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