A) 0.12V
B) 0.06V
C) 0.09V
D) 0.16V
Correct Answer: A
Solution :
\[{{E}_{cell}}={{E}^{o}}-\frac{0.059}{n}\log \frac{\left[ Product\text{ }cone. \right]}{\left[ Reactant\text{ }cone. \right]}\] \[{{({{E}_{cell}})}_{1}}=0-\frac{0.059}{1}\log \frac{0.1}{1}\] ... (i) \[{{({{E}_{cell}})}_{2}}=0-\frac{0.059}{1}\log \frac{0.01}{1}\] ... (ii) \[0.06=-\frac{0.059}{1}\log \frac{1}{10}\] ... (i) \[{{({{E}_{cell}})}_{2}}=-\frac{0.059}{1}\log \frac{1}{10}\] ... (ii) Dividing Eq. (i) by (ij), we have \[\frac{0.06}{{{({{E}_{cell}})}_{2}}}=\frac{\log 1-\log 10}{\log 1-\log 100}\] \[\frac{0.06}{{{({{E}_{cell}})}_{2}}}=\frac{0-1}{0-2\log 10}\] \[\frac{0.06}{{{({{E}_{cell}})}_{2}}}=\frac{-1}{-2}=\frac{1}{2}\] \[{{({{E}_{cell}})}_{2}}=2\times 0.06=0.12\,V\]
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