A) 0.059 V
B) 0.59 V
C) 0.118 V
D) 1.18 V
Correct Answer: B
Solution :
[b] 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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