Saved Bookmarks
| 1. |
Calcualte the emf of the following cell at 25^(@) C Ag(s) | Ag^(+) (10^(-3) M) || Cu^(2+) (10^(-1) M) | Cu(s) Given: E_("cell")^(@) = +0.46 V and log 10^(n) =n. |
|
Answer» Solution :The cell reaction is : `Cu + 2Ag^(+) to Cu^(2+) + 2Ag` EMF of the cell can be obtained by using the NERNST EQUATION: `E_("cell") = E_("cell")^(@) -0.059/2 LOG ([Cu^(2+)])/([Ag^(+)]^(2))` Substituting the values, we get `E_("cell") = 0.46 V - 0.059/2 log ([10^(-1)])/([10^(-3)])^(2)` `=0.46 - 0.0295 log 10^(5)` `=0.46 - 0.1475` or `E_("cell") = 0.3125` V |
|