Question

Can you walk through the steps for balancing `"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"`?

Answer 1

Here are the detailed steps.

Explanation:

Your equation is

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

1. Identify the oxidation number of every atom.

Left hand side: `"Cr = +6"`; `"O = -2"`; `"S = +4"`; `"H = +1"`
Right hand side: `"Cr = +3"`; `"S = +6"`; `"O = -2"`; `"H = +1"`

2. Determine the change in oxidation number for each atom that changes.

`"Cr: +6 → +3"`; Change = -3
`"S: +4 → +6"`; Change = +2

3. Make the total increase in oxidation number equal to the total decrease in oxidation number.

We need 2 atoms of `"Cr"` for every 3 atoms of `"S"`. This gives us total changes of -6 and +6.

4. Put the appropriate coefficients in front of the formulas containing those atoms.

`color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + "H"_2"O"`

5. Balance all remaining atoms other than `"H"` and `"O"`.

Done.

6. Balance O.

We have fixed 16 atoms of `"O"` on the left, so we need 16 atoms of `"O"` on the right.

And we have fixed 12 atoms of `"O"` on the right, so we need 4 more.

Put a `color(blue)(4)` in front of the `"H"_2"O"`.

`color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O"`

7. Balance H.

We have fixed 8 atoms of `"H"` on the right, so we need 8 on the left.

Put an `color(green)(8)` in front of the `"H"^+`.

`color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + color(green)(8)"H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O"`

8. Check that everything is balanced.

(a) Atoms

On the left: `"2Cr; 16 O; 3 S; 8 H"`
On the right:`"2Cr; 3 S; 16 O; 8 H"`

(b) Charge

On the left: 0
On the right: 0

Everything checks!

The final balanced equation is

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