Question 15.12: 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

H _3 AsO _3 + MnO {_4} ^-  \longrightarrow  H _3 AsO _4 + Mn ^{2+}

+1 +3 -2     +7-2                  + 1 +5 -2   +2

Arsenic is oxidized, increasing in oxidation number from +3 to +5. Manganese is reduced, decreasing in oxidation number from +7 to +2. The skeletal half-reactions for oxidation and reduction are

Oxidation:          H _3 AsO _3  \longrightarrow  H _3 AsO _4

Reduction:         MnO {_4} ^- \longrightarrow  Mn ^{2+}

Step 2 Balance the individual half-reactions.

a. In both half-reactions, the element being oxidized or reduced is already balanced-one atom of As on both sides in the oxidation half-reaction and one atom of Mn on both sides in the reduction half-reaction.

Oxidation:          H _3 AsO _3  \longrightarrow  H _3 AsO _4

Reduction:         MnO {_4} ^- \longrightarrow  Mn ^{2+}

b. The oxidation number increase for As is +2, which corresponds to the loss of
two electrons.

Oxidation:          H _3 AsO _3  \longrightarrow  H _3 AsO _4 + 2 e ^-

The oxidation number decrease for Mn is -5, which corresponds to the gain of five electrons.

Reduction:         MnO {_4} ^- + 5 e ^- \longrightarrow  Mn ^{2+}

c. Since this reaction occurs in an acidic solution, charge balance is achieved by adding H ^+ ions. In the oxidation half-reaction there is a charge of zero on the left side and a charge of -2 (two electrons) on the right side. Charge balance, at zero, is achieved by adding two H ^+ ions to the right side of the equation.

Oxidation:          H _3 AsO _3  \longrightarrow  H _3 AsO _4 + 2 e ^- + 2 H ^+

In the reduction half-reaction there is a charge of -6 on the left side (one MnO {_4} ^- ion and five electrons) and a charge of +2 on the right side. Charge balance, at a +2, is achieved by adding eight H ^+ ions to the left side of the equation.

Reduction:         MnO {_4} ^- + 5 e ^- + 8 H ^+  \longrightarrow  Mn ^{2+}

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

Oxidation:          H _3 AsO _3 + H _2 O   \longrightarrow  H _3 AsO _4 + 2 e ^- + 2 H ^+

Hydrogen balance is obtained in the reduction half-reaction by adding four H _2 O molecules to the right side of the equation.

Reduction:         MnO {_4} ^- + 5 e ^- + 8 H ^+  \longrightarrow  Mn ^{2+} + 4 H _2 O

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

Oxidation:          H _3 AsO _3 + H _2 O   \longrightarrow  H _3 AsO _4 + 2 e ^- + 2 H ^+

Reduction:         MnO {_4} ^- + 5 e ^- + 8 H ^+  \longrightarrow  Mn ^{2+} + 4 H _2 O

Step 3 Equalize electron loss and electron gain.

The lowest common multiple for an electron loss of 2 and an electron gain of 5 is 10 electrons. Thus, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

Oxidation:          5(H _3 AsO _3 + H _2 O   \longrightarrow  H _3 AsO _4 + 2 e ^- + 2 H ^+ )

Reduction:         2(MnO {_4} ^- + 5 e ^- + 8 H ^+  \longrightarrow  Mn ^{2+} + 4 H _2 O)

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

Adding the two half-reactions together, we get

Both H ^+ ion and H _2 O are on both sides of the equation. We can cancel 5 H _2 O molecules from each side and 10 H ^+ ions from each side. The final balanced equation becomes

Answer Double Check: Were H _2 O and H ^+ added to the equation in the appropriate manner to achieve H and O balance? Yes. There are 21 H atoms on each side of the equation and 23 O atoms on each side of the equation.