3) A redox titration: 160.00 mL of 0.0500 M Sn2+ solution was titrated with 0.200 M Ce according to thereaction below.Buret containing CeCalomelreferenceelectrode.0.00500 M SnMagneticstirring barTitration reaction:2 Ce++ Sn2+-------> 2 Ce3+ + Sn4+EºCee-------> Ce3+Sn+ + 2e------> Sn²+1.470 V0.139 Va) What is V. (the equivalence volume) (of Ce") needed to reach the endpoint?b) Calculate the potential (voltage), when 10.00 mL Ce" is added.c) What is the cell potential at the equivalence point?d) What is the cell potential when 60 mL Ce4+ is added?SCE = +0.241 V

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Answered: 3) A redox titration: 160.00 mL of…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.40QAP

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A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139Va. What is the VE (the equivalence volume ) (of Ce4+) needed to reach the endpoint?
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VCalculate the potential (voltage), when 10.00mL of Ce4+ is added. (SCE= +0.241 V)
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VWhat is the cell potential at the equivalence point?
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139V\What is the cell potential when 60mL Ce4+ is added?
please answer 2,3 and 4
+ H (a) CuCN (b) CaCO3 (c) Ag₂CO3 (d) PbBr₂ (e) Y(OH)3 Periodic Table B W Chapter... Supporting Materials Additional Materiala [Ca²+] = 5.2 x 104 M, [CO32- ] = 6.5x 10-6 M * G Please use the values in the resources listed below instead of the textbook values. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated. [Cu] = 1.9x10-10 M, [CN] = 1.9x10-¹0 M [Ag+] = 1.8x10-5 M, [CO32] = 2.7x10-² M [Pb²+] = 1.2x 10-2 M, [Br] = 2.4x10-2 M [Y³+] = 3.5x 10-16 M, [OH-] = 6.6x 10-³ M Constants and Factors №. webassign.net ● F 893 Supplemental Data L C C a Ⓒ + D r
A 1.000-g sample containing KHC204, H2C204, and impurities required 38.22 mL of 0.1000 M NaOH for acid-base titration. Another 1.000-g sample required 27.02 mLof 0.04000 M KMNO, for redox titration in acid solution, according to the following reaction: Mno, + 5C,0,2- + 16H* 2 2MN2* + 10co, + 8H,0 How many mmols of KHC,0, are there in the sample? 1.351 mmols KHC204 1.582 mmols KHC204 O 1.120 mmols KHC204 1.911 mmols KHC204
A 20-tablet sample of soluble saccharin was treated with 20.00 mL of 0.08181 M AgNO3 . The reaction is : • R-N-Na + Ag+  R-N-Ag(s) + Na+ • After removal of the solid, titration of the filtrate and washings required 2.81 mL of 0.04124 M SCN- . Ag+ + SCN−  AgSCN(s) (Ag+ titrated with SCNrepresents the unreacted Ag+ in the above solution.) • Calculate the average number of milligrams of saccharin (205.17) in each tablet.
3
5. The sulfate in 186.5 mg sample was precipitated as BaSO, by addition of 32.00 mL of 0.0340 M BaCI,. The precipitate was removed by filtration and the remaining BaCl, consumed 37.5 mL of 0.0200 M EDTA for titration to the end point. Calculate the % SO, in the sample. (MW of SO, is 80.06 g/mol)
A 0.64 g sample containing KCl ( mw = 74.6 ) is dissolved in 50mL of water and titrated to the Ag2CrO4 end point, requiring 26.2 mL of 0.15 M AgNO3 .The %w/w KCl in the sample is ?
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