(a)
Interpretation:
A concentration cell having Copper electrodes at different concentration is given. The cell potential for the given concentration cell at the given concentration of ions and the value of new cell potential after enough
Concept introduction:
A cell consisting of electrodes of same type which have different concentration of ions is called concentration cell. The difference in ion concentration in different compartments is the driving force for the formation of the concentration cell.
To determine: The potential for the given concentration cell.
The value of cell potential for the given cell is
(b)
Interpretation:
A concentration cell having Copper electrodes at different concentration is given. The cell potential for the given concentration cell at the given concentration of ions and the value of new cell potential after enough
Concept introduction:
A cell consisting of electrodes of same type which have different concentration of ions is called concentration cell. The difference in ion concentration in different compartments is the driving force for the formation of the concentration cell.
To determine: The value of new cell potential after enough
The value of new cell potential for the given cell is
Want to see the full answer?
Check out a sample textbook solutionChapter 17 Solutions
Chemistry: An Atoms First Approach
- You have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 1.0 104M (left side). a. Calculate the potential for this cell at 25C. b. The Cu2+ ion reacts with NH3 to form Cu(NH3)2+ by the following equation: Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)K=1.01013 Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium [NH3] = 2.0 M.arrow_forwardAn 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=1.01013arrow_forwardA voltaic cell is constructed in which one half-cell consists of a silver wire in an aqueous solution of AgNO3.The other half cell consists of an inert platinum wire in an aqueous solution containing Fe2+(aq) and Fe3+(aq). (a) Calculate the cell potential, assuming standard conditions. (b) Write the net ionic equation for the reaction occurring in the cell. (c) Which electrode is the anode and which is the cathode? (d) If [Ag+] is 0.10 M, and [Fe2+] and [Fe3+] are both 1.0 M, what is the cell potential? Is the net cell reaction still that used in part (a)? If not, what is the net reaction under the new conditions?arrow_forward
- 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)?arrow_forwardAn 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.)arrow_forwardConsider a battery made from one half-cell that consists of a capper electrode in 1 M CuSO4 solution and another half—cell that consists of a lead electrode in 1 M Pb(NO3)2 solution. (a) What are the reactions at the anode, cathode, and the overall reaction? (b) What is the standard cell potential for the battery? (c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this tell he used to make a battery that could replace a dry-cell battery? Why or why not. (d) Suppose sulfuric acid is added to the half—cell with the lead electrode and some PbSO4(s) forms. Would the cell potential increase, decrease, or remain the same?arrow_forward
- 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?arrow_forwardConsider 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.arrow_forwardManganese may play an important role in chemical cycles in the oceans. Two reactions involving manganese (in acid solution) are the reduction of nitrate ions (to NO) with Mn2+ ions and the oxidation of ammonium ions (to N2) with MnO2. (a) Write balanced chemical equations for these reactions (in acid solution). (b) Calculate Ecell for the reactions. (One half-reaction potential you need is for the reduction of N2 to NH4+, E = 0.272 V.)arrow_forward
- Consider the following reduction potentials: Co3++ 3e Co = 1.26 V Co2++ 2e Co = 0.28 V a. When cobalt metal dissolves in 1.0 M nitric acid, will Co3+ or Co2+ be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of HNO3 to get a different result in part a? Concentrated HNO3 is about 16 M.arrow_forwardAn 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 immersed in a solution with [Al3+] = 1.0 M. Sodium hydroxide is added to the aluminum compartment, causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3 has ceased, the concentration of OH is 1.0 104 M and the measured cell potential is 1.82 V. Calculate the Ksp value for Al(OH)3. Al(OH)3(s)Al3+(aq)+3OH(aq)Ksp=?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
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning