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 \[\underline{2{{H}^{+}}(1M)+2{{e}^{-}}\xrightarrow[{}]{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}(g)}}\,}\] \[\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{{{p}_{1}}}{\mathop{{{H}_{2}}(g)}}\,\xrightarrow{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}(g)}}\,}\] \[K=\frac{{{p}_{2}}}{{{p}_{1}}},\] \[E_{cell}^{o}=0.00V,n=2\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{RT}{nF}{{\log }_{e}}K\] \[=0-\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\] \[{{E}_{cell}}=\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]You need to login to perform this action.
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