The standard cell potential for the galvanic cell Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s) is E°cell = 1.10 V. Write the spontaneous reaction and confirm E°cell sign.

• A. Zn(s) + Cu2+ → Zn2+ + Cu(s); E°cell = +1.10 V
• B. Cu(s) + Zn2+ → Cu2+ + Zn(s); E°cell = -1.10 V
• C. Zn(s) + Cu(s) → Zn2+ + Cu2+; E°cell = 0 V
• D. Zn2+ + Cu2+ → Zn(s) + Cu(s); E°cell = +1.10 V

Answer: (A)

Spontaneity in a galvanic cell is determined by the sign of the standard cell potential: a positive E°cell means the reaction as written proceeds on its own. Here, the zinc electrode has the more negative tendency to gain/lose electrons, so it acts as the anode and is oxidized: Zn(s) -> Zn2+ + 2 e-. The copper electrode has the higher tendency to gain electrons, so it is the cathode and is reduced: Cu2+ + 2 e- -> Cu(s).

When you combine these half-reactions, the overall spontaneous reaction is Zn(s) + Cu2+ -> Zn2+ + Cu(s). The standard potentials give E°cell = E°cathode - E°anode = (+0.34 V) - (-0.76 V) = +1.10 V, which matches the given positive value and confirms spontaneity in this direction.

The other proposed reactions would require reversing the direction of electron flow or involve an implausible redox pairing, so they do not describe the spontaneous galvanic cell shown.