Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 21, Problem 21.152P

(a)

Interpretation Introduction

Interpretation:

For the given voltaic cell the Ecell,ΔGandK values have to be calculated.

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

Δ°G=-nFE°cellwhere,n - Number of electrons involved per equivalent of the net redox reaction in the cellF - Faraday’s Constant (96500 C) cell- Standard cell potential.

The Nernst equation depicts the relationship between Ecell and Eocell as follows,

Ecell=Eocell0.0592VnlogQwhere,Ecell=cellpotentialEocell=Standardcellpotentialn=No.ofelectronsQ=ReactionQuotient

(a)

Expert Solution
Check Mark

Explanation of Solution

The half reactions for given cell with their Eo values are noted by using Appendix D as follows,

Cd(s)Cd2+(aq)+2e E(Oxidation)=0.40VCu2+(aq)+2eCu(s) E(Reduction)=0.34V_Cd(s)+Cu2+(aq)Cu(s)+Cd2+(aq)

Observing the standard reduction potentials values it is clear that Cd oxidizes and Cu2+ reduces since, Cd is better reducing agent.

The Ecell value is calculated by using the equation cell= E°cathode - E°anode.

cell= E°cathode - E°anode=0.34V-(-0.40V)=0.74V

Next, using the Δ°G=-nFE°cell equation, the Δ°G value is calculated.

Δ°G=-nFE°cell= -(96485C/mole-)×(0.74J/C)=1.427978×105=1.4×105J

Finally, using the number of electrons and the cell potential the K value is determined.

logK=nE°cell0.0592logK= (0.74)0.0592=25K=1×1025

(b)

Interpretation Introduction

Interpretation:

The Ecell,andΔG values have to be calculated when the given voltaic cell started to operate with increased [Cd2+].

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

Δ°G=-nFE°cellwhere,n - Number of electrons involved per equivalent of the net redox reaction in the cellF - Faraday’s Constant (96500 C) cell- Standard cell potential.

The Nernst equation depicts the relationship between Ecell and Eocell as follows,

Ecell=Eocell0.0592VnlogQwhere,Ecell=cellpotentialEocell=Standardcellpotentialn=No.ofelectronsQ=ReactionQuotient

(b)

Expert Solution
Check Mark

Explanation of Solution

The half reactions for given cell with their Eo values are noted by using Appendix D as follows,

Cd(s)Cd2+(aq)+2e E(Oxidation)=0.40VCu2+(aq)+2eCu(s) E(Reduction)=0.34V_Cd(s)+Cu2+(aq)Cu(s)+Cd2+(aq)

Observing the standard reduction potentials values it is clear that Cd oxidizes and Cu2+ reduces since, Cd is better reducing agent.

Ecell=Eocell0.0592VnlogQEcell=0.74V0.0592V2log[Cd2+][Cu2+] (since,Q=ProductconcentrationReactantconcentration)

Observing the cell equation shows that the moles ratios of given ions is 1:1 and cell operates the [Cd2+] concentration increases with equal decrease in Cu2+ concentration. Therefore, the reduced [Cu2+] is 0.05M since, [Cd2+] increases from 1 to 1.95M.

Ecell=0.74V0.0592V2log[1.95][0.05]=0.69V

(c)

Interpretation Introduction

Interpretation:

For the given voltaic cell the Ecell,andΔG values have to be calculated for [Cu2+] at equilibrium.

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

Δ°G=-nFE°cellwhere,n - Number of electrons involved per equivalent of the net redox reaction in the cellF - Faraday’s Constant (96500 C) cell- Standard cell potential.

The Nernst equation depicts the relationship between Ecell and Eocell as follows,

Ecell=Eocell0.0592VnlogQwhere,Ecell=cellpotentialEocell=Standardcellpotentialn=No.ofelectronsQ=ReactionQuotient

(c)

Expert Solution
Check Mark

Explanation of Solution

The half reactions for given cell with their Eo values are noted by using Appendix D as follows,

Cd(s)Cd2+(aq)+2e E(Oxidation)=0.40VCu2+(aq)+2eCu(s) E(Reduction)=0.34V_Cd(s)+Cu2+(aq)Cu(s)+Cd2+(aq)

Observing the standard reduction potentials values it is clear that Cd oxidizes and Cu2+ reduces since, Cd is better reducing agent.

Ecell=Eocell0.0592VnlogQ0V=0.74V0.0592V2log[2-x][x]25=log[2][x]1×1025=[2][x]x=2×1025MCu2+

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Chapter 21 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 21.4 - Prob. 21.6AFPCh. 21.4 - Prob. 21.6BFPCh. 21.4 - Prob. 21.7AFPCh. 21.4 - Prob. 21.7BFPCh. 21.4 - Prob. 21.8AFPCh. 21.4 - Prob. 21.8BFPCh. 21.7 - The most ionic and least ionic of the common...Ch. 21.7 - Prob. 21.9BFPCh. 21.7 - Prob. 21.10AFPCh. 21.7 - Prob. 21.10BFPCh. 21.7 - Prob. 21.11AFPCh. 21.7 - Prob. 21.11BFPCh. 21.7 - In the final steps of the ETC, iron and copper...Ch. 21.7 - Prob. B21.2PCh. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.3PCh. 21 - Water is used to balance O atoms in the...Ch. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Aqua regia, a mixture of concentrated HNO3 and...Ch. 21 - Consider the following general voltaic...Ch. 21 - Why does a voltaic cell not operate unless the two...Ch. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Consider the following voltaic cell: In which...Ch. 21 - Consider the following voltaic cell: In which...Ch. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - A voltaic cell is constructed with an Fe/Fe2+...Ch. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - What does a negative indicate about a redox...Ch. 21 - Prob. 21.37PCh. 21 - In basic solution, Se2− and ions react...Ch. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Use the emf series (Appendix D) to arrange each...Ch. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Consider the following general electrolytic...Ch. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - In the electrolysis of molten NaBr: What product...Ch. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Write a balanced half-reaction for the product...Ch. 21 - Prob. 21.99PCh. 21 - Prob. 21.100PCh. 21 - Prob. 21.101PCh. 21 - Prob. 21.102PCh. 21 - Prob. 21.103PCh. 21 - Prob. 21.104PCh. 21 - Prob. 21.105PCh. 21 - Prob. 21.106PCh. 21 - Prob. 21.107PCh. 21 - Prob. 21.108PCh. 21 - Prob. 21.109PCh. 21 - Prob. 21.110PCh. 21 - Prob. 21.111PCh. 21 - Prob. 21.112PCh. 21 - Prob. 21.113PCh. 21 - Prob. 21.114PCh. 21 - Prob. 21.115PCh. 21 - Prob. 21.116PCh. 21 - Prob. 21.117PCh. 21 - Prob. 21.118PCh. 21 - Prob. 21.119PCh. 21 - Prob. 21.120PCh. 21 - To examine the effect of ion removal on cell...Ch. 21 - Prob. 21.122PCh. 21 - Prob. 21.123PCh. 21 - Prob. 21.124PCh. 21 - Prob. 21.125PCh. 21 - Prob. 21.126PCh. 21 - Commercial electrolytic cells for producing...Ch. 21 - Prob. 21.129PCh. 21 - Prob. 21.130PCh. 21 - The following reactions are used in...Ch. 21 - Prob. 21.132PCh. 21 - Prob. 21.133PCh. 21 - Prob. 21.134PCh. 21 - Prob. 21.135PCh. 21 - If the Ecell of the following cell is 0.915 V,...Ch. 21 - Prob. 21.137PCh. 21 - Prob. 21.138PCh. 21 - Prob. 21.139PCh. 21 - Prob. 21.140PCh. 21 - Prob. 21.141PCh. 21 - Prob. 21.142PCh. 21 - Prob. 21.143PCh. 21 - Prob. 21.144PCh. 21 - Prob. 21.145PCh. 21 - Prob. 21.146PCh. 21 - Prob. 21.147PCh. 21 - Both Ti and V are reactive enough to displace H2...Ch. 21 - For the reaction ∆G° = 87.8 kJ/mol Identity the...Ch. 21 - Two voltaic cells are to be joined so that one...Ch. 21 - Prob. 21.152PCh. 21 - Prob. 21.153P
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