InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Calculate the standard cell potentials of galvanic cells in which the following reactions take place: (i) `2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)` (ii) `Fe^(2+)(aq)+Ag^(2+)(aq)toFe^(3+)(aq)+Ag(s)` Given `E_(Cr^(3+),Cr)^(@)=-0.74V,E_(Cd^(2+),Cd)^(@)=-0.40V,E_(Ag^(+),Ag)^(@)=0.80V,E_(Fe^(3+),Fe^(2+))^(@)=0.77V` Also calculate `Delta_(r)G^(@)` and equilibrium constant of the reaction. |
|
Answer» (i) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=-0.40V-(-0.74V)=+0.34V` `Delta_(r)G^(@)=-nFE_(cell)^(@)=-6molxx96500" C "mol^(-1)xx0.34V` `=-196860" CV "mol^(-1)=-196860J" "mol^(-1)=-196.86" kJ "mol^(-1)` `-Delta_(r)G^(@)=2.303" RT "logK` `196860=2.303xx8.314xx298logK` or `logK=34.5014` K=Antilog 34.5014=3.192`xx10^(34)` (ii) `E_(cell)^(@)=+0.80V-0.77V=+0.03V` `Delta_(r)G^(@)=-nFE_(cell)^(@)=-(1mol)xx(96500" C "mol^(-1))xx(0.03V)` `=-2895" CV "mol^(-1)=-2895" J "mol^(-1)` ltBrgt `=-2.895" kJ "mol^(-1)` `Delta_(r)G^(@)=-2.303" RT "logK` `-2895=-2.303xx8.314xx298xxlogK` or log K`=0.5974` or K=Antilog (0.5974)=3.22. |
|