Consider the thermodynamic data:
#sf(" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "E^@" "(V))#
#sf(Fe^(2+)" "" " +" "" "2e" "rightleftharpoons" "" "" "Fe" "" "" "" "-0.44)
#
#sf(SO_4^(2-)+4H^++2e" "" "rightleftharpoons" "SO_2+2H_2O" "color(white)(..)+0.17)#
#sf(Fe^(3+)" "+" "" "e" "" "color(white)(.)rightleftharpoons" "" "" "Fe^(2+)" "" "" "+0.77)#
The most powerful oxidisers are to the bottom left. The most powerful reducers are at the top right when the #sf(E^@)# values are listed like this.
We would expect #sf(Fe^(3+)# to oxidise #sf(SO_2)# as it is the more powerful oxidiser so will take in electrons which are given out by #sf(SO_2)#.
The two half equations are:
#sf(SO_2+2H_2OrarrSO_4^(2-)+4H^++2e" "" "" "" "color(red)((1)))#
#sf(Fe^(3+)+erarrFe^(2+)" "" "" "" "" "" "" "" "" "" "color(white)(.)color(red)((2)))#
To get the electrons to balance you #xx# #sf(color(red)((2)))# by 2 then add:
#sf(2Fe^(3+)+SO_2+2H_2Orarr2Fe^(2+)+SO_4^(2-)+4H^+)#
The data indicates that #sf(SO_2)# is not a powerful enough reducer to reduce #sf(Fe^(2+)# down to #sf(Fe)#.
If you were to bubble #sf(SO_2)# gas through an acidified solution of iron(III) chloride you would observe the brown/yellow colour change to green showing the presence of iron(II).
