Two electrochemical cells are assembled in which the following reactions occur `:`
`V^(2)=VO^(2+)+2H^(o+)rarr 2V^(3+)+H_(2)O`
`V^(3+)+Ag^(o+)+H_(2)O rarr VO^(2+)+2H^(o+)+Ag(s)`
Calculate `E^(c-)` for half reaction `V^(3+)+e^(-)rarr V^(2+)`
Given `: E^(c-)._((Ag^(o+)|Ag))=0.799`
`E^(c-)=E^(c-)._(V^(4+)|V^(3+))-E^(c-)._(V^(3+)|V^(2+))=0.616V`
`E^(c-)=E^(c-)._(Ag^(o+)|Ag)-E^(c-)._(V^(4+)|V_(3+))=0.439V`

Explanation

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