What is the equilibrium constant for the reaction Fe2+ + Ag+ ⇌ Fe3+ + Ag?

• A. 0.30
• B. 1
• C. 3.22
• D. 10

Answer: (C)

This question tests how standard redox potentials translate into an equilibrium constant for a redox reaction. In this reaction, silver ions are reduced to solid silver, while iron(II) is oxidized to iron(III). So identify the relevant half-reactions: Ag+ + e− → Ag(s) with E° ≈ +0.7996 V, and Fe3+ + e− → Fe2+ with E° ≈ +0.771 V. Since Fe2+ is being oxidized, use the reduction potential for iron as the anode value but subtract it when calculating the cell potential.

Compute the standard cell potential: E°cell = E°cathode − E°anode = 0.7996 V − 0.7710 V ≈ 0.0286 V. With one electron transferred (n = 1), convert this to the equilibrium constant using log10 K = (n E°cell) / 0.05916 (at 25 °C). So log10 K ≈ 0.0286 / 0.05916 ≈ 0.48, giving K ≈ 10^0.48 ≈ 3.0–3.3.

Thus the equilibrium constant is about 3.2, indicating the reaction favors formation of Fe3+ and solid Ag to a modest extent.