Questions: Electrochemical Cells

Question 1

A galvanic cell is constructed from a Zn²⁺/Zn half-cell (E° = −0.76 V) and a Cu²⁺/Cu half-cell (E° = +0.34 V), with Zn as the anode. What is E°cell?

• A. −1.10 V
• B. −0.42 V
• C. +0.42 V
• D. +1.10 V

Answer: (D)

E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V. The copper half-cell is the cathode (reduction) and zinc is the anode (oxidation). A common error is to simply add the two values algebraically without recognizing that the anode potential must be subtracted, and another is to reverse the cathode/anode assignment.

Question 2

In a galvanic cell, both electrons in the external wire and cations in the salt bridge move toward the cathode, so they are doing the same job of carrying positive charge in the same direction.

• T. True
• F. False

Answer: False

Electrons flow from anode to cathode through the external wire — they are negatively charged moving in one direction. Cations in the salt bridge migrate toward the cathode to maintain charge neutrality in the cathode compartment, which becomes depleted of positive ions as reduction consumes cations. These are two distinct charge carriers serving complementary roles; conflating them misrepresents how the circuit is completed.

Question 3

Using the Nernst equation, explain why a battery's voltage decreases as it discharges.

Model answer: As a battery discharges, reactant concentrations fall and product concentrations rise, increasing the reaction quotient Q. The Nernst equation (E = E° − (RT/nF)ln Q) shows that as Q increases, cell potential E decreases below E°.

At standard conditions Q = 1 and E = E°. As the cell operates, it consumes reactants and generates products, so Q grows larger than 1. Because ln Q becomes a larger positive number, the term (RT/nF)ln Q subtracted from E° grows, lowering E. When Q equals the equilibrium constant K, E = 0 — the battery is dead and ΔG = 0.