Concept explainers
Interpretation:
The EMF of given voltaic cell should be calculated by using standard free energy changes of reactants and products.
Concept introduction:
Free energy change:
In
Where,
Free energy change:
The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.
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Chapter 19 Solutions
OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
- Calculate 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_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_forwardCalculate the standard cell potential of the following cell at 25C. Sn(s)Sn2+(aq)I2(aq)I(aq)arrow_forward
- A 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_forwardAnother type of battery is the alkaline zinc-mercury cell, in which the cell reaction is Zn(s) + HgO(s) Hg() + ZnO(s) E = + 1.35 V (a) What is the standard free energy change for this reaction? (b) The standard free energy change in a voltaic cell is the maximum electrical energy that the cell can produce. If the reaction in a zinc-mercury cell consumes 1.00 g mercury oxide, what is the standard free energy change? (c) For how many hours could a mercury cell produce a 10-mA current if the limiting reactant is 3.50 g mercury oxide?arrow_forwardAn electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, Hg2Cl, containing 1.00 M Cl . The cell potential of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is 0.268 V. What is the solubility product of mercury(I) chloride?arrow_forward
- Consider the following cell reaction at 25C. 2Cr(s)+3Fe2+(aq)2Cr3+(aq)+3Fe(s) Calculate the standard cell potential of this cell from the standard electrode potentials, and from this obtain G for the cell reaction. Use data in Appendix C to calculate H; note that Cr(H2O)63+(aq) equals Cr3+(aq). Use these values of H and G to obtain S for the cell reaction.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_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_forward
- 1. If you wish to convert 0.0100 mol of Au3+ (aq) ions into Au(s) in a “gold-plating” process, how long must you electrolyze a solution if the current passing through the circuit is 2.00 amps? 483 seconds 4.83 104 seconds 965 seconds 1450 secondsarrow_forwardIt 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 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
- 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 LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning