A) \[p({{H}_{2}})=2\,atm\,and\,[{{H}^{+}}]=2.0M\]
B) \[p({{H}_{2}})=1\,atm\,and\,[{{H}^{+}}]=2.0M\]
C) \[p({{H}_{2}})=1\,atm\,and\,[{{H}^{+}}]=1.0M\]
D) \[p({{H}_{2}})=2\,atm\,and\,[{{H}^{+}}]=1.0M\]
Correct Answer: D
Solution :
| [d] For \[{{H}^{+}}(aq)+{{e}^{-}}\to (1/2){{H}_{2}}(g),\] we have |
| \[E=-\frac{RT}{F}\ln \frac{{{\left( p{{H}_{2}}/atm \right)}^{1/2}}}{\left( [{{H}^{+}}]/M \right)}\] |
| For E to be negative, the ratio \[\frac{{{\left( p{{H}_{2}}/atm \right)}^{1/2}}}{\left( [{{H}^{+}}]/M \right)}\] should be greater than 1. Only the choice ([d] produces this result. |
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