Nernst Equation
For the cell Zn|Zn²⁺(0.1M)||Cu²⁺(0.01M)|Cu at 298 K, if E°cell = 1.10 V, the cell potential is: (Take 2.303RT/F = 0.06)
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About This Question
- Subject
- chemistry
- Chapter
- redox reactions and electrochemistry
- Topic
- nernst equation
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
1.07 V
Using the Nernst Equation for the cell reaction Zn+Cu2+→Zn2++Cu (n=2):
- Formula: Ecell=Ecell∘−n0.06log[Cu2+][Zn2+]
- Values: Ecell∘=1.10 V, n=2, [Zn2+]=0.1 M, [Cu2+]=0.01 M.
- Calculation: Ecell=1.10−20.06log(0.010.1) Ecell=1.10−0.03log(10) Ecell=1.10−0.03(1)=1.07 V.
This hard difficulty chemistry question is from the chapter redox reactions and electrochemistry, covering the topic of nernst equation. It appeared in the 2025 exam.
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