A) \[1.0\times {{10}^{30}}\]
B) \[1.0\times {{10}^{5}}\]
C) \[1.0\times {{10}^{10}}\]
D) \[1.0\times {{10}^{1}}\]
Correct Answer: C
Solution :
[c] \[{{E}_{cell}}={{E}^{o}}_{cell}-\frac{0.059}{n}\log Q\] At equilibrium, \[{{E}_{cell}}=0\]and \[Q={{K}_{c}}\] \[{{E}^{o}}_{cell}=\frac{0.059}{1}\log {{K}_{c}}\] \[\Rightarrow 0.591=\frac{0.059}{1}\log {{K}_{c}}\] \[\therefore {{K}_{c}}=1\times {{10}^{10}}\]
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