A) \[1.0\times {{10}^{10}}\]
B) \[1.0\times {{10}^{5}}\]
C) \[1.0\times {{10}^{1}}\]
D) \[1.0\times {{10}^{30}}\]
Correct Answer: A
Solution :
\[{{E}_{Cell}}=E_{Cell}^{0}-\frac{0.0591}{n}\log {{K}_{c}}\] At 298 K \[{{E}_{Cell}}=0\] \[O=0.591-\frac{0.0591}{n}\log {{K}_{c}}\] \[\log {{K}_{c}}=\frac{0.591\times 1}{0.0591}=10\]; \[{{K}_{c}}=\text{Anti }\log 10=1\times {{10}^{10}}\].
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