Ask Me Help Desk - Balance equations using the redox method?

Quote:

a. Fe + O2 ---> Fe2O3

b. KClO3 ---> KCl + O2

The four steps to using the redox method (using half-reactions).

1. Write each half reaction using the species that is either oxidized or reduced.
2. Add water to balance the oxygen atoms
3. Add hydrogen ion to balance the hydrogen atoms
4. Add electrons to balance the charge.
5. If it's in basic solution, add the same number of hydroxide ions as you have hydrogen ions to both sides. In your head, mix the hydrogen atoms with the hydroxide atoms and write them as H2O. Cancel redundant water from both sides of the equation so that only one side has water in it.

Basically, you need to know, figure out, or look up the half reactions. In "a", above, these would be the half reactions:

$Fe \,\rightleftharpoons\, Fe^{3+} + 3e^-$

$4e^- + O2 \,\rightleftharpoons\, 2O^{2-}$

Note that you can reverse the reactions. The first thing to do is to make sure the electrons are on opposite sides of the harpoons. This is so that they will cancel out when you add the equations. The second thing to do is to multiply the two equations with constants, if required, to get the number of electrons on both sides of the equation to match. We'll multiply the first equation by 4 and the second equation by 3:

$4Fe \,\rightleftharpoons\, 4Fe^{3+} + 12e^-$

$12e^- + 3O2 \,\rightleftharpoons\, 6O^{2-}$

Now add the two equations:

$4Fe + 3O2 \rightleftharpoons 4Fe^{3+} + 6O^{2-}$

Double-check to make sure the number of atoms on both sides is the same. It will always work if you did it correctly.

In this case, we have one more step, and that's to combine the iron and oxygen into a compound:

$4Fe + 3O2 \,\rightleftharpoons\, 4Fe^{3+} + 6O^{2-} \rightarrow 2Fe_2O_3$

Example B is similar, but not identical. In this case, the potassium ion is a "spectator ion". It isn't oxidized or reduced. We will eliminate it for the half reactions.

$ClO_3^- \rightarrow Cl-$
$O^{2-} \rightarrow O_2$

Step 1: Balance the oxidized or reduced atom count:

$ClO_3^- \rightarrow Cl-$
$2O^{2-} \rightarrow O_2$

Step 2: Add water to balance the oxygen

$ClO_3^- \rightarrow Cl^- + 3H_2O$
$2O^{2-} \rightarrow O_2$

Step 3: Add hydrogen ion to balance the hydrogen:

$ClO_3^- + 6H^+ \rightarrow Cl^- + 3H_2O$
$2O^{2-} \rightarrow O_2$

Step 4: Add electrons to balance the charge. In the first equation, there is a net of 5+ on the left and 1- on the right. We need to add 6 electrons on the left. In the second equation, we simply need to add four electrons to the right.

$6e^- + ClO_3^- + 6H^+ \rightarrow Cl^- + 3H_2O$
$2O^{2-} \rightarrow O_2 + 4e^-$

Now, multiply the first equation by 2, the second equation by 3 so the number of electrons is the same on both sides.

$12e^- + 2ClO_3^- + 12H^+ \rightarrow 2Cl^- + 6H_2O$
$6O^{2-} \rightarrow 3O_2 + 12e^-$

And, add the equations:

$2ClO_3^- + 12H^+ + 6O^{2-} \rightarrow 3O_2 + 2Cl^- + 6H_2O$

Now, recognize that 12H(+) and 6O(2-) would immediately form 6 H2O so

$2ClO_3^- + 6H_2O \rightarrow 3O_2 + 2Cl^- + 6H_2O$

Also note that you have 6 H2O on both sides of the equation so you can eliminate them

$2ClO_3^- \rightarrow 3O_2 + 2Cl^-$

Now you can add the K(+) back in

$2KClO_3 \rightarrow 3O_2 + 2KCl$