Chemistry

Chemistry

Problem

Consider a galvanic cell based on the overall reaction Sn2+^{2+} (aq) + 2 Fe3+^{3+} (aq) → Sn4+^{4+} (aq) + 2 Fe2+^{2+} (aq). Both half-cells initially contain 1.0 M solutions of the indicated ions. After the cell has been assembled, pure water is added to the cathode compartment only, doubling its volume but leaving the anode unchanged. Which statement best describes the immediate effect on the cell potential? a) The potential increases because dilution lowers QQ, moving the system farther from equilibrium. b) The potential is essentially unchanged because the ratio [Fe2+]/[Fe3+]\left[\text{Fe}^{2+}\right]/\left[\text{Fe}^{3+}\right] is unaffected, leaving QQ nearly the same. c) The potential decreases because the lower ionic concentrations reduce the activities of the cathode species, diminishing the driving force for electron flow even if their ratio is unchanged. d) The potential changes sign because diluting the cathode solution removes enough oxidizing power from Fe3+\text{Fe}^{3+} to force the cell reaction to run in reverse.
Intuitive explanation

This problem tests your understanding of how concentration changes affect cell potential through the Nernst equation. The key insight is that diluting only the cathode compartment changes the concentrations of the ions involved in the cathode half-reaction, which directly shifts the reduction potential at that electrode. Since cell potential depends on both the anode and cathode potentials, you need to track how dilution affects each separately—and recognize that the cathode potential is more sensitive to concentration changes when ions appear in the half-reaction with higher stoichiometric coefficients.

Intuitive explanation

This problem tests your understanding of how concentration changes affect cell potential through the Nernst equation. The key insight is that diluting only the cathode compartment changes the concentrations of the ions involved in the cathode half-reaction, which directly shifts the reduction potential at that electrode. Since cell potential depends on both the anode and cathode potentials, you need to track how dilution affects each separately—and recognize that the cathode potential is more sensitive to concentration changes when ions appear in the half-reaction with higher stoichiometric coefficients.

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