A) \[E_{cell}^{o}=\frac{n}{0.059}\log {{K}_{c}}\]
B) \[E_{cell}^{o}=\frac{0.059}{n}\log {{K}_{c}}\]
C) \[E_{cell}^{o}=0.059\,n\,\log {{K}_{c}}\]
D) \[E_{cell}^{o}=\frac{\log {{K}_{c}}}{n}\]
Correct Answer: B
Solution :
\[E_{cell}^{o}=\frac{0.0591}{n}\log {{K}_{c}}\]
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