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= 0.139Va. What is the VE (the equivalence volume ) (of Ce4+) needed to reach the endpoint?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

See similar textbooks

Question

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 E^o for each part:
Ce⁴⁺ + e^- -------> Ce³⁺ E^o= 1.470V
Sn⁴⁺ + 2e^- -------> Sn²⁺ E^o= 0.139V

a. What is the V_E (the equivalence volume ) (of Ce⁴⁺) needed to reach the endpoint?

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

What cathode potential (versus SCE) would be required to lower the total Hg(II) concentration of the following solutions to 1.00 10-6 M (assume reaction product in each case is elemental Hg): (a) an aqueous solution of Hg2+? (b) a solution with an equilibrium SCN- concentration of 0.100 M? Hg2+ + 2SCN- Hg(SCN)2(aq) = Kf = 1.8 107 c) a solution with an equilibrium Br- concentration of 0.100 M? HgBr42++ 2e- Hg(l) + 4Br- E0= 0.223 V
Sulfide ion (S2- ) is formed in wastewater by the action of an aerobic bacteria on organic matter. Sulfide can be readily protonated to form volatile, toxic H2S. In addition to the toxicity and noxious odor, sulfide and H2S cause corrosion problems because they can be easily converted to sulfuric acid when conditions change to aerobic. One common method to determine sulfide is by coulometric titration with generated silver ion.At the generator electrode, the reaction is Ag Ag+ + e-. The titration reaction is S2- + 2Ag+ Ag2S(s). (a) A digital chloridometer was used to determine the mass of sulfide in a wastewater sample. The chloridometer reads out directly in ng Cl-.In chloride determinations, the same generator reaction is used,but the titration reaction is Cl- + Ag+ AgCI(s). Derive an equation that relates the desired quantity, mass S2- (ng), to the chloridometer readout in mass Cl- (ng). (b) A particular wastewater standard gave a reading of 1689.6 ng Cl-. What total charge in coulombs was required to generate the Ag+ needed to precipitate the sulfide in this standard? (c) The following results were obtained on 20.00-mL samples containing known amounts of sulfide.17 Each standard was analyzed in triplicate and the mass of chloride recorded. Convert each of the chloride results to mass S2- (ng). (d) Determine the average mass of S2- (ng), the standard deviation, and the %RSD) of each standard. (e) Prepare a plot ofthe average mass of S2- determined (ng) versus the actual mass (ng). Determine theslope, the intercept, the standard error, and the R2 value. Comment on the fit of the data to a linear model. (f) Determine the detection limit (ng) and in parts per million using a k factor of 2 (see Equation 1-12). (g) An unknown wastewater sample gave an average reading of 893.2 ng Cl. What is the mass of sulfide (ng)? If 20.00 mL of the wastewater sample was introduced into the titration vessel, what is the concentration of S2- n parts per million?
4) A redox titration: 20.00 mL of 0.00500 M Sn²+ solution was titrated with 0.0200 M Ce+*+ according to the reactio below. Titration reaction: Burst coreaning Co Ce4+ + e-------> Ce³+ Sn²+ + 2e ------> Sn²+ appy 0.00500 M Sn Calomal Pt wire Magneto stirring bar 2 Ce4+ + Sn²+ --> 2 Ce³+ + Sn¹+ Eº 1.470 V 0.139 V a) What is VE (the equivalence volume) (of Ce4+) needed to reach the endpoint? b) Calculate the potential (voltage) relative to SCE, when 5.00 mL Ce4+ is added. SCE = +0.241 V
The titration solution in the cell below had a total volume of 50.0 mL and contained 0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00 S.C.E. || Zn(s) | titration solution Zn2* + 2e- Zn(s) E = -0.762 V MgY2-: Kf = 6.2 x 108; ZnY2-: Kf = 3.2 x 1016. What will be the cell voltage when 10.0 mL of 0.100 MEDTA ha been added? (Hint: See Exercise 15-B.) i) Total mmoles of Mg+2 ii) mmoles of EDTA added is iii) mmoles of Mg*2 reacting with EDTA iv) mmoles of MgY2- v) Concentration of MgY2- in M vi) mmoles of excess unreacted Mg*2 vii) Concentration of Mg*2 in M viii) Concentration of EDTA (Y4-) in M ix) Total mmoles of Zn*2{1:NUMERICAL:%100%0.500:0.15} x) mole of ZnY2- in M xi) Concentration of ZnY2- in M
4
W
Suitable treatment of a 2.566 g soil sample converts all arsenic to AsO43-. 50.00 mL of 0.082 M Ag+ was added to form Ag3AsO4 precipitate. The excess Ag+ was back-titrated with 21.8 mL of 0.024 M SCN- to reach the equivalence point Ag+ + SCN- → AgSCN(s) Solve for the % As (MW = 74.92 g/mol) in the sample.
A soluion of 10.0 mL of 0.0500 mol/L AgNO3 is titrated with 0.0250 mol/L NaBr in the cell: VKE||onbekende oplossing||Ag(s) Calculate the potential of the cell after adding 0.1 mL, 10.0 mL, 20.0 mL and 30.0 ml of titrant. E\deg (Ag+, Ag = 0.799 V; E(VKE = 0.244 V; Ksp(ABr) = 5.0*10-13
A solution of HClO4 was standardized by dissolving 0.4008 g of primary-standardgrade HgO in a solution of KBr: HgO(s) + 4Br- + H2O --> HgBr42- + 2OH-. The liberated OH- consumed 43.75 mL of the acid. Calculate the molar concentration of the HClO4.
Given the titration reaction: Tl3+ + 2Fe2+ « Tl+ + 2Fe3+. Tl3+ + 2e- ↔ Tl+ E0 = 1.25 V Fe3+ + e- ↔ Fe2+ E0 = 0.771 VWhat is potential of 25.00 mL of 0.100 M Tl3+ after adding the following volumes of 0.100 M Fe2+:a. 10mLb. Equivalence pointc. 65mL
A 10.0 mL solution of 0.050 M AgNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E || titration solution | Ag(s) Find the cell voltage for 30.0 mL of titrant. (A) +0.093 V (B) -0.039 V (C) -0.093 V (D) +0.039 V
Suitable treatment of a 1.579 g soil sample converts all arsenic to AsO4³-. 50.00 mL of 0.07 M Ag+ was added to form Ag3AsO4 precipitate. The excess Ag+ was back-titrated with 38.97 mL of 0.021 M SCN to reach the equivalence point Ag+SCN → AgSCN (s) Solve for the % As (MW = 74.92 g/mol) in the sample. Question 26 A 0.98 g sample containing Cu (63.546 g/mol) is analyzed by reducing copper(II) to copper (1) by the reaction of iodide ions. 2Cu²++41 2Cul + 12 The sample required 18 mL of 0.1000 M Na2S2O3 to react to all the liberated 12 completely. 25203² + 1221 +S406²- What is the %Cu in the sample?