## Balancing chemical equations

We are **balancing chemical equations** mainly by ion-electron and oxidation number process. Oxidation and reduction always go hand in hand during redox chemical equations.

Thus during the redox reaction, some elements or compounds are oxidized, other elements or compounds must be reduced.

### Balancing chemical equations by ion-electron method

- Certain the reactants and products, and their chemical formulas.
- Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.
- If the reaction occurs in acid solution use a requisite number of hydrogen ion for
**balancing**the number of atoms involved in the partial equation and for alkaline solution use hydroxyl ions. **Balancing chemical**partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation and reduction half-reactions.- Multiply each partial equation by a suitable number so each of the two partial equations involves the same number of electrons.
- Add the partial equations and cancel out species that appear on both sides of the
**chemical equations**.

### Balancing chemical equations examples

#### Oxidation of iron by potassium permanganate

KMnO_{4} in dilute H_{2}SO_{4} oxidizes iron ferrous state to the ferric state. In this chemical equation permanganate ions are the oxidant and ferrous ion the reductant.

MnO_{4}^{–} + H^{+} + Fe^{+2} ⇆ Mn^{+2} + Fe^{+3} + H_{2}O

The left-hand side of the ultimate equation will carry, MnO_{4}^{–} (permanganate ion), H^{+} (hydrogen ion) and Fe^{+2} (ferrous sulfate).

The Right-hand side will have as products, Mn^{+2} (manganese sulfate), water and Fe^{+3} (ferric sulfate). Thus the partial equation representing the reduction of the oxidant

MnO_{4}^{–} → Mn^{+2}

This reaction occurs in an acid solution, thus we utilize eight H^{+} to balance the four oxygen atoms in permanganate ion.

MnO_{4}^{–} + 8H^{+} → Mn^{+2} + 4H_{2}O

But the above partial equation still unbalanced from the viewpoint of charge, the equation balanced by bringing in five electrons.

MnO_{4}^{–} + 8H^{+} + 5e ⇆ Mn^{+2} + 4H_{2}O

The partial equation representing the oxidation of the reductant.

Fe^{+2} ⇆ Fe^{+3} + e

So this equation multiplies by 5 and then adding to the partial equation representing the reduction of the oxidant.

MnO_{4}^{–} + 8H^{+} + 5e ⇆ Mn^{+2} + 4H_{2}O |

5×( Fe^{+2} ⇆ Fe^{+3} + e) |

MnO_{4}^{–}+8H^{+}+5Fe^{+2}→Mn^{+2}+5Fe^{+3}+4H_{2}O |

#### Cr_{2}O_{7}^{-2} + I^{–} + H^{+} ⇆ Cr^{+3} + I_{2} + H_{2}O

Taking the dichromate side of the partial equation of reduction of oxidant and balancing the atoms and electrons.

Cr_{2}O_{7}^{-2} + 14H^{+} + 6e ⇆ 2Cr^{+3} + 7H_{2}O

But the partial equation representing the oxidation of the reductant.

2I^{–} ⇆ I_{2} + 2e

Thus this equation multiplies by 3 and adding with partial equations of the dichromate side.

Cr_{2}O_{7}^{-2}+14H^{+}+6e ⇆ 2Cr^{+3}+7H_{2}O |

3 (2I^{–} ⇆ I_{2} + 2e) |

Cr_{2}O_{7}^{-2}+14H^{+}+6I^{–}→2Cr^{+3}+3I_{2}+7H_{2}O |

#### Permanganate ion in alkaline solution

Permanganate ion in alkaline solution oxidizes Na_{2}SnO_{2} to sodium stannate Na_{2}SnO_{3}.

Thus the partial equation representing the reduction of oxidant in alkaline solution and balancing the chemical equations with a requisite number of OH^{–} ion.

MnO_{4}^{–} + 2H_{2}O → MnO_{2} + 4OH^{–}

But the above partial equation still unbalanced from the viewpoint of charge, the equation balanced by bringing in three electrons.

MnO_{4}^{–} + 2H_{2}O + 3e → MnO_{2} + 4OH^{–}

Again the partial equation representing the oxidation of the reductant.

SnO_{2}^{-2} + 2OH^{–} ⇆ SnO_{3}^{-2} + H_{2}O + 2e

Hence balancing these chemical equations, first partial equation multiplies 2 and the second partial equation multiplies 3 and adding to the given final balanced chemical equation.

2MnO_{4}^{–}+ 4H_{2}O + 6e ⇆ 2MnO_{2} + 8OH^{–} |

3SnO_{2}^{-2}+ 6OH^{–} ⇆ 3SnO_{3}^{-2}+ 3H_{2}O + 6e |

2MnO_{4}^{–}+3SnO_{2}^{-2}+H_{2}O→2MnO_{2}+3SnO_{3}^{-2}+2OH^{–} |

Problem

Balancing the chemical equation of permanganate to manganous by hydrogen peroxide in acid solution.

Solution

The partial equation for the reduction of oxidant

MnO_{4}^{–} + 8H^{+} + 5e ⇆ Mn^{+2} + 4H_{2}O

Hydrogen peroxide will give oxygen to an acid solution.

H_{2}O_{2} ⇆ 2H^{+} + O_{2} +2e

The first equation is multiplying by 2 and the second is 5 for balancing the chemical equation.

2MnO_{4}^{–} + 16H^{+} + 10e ⇆ 2Mn^{+2} + 8H_{2}O |

5H_{2}O_{2} ⇆ 10H^{+} + 5O_{2} +10e |

2MnO_{4}^{–}+5H_{2}O_{2}+6H^{+}⇆2Mn^{+2}+8H_{2}O+5O_{2} |

#### Mn^{+2} + BiO_{3}^{–} + H^{+} → MnO_{4}^{–} + BiO^{+} + H_{2}O

NaBiO₃ oxidizes manganese (II) to permanganate ion in sulfuric acid solution with the formation of BiO^{+} ion. Balancing chemical equation for this reaction.

2Mn^{+2} + 8H_{2}O → 2MnO_{4}^{–} + 16H^{+} + 10e |

5BiO_{3}^{–} + 20H^{+} + 10e → 5BiO^{+} + 10H_{2}O |

2Mn^{+2}+5BiO_{3}^{–}+4H^{+}→MnO_{4}^{–}+5BiO^{+} 2H_{2}O |

Problem

Balancing the chemical equation of reduction of nitrate ion to ammonia by aluminum in aqueous sodium hydroxide.

Answer

The partial equation for the reduction of oxidant in the sodium hydroxide solution.

NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

Metallic aluminum will go over to aluminate ion in alkaline solution

Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O + 3e

Multiplying by right factors for electron balancing the chemical equation.

3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻

8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O + 24e

3NO₃⁻ + 8Al + 2H₂O + 5OH⁻ → 3NH₃ + 8AlO₂⁻

### Oxidation number method

The oxidation number of oxygen remains unchanged manganese in permanganate has an oxidation number +7. Decreases and increases in oxidation numbers provide an idea about oxidation-reduction reactions.

Putting the right factors the decreases and increases in oxidation numbers and balancing chemical reactions.

MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³

Acidic solution the changes are not equal on the two sides of the above expression, hydrogen ions are added and the requisite number of water written.

MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn⁺² + 5Fe⁺³ + 4H₂O

#### Iodide ion in dilute sulfuric acid

Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂

The oxidation number of chromium decreases 2 and the oxidation number of Sn increases 1. Equalizing the increase and decrease in oxidation number.

Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂

Ionic charges are not equal on the two sides in an acid solution of the above expression. Fourteen hydrogen ions are added to the left-hand side and the requisite number of water added to the right-hand side.

Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ + 7H₂O

#### Sodium stannite to stannate in alkaline solution

MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

The oxidation number of manganese decreases 3 and the oxidation number of Sn increases 2. Equalizing the increase and decrease in oxidation number.

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²

Ionic charges are not equal on the two sides in the alkaline solution of the above expression. Two hydroxyl ions are added to the right-hand side.

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

There are two hydrogen atoms on the right and none on the left, water is added to balancing the chemical equations.

2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

#### Iodide ion and iodate ion in acid solution

I⁻ + IO₃⁻ → I₂

Change of oxidation number of iodine in the above chemical reaction.

I⁻(-1) + IO₃⁻(+5) → I₂(0)

I⁻ (-1) oxidation number of iodide increases 1 and IO₃⁻(+5) oxidation number decreases 5. Putting the right factors the decrease and increase in oxidation number for balancing equation.

5I⁻ + IO₃⁻ → 3I₂

Charges are unequal on two sides of the above expression in acid medium six hydrogen ion added on the left-hand side and write three water on the right-hand side.

I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O

#### Sulfurous acid and dichromate in acid solution

SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

Oxidation of two chromium decreases 2 × (+3) = 6 and the oxidation number of sulfur increases 2. equalizing the above chemical equation.

3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

Ionic charges are not equal on the two sides in acid medium, eight hydrogen ions are added to the left-hand side.

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³

There are eight hydrogen ion on the left and none on the right, four water added to balance the chemical equation.

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³ + 4H₂O