Consider the titration of 100.0 mL of 0.010 0 M Ce4+ in 1 M HClO4 with 0.040 0 M Cu+ to give Ce3+ and Cu2+. Calculate the potential of the indicator electrode after adding 24.5 mL of Cu+.
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- The following cell was found to have a potential of —0.492 V: Ag|AgCl(sat’d)||HA(0.200 M),NaA(0.300 M)|H2(1.00 atm),Pt Calculate the dissociation constant of HA, neglecting the junction potential.Calculate the electrode potentials for the following systems: (a) Cr2O72-(5.00 10-3 M),Cr3+(2.50 10-2 M),H+ (0.100 M)|Pt (b) UO22+(0.100 M),U4+ (0.200 M),H+ (0.600 M)|PtGalvanic cells harness spontaneous oxidationreduction reactions to produce work by producing a current. They do so by controlling the flow of electrons from the species oxidized to the species reduced. How is a galvanic cell designed? What is in the cathode compartment? The anode compartment? What purpose do electrodes serve? Which way do electrons always flow in the wire connecting the two electrodes in a galvanic cell? Why is it necessary to use a salt bridge or a porous disk in a galvanic cell? Which way do cations flow in the salt bridge? Which way do the anions flow? What is a cell potential and what is a volt?
- The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at 25 00C is given by the equation where contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that = 0.250 V and that a. What is the uncertainty in the values of pH and [H+] if the uncertainty in the measured potential is 1 m V ( 0.001 V)? b. To what precision must the potential be measured for the uncertainty in pH to be 0.02 pH unit?At what pH does Ecell = 0.00 V for the reduction of dichromate by iodide ion in acid solution, assuming standard-state concentrations of all species except H+ ion?A 30.0 mL solution of 0.0180 M Rh3+ in 1 M HClO4 was titrated with 0.0900 M Ce4+ to give Rh6+ and Ce3+. Calculate the potential (vs. Ag | AgCl) at the equivalence point. At what volume of titrant will the potential be equal to the standard potential of the Ce half-reaction (ignoring the Ag | AgCl electrode)?
- Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M Ce^4+.Given the formal potential for the reduction E^0 of Fe^3+ to Fe^2+ is +0.77V, and the formal potential for the reduction of E^0 Ce^4+ to Ce^3+ is 1.45V.A titration of 50.0 mL of 0.10 M Sn2+ with 0.2 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and calomel electrodes. Assuming the standard potential for Sn2+/Sn4+ = 0.139V ; calomel, 0.241 V; Fe2+/Fe3+ = 0.732 V. Calculate the voltage (E) at 25 ml, 50 ml, and 70 ml.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…
- Calculate the potential of the solution in the titration of 50.0 mL 0.100 M Fe2+ in 1.00 M HClO4 with 0.0167 M Cr2O72- at 10.00 mL titrant added.Calculate the indicator electrode potential after adding 50.0 mL of 0.10 M Ce4+ to 5.0 mL of 0.10 M Fe2+ in a medium of 1 M H2SO4. use your Eo Tables. O 0.721 V O 1.399 V O 0.736 V 1.496 V O 0.640 V O 0.624 V O 1.386 VCalculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0.197 V , E0 Ag+/Ag = 0.799 V, Kçç (AgCl) = 1.8 x10-8 A. 0.704v B. 0.507V C.0.238v D. 0.435V
