Concept explainers
Interpretation:
The value of
Concept introduction:
In an
If oxidation takes place on an electrode, that electrode is called anode. The removal of electrons takes place from the species present in anode.
If reduction takes place on an electrode, that electrode is called cathode. The addition of electrons takes place to the species present in cathode. The electrode potential of cell is calculated from reduction electrode potential of cathode and anode as follows:
Nernst equation is as follows:
Z = number of moles of electrons transferred in the cell.
Q is ratio of concentration of product to reactant.
Want to see the full answer?
Check out a sample textbook solutionChapter 19 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
- What 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_forwardWhat is the cell potential (Ecell) of a spontaneous cell that is run at 25C and contains [Cr3+] = 0.10 M and [Ag+] = 1.0 104 M?arrow_forward
- 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)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_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_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_forwardConsider a galvanic cell based on the following half-reactions: a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at 25C with [Fe2+] = 2.00 104 M and [La3+] = 3.00 103 M, what is the expected cell potential?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_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_forwardFour voltaic cells are set up. In each, one half-cell contains a standard hydrogen electrode. The second half-cell is one of the following: (i) Cr3+(aq, 1.0 M)|Cr(s) (ii) Fea+(aq, 1.0M)|Fe(s) (iii) Cu2+(aq, 1.0M)|Cu(s) (iv) Mg2+(aq, 1.0M)|Mg(s) (a) In which of the voltaic cells does the hydrogen electrode serve as the cathode? (b) Which voltaic cell produces the highest potential? Which produces the lowest potential?arrow_forwardCalculate the cell potential of a cell operating with the following reaction at 25C, in which [Cr2O32] = 0.020 M, [I] = 0.015 M, [Cr3+] = 0.40 M, and [H+] = 0.60 M. Cr2O72(aq)+6I(aq)+14H+(aq)2Cr3+(aq)+3I2(s)+7H2O(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: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning