A) \[10\]
B) \[2\times {{10}^{2}}\]
C) \[3\times {{10}^{2}}\]
D) \[2\times {{10}^{5}}\]
Correct Answer: A
Solution :
\[E_{cell}^{o}=\frac{0.0591}{n}{{\log }_{10}}K\] \[n=2,\,E_{cell}^{o}=0.0295V\] \[0.0295=\frac{0.0591}{2}{{\log }_{10}}K\] K = antilog \[0.998\] \[K=9.96\approx 10\]
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