Question

Cr2O7^2- +SO3^2-=Cr^3+ +SO4^2- Balance the ionic equation?

Answer 1

`Cr_2O_7^(2-) +3SO_3^(2-) +8H^+ rarr 2Cr^(3+) +3SO_4^(2-)+4H_2O(l)`

Explanation:

We gots a redox equation....so we write half equation...

`"Dichromate is reduced:"`
`Cr_2O_7^(2-) +14H^+ + 6e^(-) rarr 2Cr^(3+) +7H_2O(l)`

`"Sulfite is oxidized:"`
`SO_3^(2-) +H_2O rarr SO_4^(2-)+2H^+ +2e^(-)`

We adds THREE of the latter to one of the former to eliminate the electrons....

`Cr_2O_7^(2-) +3SO_3^(2-) +cancel(3H_2O)+cancel(14)8H^+ + cancel(6e^(-)) rarr 2Cr^(3+) +cancel(7)4H_2O(l)+3SO_4^(2-)+cancel(6H^+ +6e^(-))`

....to get finally...

`underbrace(Cr_2O_7^(2-))_"orange red" +3SO_3^(2-) +8H^+ rarr underbrace(2Cr^(3+))_"green" +3SO_4^(2-)+4H_2O(l)`

..the which is balanced with respect to mass and charge as required....

Answer 2

The balanced equation is

`"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"`

Explanation:

Step 1: Separate the skeleton equation into two half-reactions.

`"Cr"_2"O"_7^"2-" → "Cr"^"3+"`
`"SO"_3^"2-" → "SO"_4^"2-"`

Step 2: Balance all atoms other than `"H"` and `"O"`.

`"Cr"_2"O"_7^"2-" → "2Cr"^"3+"`
`"SO"_3^"2-" → "SO"_4^"2-"`

Step 3: Balance `"O"`.

Add enough `"H"_2"O"` molecules to balance `"O"`.

`"Cr"_2"O"_7^"2-" → "2Cr"^"3+" + 7"H"_2"O"`
`"SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-"`

Step 4: Balance `"H"`.

Add enough `"H"^"+"` ions to balance `"H"`.

`"Cr"_2"O"_7^"2-" + "14H"^"+" → "2Cr"^"3+" + 7"H"_2"O"`
`"SO"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+"`

Step 5: 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"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+" + 2"e"^"-"`

Step 6: 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"_3^"2-" + "H"_2"O" → "SO"_4^"2-" + "2H"^"+" + 2"e"^"-"]`

Step 7: Add the two half-reactions.

`"Cr"_2"O"_7^"2-" + stackrelcolor(blue)(8)(color(red)(cancel(color(black)(14))))"H"^"+" + color(red)(cancel(color(black)(6"e"^"-"))) → "2Cr"^"3+" + stackrelcolor(blue)(4)(color(red)(cancel(color(black)(7))))"H"_2"O"`
`ul(3"SO"_3^"2-" + color(red)(cancel(color(black)(3"H"_2"O"))) → "3SO"_4^"2-" + color(red)(cancel(color(black)("6H"^"+"))) + color(red)(cancel(color(black)(6"e"^"-")))color(white)(mmmm))`
`"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"`

Step 8: Check mass balance.

`ulbb("Atom"color(white)(m)"On the left"color(white)(m)"On the right")`
`color(white)(ml)"Cr"color(white)(mmmmll)2color(white)(mmmmmmm)2`
`color(white)(ml)"O"color(white)(mmmmll)16color(white)(mmmmmml)16`
`color(white)(ml)"S"color(white)(mmmmmll)3color(white)(mmmmmmll)3`
`color(white)(ml)"H"color(white)(mmmmlm)8color(white)(mmmmmmll)8`

Step 9. Check charge balance

`ulbb(color(white)(m)"On the left"color(white)(m))color(white)(mll)ulbb("On the right")`
`- 2 -6 +8 =0color(white)(mm)+6 - 6 = 0`

The balanced equation is

`"Cr"_2"O"_7^"2-" + "3SO"_3^"2-" + "8H"^"+" → "2Cr"^"3+" + "3SO"_4^"2-" +"4H"_2"O"`