Saved Bookmarks
| 1. |
Write the nernst equation and the e.m.f. of the following cells at 298K: (i) Mg(s)|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu(s) (ii) Fe(s)|Fe^(2+)(0.001M)||H^(+)(1M)|H_(2)(g)(1" bar")|Pt(s) (iii) Sn(s)|Sn^(2+)(0.050M)||H^(+)(0.020M)|H_(2)(g)(1" bar")|Pt(s) (iv) Pt(s)|Br_(2)(l)|Br^(-)(0.010M)|H^(+)(0.030M)|H_(2)(g)(1" bar")|Pt(s) Given E_(Mg^(2+)//Mg)^(@)=-2.37V,E_(Cu^(2+)//Cu)^(@)=+0.34V,E_(Fe^(2+),Fe)^(@)=-0.44V E_(Sn^(2+)//Sn)^(@)=-0.14V,E_(1//2Br_(2),Br^(-))^(@)=+1.08V. |
|
Answer» SOLUTION :(i) Cell reaction: `Mg+Cu^(2+)toMg^(2+)+Cu(n=2)` <BR> nernst eqn.: `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"LOG"([Mg^(2+)])/([Cu^(2+)])` `E_(cell)=0.34-(-2.37)-(0.0591)/(2)"log"(10^(-3))/(10^(-4))=2.71-0.02955=2.68V` (ii) Cell reaction: `Fe+2H^(+)toFe^(2+)+H_(2)(n=2)` Nernst eqn.: `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Fe^(2+)])/([H^(+)]^(2))` `thereforeE_(cell)=0-(-0.44)-(0.0591)/(2)"log"(10^(-3))/((1)^(2))=0.44-(0.0591)/(2)xx(-3)=0.44+0.0887=0.523V`. (iii) Cell reaction: `Sn+2H^(+)toSn^(2+)+H_(2)(n=2)` Nernst eqn. : `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Sn^(2+)])/([H^(+)]^(2))=0-(-0.14)-(0.591)/(2)"log"(0.05)/((0.02)^(2))` `=0.14-(0.0591)/(2)log125=0.14-(0.0591)/(2)(2.0969)=0.078V` (iv) Cell reaction: `2Br^(-)+2H^(+)toBr_(2)+H_(2)` (Note carefully) ltrgt Nernst eqn. : `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"(1)/([Br^(-)]^(2)[H^(+)]^(2))` `therefoerE_(cell)=(0-1.08)-(0.0591)/(2)"log"(1)/((0.01)^(2)(0.03)^(2))` `=-1.08-(0.0591)/(2)log(1.111xx10^(7))=-1.08-(0.0591)/(2)(7.0457)=-1.08-0.208=-1.288V` Thus, oxidation will OCCURS at the HYDROGEN electrode and reduction on the `Br_(2)` electrode. `E_(cell)=1.288V`. |
|