Saved Bookmarks
| 1. |
Calculate the EMF of the cell for the reaction Mg_((s))+2Ag_((aq))^(+)rarr Mg_((aq))^(2+)+2Ag_((s)). ("Given " :E^(@)Mg^(2+)//Mg=-2.37V, E^(@)Ag^(+)//Ag=0.80V, [Mg^(2+)]=0.001M[Ag^(+)]=0.001M and log10^(5)=5). |
|
Answer» SOLUTION :`E_("CELL")^(@)=E_("cathode")^(@)-E_("anode")^(@)` `E_("cell")^(@)=E_(Ag^(+)//Ag)^(@)-E_(Mg^(2+)//Mg)^(@)` `E_("cell")^(@)=0.80-(-2.37)` `E_("cell")^(@)=3.17V` `E_("cell")=E_("cell")^(@)-(0.059)/(n)log""([Mg^(2+)])/([Ag^(+)]^(2))` `E_("cell")=3.17-(0.059)/(2)log""([0.001])/([0.0001]^(2))` `E_("cell")=3.17-(0.059)/(2)LOG10^(5)` `E_("cell")=3.17-(0.059)/(2)xx5` `E_("cell")=3.022V` |
|