A) 1010
B) 1020
C) 1040
D) 1060
Correct Answer: C
Solution :
\[A(s)+2{{B}^{+}}(aq)\xrightarrow{{}}{{A}^{2+}}(aq)+2B(s)\] Oxidation \[A(s)\xrightarrow{{}}{{A}^{2+}}(aq)+2{{e}^{-}}\] Reduction \[2{{B}^{+}}(aq)+2{{e}^{-}}\xrightarrow{{}}2B(s)\] Here, \[n=2\] \[E{}^\circ =1.18V\] \[{{E}_{eq}}=?\] \[\because \] \[\log {{K}_{eq}}=\frac{nE{}^\circ F}{RT}\] \[\left[ \frac{F}{RT}=\frac{1}{0.059} \right]\] \[\therefore \] \[\log {{K}_{eq}}=\frac{2\times 1.18\times 1}{0.059}\] \[\therefore \] \[\log {{K}_{eq}}=40\]or \[{{K}_{eq}}={{10}^{40}}\]You need to login to perform this action.
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