K2Cr2O7 + H2SO4 + SO2 = K2SO4 + Cr2(SO4)3 + H2O balance by partial equation method??

Please help me to balance this equation by PARTIAL EQUATION METHOD...
NOT IN ANY OTHER METHOD!!!

1 Answer
May 5, 2018

Warning! Long Answer. Here's what I get.

Explanation:

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"K2Cr2O7+3SO2+H2SO4Cr2(SO4)3+K2SO4+H2O

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"K2Cr2O7+H2SO4+SO2K2SO4+Cr2(SO4)3+H2O

Strip the equation of all common ions, also "H"^"+", "OH"^"-"H+,OH- and "H"_2"O"H2O (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"