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arning A current of 3.77 A is passed through a Pb(NO3)₂ solution for 1.70 h. How much lead is plated out of the solution?

arning A current of 3.77 A is passed through a Pb(NO3)₂ solution for 1.70 h. How much lead is plated out of the solution?

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter17: Electrochemistry And Its Applications
Section: Chapter Questions
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Macmillan Learning Ⓒ A current of 3.77 A is passed through a Pb(NO3)2 solution for 1.70 h. How much lead is plated out of the solution? mass of lead: 33.765 Incorrect g
Transcribed Image Text:Macmillan Learning Ⓒ A current of 3.77 A is passed through a Pb(NO3)2 solution for 1.70 h. How much lead is plated out of the solution? mass of lead: 33.765 Incorrect g
Quantitative Chemical Analysis Harris | Lucy TENTH EDITION Ti³+ is to be generated in 0.10 M HCIO, for coulometric reduction of azobenzene. TiO²+ + 2H+ + e Ti³+ + H₂O E = 0.100V 4 Ti³+ + C₂H₂N=NCH₂ + 4H₂O2CH₂NH₂ +4 TiO²+ + 4H+ Azobenzene current density: At the counter electrode, water is oxidized, and O, is liberated at a pressure of 0.201 bar. Both electrodes are made of smooth Pt, and each has a total surface area of 1.00 cm². The rate of reduction of the azobenzene is 25.9 nmol/s, and the resistance of the solution between the generator electrodes is 45.7 92. Calculate the current density (A/m²) at the electrode surface. overpotential: Use Table 17-1 to estimate the overpotential for O, libertation. Table 17-1 Overpotential (V) for gas evolution at various current densities (A/m²) at 25°C 10 A/m² 100 A/m² 1 000 A/m² 10 000 A/m² Electrode Platinized Pt Smooth Pt E(cathode) = 24.9 Incorrect E(anode) = E = Incorrect Incorrect Calculate the cathode potential (versus S.H.E.) assuming that [TiO²+] surface = [TiO²+]bulk = 0.0430 M and [Ti³+] surface = 0.134 M. Aniline Incorrect Incorrect presented by Macmillan Learning H₂ 0₂ H₁₂ 0₂ 0.0154 0.398 0.0300 0.521 0.024 0.721 0.068 0.85 Calculate the anode potential(versus S.H.E.). Refer to the table of standard reduction potentials as needed. What should be the applied voltage? H₂ 0₂ H₂ 0.0405 0.638 0.0483 0.288 1.28 0.676 0₂ 0.766 1.49 V V V A/m²
Transcribed Image Text:Quantitative Chemical Analysis Harris | Lucy TENTH EDITION Ti³+ is to be generated in 0.10 M HCIO, for coulometric reduction of azobenzene. TiO²+ + 2H+ + e Ti³+ + H₂O E = 0.100V 4 Ti³+ + C₂H₂N=NCH₂ + 4H₂O2CH₂NH₂ +4 TiO²+ + 4H+ Azobenzene current density: At the counter electrode, water is oxidized, and O, is liberated at a pressure of 0.201 bar. Both electrodes are made of smooth Pt, and each has a total surface area of 1.00 cm². The rate of reduction of the azobenzene is 25.9 nmol/s, and the resistance of the solution between the generator electrodes is 45.7 92. Calculate the current density (A/m²) at the electrode surface. overpotential: Use Table 17-1 to estimate the overpotential for O, libertation. Table 17-1 Overpotential (V) for gas evolution at various current densities (A/m²) at 25°C 10 A/m² 100 A/m² 1 000 A/m² 10 000 A/m² Electrode Platinized Pt Smooth Pt E(cathode) = 24.9 Incorrect E(anode) = E = Incorrect Incorrect Calculate the cathode potential (versus S.H.E.) assuming that [TiO²+] surface = [TiO²+]bulk = 0.0430 M and [Ti³+] surface = 0.134 M. Aniline Incorrect Incorrect presented by Macmillan Learning H₂ 0₂ H₁₂ 0₂ 0.0154 0.398 0.0300 0.521 0.024 0.721 0.068 0.85 Calculate the anode potential(versus S.H.E.). Refer to the table of standard reduction potentials as needed. What should be the applied voltage? H₂ 0₂ H₂ 0.0405 0.638 0.0483 0.288 1.28 0.676 0₂ 0.766 1.49 V V V A/m²
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