A) 1.345V
B) 1.724V
C) 2.146V
D) 2.487V
Correct Answer: D
Solution :
\[A{{l}_{2}}{{O}_{3}}(2A{{l}^{3+}}+3{{O}^{2-}})\xrightarrow[{}]{{}}2Al+\frac{3}{2}{{O}_{2}};\]\[n=6{{e}^{-}}\] \[\frac{2}{3}A{{l}_{2}}{{O}_{3}}\xrightarrow[{}]{{}}\frac{4}{3}Al+{{O}_{2}};\] \[n=4{{e}^{-}}\] \[\Delta G=+960\,kJ=+960000J\] \[\Delta G=-nFE\] \[960000=-4\times 96500\times E\] \[E=-2.487\,V\] Minimum potential difference required to reduce\[A{{l}_{2}}{{O}_{3}}\]is 2.487 V.You need to login to perform this action.
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