Categories > electrochemistry > Corrosion and miscellaneous questions > On the basis of the information available from the reaction $\frac{4}{3}Al+{{O}_{2}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}},\,\Delta G=-827\,kJ\,mo{{l}^{-1}}$ of ${{O}_{2}}$. The minimum emf required to carry out an electrolysis of $A{{l}_{2}}{{O}_{3}}$ is $(F=96500\,\,C\,mo{{l}^{-1}})$

electrochemistry

Question: On the basis of the information available from the reaction $\frac{4}{3}Al+{{O}_{2}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}},\,\Delta G=-827\,kJ\,mo{{l}^{-1}}$ of ${{O}_{2}}$. The minimum emf required to carry out an electrolysis of $A{{l}_{2}}{{O}_{3}}$ is $(F=96500\,\,C\,mo{{l}^{-1}})$


1) 8.56 V
2) 2.14 V
3) 4.28 V
4) 6.42 V
Solution: Explanation: No Explanation
Corrosion and miscellaneous questions

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