# 18.2 Balancing Redox Reactions

## Learning Objectives

By the end of this section, you will be able to:

• Balance chemical equations for redox reactions using the half-reaction method

## Balancing Redox Reactions via the Half-Reaction Method

Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:

### Balancing Redox Reactions

1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
1. Add OH ions to both sides of the equation in numbers equal to the number of H+ ions.
2. On the side of the equation containing both H+ and OH ions, combine these ions to yield water molecules.
3. Simplify the equation by removing any redundant water molecules.
9. Finally, check to see that both the number of atoms and the total charges are balanced.

### Example 18.2a

#### Balancing Redox Reactions in Acidic Solution

Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.

Cr2O72− + Fe2+ → Cr3+ + Fe3+

#### Solution

Step 1. Write the two half-reactions. Each half-reaction will contain one reactant and one product with one element in common.

Fe2+ → Fe3+
Cr2O72− → Cr3+

Step 2. Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

Fe2+ → Fe3+

Cr2O72−2 Cr3+

Step 3. Balance oxygen atoms by adding H2O molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.

Fe2+ → Fe3+

Cr2O72− → 2 Cr3+ + 7 H2O

Step 4. Balance hydrogen atoms by adding H+ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.

Fe2+ → Fe3+

Cr2O72− + 14 H+ → 2 Cr3+ + 7 H2O

Step 5. Balance charge by adding electrons. The iron half-reaction shows a total charge of 2+ on the left side (1 Fe2+ ion) and 3+ on the right side (1 Fe3+ ion). Adding one electron to the right side brings that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved. The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side (1 Cr2O72− ion and 14 H+ ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr3+ ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6) = 6+, and charge balance is achieved.

Fe2+ → Fe3+ + e

Cr2O72− + 14 H+ + 6 e → 2 Cr3+ + 7 H2O

Step 6. Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.

6 Fe2+6 Fe3+ +6 e

Cr2O72− + 6 e + 14 H+ → 2 Cr3+ + 7 H2O

Step 7. Add the balanced half-reactions and cancel species that appear on both sides of the equation.

6 Fe2+ + Cr2O72− + 6 e + 14 H+ → 6 Fe3+ + 6 e + 2 Cr3+ + 7 H2O

Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

6 Fe2+ + Cr2O72− + 14 H+ → 6 Fe3+ + 2 Cr3+ + 7 H2O

A final check of atom and charge balance confirms the equation is balanced.

Component Reactants Products
Fe 6 6
Cr 2 2
O 7 7
H 14 14
charge 24+ 24+

### Exercise 18.2a

In acidic solution, hydrogen peroxide reacts with Fe2+ to produce Fe3+ and H2O. Write a balanced equation for this reaction.

### Example 18.2b

#### Balancing Equations for Redox Reactions in Basic Solutions

Write the balanced equation representing reaction between aqueous permanganate ion, MnO4, and solid chromium(III) hydroxide, Cr(OH)3, to yield solid manganese(IV) oxide, MnO2, and aqueous chromate ion, CrO42−. The reaction takes place in a basic solution.

#### Solution

Following the steps of the half-reaction method:

Step 1. Write skeletal equations for the oxidation and reduction half-reactions.

oxidation: Cr(OH)3(s) → CrO42−(aq)
reduction: MnO4(aq) → MnO2(s)

Step 2. Balance each half-reaction for all elements except H and O.

oxidation: Cr(OH)3(s) → CrO42−(aq)
reduction: MnO4(aq) → MnO2(s)

Step 3. Balance each half-reaction for O by adding H2O.

oxidation: H2O(l) + Cr(OH)3(s) → CrO42−(aq)
reduction: MnO4(aq) → MnO2(s) + 2 H2O(l)

Step 4. Balance each half-reaction for H by adding H+.

oxidation: H2O(l) + Cr(OH)3(s) → CrO42−(aq) + 5 H+(aq)
reduction: 4 H+(aq) + MnO4(aq) → MnO2(s) + 2 H2O(l)

Step 5. Balance each half-reaction for charge by adding electrons.

oxidation: H2O(l) + Cr(OH)3(s) → CrO42−(aq) + 5 H+(aq) + 3 e
reduction: 3 e + 4 H+(aq) + MnO4(aq) → MnO2(s) + 2 H2O(l)

If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.  This step is not necessary since the number of electrons is already in balance.

Add the two half-reactions and simplify.

H2O(l) + Cr(OH)3(s) + 3e + 4H+(aq) + MnO4(aq) → CrO42−(aq) + 5H+(aq) + 3e + MnO2(s) + 2H2O(l)
Cr(OH)3(s) + MnO4(aq) → CrO42−(aq) + H+(aq) + MnO2(s) + H2O(l)

If the reaction takes place in a basic medium, add OH ions the equation obtained in step 7 to neutralize the H+ ions (add in equal numbers to both sides of the equation) and simplify.

OH(aq) + Cr(OH)3(s) + MnO4(aq) → CrO42−(aq) + H+(aq) + OH(aq) + MnO2(s) + H2O(l)
OH(aq) + Cr(OH)3(s) + MnO4(aq) → CrO42−(aq) + MnO2(s) + 2 H2O(l)

### Exercise 18.2b

Aqueous permanganate ion may also be reduced using aqueous bromide ion, Br, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, BrO3. Write the balanced equation for this reaction occurring in a basic medium.