None 12. Calculate E° cell for a silver-aluminum cell in which the cell reaction isAl(s)+3Ag+(aq) → Al3+(aq) + 3Ag(s)A. -2.46 VB. +0.86 VC. -0.86 VD. +2.46 V(1) Al3+(aq) +3e → Al(s)(2) Al(OH)4(aq) + 3e → Al(s) + 40H(aq)(3) Ag (aq) + e→ Ag(s)(4) Ag2+(aq) + e→ Ag+(aq)E. None of theseE° = -1.66 VE°= -2.35 VE° = +0.80 VE° = +2.00 V

Reversing half cell reaction (1), Eo sign will be reversedAl(s)→Al3+(aq)+3e−,Eo=1.66VMultiplying by…

Answered: 12. Calculate E° cell for a…

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter19: Principles Of Chemical Reactivity: Electron Transfer Reactions

Section19.5: Electrochemical Cells Under Nonstandard Conditions

Problem 1RC

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An electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag+] = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 M NH3 that is also 0.010 M in Cu(NH3)42+, what is the cell potential at 25C? Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)K=1.01013
Calculate 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)
Consider the following galvanic cell: A 15 0-mole sample of NH is added to the Ag compartment (assume 1.00 L of total solution after the addition). The silver ion reacts with ammonia to form complex ions as shown: Ag+(aq)+NH3(aq)AgNH3+(aq)K1=2.1103AgNH3+(aq)+NH3(aq)Ag(NH3)2+(aq)K2=8.2103 Calculate the cell potential after the addition of 15.0 moles of NH3.
An 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)?
An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at 25C if the aluminum electrode is placed in a solution in which [Al3+] = 7.2 103 M? b. When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25C is 1.62 V. Calculate [Al3+] in the unknown solution. (Assume Al is oxidized.)
Consider the cell described below: Zn|Zn2+(1.00M)||Cu2+(l.00M)|Cu Calculate the cell potential after the reaction has operated long enough for the [Zn2+] to have changed by 0.20 mol/L. (Assume T=25C.)
What is the voltage of a concentration cell of Fe2+ ions where the concentrations are 0.0025 and 0.750 M? What is the spontaneous reaction?
An 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?
A 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.
What is the maximum work you can obtain from 30.0 g of nickel in the following cell when the cell potential is 0.97 V? Ni(s)Ni2+(aq)Ag+(aq)Ag(s)
Consider the cell described below: Al|Al3+(1.00M)||Pb2+(1.00M)|Pb Calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.60 mol/L. (Assume T = 25C.)
An electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag+] = 1.00 M separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00 M NH3 that also contains 2.4 103 M Cu(NH3)42+. The equilibrium between Cu2+ and NH3 is: Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)K=1.01013 and the two cell half-reactions are: Assuming Ag+ is reduced, what is the cell potential at 25C?

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