of some concentration cells. The laboratory manual asked you to construct these cells: a) Cu(s) | Cu(NO3)2(0.010M) || Cu(NO3)2(0.10M)|Cu(s) b) C(graphite) | FeSO4(0.010M), Fe(NO3)3(0.10M) || Cu(NO3)2(0.10M) | Cu(s) c) C(graphite) | FeSO4(0.10M), Fe(NO3)3(0.010M) || Cu(NO3)2(0.10M) | Cu(s) When the red lead for the voltmeter was connected to the right-hand electrode of the cells as diagrammed above, which of these cells showed a positive algebraic sign for the measured voltage? Your Register to View AnswerCu(s) | Cu(NO3)2(0.010M) || Cu(NO3)2(0.10M)| You Scored 3 points out of 3 Possible Cu(s) 32 Part II analysis. ) Scoring Scheme: 3-2-1-1 When a measured cell voltage is negative, the electrode to which the red lead is attached is actually the anode rather than the cathode (i.e., you should enter a positive voltage). The anode is the electrode at which oxidation takes place, so it is surrendering electrons to the external circuit. For the two cells involving the iron and copper species, the Cu(II)/Cu(s) couple is the anode. Which of the following half-cell reactions is taking place at the Cu electrode in the anode compartment of these two cells? Your Register to View AnswerCu(s) -> Cu2+(aq) + 2eYou Scored 2 points out of 3 Possible 34 Part II analysis. ) Scoring Scheme: 3-2-1-1 For the overall cell reaction in cells #3, #5, & #6, Cu(s) + 2Fe3+(aq) -> 2Fe2+(aq) + Cu2+(aq), which of the following is the correct expression for Q, the reaction quotient in the Nernst equation: Your Register to View AnswerQ = [Fe2+]2[Cu2+]/[Fe3+]2 You Scored 3 points out of 3 Possible 35 Part II analysis. ) Scoring Scheme: 3-2-1-1 The Nernst equation for the cell potential of cells #3, #5, & #6, is Ecell = Eocell - (0.0257/2)ln(([Cu2+][Fe2+]2)/ [Fe3+]2). What is the expression, then, for the cell voltage difference between cells #5 and #3, Ecell#5 - Ecell#3? Your Register to View Answer-(0.0257/2)ln(([Fe2+(cell #5)]2[Fe3+(cell #3)]2)/ You Scored 3 points out of 3 Possible ([Fe3+(cell #5)]2[Fe2+(cell #3)]2)) 36 Scoring Scheme: 3-3-2-1 ) Using the ion concentration values in cells #3 & #5 in the expression, Ecell#5 - Ecell#3 = -(0.0257/2)ln{([Fe2+ (cell#5)]2[Fe3+(cell#3)]2)/([Fe3+(cell #5)]2[Fe2+(cell#3)]2)}, calculate the theoretical cell voltage difference between cells #5 and #3. Cell Voltage Difference = Your Answer: 0.0590 You Scored 3 points out of 3 Possible 37 Scoring Scheme: 3-3-2-1 ) Using the ion concentration values in cells #3 & #6 in the expression, Ecell#6 - Ecell#3 = -(0.0257/2)ln{([Fe2+ (cell#6)]2[Fe3+(cell#3)]2)/([Fe3+(cell #6)]2[Fe2+(cell#3)]2)}, calculate the theoretical cell voltage difference between cells #6 and #3. Cell Voltage Difference = Your Answer: -0.0590 You Scored 3 points out of 3 Possible 39 Part III analysis. ) Data Entry - No Scoring Please enter the value of Ecell you measured in volts for the spontaneous cell reaction for cell Pb(s) | PbSO4(s), NH4+ (3.0M), SO42-(1.50M), NO3- (0.10M), Pb2+(? M) || Pb(NO3)2(0.05 M) | Pb(s). Ecell = Your Answer: 0.06 No Points Possible 40 Part III analysis. ) Scoring Scheme: 3-3-2-1 Because the standard potentials are the same for both the anode and the cathode reactions for a concentration cell, the Nernst equation for a concentration cell becomes simply, Ecell = -(0.0257/n)lnQ, where Q is the expression for the spontaneous cell reaction. For the concentration cell, Pb(s) | PbSO4(s), NH4+ (3.0M), SO42-(1.50M), NO3- (0.10M), Pb2+(? M) || Pb(NO3)2(0.05 M) | Pb(s) which of the following is the correct expression for Q for the spontaneous overall cell reaction, Pb2+(0.05M) -> Pb2+(?): Your Register to View AnswerQ = x/(0.05M) You Scored 3 points out of 3 Possible 41 Part III analysis. ) Scoring Scheme: 3-3-2-1 For the concentration cell, Pb(s) | PbSO4(s), NH4+ (3.0M), SO42-(1.50M), NO3- (0.10M), Pb2+(? M) || Pb(NO3)2(0.05 M) | Pb(s), Ecell = -(0.0257/2)ln(x/0.05), what is the expression for x in terms of Ecell ? {The notation used here is exp[x] = ex where "e" is the base of natural logarithm.} Your Register to View Answerx = 0.05exp[-2Ecell/0.0257] You Scored 3 points out of 3 Possible 42 Part III analysis. ) Scoring Scheme: 3-3-2-1 Using your measured cell potential for Pb(s) | PbSO4(s), NH4+ (3.0M), SO42-(1.50M), NO3- (0.10M), Pb2+(? M) || Pb(NO3)2(0.05 M) | Pb(s) in the expression, x = 0.05exp[-Ecell/0.01285], calculate the molarity of Pb2+ in the anode compartment. {The notation used here is exp[x] = ex where "e" is the base of natural logarithm.} Please put your answer in scientific notation. Molarity = Your Answer: 0.000470 You Scored 3 points out of 3 Possible 43 Part III analysis. ) Scoring Scheme: 3-3-2-1 Which of the following is the correct solubility product expression, Ksp, for PbSO4? Your Register to View AnswerKsp = [Pb2+][SO42-] You Scored 3 points out of 3 Possible 44 Part III analysis. ) Scoring Scheme: 3-3-2-1 Using the expression, Ksp = [Pb2+][SO42-], the concentration of sulfate in the anode compartment, and determined value Pb2+ concentration in the anode compartment, calculate Ksp. Please put your answer in scientific notation. Ksp = Your Answer: 0.000690 You Scored 3 points out of 3 Possible 45) Concluding Remarks: Briefly discuss interpretations of your observations and results. Include in your discussion, any conclusions drawn from the results and any sources of error in the experiment. Your Answer: In this lab we constructed a galvanic cell in Text Answers to be Scored by your TA order to measure the voltage of a number of different half cells so we would be able to get there standard reduction potentials. In part II, we examined the dependence of cell potential on concentration and found that it is possible to generate a potential between two half cells of the same composition but differing concentrations. In part 3 we estimated the solubility product of PbSO4 by measuring a lead concentration cell. Possible sources of error in this experiment could have been not diluting and mixing chemicals properly.

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Part 1 of 1 -5.0 PointsQuestion 1 of 10Whenanendergonicreactionreachesequilibrium0.5 PointsA. There is a higher concentration of reactants than products B. There is no net energy release or consumption. C. The rate of the forward reaction equals the
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Question 1 of 10Whatdeterminestherateofareaction?0.5 PointsA. The free energy difference between the reactants and the products B. The relative concentrations of reactants and products C. The energy of activation D. Both B and C E. All of the choices a
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Question 1 of 10onsareunabletofreelycrossmembranes.Whenchargedatoms(ions)are0.5/ 0.5 Points Ipumpedacrosscellmembranesbyspecializedproteins,anelectrochemical gradientisformed.Electrochemicalgradientsareaformoforder,because moreofaparticularionisfoundon
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Question 1 of 100.5/ 0.5 PointsWhichofthefollowingproducesthemostATPwhenarespiringcellcompletely oxidizesglucose(C6H12O6)tocarbondioxide(CO2)?A. citric acid cycle B. glycolysis C. oxidative phosphorylation (chemiosmosis) D. fermentation E. oxidation of
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Question 1 of 18Foranimationstohelpyouvisualizetheconceptsinthisquiz,goto movementof0.5/ 0.5 Pointshttp:/www.biologycorner.com/bio1/diffusion.html#Question1:SimplediffusionisdefinedastheA. molecules from areas of higher concentration to areas of lower
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Question 1 of 10Whatrole(s)doesthecytoskeletonplayinalivingcell?0.5/ 0.5 PointsA. maintaining cell shape B. movement C. contraction D. all of theseAnswerKey:DQuestion 2 of 100.5/ 0.5 PointsOfthefollowing,whatdobothmitochondriaandchloroplastshaveinc
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Question 1 of 9Theimageillustrateswhatphaseofmitosis0.5/ 0.5 PointsA. Anaphase B. Teleophase C. Metaphase D. ProphaseAnswerKey:AQuestion 2 of 90.5/ 0.5 PointsAcellthathas20chromosomesundergoesmitosis.Whichofthefollowingis true?A. two daughter cell
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Question 1 of 8Crossingoveroccursduring:0.5/ 0.5 PointsA. anaphase 1 B. metaphase 1 C. prophase 1 D. prophase 2AnswerKey:CQuestion 2 of 8Thepicturedepictswhatphaseofmeiosis0.5/ 0.5 PointsA. prophase 1 B. prophase 2 C. anaphase 1 D. anaphase 2Answ
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Question 1 of 8Apromoteris0.5/ 0.5 PointsA. A piece of DNA that can bind RNA polymerase B. A piece of RNA that is made by RNA polymerase C. A start codonAnswerKey:AQuestion 2 of 80.5/ 0.5 PointsSomepromotersareconsideredtobemoreattractivetoRNApolym
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Question 1 of 9Anoperon0.5/ 0.5 PointsA. Consists of more than one gene following a single promoter B. Allows all the genes for a pathway to be turned on and off synchronously C. Produces a polycistronic mRNA, in which start and stop codons allow for p
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Question 1 of 120.5/ 0.5 PointsThefunctionofbothalcoholfermentationandlacticacidfermentationistoA. reduce NAD+ to NADH B. reduce FADH2 to FAD+ C. reduce FAD+ to FADH2 D. oxidize NADH to NAD+ E. none of the aboveAnswerKey:DQuestion 2 of 120.5/ 0.5 Po
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Question 1 of 10Theactivesiteofanenzymeistheregionthat0.5/ 0.5 PointsA. is involved in the catalytic reaction of the enzyme. B. binds the products of the catalytic reaction. C. is inhibited by the presence of a coenzyme or a cofactor. D. both A and BA
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WhenDNAisreplicated,newnucleotidesareaddedontowhichendofthegrowingstrand?A. New nucleotides are added onto both ends. B. New nucleotides are added onto only the 5' end (where a phosphate group, PO4-, is exposed). C. New nucleotides are added onto only th
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Question 1 of 15HowcanDNAdamagebedifferentiatedfromamutation?0.5/ 0.5 PointsA. Damage and mutation are two words for the same phenomenon. B. Mutated DNA has a normal Watson-Crick structure, but different information. C. Damaged DNA has a normal Watson-
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Question 1 of 160.5/ 0.5 PointsThesugarribosecanbedifferentiatedfromthesugardeoxyriboseby determiningwhetherahydroxyl(OH)isattachedtowhichcarbonofthesugar?A. a. Carbon #1 Feedback: Correct! Examine figure 11.1 in the text. Oxygen forms one B. corner of
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Copyright 1994 Scientific American, Inc.The Origin of Life on the EarthGrowing evidence supports the idea that the emergence of catalytic RNA was a crucial early step. How that RNA came into being remains unknownby Leslie E. OrgelWhen the earth forme