A) \[0.59\] V
B) \[0.118\] V
C) \[1.18\] V
D) \[0.059\] V
Correct Answer: A
Solution :
\[\underset{1\,atm}{\mathop{{{H}_{2}}}}\,\,\xrightarrow{\,}\,\underset{{{10}^{-10}}}{\mathop{2{{H}^{+}}}}\,+2{{e}^{-}}\] \[{{E}_{{{H}_{2}}/{{H}^{+}}}}\,=0-\frac{0.059}{2}\,\log \,\frac{{{({{10}^{-10}})}^{2}}}{1}\] \[{{E}_{{{H}_{2}}/{{H}^{+}}}}\,=+0.59\,V\]You need to login to perform this action.
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