A zinc sacrificial anode is attached to a steel structure in seawater. Treating it as a cell under standard conditions with standard electrode potentials of −0.44 V for Fe²⁺/Fe and −0.76 V for Zn²⁺/Zn, which correctly gives the emf and the metal that dissolves?

Answer: A. About 0.32 V; zinc dissolves and the steel is protected

The emf is the potential of the cathode side (the more noble iron) minus that of the anode side (the more active zinc): E = −0.44 − (−0.76) = 0.32 V. The active zinc becomes the anode and dissolves as Zn → Zn²⁺ + 2e⁻, and the electrons flow into the steel, making it the cathode, so dissolution of the steel is suppressed. This is cathodic protection by sacrificial anodes (the galvanic anode method), used for ship hulls, harbor steel structures and buried pipelines; the anodes are consumed and are replaced periodically. About 1.20 V is the mistake of adding the magnitudes of the two potentials. The choice of about 0.32 V with the steel dissolving reverses noble and active. Actual potentials in seawater differ from standard electrode potentials, but zinc is still more active than steel.