Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393600681
Author: Gilbert
Publisher: W. W. Norton & Company
Question
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Chapter 17, Problem 17.42QA
Interpretation Introduction

To find:

The value of  Ecell0 and G0 for the given two redox reactions by using the standard reduction potential listed in Appendix 6 of the text.

 

Expert Solution & Answer
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Answer to Problem 17.42QA

Solution:

a) Ecell0= 0.789 V; G0=-152 kJ

b) Ecell0=-0.03VG0= 2.90 kJ

 

Explanation of Solution

1) Concept:

To calculate the value of G0 and Ecell0  for the two given reactions, we can use the appropriate standard potentials listed in Appendix 6.  First, we will calculate the EOx0 and Ered0 from the balanced reaction. Then, we will calculate the Ecell0 of the reaction by using EOx0 and Ered0 values.

Form the Ecell0 and G0 formula we can calculate G0.

2) Formula:

i) Ecell0= Ecathod0-Eanode0    

ii) G0= -nFEcell         

      

3) Given:

i) Fe(s)+Cu(aq)2+ Fe(aq)2++ Cu(s)    

ii) Ag(s)+Fe(aq)2+ Ag(aq)++ Fe(aq)2+

4) Calculations:

a) The half reaction at the cathode is based on the reduction of Cu2+ to Cu. The appropriate half-reaction in Table A6.1 is

Cu(aq)2++ 2e-  Cu(s)  Eanode0= 0.342 V

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1 in which Fe2+ is the reactant and Fe is the product:

Fe(aq)2++ 2e-  Fe(s)  Eanode0= -0.447 V

Reversing the Fe2+half reaction and adding it to the Cu2+ half reaction, we get

Cu(aq)2++ 2e-  Cu(s) 

Fe(s)  Fe(aq)2++ 2e- 

Chemistry: An Atoms-Focused Approach, Chapter 17, Problem 17.42QA , additional homework tip  1 

Fe(s)+Cu(aq)2+ Fe(aq)2++ Cu(s)    

Calculating Ecell0,

Ecell0= Ecathod0-Eanode0= 0.342 V--0.447 V= 0.789 V

Calculating G0,

G0= -nFEcell         

Gcell0=-2 mol  × 9.65 ×104Cmol ×0.789 V

Gcell0= -15.2277× 104C.V=-15.2277× 104J= -152 kJ

b) The half reaction at the cathode is based on the reduction of Fe3+ to Fe2+. The appropriate half-reaction in Table A6.1 is

Fe(aq)3++ e-  Fe2+(s)  Ecathod0= 0.770 V

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1 in which Ag+ is the reactant and Ag is the product:

Ag(aq)++ e-  Ag(s)  Eanode0= 0.800 V

Reversing the Ag+half reaction and adding it to  Fe3+  half reaction, we get

Fe(aq)3++ e-  Fe2+(s)               

Ag(aq)++ e-  Ag(s) 

Chemistry: An Atoms-Focused Approach, Chapter 17, Problem 17.42QA , additional homework tip  2

Ag(s)+Fe(aq)2+ Ag(aq)++ Fe(aq)2+

Calculating Ecell0,

Ecell0= Ecathod0-Eanode0=  0.770 V-0.800 V= -0.03V

Calculating G0,

G0= -nFEcell         

Gcell0=-1 mol  × 9.65 ×104Cmol ×-0.03V

Gcell0= 0.2895× 104C.V=0.2895× 104J= 2.90 kJ

Conclusion:

We calculate the G0 and Ecell0 by using the standard reduction potential values and formula between G0 and Ecell0.

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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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