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