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 17.41QA
Interpretation Introduction

To find:

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

Expert Solution & Answer
Check Mark

Answer to Problem 17.41QA

Solution:

a)  Ecell0= -0.478 V; G0=92.2 kJ

b)  Ecell0=0.505 VG0= -97.5kJ

Explanation of Solution

1) Concept:

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

From the Ecell0 and G0 formula, we can calculate G0.

2) Formula:

i)   Ecell0= Ecathod0-Eanode0    

ii) G0= -nFEcell         

3) Given:

i)   Cu(s)+Sn(aq)2+ Cu(aq)2++ Sn(s)    

ii)   Zn(s)+Ni(aq)2+ Zn(aq)2++ Ni(s)

4) Calculations:

a) The half reaction at the cathode is based on the reduction potential value of Sn2+ to Sn. The appropriate half-reaction in Table A6.1 is

Sn(aq)2++ 2e-  Sn(s)  Ecathode0= -0.136 V

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1, in which Cu2+ is the reactant and Cu is the product.

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

Reversing the Cu2+half reaction and adding it to the Sn2+ half reaction, we get

Sn(aq)2++ 2e-  Sn(s)         

Cu(s)  Cu(aq)2++ 2e- 

Chemistry: An Atoms-Focused Approach, Chapter 17, Problem 17.41QA , additional homework tip  1

Cu(s)+Sn(aq)2+ Cu(aq)2++ Sn(s)    

Calculating Ecell0:

Ecell0= Ecathode0-Eanode0=  -0.136 V-0.342 V= -0.478V

Calculating G0:

G0= -nFEcell         

Gcell0=-2 mol  × 9.65 ×104 Cmol ×(-0.478 V)

Gcell0= 9.2254× 104 C.V=9.2254× 104J= 92.2 kJ

b) The half reaction at the cathode is based on the reduction of Ni2+ to Ni. The appropriate half-reaction in Table A6.1 is

Ni(aq)2++ 2e-  Ni(s)  Ecathode0= -0.257 V

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1, in which Zn2+ is the reactant and Zn is the product.

Zn(aq)2++ 2e-  Zn(s)  Eanode0= -0.762 V

Reversing the Zn2+half reaction and adding it to the Ni2+  half reaction, we get

Ni(aq)2++ 2e-  Ni(s) 

Zn(s)   Zn(aq)2++ 2e-

Chemistry: An Atoms-Focused Approach, Chapter 17, Problem 17.41QA , additional homework tip  2

Zn(s)+Ni(aq)2+ Zn(aq)2++ Ni(s)

Calculating Ecell0:

Ecell0= Ecathode0-Eanode0=  -0.257 V-(-0.762) V= 0.505V

Calculating G0:

G0= -nFEcell         

 Gcell0=-2 mol  × 9.65 ×104Cmol ×0.505V

Gcell0= -9.7465× 104 C.V=-9.7465× 104 J= -97.5 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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