A) \[{{10}^{25}}\]
B) \[{{10}^{20}}\]
C) \[{{10}^{15}}\]
D) \[{{10}^{30}}\]
Correct Answer: D
Solution :
\[E_{cell}^{o}=\frac{0.0591}{n}{{\operatorname{logK}}_{c}}\] \[0.59=\frac{0.0591}{3}\log {{K}_{c}}\] \[\frac{0.59\times 3}{0.059}=\log {{K}_{c}}\] \[\therefore \] \[\log {{K}_{c}}=30\] \[{{K}_{c}}=\text{antilog 30}\] \[={{10}^{30}}\]
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