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Wright CHM 1220 - CHM 1220 SI_Electrochemistry_Rev 2_ANSWERS

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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. Ⓑ 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.Chapter 17 SI Activity: Electrochemistry CHM 1220Problems2. 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 half-reaction and +1.68 V for the manganese half-reaction. Pb2+ (aq, 0.10 M) + 2 e–  Pb (s) MnO4– (aq, 1.50 M) + 4 H+ (aq) + 3 e–  MnO2 (s) + 2 H2O (l)3. The redox reaction Aa+ + 2 B  A + 2 Bb+ with a Gibbs energy change of –552 kJ produces a standard cell po-tential of 1.43 V. What is the charge on the ion Aa+? On Bb+?ANSWERSQuestions? Comments? Suggestions? Email me at steven [email protected] from Dan Harris’ Quantitative Chemical Analysis, 9/e, ©2015 by W. H. Freeman and Company, an imprint of Macmillian Education.Calculate Cell Potential:Cd (s) | Cd2+ (aq) || Ag+ (aq) | Ag (s)EE Ecell cathode anode V V V¥¥ ¥080040 120...Determine overall rxn:Need number of electrons transferred:From stoichiometry:Need Nernst Eqn: 3 Pb (s) + 2 MnO4– (aq, 1.50 M) + 8 H+ (aq)  3 Pb2+ (aq, 0.10 M) + 2 MnO2 (s) + 4 H2O (l)–3[ ]2[ ]EERTnFQERTnFcell cell cellPbMnOH ¥¥ln ln2342811680138 31447 95 0 273 156965 1001043......ln.115010179228.. VrrJmolCmol VGnFE nGFE¥¥¥¥ 552 109651014334.. 4A and B42++Rev. 1 – Fall 2017 – J. Stevens4. Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dis-solve, 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 dis-solution 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– .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?6. Calculate E¥ , ΔrG¥ , and K for the reaction as written below.2 Cu2+ (aq) + 2 I– (aq) + OHHO (aq)  2 CuI (s) + 2 H+ (aq) + OO (aq)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) E¥red = +0.339 V CuI (s) + e–  Cu (s) + I– (aq) E¥red = –0.185 V OO (aq) + 2 H+ (aq) + 2 e–  OHHO (aq) E¥red = +0.700 Vhydroquinone quinoneRelevant Half-Reactions & their Summation:Electrolysis Dimensional Analysis: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.7601120024136001.. A A days hrs days hCsrrmol e Cmol Cr3 mol e51.996 965104.gg Crmol Cr g 235 8.12 22220 185 35 701... Cu I CuI e krsaqs GF  ¥JJ Cu e Cu kJhr22 42 40339 130 83322.... aqsGF ¥yydro quinone 2 H e kJrr aq GFG220 700 135 083¥..4412331 4¥ ¥¥¥rrr kJGGG .EGF4420 163¥¥ r V.Ke eGRT  r431 40 008314 298 1532 105¥....Au ClAuCl eNO H e NO saqaqaq aq g434343244234 H OAu Cl HNOAuCl NO saqaqaqg22 H O


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Wright CHM 1220 - CHM 1220 SI_Electrochemistry_Rev 2_ANSWERS

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