A) In \[{{K}_{C}}=\frac{nFE_{cell}^{''}}{RT}\]
B) \[{{K}_{C}}=\frac{nFE_{cell}^{o}}{RT}\]
C) \[E_{cell}^{o}\frac{-RT}{nF}\] In \[{{K}_{C}}\]
D) \[{{K}_{C}}=\frac{RT}{nF}\] In \[E_{cell}^{o}\]
Correct Answer: A
Solution :
\[e_{cell}^{o}=\frac{RT}{nF}\] In \[{{K}_{C}}\] In \[{{K}_{C}}=\frac{nFE_{cell}^{o}}{RT}\]You need to login to perform this action.
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