A) 4.28 V
B) 6.42 V
C) 8.56 V
D) 2.14 V
Correct Answer: D
Solution :
[d] \[\frac{4}{3}Al+{{O}_{2}}\xrightarrow[{}]{}\frac{2}{3}A{{l}_{2}}{{O}_{3}};\]\[\Delta G=-827kJmo{{l}^{-1}}\] Number of electrons involved is 4 \[\left( \frac{2}{3}\times 6=4 \right)\] \[\Delta G{}^\circ =-nFE{}^\circ \] or \[\frac{827000}{4\times 96500}=E{}^\circ \] \[\therefore \]\[E{}^\circ =2.14V\]You need to login to perform this action.
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