A) \[29.5\times {{10}^{-2}}\]
B) 10
C) \[1\times {{10}^{10}}\]
D) \[2.95\times {{10}^{-10}}\]
Correct Answer: C
Solution :
[c] Using the relation, \[E{{{}^\circ }_{cell}}=\frac{2.303RT}{nF}\log {{K}_{eq}}=\frac{0.0591}{n}\log {{K}_{eq}}\] \[\therefore 0.295V=\frac{0.0591}{2}\log {{K}_{eq}}\] Or \[\log {{K}_{eq}}=\frac{2\times 0.295}{0.0591}=10\] Or \[{{K}_{eq}}=1\times {{10}^{10}}\]You need to login to perform this action.
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