A) 0.540V
B) 0.4810V
C) 0.5696V
D) 0.5105V
Correct Answer: D
Solution :
\[{{E}_{cell}}=E_{cell}^{o}+\frac{0.059}{2}\log \frac{[N{{i}^{2+}}]}{[Z{{n}^{2+}}]}\] \[0.5105=E_{cell}^{o}+\frac{0.059}{2}\log \frac{1}{1}\] \[E_{cell}^{o}=0.5105\] \[(\because \log 1=0)\]You need to login to perform this action.
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