Determine the standard potential for each cell and identify the reactions as spontaneous or nonspontaneous as written. 2Al3+(aq) + 3Cu(s) → 3Cu2+(aq) + 2Al(s)

Answer 1

2Al³⁺(aq) + 3Cu(s) ==> 3Cu²⁺(aq) + 2Al(s)

This reactions has the Al³⁺ being reduced to Al(s) and the Cu(s) being oxidized to Cu²⁺(aq)

Looking at the 2 standard reduction potentials, we have...

Al³⁺ + 3e- ==> Al(s) Eº= -1.66 V

Cu²⁺ ==> Cu(s) Eº = 0.34 V

As written, the Eº_cell = -1.66 - 0.34 = -2.00 V

This would be nonspontaneous because the voltage is negative. You could tell that from the original Eº values because to be spontaneous, the cathode should have been Cu and the anode should be Al but this reaction has it reversed.