A) \[p({{H}_{2}})=2\] atm and \[[{{H}^{+}}]=2.0\] M
B) \[p({{H}_{2}})=1\] atm and \[[{{H}^{+}}]=2.0\] M
C) \[p({{H}_{2}})=1\] atm and \[[{{H}^{+}}]=1.0\] M
D) \[p({{H}_{2}})=2\] atm and \[[{{H}^{+}}]=1.0\] M
Correct Answer: D
Solution :
\[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] Apply Nernst equation \[E=0-\frac{0.059}{1}\log \frac{{{P}_{{{H}_{2}}}}^{\frac{1}{2}}}{[{{H}^{+}}]}\] \[E=-\frac{0.059}{1}\log \frac{{{2}^{1/2}}}{1}\] Therefore E is negative.
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