Two electrochemical cells are assembled in which the following reaction occur.
`V^2+VO^2+2H^+ to 2V^3+H_2O, E_(cell) ^@=0.616V`
`V^3+Ag^+ +H_2O to VO^2+2H^+ + Ag(s), E_(cell)^@=0.439V`
calculate `E^@` for the half cell reaction
`V^(3+) +e to V^(2+) (E_(Ag)^@, Ag=0.799` volt )

Explanation

Let us find the number of electrons involved in each reaction
`V^(2+)+VO^(2+)+2H^+ to V^(3+)+V^(3+)+H_2O....(I)`

`V^(2+)+Ag^+ +H_2O to Ag(s)+VO^(2+)+2H^+`

Adding (1) and (2) we get
`V^(2+)+Ag^+ to V^(3+) +Ag(s), Delta G^@=-(1 times F times 0.616+1 times F times 0.439)`
`=1.055F`
`therefore E_(cell)^@= - (DeltaG^@)/(nF)= (1.055 F)/F=1.055V`
Furhter for the above cell
`E_(cell)^@=E^@Ag^+,Ag-E_(v^(3+))^@,V^(2+)`
`1.055=0.799-E_(v^(3+))^@,V^(2+)`
or `E_(v^(3+))^@,V^(2+)=0.799-1.055`
`=-0.256` volt