A) 0.059 V
B) 0.59 V
C) 0.118 V
D) 1.18 V
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, \[{{E}_{cell}}={{E}^{o}}_{cell}=\frac{0.0591}{2}\log \frac{{{[{{H}^{+}}]}^{2}}}{{{p}_{{{H}_{2}}}}}\] For hydrogen electrode \[{{E}^{o}}_{cell}=0\] \[{{E}_{cell}}=-\frac{0.0591}{2}\log \frac{{{({{10}^{-10}})}^{2}}}{1}\] \[=+\frac{0.0591\times 2}{2}\log \frac{1}{{{10}^{-10}}}\] \[=0.0591\,\log \,{{10}^{10}}\] \[=0.059\times 10=0.591\,V\]
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