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 )

Two electrochemical cells are assembled in which the following reactio

1.	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 )
Answer» Solution :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