Choose the balanced equation for the reaction. Au3+(aq) + Cu+(aq) → Au(s) + Cu2+(aq) Au(s) + Cu2+(aq) → Au3+(aq) + Cu+(aq) Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq) Au(s) + 3Cu2+(aq) → Au3+(aq) + 3Cu+(aq) b Use the standard reduction potentials for the galvanic cell to determine E°.

Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)

1.35 V

Explanation:

Given that the reduction potential of

Cu^2+(aq) + e- > Cu^+(aq) is +0.15 V

While the reduction potential of

Au3+(aq) + 3 e- > Au(s) is + 1.50V

It is clear that the Cu^+(aq)/Cu^2+(aq) system is the anode while Au^3+(aq)/Au(s) system is the cathode based on the reduction potentials shown above. The number of electrons transferred (n) =3

E°cell = E°cathode - E°anode

E°cell= 1.50-0.15

E°cell= 1.35 V