Nernst Equation
For the cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), the Nernst equation at 298 K gives E = E° − (0.0591/n) log Q. If [Zn²⁺] = 0.1 M and [Cu²⁺] = 1.0 M, and E° = 1.10 V, the cell potential E is approximately:
Select the correct option:
Solution
1.13 V
Using the Nernst equation: E = E° − (0.0591/n) log([Zn²⁺]/[Cu²⁺]). Here n = 2 (two electrons transferred), [Zn²⁺] = 0.1 M, and [Cu²⁺] = 1.0 M. So Q = 0.1/1.0 = 0.1. E = 1.10 − (0.0591/2) × log(0.1) = 1.10 − (0.02955) × (−1) = 1.10 + 0.02955 ≈ 1.13 V. Since Q < 1 (product concentration is low relative to reactant), the cell potential increases beyond E°. This demonstrates that a galvanic cell delivers higher voltage when it is far from equilibrium with excess reactants.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More nernst equation Practice Questions
About This Question
- Subject
- chemistry
- Chapter
- redox reactions and electrochemistry
- Topic
- nernst equation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.13 V
Using the Nernst equation: E = E° − (0.0591/n) log([Zn²⁺]/[Cu²⁺]). Here n = 2 (two electrons transferred), [Zn²⁺] = 0.1 M, and [Cu²⁺] = 1.0 M. So Q = 0.1/1.0 = 0.1. E = 1.10 − (0.0591/2) × log(0.1) = 1.10 − (0.02955) × (−1) = 1.10 + 0.02955 ≈ 1.13 V. Since Q < 1 (product concentration is low relative to reactant), the cell potential increases beyond E°. This demonstrates that a galvanic cell delivers higher voltage when it is far from equilibrium with excess reactants.
This medium difficulty chemistry question is from the chapter redox reactions and electrochemistry, covering the topic of nernst equation. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse redox reactions and electrochemistry questions on RankGuru.