A) \[{{E}_{cell}}=E_{cell}^{\text{o}}-\frac{0.592}{n}{{\log }_{10}}[F{{e}^{2+}}]\,{{[{{I}^{-}}]}^{2}}\]
B) \[{{E}_{cell}}=E_{cell}^{\text{o}}-\frac{0.0592}{n}{{\log }_{10}}[F{{e}^{2+}}]\,\,{{[{{I}^{-}}]}^{2}}\]
C) \[{{E}_{cell}}=E_{cell}^{\text{o}}-\frac{0.0592}{n}{{\log }_{10}}[F{{e}^{2+}}]\,\,[{{I}^{-}}]\]
D) \[{{E}_{cell}}=E_{cell}^{\text{o}}-\frac{0.0592}{n}{{\log }_{10}}\frac{[F{{e}^{2+}}]\,\,{{[{{I}^{-}}]}^{2}}}{[Fe]\,\,[{{I}_{2}}]}\]
Correct Answer: B
Solution :
Cell reaction of a given cell is \[\begin{align} & \underline{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,Fe\xrightarrow{{}}F{{e}^{2+}}+2{{e}^{-}} \\ & {{I}_{2}}+2{{e}^{-}}\xrightarrow{{}}2{{I}^{-}} \\ \end{align}} \\ & \,\,Fe+{{I}_{2}}\xrightarrow{{}}F{{e}^{2+}}+2{{I}^{-}} \\ \end{align}\] According to Nernst equation \[{{E}_{cell}}=E_{cell}^{\text{o}}+\frac{0.0592}{n}{{\log }_{10}}\frac{[Fe]\,[{{I}_{2}}]}{[F{{e}^{2+}}]\,{{[{{I}^{-}}]}^{2}}}\] \[[Fe]=[{{I}_{2}}]=1\] \[{{E}_{cell}}=E_{cell}^{\text{o}}-\frac{0.0592}{n}{{\log }_{10}}[F{{e}^{2+}}]\,{{[{{I}^{-}}]}^{2}}\]
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