The balanced equation is
#"K"_2"Cr"_2"O"_7 + "3SO"_2 + "H"_2"SO"_4 → "Cr"_2("SO"_4)_3 + "K"_2"SO"_4 + "H"_2"O"#
I think you are referring to the ion-electron method or the half-reaction method.
Step 1. Write the skeleton equation
The molecular equation is
#"K"_2"Cr"_2"O"_7 + "H"_2"SO"_4 + "SO"_2 → "K"_2"SO"_4 + "Cr"_2("SO"_4)_3 + "H"_2"O"#
Strip the equation of all common ions, also #"H"^"+", "OH"^"-"# and #"H"_2"O"# (these come back in during the balancing procedure).
#color(red)(cancel(color(black)("K"^"+"))) + "Cr"_2"O"_7^"2-" + color(red)(cancel(color(black)("H"^"+"))) + color(red)(cancel(color(black)("SO"_4^"2-"))) + "SO"_2 → color(red)(cancel(color(black)("K"^"+"))) + color(red)(cancel(color(black)("SO"_4^"2-"))) + "Cr"^"3+" + color(red)(cancel(color(black)("SO"_4^"2-"))) + color(red)(cancel(color(black)("H"_2"O")))#
The above equation simplifies to
#"Cr"_2"O"_7^"2-" + "SO"_2 → "Cr"^"3+"#
Step 2: Separate the skeleton equation into two half-reactions.
#"Cr"_2"O"_7^"2-" → "Cr"^"3+"#
#"SO"_2 → ?#
Uh, oh! We have no product for the second half-reaction. It must have been one of those sulfate ions we deleted. We write
#"Cr"_2"O"_7^"2-" → "Cr"^"3+"#
#"SO"_2 → "SO"_4^"2-"#
Step 3: Balance all atoms other than #"H"# and #"O"#.
#"Cr"_2"O"_7^"2-" → "2Cr"^"3+"#
#"SO"_2 → "SO"_4^"2-"#
Step 4: Balance #"O"#.
Add enough #"H"_2"O"# molecules to balance #"O"#.
#"Cr"_2"O"_7^"2-" → "2Cr"^"3+" + 7"H"_2"O"#
#"SO"_2 + 2"H"_2"O" → "SO"_4^"2-"#
Step 5: Balance #"H"#.
Add enough #"H"^"+"# ions to balance #"H"#.
#"Cr"_2"O"_7^"2-" + "14H"^"+" → "2Cr"^"3+" + 7"H"_2"O"#
#"SO"_2 + 2"H"_2"O" → "SO"_4^"2-" + "4H"^"+"#
Step 6: Balance charge.
Add electrons to the side that needs more negative charge.
#"Cr"_2"O"_7^"2-" + "14H"^"+" +6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"#
#"SO"_2 + 2"H"_2"O" → "SO"_4^"2-" + "4H"^"+" + 2"e"^"-"#
Step 7: Equalize electrons transferred.
Multiply each half-reaction by numbers to get the lowest common multiple of electrons transferred.
#1 ×["Cr"_2"O"_7^"2-" + "14H"^"+" +6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"]#
#3 × ["SO"_2 + 2"H"_2"O" → "SO"_4^"2-" + "4H"^"+" + 2"e"^"-"]#
Step 8: Add the two half-reactions.
#"Cr"_2"O"_7^"2-" + stackrelcolor(blue)(2)(color(red)(cancel(color(black)(14))))"H"^"+" + color(red)(cancel(color(black)(6"e"^"-"))) → "2Cr"^"3+" + stackrelcolor(blue)(1)(color(red)(cancel(color(black)(7))))"H"_2"O"#
#ul(3"SO"_2 + color(red)(cancel(color(black)(6"H"_2"O"))) → "3SO"_4^"2-" + color(red)(cancel(color(black)("12H"^"+"))) + color(red)(cancel(color(black)(6"e"^"-")))color(white)(mmmm))#
#"Cr"_2"O"_7^"2-" + "3SO"_2 + "2H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"H"_2"O"#
Step 9. Replace the spectator ions
#"Cr"_2"O"_7^"2-" + "3SO"_2 + "2H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" + color(white)(mmmmll)"H"_2"O"#
#ul("2K"^"+" +color(white)(mmmmml)"SO"_4^"2-" → color(white)(mmmmmmmm)"2K"^"+" + "SO"_4^"2-"color(white)(mm))#
#"K"_2"Cr"_2"O"_7 + "3SO"_2 + "H"_2"SO"_4 → "Cr"_2("SO"_4)_3 + "K"_2"SO"_4 + "H"_2"O"#
Step 10: Check mass balance.
#ulbb("Atom"color(white)(m)"On the left"color(white)(m)"On the right")#
#color(white)(ml)"K"color(white)(mmmmm)2color(white)(mmmmmmm)2#
#color(white)(ml)"Cr"color(white)(mmmmll)2color(white)(mmmmmmm)2#
#color(white)(ml)"O"color(white)(mmmmll)17color(white)(mmmmmml)17#
#color(white)(ml)"S"color(white)(mmmmmll)3color(white)(mmmmmmll)3#
#color(white)(ml)"H"color(white)(mmmmlm)2color(white)(mmmmmmll)2#
∴ The balanced equation is
#"K"_2"Cr"_2"O"_7 + "3SO"_2 + "H"_2"SO"_4 → "Cr"_2("SO"_4)_3 + "K"_2"SO"_4 + "H"_2"O"#
