In a galvanic cell using Zn2+/Zn and Cu2+/Cu with standard reduction potentials E°(Zn2+/Zn) = -0.76 V and E°(Cu2+/Cu) = +0.34 V, what are E°cell and which metal serves as the anode and cathode?

In a galvanic cell using Zn2+/Zn and Cu2+/Cu with standard reduction potentials E°(Zn2+/Zn) = -0.76 V and E°(Cu2+/Cu) = +0.34 V, what are E°cell and which metal serves as the anode and cathode?

• A. E°cell = 1.10 V; Anode: Cu; Cathode: Zn
• B. E°cell = -1.10 V; Anode: Zn; Cathode: Cu
• C. E°cell = 1.34 V; Anode: Cu; Cathode: Zn
• D. E°cell = 1.10 V; Anode: Zn; Cathode: Cu

Answer: (D)

In a galvanic cell, the substance with the higher reduction potential gets reduced at the cathode, while the other is oxidized at the anode. Here, copper has the higher reduction potential (+0.34 V) compared with zinc (-0.76 V). So Cu2+ will gain electrons at the cathode, and Zn will lose electrons at the anode. Calculate the standard cell potential using E°cell = E°cathode − E°anode: 0.34 − (−0.76) = +1.10 V. A positive value means a spontaneous, favorable reaction in this arrangement. Therefore, the anode is zinc and the cathode is copper, giving the overall reaction Zn + Cu2+ → Zn2+ + Cu.

In a galvanic cell, the substance with the higher reduction potential gets reduced at the cathode, while the other is oxidized at the anode. Here, copper has the higher reduction potential (+0.34 V) compared with zinc (-0.76 V). So Cu2+ will gain electrons at the cathode, and Zn will lose electrons at the anode.

Calculate the standard cell potential using E°cell = E°cathode − E°anode: 0.34 − (−0.76) = +1.10 V. A positive value means a spontaneous, favorable reaction in this arrangement. Therefore, the anode is zinc and the cathode is copper, giving the overall reaction Zn + Cu2+ → Zn2+ + Cu.