A galvanic cell is constructed with the following half-cells at 298 K: Cathode…
CHEMISTRY · Multi-part · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry
Problem
A galvanic cell is constructed with the following half-cells at 298 K:
Cathode compartment: Pt(s) | Fe3+(aq, 1.0 M), Fe2+(aq, 1.0 M)
Anode compartment: Pt(s) | Sn4+(aq, 1.0 M), Sn2+(aq, 1.0 M)
Standard reduction potentials are
Fe3+(aq)+e−→Fe2+(aq) Eo=+0.77V
Sn4+(aq)+2e−→Sn2+(aq) Eo=+0.15V
The cell is allowed to operate only briefly so that concentrations remain essentially at their initial values. Three separate modifications are then performed in sequence, allowing the cell to reach a new instantaneous potential after each step:
Step 1: Solid FeCl3 is added to the cathode compartment until [Fe3+] becomes 2.0 M (all other species unchanged).
Step 2: After the system reaches the new potential from Step 1, solid Na2SnO2 is added to the anode compartment until [Sn2+] becomes 2.0 M (all other species at the values present after Step 1).
Step 3: An equal volume of distilled water is added to BOTH compartments, doubling the total volume in each half-cell without removing any material.
(a) For Step 1, predict whether the cell potential immediately after the addition of FeCl3 is greater than, less than, or equal to the original potential. Provide a clear chemical justification.
(b) For Step 2, predict whether the new cell potential is greater than, less than, or equal to the potential obtained at the end of Step 1. Explain your reasoning.
(c) For Step 3, predict whether the final cell potential is greater than, less than, or equal to the potential obtained at the end of Step 2. Justify your answer in terms of how the dilution affects the reaction quotient.
(d) Rank the magnitudes of the four cell potentials (original, after Step 1, after Step 2, after Step 3) from largest to smallest, and summarize the key reasoning steps that support your ranking.