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The EMF of a Daniel cell at 298K is E(1)...

The EMF of a Daniel cell at 298K is `E_(1)`. The cell is: `Zn(s)abs( ZnSO_(4) (0.01M)) | abs(CuSO_(4)(1.0M)) Cu(s).` When the concentration of `ZnSO_(4)` is 1.0 M and that of `CuSO_(4)` is `0.01` M the emf changes to `E_(2)`. What is the relationship between `E_(1) and E_(2)`?

A

`E_(2) = 0 ne E_(1)`

B

`E_(1) gt E_(2)`

C

`E_(1) lt E_(2)`

D

`E_(1) = E_(2)`

Text Solution

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The correct Answer is:
B
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