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Find the Gibbs energy change of the cell at 298K. 3 Cr |Cr3+ (0.1M)|| Cu2+(0.01M) + |Cu. Given E° Cr3+ / Cr = – 0.75V, Cu2+/Cu = +0.34V, F = 96500C.

Find the Gibbs energy change of the cell at 298K. 3
Cr |Cr3+ (0.1M)|| Cu2+(0.01M) + |Cu.
Given E° Cr3+ / Cr = – 0.75V, Cu2+/Cu = +0.34V, F = 96500C.

Grade:9

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago

Cr ------Cr+3 +3e-
Cu+2 +2e- ----Cu

E=E0 +.059/n log(Cu+2/Cr+3)
=.34+.75 +.059/6log(.01/.1)
=1.09+.059/6log.1
=1.08

deltaG=-nFE
=-6*96500*1.08
=-625.32KJ

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