Two electrochemical cells are assembled in which the following reactions occur V^(2+)+VO^(2+)+2H^(+) to 2V^(3+)+H_(2),E_(cell)^(@)=0.616V V^(3+)+Ag^(+)+H_(2)toVO^(2+)+2H^(+)+Ag(s),E_(cell)^(@)=0.439V Calculate the E^(@) value for the half-cell reaction V^(3+)+e^(-)toV^(2+). (Given: E_(Ag^(+)//Ag)^(@)=0.799volt) lt

Two electrochemical cells are assembled in which the following reactio

1.	Two electrochemical cells are assembled in which the following reactions occur V^(2+)+VO^(2+)+2H^(+) to 2V^(3+)+H_(2),E_(cell)^(@)=0.616V V^(3+)+Ag^(+)+H_(2)toVO^(2+)+2H^(+)+Ag(s),E_(cell)^(@)=0.439V Calculate the E^(@) value for the half-cell reaction V^(3+)+e^(-)toV^(2+). (Given: E_(Ag^(+)//Ag)^(@)=0.799volt) lt
Answer» Solution :Two half CELLS for the 1st cell reaction will be (i) `V^(2+)toV^(3+)+e^(-),E_(1)^(@)(O x)` (ii) `VO^(2+)+2H^(+)+e^(-)toV^(3+)+H_(2)O,E_(2)^(@)(red.)` Two half-cells for the 2ND cell reaction will be (iii) `V^(3+)+H_(2)OtoVO^(2+)+2H^(+)+e^(-),E_(3)^(@)(O x),` (iv) `AG^(+)+e^(-)toAg,E_(4)^(@)(Red).0.799V` (given) `E_(3)^(@)+E_(4)^(@)=0.439V` `thereforeE_(3)^(@)(O x.)=0.439-E_(4)^(@)=0.439-0.799=-0.360V` As reaction (ii) is reverse of reaction (iii) `E_(2)^(@)(Red.)=-E_(3)^(@)(O x.)=+0.360V` Further, `E_(1)^(@)(O x)+E_(2)^(@)(Red.)=0.616` or `E_(1)^(@)(O x.)=0.616-E_(2)^(@)(Red.)=0.616-0.360=0.256V` As required reaction is reverse of reaction (i) `E^(@)`(Reqd. reaction)=-0.256V