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 :
\[Cu(s)+2A{{g}^{2+}}(ag)\xrightarrow[{}]{{}}C{{u}^{2+}}(aq)+2Ag(s)\] \[{{E}^{o}}=0.46\,V\,at\,298\,K\] \[RT\,\ln \,K=nF{{E}^{o}}\] \[\ln \,K=\frac{nF{{E}^{o}}}{RT}\] \[=\frac{2\times 0.46}{0.0591}\] \[K=4\times {{10}^{15}}\]You need to login to perform this action.
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