# 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"?

Oct 22, 2015

Here are the detailed steps.

#### Explanation:

$\text{Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O}$

1. Identify the oxidation number of every atom.

Left hand side: $\text{Cr = +6}$; $\text{O = -2}$; $\text{S = +4}$; $\text{H = +1}$
Right hand side: $\text{Cr = +3}$; $\text{S = +6}$; $\text{O = -2}$; $\text{H = +1}$

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

$\text{Cr: +6 → +3}$; Change = -3
$\text{S: +4 → +6}$; Change = +2

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

We need 2 atoms of $\text{Cr}$ for every 3 atoms of $\text{S}$. This gives us total changes of -6 and +6.

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

$\textcolor{red}{1} \text{Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + "H"_2"O}$

5. Balance all remaining atoms other than $\text{H}$ and $\text{O}$.

Done.

6. Balance O.

We have fixed 16 atoms of $\text{O}$ on the left, so we need 16 atoms of $\text{O}$ on the right.

And we have fixed 12 atoms of $\text{O}$ on the right, so we need 4 more.

Put a $\textcolor{b l u e}{4}$ in front of the $\text{H"_2"O}$.

$\textcolor{red}{1} \text{Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O}$

7. Balance H.

We have fixed 8 atoms of $\text{H}$ on the right, so we need 8 on the left.

Put an $\textcolor{g r e e n}{8}$ in front of the ${\text{H}}^{+}$.

$\textcolor{red}{1} \text{Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + color(green)(8)"H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O}$

8. Check that everything is balanced.

(a) Atoms

On the left: $\text{2Cr; 16 O; 3 S; 8 H}$
On the right:$\text{2Cr; 3 S; 16 O; 8 H}$

(b) Charge

On the left: 0
On the right: 0

Everything checks!

The final balanced equation is

$\text{Cr"_2"O"_7^(2-) + 3"SO"_3^(2-) + 8"H"^+ → 2"Cr"^(3+) + 3"SO"_4^(2-) + 4"H"_2"O}$