Respuesta :

pstnonsonjoku

Answer:

2.0 V

Explanation:

For the oxidation half cell;

Al(s) -------> Al^3+(aq) + 3e.

For reduction half cell;

Cu^2+(aq) +2e ------> Cu(s).

E°cell = E°cathode - E°anode

But;

E°cathode= 0.34 V

E°anode = -1.66 V

E°cell= 0.34 -(-1.66)

E°cell= 2.0 V

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