Balancing chemical equations by ion-electron

Balancing chemical equations

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

Thus during the redox reaction, some or compounds act as an oxidizing agent, other elements or compounds must be a reducing agent.

Balancing chemical equations by ion-electron method

1. Certain the reactants and products, and their chemical formulas.
2. Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.
3. 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.
4. Balancing chemical partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation and reduction half-reactions.
5. Multiply each partial equation by a suitable number so each of the two partial equations involves the same number of electrons.
6. Add the partial equations and cancel out species that appear on both sides of the chemical equations.

Balancing chemical equations examples

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

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

Originally published at https://www.priyamstudycentre.com on April 14, 2019.