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