Wright CHM 1220 - CHM 1220 SI_Electrochemistry_Rev 2_ANSWERS (2 pages)

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CHM 1220 SI_Electrochemistry_Rev 2_ANSWERS



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CHM 1220 SI_Electrochemistry_Rev 2_ANSWERS

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Wright State University
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Chm 1220 - General Chemistry II
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Questions Comments Suggestions Email me at stevens 161 wright edu Chapter 17 SI Activity Electrochemistry ANSWERS CHM 1220 Problems 1 Consider an electrochemical cell constructed of cadmium and silver electrodes with nitrate salts of both metals as charge balancing electrodes and a potassium nitrate salt bridge Standard reduction potentials are provided below for your convenience Cd2 aq 2 e Cd s E red 0 40 V Ag aq e Ag s E red 0 80 V Represent this in cell shorthand notation Cd s Cd2 aq Ag aq Ag s Sketch the full cell in the space below Be sure to include anode cathode labels indicate the flow of electrons and ions and calculate the standard cell reduction potential Calculate Cell Potential Ecell Ecathode Eanode 0 80 V 0 40V 1 20 V Figure from Dan Harris Quantitative Chemical Analysis 9 e 2015 by W H Freeman and Company an imprint of Macmillian Education 2 An electrochemical cell is constructed from the two half reactions below and is buffered at a pH of 2 000 Calculate the cell potential at 95 C given that the standard reduction potential is 0 13 V for the lead halfreaction and 1 68 V for the manganese half reaction 3 2 aq 0 10 M 2 e Pb s Pb MnO 2 4 aq 1 50 M 4 H aq 3 e MnO2 s 2 H2O l Determine overall rxn 3 Pb s 2 MnO4 aq 1 50 M 8 H aq 3 Pb2 aq 0 10 M 2 MnO2 s 4 H2O l 2 3 Need Nernst Eqn Pb RT RT Ecell Ecell ln Q Ecell ln nF nF MnO 2 H 8 4 8 31447 95 0 273 15 ln 0 10 1 68 0 13 8 2 6 9 65 10 4 1 50 10 2 3 1 79 V 3 The redox reaction Aa 2 B A 2 Bb with a Gibbs energy change of 552 kJ produces a standard cell potential of 1 43 V What is the charge on the ion Aa On Bb J 552 103 mol r G Need number of electrons transferred r G nFE n 4 FE 9 65 104 molC 1 43 V From stoichiometry A 4 and B2 4 Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid It will dissolve however in aqua regia a mixture of the two concentrated acids The dissolution products are the AuCl4 complex and gaseous NO Using the method of half reactions write a balanced chemical equation for the dissolution of gold in aqua regia If the AuCl4 half reaction has a standard reduction potential of 1 00 V suggest a substance that could recover the gold metal from AuCl4 Relevant Half Reactions their Summation Au s 4 Cl aq NO3 aq 4 H aq 3 e AuCl 4 aq 3 e NO g 2 H 2O Au s 4 Cl aq 4 H aq NO3 AuCl 4 aq NO g 2 H 2 O Since Au is oxidized from Au0 to Au3 the tetrachloroaurate anion would need to be reduced to recover the gold Therefore substance with a reduction potential LOWER THAN 1 00 V would do the trick 5 A Cr3 aq solution is electrolyzed using a current of 7 60 A What mass of Cr s is plated out after 2 00 days Electrolysis Dimensional Analysis 24 hrs 3600 s mol e mol Cr 51 996 g Cr 1 Cs 2 00 days 235 8 g 4 1 A 1 day 1 hr 9 65 10 C 3 mol e mol Cr 7 60 A 6 Calculate E rG and K for the reaction as written below 2 Cu2 aq 2 I aq HO OH aq 2 CuI s 2 H aq O hydroquinone O aq quinone All or part of the list of half reactions below may prove useful in your progress toward a solution Cu aq e Cu s E red 0 518 V Cu2 aq 2 e Cu s Ered 0 339 V CuI s e Cu s I aq Ered 0 185 V O O aq 2 H aq 2 e HO 1 2 Cu s 2 I aq 2 CuI s 2 e 2 2 Cu 2 aq 4 e 2 Cu s 3 hydro quinone 2 H aq 2 e r G4 r G1 r G2 r G3 31 4 kJ E4 r G4 0 163 V 2F Rev 1 Fall 2017 J Stevens E red 0 700 V OH aq K e r G1 2 F 0 185 35 70 kJ r G2 4 F 0 339 130 83 kJ r G3 2 F 0 700 135 08 kJ r G4 31 4 RT e 0 008314 298 15 3 2 105


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