A) 0.0059 V
B) 0.059 V
C) 5.9 V
D) 0.59 V
Correct Answer: B
Solution :
\[Ag|A{{g}^{+}}(.1m)||A{{g}^{+}}1M|Ag|\] \[{{E}_{Cell}}=\frac{2.303RT}{nF}\log \frac{{{c}_{1}}}{{{c}_{2}}}=\frac{0.059}{1}\log \frac{1}{0.1}\] \[=0.059\log 10=0.059\ Volt\].
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