Answer:
0.900 V
Step-by-step explanation:
Oxidation half cell;
2Al(s) -----> 2Al^3+(aq) + 6e
Reduction half equation;
3Zn^2+(aq) + 6e ----> 3Zn(s)
E°anode = -1.66V
E°cathode= -0.76 V
E°cell= E°cathode - E°anode
E°cell= -0.76-(-1.66)
E°cell= 0.900 V
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