A) \[2\]
B) \[1\]
C) \[7\]
D) \[2.7\]
Correct Answer: A
Solution :
\[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] \[{{E}_{cell}}={{E}^{o}}_{cell}-0.0591\log \frac{1}{{{H}^{+}}}\] \[-0.118V=0+0.0591\,[{{H}^{+}}]\] \[-0.118V=0+0.0591\{-\log [{{H}^{+}}]\}\] \[-0.118V=-0.0591pH\] \[pH=\frac{0.118}{0.0591}=2\]You need to login to perform this action.
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