A) \[{{10}^{14}}\text{ }atm\]
B) \[{{10}^{12}}\text{ }atm\]
C) \[{{10}^{-10}}atm\]
D) \[{{10}^{-4}}atm\]
Correct Answer: D
Solution :
| [d] \[2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)\] |
| \[\therefore \] \[E={{E}^{o}}-\frac{0.0591}{2}\log \frac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\] |
| \[0=0-0.0295\,\,\log \frac{{{P}_{{{H}_{2}}}}}{{{[{{10}^{-7}}]}^{2}}}\] |
| \[\frac{{{P}_{{{H}_{2}}}}}{{{({{10}^{-7}})}^{2}}}=1\] |
| \[{{P}_{{{H}_{2}}}}={{10}^{-4}}\,\text{atm}\] |
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