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Topic: EMF, Nernst equation and applications· Redox Reactions and Electrochemistry

In the electrochemical cell :Zn|ZnSO_4(0.01\ M)||CuSO_4(1.0\ M)|Cu, the emf of this Daniel cell is E_1. When the concentration of ZnSO_4 is changed to 1.0 M and that of CuSO_4 changed to 0.01 M, the emf changes to E_2. From the followings, which one…

In the electrochemical cell :

\(Zn|ZnSO_4(0.01\ M)||CuSO_4(1.0\ M)|Cu\), the emf of this Daniel cell is \(E_1\). When the concentration of \(ZnSO_4\) is changed to 1.0 M and that of \(CuSO_4\) changed to 0.01 M, the emf changes to \(E_2\). From the followings, which one is the relationship between \(E_1\) and \(E_2\) ? (Given, \(\frac{RT}{F} = 0.059\))

1. \(E_1 < E_2\)

2. \(E_1 > E_2\)

3. \(E_2 = 0 \neq E_1\)

4. \(E_1 = E_2\)

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Answer: 2. \(E_1 > E_2\)

Key idea: apply the Nernst equation, \(E = E^\circ - \dfrac{0.059}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}\).

  • \(E_1\): \(\log\dfrac{0.01}{1}=-2\), so \(E_1 = E^\circ + 0.059\).
  • \(E_2\): \(\log\dfrac{1}{0.01}=2\), so \(E_2 = E^\circ - 0.059\).

Hence \(E_1 > E_2\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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