Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 17, Problem 17.35QA
Interpretation Introduction

To calculate:

The values of G0 and Ecell0 of the given reactions.   

Expert Solution & Answer
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Answer to Problem 17.35QA

Solution:

a) G0=-34.5 kJ; Ecell0=0.358 V 

b) G0=2.9 kJ; Ecell0=-0.030 V 

Explanation of Solution

1) Concept:

We are asked to calculate the value of G0 and Ecell0 for the given two reactions using appropriate standard free energies of formation from Appendix 4. Multiplying the Gf0 values by the appropriate numbers of moles and subtracting the sum of the reactants values from the sum of the product values given by the equation below will yield the value for  Grxn0.

Grxn0= nproductsGf products0-  nreactantsGf reactants0 

The relation between change in standard free energy and standard cell potential is

G0=- nFEcell0, where, the value of Faraday constant (F) is   9.65 ×104Cmol   and n is the mole of electrons transferred in the half reactions of the cell.

2) Formula:

i) G0=nproductGfproduct0-nreactantGfreactant0

ii)  G0=- nFEcell0     

3) Given:

i) 2Cu +(aq) Cu 2+(aq)+Cu (s)      

ii) Ag (s)+ Fe 3+(aq)  Ag +(aq)+Fe 2+(aq)

4) Calculations:

a)

From the Appendix 4, the values of Gf0 are as follows:

Gf0(kJ/mol)
Cu+2(aq) 65.5
Cu (s) 0
Cu+ (aq) 50.0

2Cu +(aq) Cu 2+(aq)+Cu (s)      

     

G0=nproductGfproduct0-nreactantGfreactant0

G0= 1 mol Cu2+×65.5kJmolCu2++1 mole Cu ×0kJmolCu-2 mol Cu+×50kJmol Cu+

G0=65.5kJ+0 kJ-100kJ

G0= -34.5 kJ

G0= -34.5 kJ × 103J1 kJ= -3.45 ×104 J

The number of moles of electrons transferred in the reaction is 1 mole, so n = 1.

Ecell0 is calculated as

G0=-nFEcell0     

-3.45 ×104 J=- 1 mol × 9.65 ×104Cmol × Ecell0-3.45 ×104 J = - 9.65 ×104C × Ecell0     

Ecell0=0.358 V     

b) Ag (s)+ Fe 3+(aq)  Ag +(aq)+Fe 2+(aq)

From the Appendix 4, the values of Gf0 are

Gf0(kJ/mol)
Ag (s) 0
Fe3+ (aq) -4.7
Ag+ (aq) 77.1
Fe2+aq -78.9

G0=nproductGfproduct0-nreactantGfreactant0

G0= 1 mol Ag+ ×77.1kJmol Ag+ +1 mole Fe2+ ×-78.9kJmolFe2+-1 mol Fe3+×-4.7kJmol Fe3++ 1 mole Ag ×0kJmolAg

G0=77.1 kJ-78.9 kJ+4.7 kJ-0 kJ

G0=2.9 kJ

G0= 2.9 kJ × 103J1 kJ= 2.9 ×103 J

The number of moles of electrons transferred in the reaction is 1 mole, so n = 1.

Ecell0 is calculated as

G0=-nFEcell0     

2.9 ×103 J=- 1 mol × 9.65 ×104Cmol × Ecell0     

2.9 ×103 J = - 9.65 ×104C × Ecell0     

Ecell0=-0.030 V     

Conclusion:

When the cell reaction is spontaneous, the value of G0 is negative and the value of Ecell0 is positive. When the cell reaction is non-spontaneous, the value of G0 is positive and the value of Ecell0 is negative.

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Chapter 17 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 17 - Prob. 17.11QACh. 17 - Prob. 17.12QACh. 17 - Prob. 17.13QACh. 17 - Prob. 17.14QACh. 17 - Prob. 17.15QACh. 17 - Prob. 17.16QACh. 17 - Prob. 17.17QACh. 17 - Prob. 17.18QACh. 17 - Prob. 17.19QACh. 17 - Prob. 17.20QACh. 17 - Prob. 17.21QACh. 17 - Prob. 17.22QACh. 17 - Prob. 17.23QACh. 17 - Prob. 17.24QACh. 17 - Prob. 17.25QACh. 17 - Prob. 17.26QACh. 17 - Prob. 17.27QACh. 17 - Prob. 17.28QACh. 17 - Prob. 17.29QACh. 17 - Prob. 17.30QACh. 17 - Prob. 17.31QACh. 17 - Prob. 17.32QACh. 17 - Prob. 17.33QACh. 17 - Prob. 17.34QACh. 17 - Prob. 17.35QACh. 17 - Prob. 17.36QACh. 17 - Prob. 17.37QACh. 17 - Prob. 17.38QACh. 17 - Prob. 17.39QACh. 17 - Prob. 17.40QACh. 17 - Prob. 17.41QACh. 17 - Prob. 17.42QACh. 17 - Prob. 17.43QACh. 17 - Prob. 17.44QACh. 17 - Prob. 17.45QACh. 17 - Prob. 17.46QACh. 17 - Prob. 17.47QACh. 17 - Prob. 17.48QACh. 17 - Prob. 17.49QACh. 17 - Prob. 17.50QACh. 17 - Prob. 17.51QACh. 17 - Prob. 17.52QACh. 17 - Prob. 17.53QACh. 17 - Prob. 17.54QACh. 17 - Prob. 17.55QACh. 17 - Prob. 17.56QACh. 17 - Prob. 17.57QACh. 17 - Prob. 17.58QACh. 17 - Prob. 17.59QACh. 17 - Prob. 17.60QACh. 17 - Prob. 17.61QACh. 17 - Prob. 17.62QACh. 17 - Prob. 17.63QACh. 17 - Prob. 17.64QACh. 17 - Prob. 17.65QACh. 17 - Prob. 17.66QACh. 17 - Prob. 17.67QACh. 17 - Prob. 17.68QACh. 17 - Prob. 17.69QACh. 17 - Prob. 17.70QACh. 17 - Prob. 17.71QACh. 17 - Prob. 17.72QACh. 17 - Prob. 17.73QACh. 17 - Prob. 17.74QACh. 17 - Prob. 17.75QACh. 17 - Prob. 17.76QACh. 17 - Prob. 17.77QACh. 17 - Prob. 17.78QACh. 17 - Prob. 17.79QACh. 17 - Prob. 17.80QACh. 17 - Prob. 17.81QACh. 17 - Prob. 17.82QACh. 17 - Prob. 17.83QACh. 17 - Prob. 17.84QACh. 17 - Prob. 17.85QACh. 17 - Prob. 17.86QACh. 17 - Prob. 17.87QACh. 17 - Prob. 17.88QACh. 17 - Prob. 17.89QACh. 17 - Prob. 17.90QACh. 17 - Prob. 17.91QACh. 17 - Prob. 17.92QACh. 17 - Prob. 17.93QACh. 17 - Prob. 17.94QACh. 17 - Prob. 17.95QACh. 17 - Prob. 17.96QACh. 17 - Prob. 17.97QACh. 17 - Prob. 17.98QACh. 17 - Prob. 17.99QACh. 17 - Prob. 17.100QACh. 17 - Prob. 17.101QACh. 17 - Prob. 17.102QA
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