Answer:
Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)
1.35 V
Step-by-step 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