A) 0.059V
B) 0.59V
C) 0.118V
D) 1.18V
Correct Answer: B
Solution :
For hydrogen electrode, oxidation half reaction is \[\underset{(1\,atm)}{\mathop{{{H}_{2}}}}\,\xrightarrow{{}}\underset{(At\,pH\,10)}{\mathop{2{{H}^{+}}}}\,+2{{e}^{-}}\] If \[pH=10\] \[{{H}^{+}}=1\times {{10}^{-pH}}=1\times {{10}^{-10}}\] From Nernst equation, \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{{{[{{H}^{+}}]}^{2}}}{{{p}_{{{H}_{2}}}}}\] For hydrogen electrode, \[E_{cell}^{o}=0\] \[{{E}_{cell}}=-\frac{0.0591}{2}\log \frac{{{({{10}^{-10}})}^{2}}}{1}\] \[=+\frac{0.0591\times 2}{2}\log \frac{1}{{{10}^{-10}}}\] \[=0.0591\,\log {{10}^{10}}\] \[=0.0591\times 10=0.591\,V\]
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