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)` ?

Options

A. `1.10 V `
B. `1.04 V `
C. `1.16 V `
D. `1.07 V`

Answer

D

Explanation

`E = E^(Theta) - (0.059)/(2) "log"([Zn^(2+)])/([Cu^(2+)])`
`E^(Theta) = E_((Cu^(2+)|Cu))^(Theta) - E_((Zn^(2+) |Zn))^(Theta)`
`E^(Theta) = 0.34 - (-0.76) = 1.10 V `
`therefore E = 1.10 - 0.295 = 1.0705 V `.