A) 4.5 V
B) 3.0 V
C) 5.0 V
D) 2.5 V
Correct Answer: B
Solution :
In the reaction \[\frac{2}{3}A{{l}_{2}}{{O}_{3}}\xrightarrow[{}]{{}}\frac{4}{3}Al+{{O}_{2}}\] For the oxidation half-reaction \[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow[{}]{{}}Al\] no. of electron transfered (n)= 3 \[\Delta {{G}^{o}}=-nF{{E}^{o}}\] \[940=3\times 96500\times {{E}^{o}}\] \[{{E}^{o}}=\frac{940\times {{10}^{3}}J}{3\times 96500}\]\[=3.24\approx 3V\]You need to login to perform this action.
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