A) \[1.0\times {{10}^{1}}\]
B) \[1.0\times {{10}^{5}}\]
C) \[1.0\times {{10}^{10}}\]
D) \[1.0\times {{10}^{30}}\]
Correct Answer: C
Solution :
Relation between\[{{K}_{eq}}\]and\[{{K}_{cell}}\]is \[E_{cell}^{o}=\frac{2.303RT}{nF}\log {{K}_{eq}}\] \[E_{cell}^{o}=\frac{0.0591RT}{n}\log {{K}_{eq}}\] (at 298 K) \[0.591=\frac{0.0591}{n}\log {{K}_{eq}}\] \[\therefore \] \[\log {{K}_{eq}}=10\] \[\therefore \] \[{{K}_{eq}}=1\times {{10}^{10}}\]
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