A) 0.177V
B) 0.087V
C) \[-0.177V\]
D) 0.059V
Correct Answer: C
Solution :
\[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] \[E={{E}^{o}}-\frac{0.059}{n}\log \frac{1}{[{{H}^{+}}]}\] \[=0-\frac{0.059}{1}pH\] \[E=-0.059\times 3=-0.177V\]
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