Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 19, Problem 9PE
Interpretation Introduction
Interpretation:
The direction in which
Concept introduction:
Redox reactions are described as the reactions in which electrons are lost by one species, and gained by another species present in the
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Chapter 19 Solutions
Introductory Chemistry: An Active Learning Approach
Ch. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Classify each of the following half-reaction...Ch. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10E
Ch. 19 - Prob. 11ECh. 19 - Identify each of the following half-reaction as...Ch. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - In this section, each equation identifies an...Ch. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - As an example of an electrolytic cell, the text...Ch. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 19.1TCCh. 19 - Prob. 19.2TCCh. 19 - Prob. 19.3TCCh. 19 - Prob. 1CLECh. 19 - Prob. 2CLECh. 19 - Prob. 3CLECh. 19 - Prob. 4CLECh. 19 - Prob. 5CLECh. 19 - Prob. 1PECh. 19 - Prob. 2PECh. 19 - Prob. 3PECh. 19 - Prob. 4PECh. 19 - Prob. 5PECh. 19 - Prob. 6PECh. 19 - Consider the reaction of copper and nitric acid:...Ch. 19 - Prob. 8PECh. 19 - Prob. 9PECh. 19 - Prob. 10PECh. 19 - Prob. 11PECh. 19 - Aqueous chromate ion, CrO42(aq), and hydrogen...Ch. 19 - Prob. 13PE
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- 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_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 potential developed by a voltaic cell using the following reaction if all dissolved species are 0.015 M. 2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) 2 Fe3+(aq) + 2 H2O()arrow_forward
- Consider a voltaic cell in which the following reaction occurs. Zn(s)+Sn2+(aq)Zn2+(aq)+Sn(s) (a) Calculate E° for the cell. (b) When the cell operates, what happens to the concentration of Zn2+? The concentration of Sn2+? (c) When the cell voltage drops to zero, what is the ratio of the concentration of Zn2+ to that of Sn2+? (d) If the concentration of both cations is 1.0 M originally, what are the concentrations when the voltage drops to zero?arrow_forwardUse the following half-equations to write three spontaneous reactions. Justify your answers by calculating E° for the cells. 1. MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2OE=+1.512V 2. O2(g)+4H+(aq)+4e2H2OE=+1.229V 3. Co2+(aq)+2eCo(s)E=0.282Varrow_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
- Calculate E° for the following voltaic cells: (a) MnO2+4H+(aq)+2I(aq)Mn2+(aq)+2H2O+I2(s) (b) H2(g)+2OH(aq)+S(s)2H2O+S2(aq) (c) an Ag-Ag+ half-cell and an Au-AuCl4- half-cellarrow_forwardConsider a voltaic cell in which the following reaction takes place. 2F2+(aq)+H2O2(aq)+2H+(aq)2Fe3+(aq)+2H2O (a) Calculate E°. (b) Write the Nernst equation for the cell. (c) Calculate E at 25°C under the following conditions: [Fe2+]=0.00813M,[H2O2]=0.914M,[Fe3+]=0.199M,ph=2.88 .arrow_forwardThe following oxidationreduction reactions are used in electrochemical cells. Write them using cell notation. (a) 2Ag+(aq)(0.50M)+Ni(s)2Ag(s)+Ni2+(aq)(0.20M) (b) Cu(s)+PtCl62-(aq)(0.10M)Cu2+(aq)(0.20M)+PtCl42-(aq)(0.10M)+2Cl(aq)(0.40M) (c) Pb(s)+SO42-(aq)(0.30M)+2AgCl(s)PbSO4(s)+2Ag(s)+2Cl(aq)(0.20M) (d) In a galvanic cell, one half-cell contains 0.010 M HCI and a platinum electrode, over which H2 is bubbled at a pressure of 1.0 atm. The other half-cell is composed of a zinc electrode in a 0.125 M solution of Zn(NO3)2.arrow_forward
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