Question 15.13: Balancing a Disproportionation Redox Equation Balance the fo......

a. Oxidation Number Method

Step 1 In assigning oxidation numbers, we immediately become aware that this is a disproportionation reaction. Nitrogen is the only element for which an oxidation number change occurs.

NO _2  \longrightarrow  NO {_3} ^- + NO

+4 -2                  +5 -2     +2 -2

Step 2 Since the species NO _2 is undergoing both oxidation and reduction, for balancing purposes we will write it twice on the reactant side of the equation. With the NO _2 in two places, brackets can then be drawn in the normal manner to connect the substances involved in oxidation and reduction.

(The NO _2 molecules will be recombined later into one location.)

Step 3 The change in oxidation number per formula unit in both cases is the same as per atom.

Step 4 By multiplying the oxidation number increase by 2, we equalize the oxidation number increase and decrease per formula unit.

Step 5 The bracket notation indicates that two N atoms undergo an increase in oxidation number for every one that undergoes a decrease in oxidation number. Translating this information into equation coefficients gives

2 NO _2 + 1 NO _2  \longrightarrow  2 NO {_3} ^- + 1 NO

Now that the equation coefficients for the substance involved in oxidation and reduction, NO _2 , have been determined, we can combine the NO _2 into one location, reversing the process carried out in step 2.

3 NO _2  \longrightarrow  2 NO {_3} ^- + 1 NO

Step 6 The only atoms left to balance are oxygen atoms.

Step 7 Since this is a net ionic equation, charge must be balanced. In an acidic solution, which is the case here, we balance the charge by adding H ^+ ion. As the equation now stands, we have a charge of -2 on the right side (2 NO {_3} ^- ions). By adding two H ^+ ions to the right side of the equation we balance the charge at zero.

3 NO _2  \longrightarrow  2 NO {_3} ^- + 1 NO + 2 H ^+

Step 8 Hydrogen atom balance is achieved through the addition of H _2 O molecules. There are no H atoms on the left side and two H atoms on the right side. Addition of one H _2 O molecule to the left side will balance the H atoms at two on each side.

1 H _2 O + 3 NO _2  \longrightarrow  2 NO {_3} ^- + 1 NO + 2 H ^+

Step 9 The oxygen atoms should automatically balance. They do, at seven atoms on each side.

H _2 O + 3 NO _2  \longrightarrow  2 NO {_3} ^- + NO + 2 H ^+

b. Half-Reaction Method

Step 1 Determine the oxidation and reduction skeletal half-reactions.

Assignment of oxidation numbers is the same as in part (a).

NO _2  \longrightarrow  NO {_3} ^- + NO

+4 -2                  +5 -2     +2 -2

Nitrogen is undergoing both oxidation and reduction. The skeleton half-reactions for oxidation and reduction are

Oxidation:          NO _2  \longrightarrow  NO {_3} ^-

Reduction:         NO _2  \longrightarrow  NO

Note how disproportionation is handled at this point. The substance undergoing disproportionation appears as a reactant in both the oxidation and reduction half-reactions.

Step 2 Balance the individual half-reactions.

a. In both half-reactions, the element being oxidized or reduced is already balanced-one atom of N in both cases.

Oxidation:          NO _2  \longrightarrow  NO {_3} ^-

Reduction:         NO _2  \longrightarrow  NO

b. The oxidation number increase for the oxidized N is +1, which corresponds to the loss of one electron.

Oxidation:          NO _2  \longrightarrow  NO {_3} ^- + 1 e ^-

The oxidation number decrease for the reduced N is -2, which corresponds to the gain of two electrons.

Reduction:         NO _2 + 2 e ^-  \longrightarrow  NO

c. Since this reaction occurs in an acidic solution, charge balance is achieved by adding H ^+ ions. In the oxidation half-reaction there is no charge on the left side of the equation and a charge of -2 on the right side. Adding two H ^+ ions to the right side of the equation will balance the charge at zero on both sides.

Oxidation:          NO _2  \longrightarrow  NO {_3} ^- + 1 e ^- + 2 H ^+

In the reduction half-reaction there is a charge of -2 on the left side of the equation and no charge on the right side. Charge balance is achieved by adding two H ^+ ions to the left side of the equation.

Reduction:         NO _2 + 2 e ^- + 2 H ^+  \longrightarrow  NO

d. Hydrogen balance is obtained in the oxidation half-reaction by adding one H _2 O to the left side of the equation.

Oxidation:          NO _2 + H _2 O   \longrightarrow  NO {_3} ^- + 1 e ^- + 2 H ^+

Hydrogen balance is obtained in the reduction half-reaction by adding one H2O to the right side of the equation.

Reduction:         NO _2 + 2 e ^- + 2 H ^+  \longrightarrow  NO + H _2 O

e. Oxygen balances at two atoms on each side in both the oxidation and reduction half-reactions. The two balanced half-reactions are

Oxidation:          NO _2 + H _2 O   \longrightarrow  NO {_3} ^- + 1 e ^- + 2 H ^+

Reduction:         NO _2 + 2 e ^- + 2 H ^+  \longrightarrow  NO + H _2 O

Step 3 Equalize electron loss and electron gain.

The oxidation half-reaction involves the loss of one electron. The reduction half-reaction involves the gain of two electrons. Multiplying the oxidation half­ reaction by a factor of 2 will cause electron loss and gain to be equal at two electrons.

Oxidation:          2(NO _2 + H _2 O   \longrightarrow  NO {_3} ^- + 1 e ^- + 2 H ^+ )

Reduction:         NO _2 + 2 e ^- + 2 H ^+  \longrightarrow  NO + H _2 O

Step 4 Add the half-reactions, and cancel identical species.

Adding the two half-reactions together, we get

Both H _2 O and H ^+ are on both sides of the equation, and some of each can be cancelled. Also, NO _2 appears in two places on the left side of the equation and needs to be combined. The final balanced equation is

Answer Double Check: Have both atom balance and charge balance been achieved for this chemical equation? Yes. Atom balance is at 3 N, 7 O, and 2 H. Charge balance is at zero. On the product side of the equation there are two -1 ions and two +1 ions, producing a net charge of zero.