Questions: Electrochemistry and the Nernst Equation

Question 1

A galvanic cell starts with all reactants and products at standard conditions (E = E°). As the cell operates, products accumulate and reactant concentrations fall. What happens to the measured cell potential?

• A. The cell potential increases, because the products are driving the reaction forward more strongly
• B. The cell potential stays the same until equilibrium, then suddenly drops to zero
• C. The cell potential decreases continuously as Q increases, approaching zero as the cell reaches equilibrium
• D. The cell potential oscillates because the Nernst equation contains a logarithm, which alternates sign as Q passes through 1

Answer: (C)

As the cell operates, products accumulate and reactants are consumed, so the reaction quotient Q increases. In the Nernst equation E = E° − (RT/nF) ln Q, a rising Q means the ln Q term grows more positive, and E falls. This continuous decrease reflects the diminishing thermodynamic driving force as the system approaches equilibrium. At equilibrium, Q = K and E = 0 — the cell can no longer do electrical work. Option A inverts the logic; more products means less driving force, not more. The drop to zero is gradual, not sudden (option B).

Question 2

For a cell reaction with n = 2 and E° = +0.50 V at 25°C, what is the relationship between E° and the equilibrium constant K?

• A. K cannot be determined from E° alone; the reaction quotient Q is also needed
• B. E° = (RT/nF) ln K, so K = exp(nFE°/RT) ≈ 10^(nE°/0.05916) ≈ 10^17, strongly favoring products
• C. E° = ΔG°/nF, so K = E° × nF, which has units of joules per mole
• D. K = 1/E° for standard conditions, so a higher voltage means a smaller equilibrium constant

Answer: (B)

At equilibrium, E = 0 and Q = K. Setting E = 0 in the Nernst equation gives 0 = E° − (RT/nF) ln K, which rearranges to E° = (RT/nF) ln K. For E° = +0.50 V and n = 2 at 25°C: ln K = nFE°/RT = 2 × 0.50/0.02569 ≈ 38.9, so K ≈ e^38.9 ≈ 10^16–17. A positive E° corresponds to K > 1; the larger E°, the more overwhelmingly products are favored. This connection between measured voltage and equilibrium thermodynamics — bridging electrochemistry and the ΔG = ΔG° + RT ln Q framework — is the deepest insight of the Nernst equation.

Question 3

A standard cell potential E° > 0 implies that the equilibrium constant K > 1 for the cell reaction, meaning products are thermodynamically favored at equilibrium.

• T. True
• F. False

Answer: True

From E° = (RT/nF) ln K, a positive E° requires ln K > 0, which means K > 1. This is also consistent with ΔG° = −nFE°: a positive E° gives a negative ΔG°, which means the reaction is spontaneous under standard conditions and products are favored at equilibrium. The quantitative connection (E° ≈ 0.05916/n × log K at 25°C) shows that even a modest positive E°, say 0.3 V with n = 1, gives K ≈ 10^5. Electrochemistry and thermodynamics are not separate frameworks — they are the same framework expressed in different units.

Question 4

The Nernst equation mainly applies to cells operating at non-standard concentrations; under standard conditions (most activities = 1), a separate equation is expected to be used.

• T. True
• F. False

Answer: False

The Nernst equation is universally valid. Under standard conditions, all activities equal 1, so Q = 1 and ln Q = 0. The Nernst equation then reduces to E = E° − 0 = E°. Standard conditions are simply a special case, not a separate regime requiring a different equation. This is analogous to ΔG = ΔG° + RT ln Q: when Q = 1, ΔG = ΔG°. The Nernst equation is the most general statement; standard conditions are the simplification you use when concentrations happen to be 1 M.

Question 5

A battery is observed to have a lower voltage than its nominal (standard) rated potential after extended use. Use the Nernst equation to explain mechanistically why this occurs.

Model answer: As a battery discharges, the cell reaction consumes reactants and produces products, continuously increasing the reaction quotient Q. In the Nernst equation E = E° − (RT/nF) ln Q, a larger Q increases the subtracted term, directly reducing E. The cell potential thus falls progressively from E° toward zero as Q rises toward the equilibrium constant K. When E = 0, Q = K and the battery is exhausted — it can no longer drive current because there is no thermodynamic driving force remaining. The battery is not 'broken'; it has simply reached chemical equilibrium.

This mechanistic explanation unifies the everyday observation (batteries run down) with the thermodynamic framework (cells approach equilibrium). It also explains why rechargeable batteries work: forcing current through the cell in reverse drives Q back below K, restoring reactant concentrations and rebuilding the voltage. The same logic explains concentration cells — two identical electrodes in solutions of different concentration generate voltage purely from the Q imbalance, with no net chemical reaction.