The redox titration we did in lab used 0.010 M Cr2072¯ to titrate 0.060 M and 20 mL Fe2+ The titration use Pt electrode to indicate the potential of the solution and use SCE as the reference electrode, calculate the cell potential at half of the equivalent ½ Ve, at 7.0 mL, Ve and 13 mL of Cr20,2-? E for SCE is 0.241 V, Eº (Cr2O¬2- +14 H*+6e ----2Cr³+ + ZH2O) = 1.36 V

The redox titration we did in lab used 0.010 M Cr2072¯ to titrate 0.060 M and 20 mL Fe2+ The titration use Pt electrode to indicate the potential of the solution and use SCE as the reference electrode, calculate the cell potential at half of the equivalent ½ Ve, at 7.0 mL, Ve and 13 mL of Cr20,2-? E for SCE is 0.241 V, Eº (Cr2O¬2- +14 H*+6e ----2Cr³+ + ZH2O) = 1.36 V

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Is titrated with 0.110 M MnO4“. Co3+ + e = Co2+ E° = 1.92 V MnO4 + e MnO42- E° = 0.56 V Calculate the ff: 1. volume at equivalence point: 45.45 mL 2. cell potential at equivalence point: +1.24 V 3. potential when the volume of titrant is 50O mL: [ Select ]
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A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is = 5.0 x 10-13. Ksp Which precipitate forms first? O TIBI O AgBr Which of the expressions shows how the cell potential, E, depends on [Ag*]? E = 0.175 V – [0.799 V – 0.05916log ([Ag+])] 1 E = |0.799 V – 0.05916log - 0.175 V [Ag*) E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V E = 0.175 V - |0.799 V – 0.05916log [Ag* * (ল) Calculate the first and second equivalence points of the titration. first equivalence point: mL second equivalence point: mL What is the cell potential after each of the given volumes of 0.720 M NaBr have been…
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