## Balancing Chemical Equations in Science

**Balancing chemical equations** by ion electron formula and change of oxidation numbers are uses to balance oxidation reduction reactions (redox) for science in chemistry online study. The ion electron method set up partials equation of oxidant and reductant and balance the charges these partial equations in acidic or basic solution, other hand oxidation number for balancing chemical reaction set up the change or calculation of oxidation number of reactant and product elements in chemical compounds.

### Balancing Chemical Equations by Ion-electron Method

**Balancing chemical **equations by the ion electron method has certain general rules which use for balance oxidation reduction reaction or redox reactions. Therefore, the usage rules for balancing these equations

- Certain the oxidizing agent and reducing agent, and their chemical formula.
- 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 the atom involved in the partial equation and for alkaline solution use hydroxyl ions.
- Balancing the changes of partial chemical equations by adding a suitable number of electrons. These electron charges involve balancing oxidation and reduction half-reactions.
- Multiply each partial equation by a suitable factor. Therefore, 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 reaction.

### Examples for Balance Chemical Equations

#### Iron in potassium permanganate solution

Potassium permanganate in dilute sulfuric acid oxidizes iron from the ferrous state to the ferric state. Therefore, permanganate ion is the oxidant and ferrous ion is 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}^{–}, H^{+} ion, and Fe^{+2} ion, and the right-hand side will have Mn^{+2}, water, and Fe^{+3}. Therefore, the partial equation representing the reduction of the oxidant, MnO_{4}^{–} → Mn^{+2}. This reaction occurs in an acid solution, thus we use hydrogen ion to balance the four oxygen atoms in permanganate ion.

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

The above partial equation still unbalanced from the viewpoint of charges and the equation balancing 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. This equation balancing with respect to charges and reacting elements. Therefore, to balancing charges of partial equations, this equation multiplies by 5 and then adding with the first equation.

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

#### Potassium iodide in acid solution of dichromate

Oxidation of potassium iodide in an acid solution of dichromate, chromium reduces from +6 state to +3 state and iodine ion oxidized to form the elementary iodine molecule. Therefore, the balance partial equations of the dichromate side and iodine side are

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

2I^{–} ⇆ I_{2} + 2e

For balencing overall equations iodine side multiplies by 3 and adding with 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 Na_{2}SnO_{3}. Therefore, 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^{–}

The above partial equation still unbalanced from the viewpoint of charge, the equation balance 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 **balance 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: Balance partial equations for the above chemical reaction are

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

H_{2}O_{2} ⇆ 2H^{+} + O_{2} + 2e

Balancing the overall charges of the chemical equation multiplying by 2 and second by 5 and adding.

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}

### Working Examples for Balencing Equations

#### Oxidation of Mn^{+2} to MnO_{4}^{–} by acidic NaBiO_{3}

NaBiO_{3} oxidizes Mn^{+2} to MnO_{4}^{–} in sulfuric acid solution with the formation of BiO^{+} ion. Balancing partials and overall chemical equations for the above chemical reaction are

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 Balance partials and overall equation for the reduction of nitrate ion to ammonia by aluminum in an aqueous sodium hydroxide are

3NO_{3}^{–} + 18H_{2}O + 24e ⇆ 3NH_{3} + 27OH^{–}

8Al + 32OH^{–} ⇆ 8AlO_{2}^{–} + 16H_{2}O + 24e

3NO_{3}^{–} + 8Al + 2H_{2}O + 5OH^{–} → 3NH_{3} + 8AlO_{2}^{–}

### Balancing Chemical Equation by Oxidation Number

Balancing oxidation reduction reactions by the change of oxidation number going hand to hand with ion electron methods for balance chemical equations. Let us illustrate this balancing method by oxidation of iron by acidic permanganate solution.

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

The oxidation number of Mn in MnO_{4}^{–} decreases by 5 and iron increases by 1 in the above equation. Therefore, putting the right factors the decreases and increases in oxidation numbers and hydrogen ion (acid solution) and water molecules for balance chemical equations.

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

#### Balancing iodide ion in acidic dichromate solution

Cr_{2}O_{7}^{-2} + I^{–} → 2Cr^{+3} + I_{2}

The oxidation number of chromium decreases by 2 and the oxidation number of iodine increases by 1. Equalizing the increase and decrease in oxidation number by putting the right factor on the above equation. Fourteen hydrogen ions and the requisite number of water molecules also added to balance the above equation.

Cr_{2}O_{7}^{-2} + 6I^{–} + 8H^{+} → 2Cr^{+3} + 3I_{2} + 7H_{2}O

#### Sodium stannite to stannate in alkaline solution

MnO_{4}^{–} + SnO_{2}^{-2} → MnO_{2} + SnO_{3}^{-2}

The oxidation number of manganese decreases by 3 and the oxidation number of tin increases by 2. Hence, equalizing the increase and decrease in oxidation number by putting the right factor and two hydroxyl ions and the requisite number of water molecules are added to balance the above chemical equation. Therefore, the balancing equation for the above reactions

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

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

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

I^{–} (-1) + IO_{3}^{–} (+5) → I_{2} (0)

Therefore, the oxidation number of iodide increases by 1, and iodate ion decreases by 5. Putting the right factors for decrease and increase in oxidation number and balancing the above equation.

I- + IO_{3}^{–} + 6H^{+} → I_{2} + 3H_{2}O

#### Balancing sulfurous scid in dichromate ion

Balancing chemical equation for the oxidation of sulfurous acid to sulfuric acid by potassium dichromate

3SO_{3}^{-2} + Cr_{2}O_{7}^{-2} + 8H^{+} → 3SO_{4}^{-2} + 2Cr^{+3} + 4H_{2}O

The oxidation number of two chromium atom decrease by 6 and the oxidation number of sulfur increases by 2. Therefore, we put the right factor and the requisite number of hydrogen ion and water to balance the above ionic reactions.

Therefore, the above oxidation reduction reactions discussion useful for balancing any chemical equations by ion electron formula and oxidation number method. For understanding, this balancing formula only balance the chemical equations of oxidation reduction (redox reaction) reactions in science.