A) \[2.38\times {{10}^{38}}\]
B) \[1.08\times {{10}^{42}}\]
C) \[1.67\times {{10}^{39}}\]
D) \[3.24\times {{10}^{44}}\]
Correct Answer: A
Solution :
\[{{E}_{cell}}={{E}_{c}}-{{E}_{a}}=1.464V\] \[{{E}_{C{{e}^{4+}}/C{{e}^{2+}}}}=E_{C{{e}^{4+}}/C{{e}^{3+}}}^{o}-\frac{0.059}{1}\log \frac{[C{{e}^{3+}}]}{[C{{e}^{4+}}]}\] \[=E_{C{{e}^{4+}}/C{{e}^{3+}}}^{o}-\frac{0.059}{1}\log \frac{10}{90}\] \[{{E}_{cell}}={{E}_{calo}}-{{E}^{o}}C{{e}^{4+}}/C{{e}^{3+}}\] \[1464={{E}_{calo}}-\left[ \left( {{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}-0.059\log \frac{1}{8} \right) \right]\] \[1464=0.28-\left( {{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}-0.059\log \frac{1}{8} \right)\] \[{{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}=-1.24V\] Now,\[2C{{e}^{3+}}+2{{H}^{+}}C{{e}^{4+}}+{{H}_{2}}\] At equilibrium, \[{{E}_{cell}}=0\] \[\therefore \]\[{{E}^{o}}=\frac{0.059}{2}\log k\] At anode \[2C{{e}^{3+}}\xrightarrow[{}]{{}}2C{{e}^{4+}}+2{{e}^{-}}\](oxidation) At cathode \[2{{H}^{+}}+2{{e}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}\](reduction) Cell reaction \[2C{{e}^{3+}}+2{{H}^{+}}\xrightarrow[{}]{{}}2C{{e}^{4+}}+{{H}_{2}}\] \[E_{cell}^{o}={{E}^{o}}\]red(cathode) \[-{{E}^{o}}\] red(anode) \[E_{red({{H}_{2}})}^{o}-E_{red}^{o}(C{{e}^{4+}}/C{{e}^{3+}})\] \[=0-(-124)=124\] \[{{E}^{o}}=\frac{0.059}{2}\log k\] \[1.24=\frac{0.059}{2}\log k\] \[\log k=42.03\] \[k=1.08\times {{10}^{42}}\]
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