The e.m.f. of a Daniell cell at 298K is E 1 Zn | array l ZnSO _ 4 (0.0 M ) array | |…
Chemistry · JEE Advanced · NTA Exams — Electrochemistry
The e.m.f. of a Daniell cell at 298K is E1
Zn \(\left|\begin{array}{l} \mathrm{ZnSO}_{4} \\ (0.0 \mathrm{M}) \end{array}\right|\) \(\left|\begin{array}{l} \mathrm{CuSO}_{4} \\ (1.0 \mathrm{M}) \end{array}\right|\) Cu
When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2 ?
Zn \(\left|\begin{array}{l} \mathrm{ZnSO}_{4} \\ (0.0 \mathrm{M}) \end{array}\right|\) \(\left|\begin{array}{l} \mathrm{CuSO}_{4} \\ (1.0 \mathrm{M}) \end{array}\right|\) Cu
When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2 ?
- E1 < E2
- E1 = E2
- E2 = 0 ≠ E1
- E1 > E2
Answer
(D) E 1 > E 2
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