CBSE Class 12-science Answered
3.57 plz
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Asked by lovemaan5500 | 19 Aug, 2019, 23:11: PM
Given:
E0Cu2+/Cu = 0.34 V
E0Ag2+/Ag = 0.80 V
R = 8.314 JK− mol−
F = 96500 C/mol
E0cell = E0cathode −E0anode
= 0.80 − 0.34
= 0.46 V
The net cell reaction,
Cu + 2Ag+ → Cu+ + 2Ag
n = 2
Nernst equation,
![straight E subscript cell space equals space straight E subscript cell superscript 0 space minus fraction numerator 2.303 RT over denominator nF end fraction log fraction numerator open square brackets Cu to the power of 2 plus end exponent close square brackets over denominator open square brackets Ag to the power of plus close square brackets squared end fraction
space space space space space space space space equals 0.46 space minus space fraction numerator 2.303 cross times 8.314 cross times 298 over denominator 2 cross times 96500 end fraction log 4 over open parentheses 0.1 close parentheses to the power of blank
space space space space space space space space space equals 0.46 space minus open parentheses 0.0295 space cross times space 2.602 close parentheses
space space space space space space space space equals space 0.46 space minus 0.0769
straight E subscript cell space equals space 0.3831 space straight V](https://images.topperlearning.com/topper/tinymce/cache/845f0153c7f7a881e82bea39cdbde6a7.png)
Cell potential is 0.3831 V
Answered by Varsha | 20 Aug, 2019, 15:49: PM
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