A) \[2.4\times {{10}^{10}}\]
B) \[2.0\times {{10}^{10}}\]
C) \[4.0\times {{10}^{10}}\]
D) \[4.0\times {{10}^{15}}\]
Correct Answer: D
Solution :
\[Cu(s)+2A{{g}^{2+}}(aq)\xrightarrow{{}}C{{u}^{2+}}(aq)+2Ag(s);\]\[{{E}^{o}}=0.46\,V\]at \[298\,K\] \[\log \,{{K}_{C}}=\frac{nF{{E}^{o}}}{RT}=\frac{2\times 0.46}{0.059}\] \[{{K}_{C}}=4\times {{10}^{15}}\]You need to login to perform this action.
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