A) \[E_{cell}^{0}=\frac{n}{0.059}\log {{K}_{c}}\]
B) \[E_{cell}^{0}=\frac{0.059}{n}\log {{K}_{c}}\]
C) \[E_{cell}^{0}=0.059\,n\,\log {{K}_{c}}\]
D) \[E_{cell}^{0}=\frac{\log {{K}_{c}}}{n}\]
Correct Answer: B
Solution :
\[E_{\text{cell}}^{o}=\frac{2.303\,\,RT}{nF}\log \,K=\frac{0.0591}{n}\log \,{{K}_{c}}\,at\,\,298K\].
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