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 \[{{E}^{o}}_{cell}=0\] |
| \[{{E}_{cell}}=-\frac{0.0591}{2}\log \frac{{{({{10}^{-10}})}^{2}}}{1}\] |
| \[=+\frac{0.0591\times 2}{2}\log \frac{1}{{{10}^{-10}}}\] |
| \[=0.0591\,\log \,{{10}^{10}}\] |
| \[=0.059\times 10=0.591\,V\] |
You need to login to perform this action.
You will be redirected in
3 sec
