A) \[-1.66V\]
B) \[+1.66V\]
C) \[-0.83\text{ }V\]
D) \[+0.83\text{ }V\]
Correct Answer: A
Solution :
\[E_{cell}^{o}=E_{2{{H}^{+}}/{{H}_{2}}}^{o}-E_{A{{l}^{3+}}/Al}^{o}\] \[1.66=0-E_{A{{l}^{3+}}/Al}^{o}\] \[E_{A{{l}^{3}}/Al}^{o}=-1.66V\]
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