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Given the standard reduction potentials: Fe2+ + 2e- -> Fe; E° = -0.41 V Ag+ + e- -> Ag; E° = +0.80 V O2 + 2H2O + 4e- -> 4OH-; E° = +0.40 V A bar of silver is joined by a wire to an iron pipe buried underground. What is the expected outcome?
- The silver would corrode, a current would flow in the wire, and O2 would be reduced at the iron pipe surface.
- The silver would corrode, a current would flow in the wire, and Fe2+ would be reduced at the iron pipe surface.
- The iron pipe would corrode, a current would flow in the wire, and Ag+ would be reduced at the silver surface.
- The iron pipe would corrode, no current would flow in the wire, and O2 would be reduced at the iron pipe surface.
Correct answer: The iron pipe would corrode, a current would flow in the wire, and Ag+ would be reduced at the silver surface.
Solution
Iron (E° = -0.41 V) is much more easily oxidized than silver (E° = +0.80 V). When the two are connected, a galvanic cell forms: iron is the anode and corrodes, while reduction occurs at the silver (cathode). Since the electrodes have different potentials, a net cell emf drives a current through the wire. At the silver cathode, the species with high reduction potential present (Ag+ from any dissolved silver, in the option's framing) is reduced. Note this is harmful — it accelerates iron corrosion, unlike a sacrificial anode (e.g., Mg or Zn) which would protect the iron.
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