A) \[-0.207\,V\]
B) \[+\,0.207\,V\]
C) \[-\,0.414\,\,V\]
D) \[+0.414\,V\]
Correct Answer: C
Solution :
For water at \[298K,\] \[[{{H}^{+}}]={{10}^{-7}}M\] Reduction reaction is \[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] \[\therefore \] \[E=\frac{RT}{F}\] In \[\frac{P_{{{H}_{2}}}^{{\scriptstyle{}^{1}/{}_{2}}}}{[{{H}^{+}}]}\] \[=-0.0591\,\log \,\frac{P_{{{H}_{2}}}^{{\scriptstyle{}^{1}/{}_{2}}}}{[{{H}^{+}}]}=-0.0591\,\,\log \frac{1}{{{10}^{-7}}}\] \[=-0.4137=-0.414\,V\]
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