A) \[{{E}_{2}}=0\ne {{E}_{1}}\]
B) \[{{E}_{1}}>{{E}_{2}}\]
C) \[{{E}_{1}}<{{E}_{2}}\]
D) \[{{E}_{1}}={{E}_{2}}\]
Correct Answer: B
Solution :
\[{{E}_{1}}={{E}_{o}}-\frac{0.0591}{2}\log \frac{0.01}{1}={{E}_{o}}+\frac{0.0591}{2}\times 2\] \[{{E}_{2}}={{E}_{o}}-\frac{0.0591}{2}\log \frac{100}{0.01}={{E}_{o}}-\frac{0.0591}{2}\times 4\] \[\therefore \,\,{{E}_{1}}>{{E}_{2}}\].
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