Oxidation Number Change Method — Balancing Redox Reactions

We're beginning our journey through electrochemistry! We started by introducing a new type of chemical reaction. That's Redox (oxidation-reduction) reactions. Then we learned that each atom in a redox reaction has an oxidation number. For the most part, the oxidation number comes from the element's periodic table group. But, there are exceptions. For example, the oxidation state of the diatomic gases is zero.

What causes a redox reaction? We need an oxidizing or reducing agent. They are atoms or molecules that readily lose or gain electrons. We identify redox reactions by the movement of electrons. Specifically, the movement of valence electrons changes the oxidation number. And we get a reaction! Oxidation number change often affects physical properties. That's why we examined oxidation state and color. Today, we'll take the next step. We'll balance redox reactions based on their oxidation number.

Why a New Method?

Way back, we learned to write and balance chemical equations. So, why do we need a new method for redox reactions? Before, finding the mole ratio also balanced the charge. So, we obeyed the conservation of mass and charge. But, not all equations are so easy. Let's look at an example.

$Au_{(aq)}^{3+} + I_{(aq)}^{-} \rightarrow Au_{(s)} + I_{2(g)}$

It seems that to balance the equation, we only need to double the iodide ion. But, check the charges. The charge of the reactants is +1. On the products' side, it's 0. Our equation isn't balanced!

Oxidation Number Change Method Steps

So, how do we balance redox reactions? We use the oxidation change method. It uses the charge (oxidation numbers) to balance the mass. It has three steps.

1. Assign oxidation numbers to each atom and ion.

2. Determine what atoms are oxidized and reduced and by how much.

3. Find the lowest common multiple of the oxidation and reduction. That will tell you how many atoms you need on each side of the equation. So, you've found your coefficients.

Examples

1. Use the redox change method to balance the equation $Fe_{2}O_{3(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}$.

$Fe: +3 \rightarrow 0$

We start with iron(III) oxide. Two iron atoms are ionically bonded to three oxygen. Oxygen has a -2 oxidation state. So, iron must start at +3. In the products, iron is in its elemental state. So, its oxidation number is 0. For iron to go from +3 to 0, it had to gain electrons. So, iron got reduced.

$O: -2 \rightarrow -2$

Oxygen is a powerful oxidizing agent. So, it very rarely changes its oxidation state. Here the oxidation state is the same in both the products and reactants. So, it was neither oxidized nor reduced.

$C: +2 \rightarrow +4$

In the reactants, we start with carbon monoxide. The carbon bonded one oxygen. In the products, we have carbon dioxide. Carbon's oxidation state increased from +2 to +4. So, carbon must have lost electrons. It was oxidized.

So now we know what was oxidized and reduced. But, how do we balance the equation? We take the least common multiple of 2 and 3. That's 6! So we need two iron atoms because $2\times3=6$. We also need three carbon atoms because $3\times 2=6$. We already have two iron atoms in the reactants. So, we only add a coefficient to elemental iron in the products. We only have one carbon for both the reactants and products. So, we'll add coefficients to both sides.

$Fe_{2}O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}$.

1. Use the redox change method to balance the equation $H_{2}S+ HNO_{3} \rightarrow S + NO + H_{2}O$.

$S: -2\rightarrow 0$

Sulfer is oxidized by +2.

$N: +5 \rightarrow +2$

Nitrogen is reduced by -3.

$O: -2\rightarrow -2$

$H: +1 \rightarrow +1$

Neither hydrogen nor oxygen gets oxidized or reduced. So, we'll only need to balance their masses.

The common multiple for 2 and 3 is 6. So we need to have three sulfur and two nitrogen on each side. We finish balancing the equation by using the water to balance hydrogen.

$3H_{2}S+ 2HNO_{3} \rightarrow 3S + 2NO + 4H_{2}O$

1. Use the redox change method to balance the equation $K + KNO_{3} \rightarrow N_{2} + K_{2}O$.

$K: 0 \rightarrow +1$

Potassium metal is oxidized by +1.

$K\, (from \, KNO_{3}): +1 \rightarrow +1$

$N: +5 \rightarrow 0$

Nitrogen is reduced by -5.

$O: -2 \rightarrow -2$

The lowest common multiple between 1 and 5 is 5. So we need five potassium atoms and one nitrogen on each side. But, there is a trick! We have one potassium that is not oxidized. So, we need to account for that. We will have one potassium atom on each side that did not change its oxidation state.

So, our balanced equation would be $5K + 1KNO_{3} \rightarrow \frac{1}{2}N_{2} + 3K_{2}O$. But, we have a problem! To only have one nitrogen on the product's side, we have a coefficient of one-half. We don't like having a fraction. So, we need to multiply everything by two. So, our balanced equation is $10K + 2KNO_{3} \rightarrow N_{2} + 6K_{2}O$.

Other Methods

You should know that there are other methods for balancing redox reactions. Some of the methods are dependent on the solution's pH. So, don't forget everything we've learned about acids and bases. I've linked the other methods below.

• Half-reaction method

• Balancing redox reaction in acidic solution.

• Balancing redox reactions in basic solutions.

(Video) Oxidation Number Method

by JFR Science

In this video, John explains the oxidation number change method. He walks you step-by-step through an example problem. He'll even show you how to convert from an acidic solution to a basic solution. If you need help with this method, this video can help!