A) \[5.08\]
B) \[3\]
C) \[4\]
D) \[6\]
Correct Answer: A
Solution :
Electrode potential \[=-0.3\text{ }V\] The electrode reaction may be given as \[2{{H}^{+}}_{(aq)}+2{{e}^{-}}\xrightarrow{{}}{{H}_{2(g)}}\] \[E={{E}^{o}}-\frac{0.059}{2}\log \frac{1}{{{[{{H}^{+}}]}^{2}}}\] \[-0.3=0-\frac{0.059}{2}\left( -2\,\log \,\left[ {{H}^{+}} \right] \right)\] \[-0.3=-0.059\,pH\] \[pH=\frac{0.3}{0.059}\] \[pH=5.085\]You need to login to perform this action.
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