Concept explainers
Balance each of the following unbalanced equations; then calculate the standard potential, E°, and decide whether each is product favored at equilibrium as written. (All reactions are carried out in acid solution.)
- (a) Sn2+(aq) + Ag(s) → Sn(s) + Ag+(aq)
- (b) Al(s) + Sn4+(aq) → Sn2+(aq) +Al3+(aq)
- (c) ClO3−(aq) + Ce3+(aq) → Cl2(g) + Ce4+(aq)
- (d) Cu(s) + NO3− (aq) → Cu2+(aq) + NO(g)
a)
Interpretation:
The
Concept introduction:
Electrochemical cells:
In this type of cells, chemical energy is converted into electrical energy.
In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.
An anode is indicated by negative sign and cathode is indicated by the positive sign.
Electrons flow in the external circuit from the anode to the cathode.
In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.
Under certain conditions a cell potential is measured it is called as standard potential
Standard potential
The
The
Answer to Problem 19PS
The value is negative, therefore it is not product favoured.
Explanation of Solution
The given reaction is as follows.
Balance the each half reaction:
Find the overall reaction.
Let’s write the half reactions:
Let’s calculate the
The value is negative, therefore it is not product favoured.
b)
Interpretation:
The
Concept introduction:
Electrochemical cells:
In this type of cells, chemical energy is converted into electrical energy.
In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.
An anode is indicated by negative sign and cathode is indicated by the positive sign.
Electrons flow in the external circuit from the anode to the cathode.
In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.
Under certain conditions a cell potential is measured it is called as standard potential
Standard potential
The
The
Answer to Problem 19PS
The value is positive, therefore it is a product favoured.
Explanation of Solution
The given reaction is as follows.
Balance the each half reaction:
Find the overall reaction.
Let’s write the half reactions:
Let’s calculate the
The value is positive, therefore it is a product favoured.
c)
Interpretation:
The
Concept introduction:
Electrochemical cells:
In this type of cells, chemical energy is converted into electrical energy.
In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.
An anode is indicated by negative sign and cathode is indicated by the positive sign.
Electrons flow in the external circuit from the anode to the cathode.
In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.
Under certain conditions a cell potential is measured it is called as standard potential
Standard potential
The
The
Answer to Problem 19PS
The value is negative, therefore it is not a product favoured
Explanation of Solution
The given reaction is as follows.
Balance the each half reaction:
The second equation is multiply with 10 .we get equal number of electrons.
Find the overall reaction.
Let’s write the half reactions:
Let’s calculate the
The value is negative, therefore it is not a product favoured.
d)
Interpretation:
The
Concept introduction:
Electrochemical cells:
In this type of cells, chemical energy is converted into electrical energy.
In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.
An anode is indicated by negative sign and cathode is indicated by the positive sign.
Electrons flow in the external circuit from the anode to the cathode.
In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.
Under certain conditions a cell potential is measured it is called as standard potential
Standard potential
The
The
Answer to Problem 19PS
The value is positive, therefore it is a product favoured.
Explanation of Solution
The given reaction is as follows.
Balance the each half reaction:
The first equation is multiplied with 3 and second equation is multiply with 2.we get equal number of electrons.
Find the overall reaction.
Let’s write the half reactions:
Let’s calculate the
The value is negative, therefore it is not product favoured.
Want to see more full solutions like this?
Chapter 19 Solutions
CHEMISTY+CHEMICAL REACTIVITY
- 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?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_forwardFor each of the reactions, calculate E from the table of standard potentials, and state whether the reaction is spontaneous as written or spontaneous in the reverse direction under standard conditions. (a) Zn(s)+Fe2+(aq)Zn2+(aq)+Fe(s) (b) AgCl(s)+Fe2+(aq)Ag(s)+Fe3+(aq)+Cl(aq) (c) Br2(l)+2Cl(aq)Cl2(g)+2Br(aq)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_forwardFor each reaction listed, determine its standard cell potential at 25 C and whether the reaction is spontaneous at standard conditions. (a) Mn(s)+Ni2+(aq)Mn2+(aq)+Ni(s) (b) 3Cu2+(aq)+2Al(s)2Al3+(aq)+3Cu(s) (c) Na(s)+LiNO3(aq)NaNO3(aq)+Li(s) (d) Ca(NO3)2(aq)+Ba(s)Ba(NO3)2(aq)+Ca(s)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
- A half-cell that consists of a copper wire in a 1.00 M Cu(NO3)2 solution is connected by a salt bridge to a solution that is 1.00 M in both Pu3+ and Pu4+, and contains an inert metal electrode. The voltage of the cell is 0.642 V, with the copper as the negative electrode. (a) Write the half-reactions and the overall equation for the spontaneous chemical reaction. (b) Use the standard potential of the copper half-reaction, with the voltage of the cell, to calculate the standard reduction potential for the plutonium half-reaction.arrow_forwardYou have 1.0 M solutions of Al(NO3)3 and AgNO3 along with Al and Ag electrodes to construct a voltaic cell. The salt bridge contains a saturated solution of KCl. Complete the picture associated with this problem by a writing the symbols of the elements and ions in the appropriate areas (both solutions and electrodes). b identifying the anode and cathode. c indicating the direction of electron flow through the external circuit. d indicating the cell potential (assume standard conditions, with no current flowing). e writing the appropriate half-reaction under each of the containers. f indicating the direction of ion flow in the salt bridge. g identifying the species undergoing oxidation and reduction. h writing the balanced overall reaction for the cell.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_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 LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningPrinciples of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
- Chemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning