Question

Consider a galvanic cell constructed from the following half-cells that are linked by a porous membrane. (1) an Au electrode dipped into 1.0M Au(NO₃)₃ and (2) an Fe electrode dipped into 1.0M FeSO₄. Answer the following questions:

a. Which electrode is the cathode?
b. Write the balanced equation for the reaction occurring while the cell is discharging.
c. What emf should the cell generate?
d. In what direction will electrons flow in the outer circuit?
e. Toward which electrode will positive ions migrate?

A. a) Au, b) (Fe²⁺ + 2e⁻ → Fe), c) 0.80 V, d) Fe to Au, e) Toward Au

B. a) Fe, b) (Fe²⁺ + 2e⁻ → Fe), c) 0.80 V, d) Au to Fe, e) Toward Fe

C. a) Au, b) (Au³⁺ + 3e⁻ → Au), c) 0.80 V, d) Au to Fe, e) Toward Au

D. a) Fe, b) (Au³⁺ + 3e⁻ → Au), c) 0.80 V, d) Fe to Au, e) Toward Fe

Answer 1

The Au electrode is the cathode, the half-reactions are Au(NO₃)₃ + 3e⁻ → Au and Fe → Fe²⁺ + 2e⁻, the emf generated is 0.80 V, electrons flow from the Fe electrode to the Au electrode in the outer circuit, and positive ions migrate toward the Au electrode.

Step-by-step explanation:

The half-reaction that occurs at each electrode in this galvanic cell is:

At the Au electrode (cathode): Au(NO₃)₃ + 3e⁻ → Au

At the Fe electrode (anode): Fe → Fe²⁺ + 2e⁻

Based on the half-reactions, the Au electrode is the cathode and the Fe electrode is the anode. In this cell, electrons will flow from the Fe electrode (anode) to the Au electrode (cathode) through the outer circuit. The positive ions will migrate toward the Au electrode.