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) lt E_(2)`B. `E_(1) gt E_(2)`C. `E_(2)=0neE_(1)`D. `E_(1)=E_(2)`

The emf of a Daniell cell at `298 K` is `E_(1)` `Zn|ZnSO_(4)(0.01 M)||

1.	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) lt E_(2)`B. `E_(1) gt E_(2)`C. `E_(2)=0neE_(1)`D. `E_(1)=E_(2)`
Answer» Correct Answer - B (b) According to Nernst equation. `E_(cell)=E_(cell)^(@)-(0.059)/(n)"log"([Zn^(2+)])/([Cu^(2+)])` `E_(1)=E_(cell)^(@)-(0.059)/(2)"log"(0.01)/(1)` `=E_(cell)^(@)-(0.059)/(2)(-2)=E_(cell)^(@)+0.059` `E_(2)=E_(cell)^(@)-(0.059)/(2)"log"(1)/(0.01)` `E_(2)=E_(cell)^(@)-(0.059)/(2)(2)=E_(cell)^(@)-0.059` Hence, `E_(1) gt E_(2)`.

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