The emf of a Daniell cell at 298 K is E 1. Zn | array c ZnSO _ 4 10.01 M , array | |…

Chemistry · JEE Main · NTA ExamsRedox Reactions and Electrochemistry

The emf of a Daniell cell at 298 K is E1.

\(\mathrm{Zn}\left|\begin{array}{c} \mathrm{ZnSO}_{4} \\ 10.01 \mathrm{M}, \end{array}\right|\left|\begin{array}{ccc} \mathrm{CuSO}_{4} & \mathrm{Cu} \\ 1.0 \mathrm{M}, 0 \end{array}\right|\)

When the concentration of ZnSO4 is 1.0M and that of CuSO4 is 0.01M, the emf changed to E2.  What is the relationship between E1 and E2?

  1. E1 = E2
  2. E2 = 0 ≠ E1
  3. E1 > E2
  4. E1 < E2

Answer

(C) E 1 > E 2

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