Nernst Equation
At 298 K a Daniell cell operates with [Zn^2+] = 0.1 M and [Cu^2+] = 0.01 M; how does its EMF compare with the standard value of 1.10 V?
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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 approximately
The Nernst equation relates the actual cell potential to the standard potential and the reaction quotient: E_cell = E°_cell - (0.0591/n) log Q at 298 K, where Q for the Daniell reaction is [Zn^2+]/[Cu^2+]. Here n = 2 because two electrons are transferred. The quotient Q = 0.1/0.01 = 10, so log Q = 1. Substituting, E_cell = 1.10 - (0.0591/2)(1) = 1.10 - 0.0296 = 1.07 V. The EMF drops slightly below the standard value because the higher product-ion concentration relative to reactant-ion concentration makes the forward reaction marginally less favourable. Option 1.10 V ignores the concentration term entirely. Option 1.13 V uses the wrong sign on the correction. Option 0.55 V wrongly halves the standard potential instead of applying the logarithmic term. This is a direct JEE application of the Nernst relationship to non-standard concentrations. Examiners frequently test whether a student can connect reaction quotient with the underlying principle rather than merely recalling an isolated fact. Working through the logic step by step, rather than memorising the result, makes it clear why concentration cell effect governs the behaviour seen here. Plausibility check: since Q > 1, the correction must be negative, so an EMF slightly under 1.10 V is the only consistent outcome.
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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