I am assuming we have vanadium(V) as #VO_2^+# and vanadium(IV) as #VO^(2+)#.
List the 1/2 equations in order least positive to most positive:
# " " "E"^@("V")#
#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#
#VO_(2(aq))^(+)+2H_((aq))^(+)+erightleftharpoonsVO_((aq))^(2+)+H_2O_((l))" "+1.00#
#Cr_2O_(7(aq))^(2-)+14H_((aq))^(+)+6erightleftharpoons2Cr_((aq))^(3+)+7H_2O_((l))" "+1.33#
#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#
Note we use the #rightleftharpoons# symbol to show that the 1/2 cells can go in either direction depending on what they are coupled with.
The 1/2 cell with the most +ve #"E"^@# is the one which will take in the electrons.
From this we can see that the 2nd 1/2 cell will be driven left to right and the 1st 1/2 cell right to left in accordance with the arrows.
So the 2 half - reactions are:
#VO_((aq))^(2+)+H_2O_((l))rarrVO_(2(aq))^(+)+2H_((aq))^(+)+e" "color(red)((1))#
#Cr_2O_(7(aq))^(2-)+14H_((aq))^(+)+6erarr2Cr_((aq))^(3+)+7H_2O_((l))" "color(red)((2))#
To get the electrons to balance you can see that we need to multiply #color(red)((1))# by #6# and add this to equation #color(red)((2))rArr#
#6VO_((aq))^(2+)+6H_2O_((l))+Cr_2O_(7(aq))^(2-)+cancel(14H_((aq))^(+))+cancel(6e)rarr6VO_(2(aq))^(+)+cancel(12H_((aq))^(+))+cancel(6e)+ 2Cr_((aq))^(3+)+7H_2O_((l)) "#
This simplifies to:
#6VO_((aq))^(2+)+6H_2O_((l))+Cr_2O_(7(aq))^(2-)+2H_((aq))^(+)rarr6VO_(2(aq))^(+)+ 2Cr_((aq))^(3+)+7H_2O_((l)) "#
To find #"E"_(cell)^@# you subtract the least +ve #"E"^@# from the most +ve #"E"^@# value.
This means that #"E"_(cell)^@# always has a +ve value which must be the case for the reaction to be spontaneous under standard conditions.
So in this case:
#"E"_(cell)^@=1.33-1.0=0.33"V"#
The advantage of this convention is that you do not have to make assumptions about the direction of the 1/2 cells and manipulate their signs.
