You are titrating 40.040.0 mL of 0.0100 M Sn2+Sn2+ in 1M HCIHCI with 0.0500 M T|3+TI3+ resulting in the formation of Sn4+Sn4+ and Tl+Tl+. A PtPt indicator electrode and a saturated Ag|AGCIAG|AgCl reference electrode are used to monitor the titration. Calculate the cell potential (EE) after each of the TI2. TI2.
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- A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Calculate the values of ? for the cell after each of the given volumes of the Cu+Cu+ titrant have been added.A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+ Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions. half‑reaction: ?∘=0.161 V half‑reaction: ?∘=0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. ?E of the Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V. ?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V ?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V ?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V E=0.767…A 10.0-mL solution of 0.050 0 M AgNO3 was titrated with 0.025 0 M NaBr in the cell S.C.E. || titration solution | Ag(s) Find the cell voltage for 0.1 and 30.0 mL of titrant.
- Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution.MnO4- + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2(g) + 8H2OCalculate the molar concentration of the KMnO4 solution.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*10-13A solution prepared by mixing 22.2 mL of 0.400 M NaCl and 22.2 mL of 0.400 M KI was titrated with 0.200 M AgNO, in a cell containing a silver indicator electrode and a saturated calomel reference electrode. What is [Ag*] when 21.2 mL of 0.200 M AgNO, has been added? Express your answer as x, where [Ag+] is a quotient having the form Ksp.Agi/x. 7.5 x10-16 x = Incorrect
- Calculate the potential at the equivalence point in the titration of 100 mL 0.100 M Fe²+ in 0.500 M H,SO, with 100 mL of 0.0200 M MnO4¯. Keg = 50 × 1062 E Fe3+/Fe2+=0.771V %3DConsider the titration of 100.0 mL of 0.010 0 M Ce4+ in 1 M HClO4 by 0.040 0 M Cu+ to give Ce3+ and Cu2+, using Pt and saturated Ag | AgCl electrodes to find the end point. (a) Write a balanced titration reaction. (b) Write two different half-reactions for the indicator electrode. (c) Write two different Nernst equations for the cell voltage. (d) Calculate E at the following volumes of Cu+: 1.00, 12.5, 24.5, 25.0, 25.5, 30.0, and 50.0 mL. Sketch the titration curve.The calcium content of a milk sample was to be determined by EDTA titration. A sample weighing 0.206 g was mixed with an aqueous buffer at pH 11.5. A few drops of EBT indicator were added and the solution required 24.25 mL of 1.538 x 10-2 M EDTA to reach the endpoint. Calculate the weight percent of calcium (FW =40) in the milk.
- What is the schematic representation of a cell with a saturated calomel electrode as the reference electrode and a silver indicator electrode to measure the p-function of [CrO4²-]? SCE | Ag2CrO4(sat'd), CrO42-(x M)| Ag O SCE || Ag2 CrO4(s), CrO4²-(x M)| Ag SCE ||Ag2 CrO4(sat'd), CrO4²-(x M) | Ag SCE || Ag2CrO4(s), CrO4²-(x M) | AgA 50.00 mL volume of 0.0600M K2CrO4 is mixed with 50.00 mL 0.0800 M AgNO3. Calculate the concentration of Ag+, CrO42-, K+, and NO3- at equilibrium. The solubility product of Ag2CrO4(s) is 1.20 x 10-12. Please show workTo determine the concentration of glucose in the packaged juice industry, 50 mL of sample is analyzed. Analysis of it (and other reducing sugars) was carried out there using triiodide back titration. An excess volume of 75.00 mL of 0.338 M triiodide standard solution in alkaline medium was added to the glucose solution. The resulting solution was acidified and the excess triiodide was titrated with 18.37 mL of standard 0.526 M thiosulfate solution. Calculate is the concentration in ppm in the sample. In alkaline medium: Glucose (C6H12O6) / Gluconate (C6H11O7-), Triiodide (I3- ) / Iodide (I-) In acid medium: Thiosulfate (S2O32-) / Tetrathionate (S4O62-), Triiodide (I3- ) / Iodide (I-)