Balancing Redox Equations

It is not always possible to balance redox equations using the simple "inspection" technique. The following unbalanced net ionic equation provides an example.

Au3+(aq) + I?(aq) ? Au(s) + I2(s)

At first glance, it seems that this equation can be balanced by placing a 2 in front of the I?.

Au3+(aq) + 2I?(aq) ? Au(s) + I2(s)

Note, however, that although the atoms are now balanced, the charge is not. The sum of the charges on the left is +1, and the sum of the charges on the right is zero, as if the products could somehow have one more electron than the reactants. To correctly balance this equation, it helps to look more closely at the oxidation and reduction that occur in the reaction. The iodine atoms are changing their oxidation number from ?1 to 0, so each iodide ion must be losing one electron. The Au3+ is changing to Au, so each gold(III) cation must be gaining three electrons. The half-reactions are:

I?(aq) ? ½I2(s) + e?

Au3+(aq) + 3e? ? Au(s) We know that in redox reactions, the number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidizing agent; thus, for each Au3+ that gains three electrons, there must be three I? ions that each lose one electron. If we place a 3 in front of the I? and balance the iodine atoms with a 3/2 in front of the I2, both the atoms and the charge will be balanced.

Au3+(aq) + 3I?(aq) ? Au(s) + 3/2I2(s)

or 2Au3+(aq) + 6I?(aq) ? 2Au(s) + 3I2(s) Balancing Redox Equations Using the Oxidation Number Method

In most situations that call for balancing an equation, you are not told whether the reaction is redox or not. In these circumstances, you can use a procedure called the oxidation number method, which is outlined below.

Sample Study Sheet: Balancing Redox Equations Using the Oxidation Number Technique

Tip-off - If you are asked to balance an equation and if you are not told whether the reaction is a redox reaction or not, you can use the following procedure.

General Steps

Step 1: Try to balance the atoms in the equation by inspection, that is, by the standard technique for balancing non-redox equations. (Many equations for redox reactions can be easily balanced by inspection.) If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3.

Step 2: Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step 3.

Step 3: If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and use them to decide whether the reaction is a redox reaction. If it is not redox, return to Step 1 and try again. If it is redox, go to Step 4.

Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (a ratio that makes the number of electrons lost equal to the number of electrons gained) Step 6: Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. (These coefficients are usually placed in front of the formulas on the reactant side of the arrow.)

Step 7: Balance the rest of the equation by inspection.

Balancing Redox Reactions Using the Oxidation Number Method

EXAMPLE 1 Balance the following redox equation using either the "inspection" technique or the "oxidation number" method. Be sure to check that the atoms and the charge are balanced.

HNO3(aq) + H3AsO3(aq) ? NO(g) + H3AsO4(aq) + H2O(l)

Solution:

Step 1: Try to balance the atoms by inspection.

Step 2: The H and O atoms are difficult to balance in this equation. You might arrive at the correct balanced equation using a "trial and error" technique, but if you do not discover the correct coefficients fairly quickly, proceed to Step 3.

Step 3: Is the reaction redox?

The N atoms change from +5 to +2, so they are reduced. This information is enough to tell us that the reaction is redox. (The As atoms, which change from +3 to +5, are oxidized.)

Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

As +3 to +5 Net Change = +2

N +5 to +2 Net Change = ?3

Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number.

As atoms would yield a net increase in oxidation number of +6. (Six electrons would be lost by three arsenic atoms.) 2 N atoms would yield a net decrease of ?6. (Two nitrogen atoms would gain six electrons.) Thus the ratio of As atoms to N atoms is 3:2.

Step 6: To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing.

2HNO3(aq) + 3H3AsO3 ? NO(g) + H3AsO4(aq) + H2O(l)

Step 7: Balance the rest of the equation by inspection.

2HNO3(aq) + 3H3AsO3(aq) ? 2NO(g) + 3H3AsO4(aq) + H2O(l)

Balancing Redox Reactions Using the Oxidation Number Method EXAMPLE 2 Balance the following redox equation using either the "inspection" technique or the "oxidation number" method. Be sure to check that the atoms and the charge are balanced. Cu(s) + HNO3(aq) ? Cu(NO3)2(aq) + NO(g) + H2O(l)

Solution:

The nitrogen atoms and the oxygen atoms are difficult to balance by inspection, so we will go to Step 3. The copper atoms are changing their oxidation number from 0 to +2, and some of the nitrogen atoms are changing from +5 to +2. These changes indicate that this reaction is a redox reaction. We next determine the changes in oxidation number for the atoms oxidized and reduced.

Cu 0 to +2 Net Change = +2

Some N +5 to +2 Net Change = ?3

We need three Cu atoms (net change of +6) for every 2 nitrogen atoms that change (net change of ?6). Although the numbers for the ratio determined in Step 5 are usually put in front of reactant formulas, this equation is somewhat different. Because some of the nitrogen atoms are changing and some are not, we need to be careful to put the 2 in front of a formula in which all of the nitrogen atoms are changing or have changed. We therefore place the 2 in front of the NO(g) on the product side. The 3 for the copper atoms can be placed in front of the Cu(s).

3Cu(s) + HNO3(aq) ? Cu(NO3)2(aq) + 2NO(g) + H2O(l)

We balance the rest of the atoms using the technique described in Chapter 4, being careful to keep the ratio of Cu to NO 3:2.

3Cu(s) + 8HNO3(aq) ? 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

EXAMPLE 3 Balancing Redox Reactions Using the Oxidation Number Method Balance the following redox equation using either the "inspection" technique or the "oxidation number" method. Be sure to check that the atoms and the charge are balanced.

NO2(g) + H2(g) ? NH3(g) + H2O(l)

Solution:

The atoms in this equation can be balanced by inspection. (You might first place a 2 in front of the H2O to balance the O's, then 7/2 in front of the H2 to balance the H's, and then multiply all the coefficients by 2 to get rid of the fraction.)

2NO2(g) + 7H2(g) ? 2NH3(g) + 4H2O(l)

We therefore proceed to Step 2. For the reaction between NO2 and H2, the net charge on both sides of the equation in Step 1 is zero. Because the charge and the atoms are balanced, the equation is correctly balanced.

Balancing Redox Equations for Reactions in Acidic Conditions Using the Half-reaction Method

Redox reactions are commonly run in acidic solution, in which case the reaction equations often include H2O(l) and H+(aq). This page will show you how to write balanced equations for such reactions even when you do not know whether the H2O(l) and H+(aq) are reactants or products. For example, you may know that dichromate ions, Cr2O72?, react with nitrous acid molecules, HNO2, in acidic conditions to form chromium ions, Cr3+, and nitrate ions, NO3?. Because the reaction requires acidic conditions, you assume that H2O(l) and H+(aq) participate in some way, but you do not know whether they are reactants or products, and you do not know the coefficients for the reactants and products. An unbalanced equation for this reaction might be written

Cr2O72?(aq) + HNO2(aq) ? Cr3+(aq) + NO3?(aq) (acidic)

In order to balance equations of this type, we need a special technique called the half-reaction method or the ion-electron method.

Sample Study Sheet: Balancing Redox Equations Run in Acidic Conditions Using the Half-reaction Technique

Tip-off - If you are asked to balance a redox equation and told that it takes place in an acidic solution, you can use the following procedure.

General Steps

Step 1: Write the skeletons of the oxidation and reduction half-reactions. (The skeleton reactions contain the formulas of the compounds oxidized and reduced, but the atoms and electrons have not yet been balanced.) See Example.

Step 2: Balance all elements other than H and O.

Step 3: Balance the oxygen atoms by adding H2O molecules where needed.

Step 4: Balance the hydrogen atoms by adding H+ ions where needed.

Step 5: Balance the charge by adding electrons, e-.

Step 6: If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost.

Step 7: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them.

Step 8: Check to make sure that the atoms and the charges balance.