A) 4.5 V
B) 3.0 V
C) 2.5 V
D) 5.0 V
Correct Answer: C
Solution :
The half cell reaction are at anode \[2{{O}^{2-}}+4{{e}^{-}}\xrightarrow{{}}{{O}_{2}}]\times 3\] At cathode \[A{{l}^{3+}}\xrightarrow{{}}Al+3{{e}^{-}}]4\] Net reaction \[4Al+6{{O}^{2-}}\xrightarrow{{}}3{{O}_{2}}+4Al\] or \[\frac{4}{3}Al+2{{O}^{2-}}\xrightarrow{{}}{{O}_{2}}+\frac{4}{3}Al\] \[\therefore \] \[n=\frac{12}{3}=4\] \[\Delta {{G}^{o}}=-nF{{E}^{o}}\] Here, \[\Delta {{G}^{o}}=+960\,kJ\,mo{{l}^{-1}}\] \[=960\times 1000\,J\,mo{{l}^{-1}}\] \[n=4\] \[F=96500\,coulomb\] \[\therefore \]\[960\times 1000=-4\times 96500\times {{E}^{o}}\] \[{{E}^{o}}=-\frac{960000}{4\times 96500}\] \[=-2.48\,V\]You need to login to perform this action.
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