A) \[{{P}_{{{H}_{2}}}}\,=\,\,1\,\,atm,\,[{{H}^{+}}]\,\,=\,\,1\,\,M\]
B) \[{{P}_{{{H}_{2}}}}\,=\,\,1\,\,atm,\,[{{H}^{+}}]\,\,=\,\,2\,\,M\]
C) \[{{P}_{{{H}_{2}}}}\,=\,\,2\,\,atm,\,[{{H}^{+}}]\,\,=\,\,1\,\,M\]
D) \[{{P}_{{{H}_{2}}}}\,=\,\,2\,\,atm,\,[{{H}^{+}}]\,\,=\,\,0.1\,\,M\]
Correct Answer: D
Solution :
\[2{{H}^{+}}+\underset{(g)}{\mathop{2{{e}^{-}}}}\,\to {{H}_{2}}\] \[{{E}_{red}}\,=\,{{E}^{{}^\circ }}_{\operatorname{Re}d}-\frac{0.059}{2}\,\,\log \,\frac{{{P}_{{{H}_{2}}}}}{{{\left[ {{H}^{+}} \right]}^{2}}}\]
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