Question:
In the electrochemical cell :
\(Zn|ZnSO_4(0.01M)||CuSO_4(1.0 M)|Cu,\)
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 following, which one is the relationship between \(E_1\) and \(E_2\)?
(Given, \(\frac {RT}{F}= 0.059\))
In the electrochemical cell :
\(Zn|ZnSO_4(0.01M)||CuSO_4(1.0 M)|Cu,\)
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 following, which one is the relationship between \(E_1\) and \(E_2\)?
(Given, \(\frac {RT}{F}= 0.059\))
\(Zn|ZnSO_4(0.01M)||CuSO_4(1.0 M)|Cu,\)
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 following, which one is the relationship between \(E_1\) and \(E_2\)?
(Given, \(\frac {RT}{F}= 0.059\))
Updated On: Apr 14, 2025
- \(E_1 = E_2\)
- \(E_1 < E_2\)
- \(E_1 > E_2\)
- \(E_2=0 ≠ E_1\)
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The Correct Option is C
Solution and Explanation
The correct option is (C): \(E_1 > E_2\).
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