Question

What is E∘cell (in units of volts) for a galvanic cell constructed from aluminum and gold electrodes?

E∘ aluminum = −1.662

E∘ gold =1.498

• Report your answer with three decimal places.

V

Answer 1

E(0)Al = -1.662 v

E(0)Au = 1.498 v

These are reduction potentials of Al and Au .More the reduction
potential more tendency to undergo reduction. And the electrode
where reduction happens is cathode.Since gold has higher
value it will be the cathode and reduction takes place.
Al will undergo oxidation and act as anode.
cell can be represented as
Al(s)|Al3+(aq)||Au+(aq)|Au(s)

E.M.F of cell= E(cathode)-E(anode)
= E(Au)-E(Al) = 1.498 V- [-1.662 V] =3.16 V
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Thank you!!