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The emf of a Daniel cell Zn|Zn^(2+)(0.00...

The emf of a Daniel cell `Zn|Zn^(2+)(0.001M)||Cu^(2+)(1M)|Cu` at 298 K is `E_(1)`. When concentration of `ZnSO_(4)` is 1 M and concentration of `CuSO_(4)` is 0.001M, emf changes to `E_(2)`. Relation between `E_(1)` and `E_(2)` is

A

`E_(1)=E_(2)`

B

`E_(1)gtE_(2)`

C

`E_(1)ltE_(2)`

D

`E_(1)=(E_(2))/2`

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BRILLIANT PUBLICATION-REDOX REACTION & ELECTROCHEMISTRY-QUESTION (ELECTROCHEMISTRY) (LEVEL -I) (HOMEWORK)
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