Question 15.11: 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-} + Cl _2  \longrightarrow  SO {_4} ^{2-} + Cl ^-

-2      0                  +6 -2    -1

Sulfur is oxidized, increasing in oxidation number from -2 to +6. Chlorine is reduced, decreasing in oxidation number from 0 to -1. The skeletal half-reactions for oxidation and reduction are

Oxidation:          S ^{2-}  \longrightarrow  SO {_4} ^{2-}
Reduction:         Cl _2  \longrightarrow  Cl ^-

Step 2 Balance the individual half-reactions.

a. In the oxidation half-reaction the S is already balanced.

Oxidation:          S ^{2-}  \longrightarrow  SO {_4} ^{2-}

To balance the Cl in the reduction half-reaction, the coefficient 2 must be added on the right side.

Reduction:         Cl _2  \longrightarrow  2 Cl ^-

b. The oxidation number increase for S is +8, which corresponds to the loss of eight electrons.

Oxidation:          S ^{2-}  \longrightarrow  SO {_4} ^{2-} + 8 e ^-

The oxidation number decrease for Cl is -1, which corresponds to a gain of one electron. Since there are two Cl atoms changing, the total electron gain is two electrons.

Reduction:         Cl _2 + 2 e^-  \longrightarrow  2 Cl ^-

c. Since this reaction occurs in a basic solution, charge balance is achieved by adding OH ^- ions. In the oxidation half-reaction there is a charge of -2 on the left side and a charge of -10 (one SO {_4} ^{2-} ion and eight electrons) on the right side. The charge is brought into balance, at a -10, by adding eight OH ^- ions

Oxidation:          S ^{2-} + 8 OH ^-  \longrightarrow  SO {_4} ^{2-} + 8 e ^-

In the reduction half-reaction, the charge is already balanced at a -2 on each side. No OH ^- ions are needed.

Reduction:         Cl _2 + 2 e^-  \longrightarrow  2 Cl ^-

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

Oxidation:          S ^{2-} + 8 OH ^-  \longrightarrow  SO {_4} ^{2-} + 8 e ^- + 4 H _2 O

Hydrogen balance is not needed in the reduction half-reaction since no hydro­ gen is present.

Reduction:         Cl _2 + 2 e^-  \longrightarrow  2 Cl ^-

e. Oxygen balances at eight atoms on each side of the equation in the oxidation half-reaction. Oxygen is not present in the reduction ha If-reaction. The two balanced half-reactions are

Oxidation:          S ^{2-} + 8 OH ^-  \longrightarrow  SO {_4} ^{2-} + 8 e ^- + 4 H _2 O

Reduction:         Cl _2 + 2 e^-  \longrightarrow  2 Cl ^-

Step 3 Equalize electron loss and electron gain.

Eight electrons are produced in the oxidation half-reaction and two electrons are gained in the reduction half-reaction. Multiplying the reduction half-reaction by 4 will cause electron loss and electron gain to be equal at eight electrons.

Oxidation:          S ^{2-} + 8 OH ^-  \longrightarrow  SO {_4} ^{2-} + 8 e ^- + 4 H _2 O

Reduction:         4(Cl _2 + 2 e^-  \longrightarrow  2 Cl ^- )

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

Adding the two half-reactions together, we get

Oxidation:          S ^{2-} + 8 OH ^-  \longrightarrow  SO {_4} ^{2-}  +  \cancel{8 \textrm{ e}^-} + 4 H _2 O

Reduction:         4 Cl _2  +  \cancel{8 \textrm{ e}^-}  \longrightarrow  8 Cl ^-

S ^{2-} + 8 OH ^-  + 4 Cl _2  \longrightarrow  SO {_4} ^{2-} + 4 H _2 O + 8 Cl ^-

There are no species to cancel other than the electrons.

Answer Double Check: Do the product and reactant sides of the equation have the same net charge? Yes. Both sides of the equation have a net charge of -10.