Question 15.10: Balancing a Net Ionic Redox Equation Using the Half-Reaction......

Step 1 Determine the oxidation and reduction skeletal half-reactions. Assigning oxidation numbers, we get

S ^{2-} + NO {_3} ^-  \longrightarrow  S + NO

-2      +5 -2              0   +2 -2

Sulfur is oxidized, increasing in oxidation number from -2 to 0. Nitrogen is reduced, decreasing in oxidation number from +5 to +2. The skeletal half-reactions for oxidation and reduction are

Oxidation:          S ^{2-}  \longrightarrow  S
Reduction:         NO {_3} ^-  \longrightarrow  NO

Step 2 Balance the individual half-reactions.

a. In both half-reactions, the element being oxidized or reduced is already balanced—one atom of S on both sides in the first half-reaction and one atom of N on both sides in the second half-reaction. There are no other elements present except oxygen.

Oxidation:          S ^{2-}  \longrightarrow  S
Reduction:         NO {_3} ^-  \longrightarrow  NO

b. The oxidation number increase for S is +2. This is caused by the loss of two electrons, which are shown on the product side of the oxidation half-reaction.

Oxidation:          S ^{2-}  \longrightarrow  S + 2 e ^-

The oxidation number decrease for N is -3. This results from the gain of three electrons, which are shown on the reactant side of the reduction half-reaction.

Reduction:  NO {_3} ^- + 3 e ^-  \longrightarrow  NO

c. Since this is an acidic solution reaction, charge balance is achieved by adding H ^+ ions. In the oxidation half-reaction there is a charge of -2 on each side of the equation. No H ^+ ions are needed since the charge is already in balance.

Oxidation:          S ^{2-}  \longrightarrow  S + 2 e ^-

In the reduction half-reaction there is a charge of -4 on the left side of the equation (-1 from the NO {_3} ^- ion and -3 from the three electrons). There is no charge on the right side of the equation. Charge balance is achieved by adding four H ^+ ions to the left side of the equation. Each side of the equation will now have zero charge.

Reduction:  NO {_3} ^- + 3 e ^- + 4 H ^+  \longrightarrow  NO

d. Water molecules are used to achieve hydrogen balance. Since no hydrogen is present in the oxidation half-reaction, no water molecules are needed.

Oxidation:          S ^{2-}  \longrightarrow  S + 2 e ^-

In the reduction half-reaction, two water molecules are added to the right side of the equation. We now have four hydrogen atoms on each side of the equation.

Reduction:  NO {_3} ^- + 3 e ^- + 4 H ^+  \longrightarrow  NO + 2 H _2 O

e. There are no oxygen atoms present in the oxidation half-reaction. In the reduction half-reaction, the oxygen balances at three atoms on each side of the equation. The two balanced half-reactions are

Oxidation:          S ^{2-}  \longrightarrow  S + 2 e ^-

Reduction:  NO {_3} ^- + 3 e ^- + 4 H ^+  \longrightarrow  NO + 2 H _2 O

Step 3 Equalize electron loss and electron gain.

Two electrons are produced in the oxidation half-reaction and three electrons are gained in the reduction half-reaction. To equalize electron loss and electron gain, we multiply the oxidation half-reaction by three and the reduction half-reaction by two. We have then an electron loss of 6 and an electron gain of 6.

Oxidation:          3(S ^{2-}  \longrightarrow  S + 2 e ^- )

Reduction:  2(NO {_3} ^- + 3 e ^- + 4 H ^+  \longrightarrow  NO + 2 H _2 O)

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

Adding the two half-reactions together, we get

There are no species to cancel other than the electrons. The electrons must always cancel. If they do not, we have made a mistake in a previous step.

Answer Double Check: Do the reactant and product sides of the equation have the same net ionic charge? Yes. There is no charge associated with the product side of the equation since none of the products is an ion. Thus, the sum of ionic charges on the reactant side of the equation must add to zero. Such is the case; 3(-2) + 2(-1) + 8(+1) = 0.