A) 0.59 V
B) 0.00 V
C) -0.59 V
D) -0.059
Correct Answer: C
Solution :
\[{{H}^{+}}(pH=10)|{{H}_{2}}(1\,atm)|Pt(s)\] Reaction: \[2{{H}^{+}}(pH=10)+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}(1\,atm)\] \[E={{E}^{o}}-\frac{0.0591}{2}\log \left( \frac{{{p}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}} \right)\] \[=0-\frac{0.0591}{2}\log \frac{1}{{{({{10}^{-10}})}^{2}}}\] \[=-\frac{0.0591}{2}\times 2\log \frac{1}{{{10}^{-10}}}\] \[=-0.0591\times 10=-0.591\] ie, \[E=-0.591\,V\]
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