A) \[+1.66V\]
B) \[-3.26V\]
C) \[3.26V\]
D) \[-1.66V\]
Correct Answer: D
Solution :
\[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[\therefore \] \[2.46=(+0.80)-E_{A{{l}^{3+}}/Al}^{o}\] or \[E_{A{{l}^{3+}}/Al}^{o}=0.80-2.46=-1.66V\]You need to login to perform this action.
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