| Given, | \[F{{e}^{2+}}+C{{e}^{4+}}F{{e}^{3+}}+C{{e}^{3+}}\] |
| \[{{E}^{o}}(C{{e}^{4+}}/C{{e}^{3+}})=1.44\,V\] | |
| \[{{E}^{o}}(F{{e}^{3+}}/F{{e}^{2+}})=0.68\,V\] |
A) \[4.52\times {{10}^{12}}\]
B) \[6.88\times {{10}^{12}}\]
C) \[7.13\times {{10}^{10}}\]
D) \[5.02\times {{10}^{10}}\]
Correct Answer: B
Solution :
\[F{{e}^{2+}}+C{{e}^{4+}}F{{e}^{3+}}+C{{e}^{3+}}\] \[{{E}^{o}}=E_{(C{{e}^{4+}}/C{{a}^{3+}})}^{o}-E_{(F{{e}^{3+}}/F{{e}^{2+}})}^{o}\] \[{{E}^{o}}=1.44-0.68\] \[=0.76\,V\] \[\because \]\[{{E}^{o}}=0.0592\,\log K\] \[\log \,K=\frac{{{E}^{o}}}{0.0592}=\frac{0.76}{0.0592}=12.83\] \[K=6.88\times {{10}^{12}}\]
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