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"#?

1 Answer
Oct 22, 2015

Answer:

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"#