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The emf of a Daniell cell at 298 K is E(...

The emf of a Daniell cell at `298 K` is `E_(1)`
`Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu`
When the concentration of `ZNSO_(4)` is `1.0 M` and that of `CuSO_(4)` is `0.01 M`, the `emf` changed to `E_(2)`. What is the relationship between `E_(1)` and `E(2)` ?

A

`E_1 gt E_2`

B

`E_1 lt E_2 `

C

`E_1 =E_2`

D

`E_2 = 0 ne E_1`

Text Solution

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The correct Answer is:
A
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The emf of a Daniell cell at 298 K is E_1 Zn underset((0.01M))|NnSO_4 |underset((1.0M))| CuSO_4 | Cu when the concentration of ZnSO_4 is 1.0 M and that of CuSO_4 is 0.01 M the emf changed to E_2 what is the relationship between E_1 and E_2 .

Calculate EMF of following cell at 298 K Zn|ZnSO_(4) (0.1 M) ||CuSO_(4) (1.0 M)|Cu (s) if E_(cell) = 2.0 V

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