A) \[\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]
B) \[\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]
C) \[\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\]
D) \[\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\]
Correct Answer: B
Solution :
LHS half cell \[\underset{{{p}_{1}}}{\mathop{{{H}_{2}}(g)}}\,\xrightarrow{{}}2{{H}^{+}}(1M)+2{{e}^{-}}\] RHS half cell \[{{p}_{1}}\] \[\begin{align} & \underline{2{{H}^{+}}(1M)+2{{e}^{-}}\xrightarrow{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}(g)}}\,} \\ & \underline{\underset{{{p}_{1}}}{\mathop{{{H}_{2}}}}\,(g)\xrightarrow{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}}}\,(g)} \\ \end{align}\] \[E_{cell}^{o}=0.00\,V,\,K=\frac{{{p}_{2}}}{{{p}_{1}}},\,n=2\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{RT}{nF}{{\log }_{e}}K=0-\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\] \[{{E}_{cell}}=\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]
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