A) \[E=E{}^\circ +\frac{RT}{nF}\log \frac{1}{[{{M}^{n+}}]}\]
B) \[E{}^\circ =E+RT\,in\,[{{M}^{n+}}]\]
C) \[E=E{}^\circ -\frac{RT}{nF}\,in\frac{1}{\,[{{M}^{n+}}]}\]
D) \[\frac{E}{E{}^\circ }=\frac{RT}{nF}\,in[{{M}^{n+}}]\]
Correct Answer: C
Solution :
For the electron reaction,\[{{M}^{n+}}(aq)+n{{e}^{-}}\xrightarrow[{}]{{}}M(s)\]Nernst equation is\[E=E{}^\circ -\frac{RT}{nF}in\frac{1}{[{{M}^{n+}}]}\]
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