Standard Cell Potential and Equilibrium Constant for Zn–Cu Cell (E° = 1.10 V)

Question statement

For the galvanic cell:

Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s)

the standard cell potential is E° = 1.10 V at 298 K. The equilibrium constant is written as
K = y × 10³⁷; find y (to two decimal places) using
F = 96485 C mol⁻¹ and R = 8.314 J K⁻¹ mol⁻¹.

Step‑by‑step solution

1. Relationship between E° and K

Thermodynamics provides two key equations linking cell potential, Gibbs free energy, and the equilibrium constant:

• ΔG° = −RT ln K
• ΔG° = −n F E°

Equating the two expressions for ΔG° gives:

−n F E° = −RT ln K ⇒ E° = (RT / nF) ln K

This is the key relationship between standard cell potential and the equilibrium constant.

2. Identify n for the Zn–Cu cell

The half‑reactions for the Zn–Cu galvanic cell are:

• Oxidation: Zn → Zn²⁺ + 2e−
• Reduction: Cu²⁺ + 2e− → Cu

Because two electrons are transferred in the overall reaction, the value of n is 2.

3. Insert numerical values

Use the data: E° = 1.10 V, T = 298 K, R = 8.314 J K⁻¹ mol⁻¹, F = 96485 C mol⁻¹, and n = 2.

ln K = (n F E°) / (R T)
= (2 × 96485 × 1.10) / (8.314 × 298)
≈ 85.49

Convert the natural logarithm to base‑10:

log₁₀ K = (ln K) / 2.303 ≈ 85.49 / 2.303 ≈ 37.12

Therefore,

K = 10^37.12 ≈ 2.57 × 10³⁷

So in the form K = y × 10³⁷, the value of y is 2.57.

Concept explanation

A positive standard cell potential E° means the cell reaction is spontaneous and has a large equilibrium constant K » 1.

In this Zn–Cu cell, E° = 1.10 V leads to K ≈ 10³⁷, which shows that the products Zn²⁺ and Cu(s) are overwhelmingly favored at equilibrium under standard conditions.