A) \[{{10}^{25}}\]
B) \[{{10}^{20}}\]
C) \[{{10}^{15}}\]
D) \[{{10}^{30}}\]
Correct Answer: D
Solution :
\[E_{cell}^{o}=\frac{0.0591}{n}\log \,\,{{K}_{c}}\] \[0.59=\frac{0.0591}{3}\log \,\,{{K}_{c}}\] \[\frac{0.059\times 3}{0.059}=\log \,\,{{K}_{c}}\] \[\therefore \] \[\log \,{{K}_{c}}=30\] \[{{K}_{c}}=\] antilog 30 \[={{10}^{30}}\]You need to login to perform this action.
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