PRINCIPLES OF MODERN CHEMISTRY-OWLV2
8th Edition
ISBN: 9781305271609
Author: OXTOBY
Publisher: CENGAGE L
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- What is the standard cell potential you would obtain from a cell at 25C using an electrode in which I(aq) is in contact with I2(s) and an electrode in which a chromium strip dips into a solution of Cr3(aq)?arrow_forwardAt 298 K, the solubility product constant for solid Ba(IO3)2 is 1.5 109. Use the standard reduction potential of Ba2+(aq) to find the standard potential for the half-reaction Ba(IO3)2(s)+2eBa(s)+2IO3(aq)arrow_forwardWhat is the standard cell potential you would obtain from a cell at 25C using an electrode in which Hg22+(aq) is in contact with mercury metal and an electrode in which an aluminum strip dips into a solution of Al3+(aq)?arrow_forward
- At 298 K, the solubility product constant for Pb(IO3)2 is 2.6 1013, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction Pb(IO3)2(s)+2ePb(s)+2IO3(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions, and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/Pb(IO3)2 electrode in a 3.5 103 M solution of NaIO3.arrow_forwardAt 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction PbC2O4(s)+2ePb(s)+C2O42(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/PbC2O4 electrode in a 0.025 M solution of Na2C2O4.arrow_forwardCalculate the cell potential of a cell operating with the following reaction at 25C, in which [MnO4] = 0.010 M, [Br] = 0.010 M. [Mn2] = 0.15 M, and [H] = 1.0 M. 2MNO4(aq)+10Br(aq)+16H+(aq)2MN2(aq)+5Br2(l)+8H2O(l)arrow_forward
- Calculate the standard cell potential of the following cell at 25C. Cr(s)Cr3(aq)Hg22(aq)Hg(l)arrow_forwardAn electrolysis experiment is performed to determine the value of the Faraday constant (number of coulombs per mole of electrons). In this experiment, 28.8 g of gold is plated out from a AuCN solution by running an electrolytic cell for two hours with a current of 2.00 A. What is the experimental value obtained for the Faraday Constant?arrow_forwardA galvanic cell is based on the following half-reactions: In this cell, the copper compartment contains a copper electrode and [Cu2+] = 1.00 M, and the vanadium compartment contains a vanadium electrode and V2+ at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M H2EDTA2, resulting in the reaction H2EDTA2(aq)+V2+(aq)VEDTA2(aq)+2H+(aq)K=? The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 mL H2EDTA2 solution added. At the stoichiometric point, was observed to be 1 .98 V. The solution was buffered at a pH of 10.00. a. Calculate before the titration was carried out. b. Calculate the value of the equilibrium constant, K, for the titration reaction. c. Calculate at the halfway point in the titration.arrow_forward
- It took 150. s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of 1+.arrow_forwardCalculate the standard cell potential of the cell corresponding to the oxidation of oxalic acid, H2C2O4, by permanganate ion. MnO4. 5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l) See Appendix C for free energies of formation: Gf for H2C2O4(aq) is 698 kJ.arrow_forwardAn aqueous solution of an unknown salt of gold is electrolyzed by a current of 2.75 amps for 3.39 hours. The electroplating is carried out with an efficiency of 93.0%, resulting in a deposit of 21.221 g of gold. a How many faradays are required to deposit the gold? b What is the charge on the gold ions (based on your calculations)?arrow_forward
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Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY