A) increase of 59 mV
B) decrease of 59 mV
C) increase of 29.5 mV
D) decrease of 29.5 mV
Correct Answer: B
Solution :
\[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{1}{[Z{{n}^{2+}}]}\] \[=E_{cell}^{0}+0.02955\log \,C\] \[E_{cell}^{1}=E_{cell}^{0}+0.02955\log \frac{C}{100}=E_{cell}^{0}\] \[+\,0.02955\,(log\,C-2)\] \[\therefore \]\[{{E}_{cell}}\]decrease by \[0.02955\times 2V=0.059V=59mV\]
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