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Balancing Chemical Equations

Balancing chemical equations practice

Balancing chemical equations by ion-electron formula and change of oxidation number are uses to balance the oxidation reduction process or redox reactions for learning chemistry.

Balancing chemical equations by ion-electron formula and oxidation number balance method in chemistry

  1. The method is set up for balancing the partials equation of oxidant and reductant and balances the charges of these partial equations in acids and bases solution.
  2. Other hand oxidation numbers use for balancing chemical equations set up by change or calculation of the oxidation number of reactant and product atoms present in chemical elements, molecules, or ions.

Balancing redox reactions by ion electron method

Balancing chemical equations by the ion-electron method has certain general rules or formulas which we use to balance redox reactions. We use rules for balancing equations by the ion-electron method may include,

  • Certain the oxidizing agent and reducing agent, and their chemical formula for balancing chemical equations.
  • Set up the partial balancing equations representing the reduction of the oxidant and the oxidation of the reductant.
  • If the reaction occurs in an acid solution, a requisite number of hydrogen ions for balancing the number of the atom involved in the partial equations. For the alkaline solution or high pH solution, we use hydroxyl ions.
  • Balancing the changes of partial chemical equations by adding a suitable number of electrons. These electron particles are involved in balancing half-reactions.
  • Multiply each partial equation by a suitable factor for balancing ionization chemicals. Each of the two partial equations involves the same number of electrons.
  • Add the partial equations and cancel out similar species that appear on both sides of the balancing chemical reaction.

Examples for balancing chemical equations

Balancing Fe in KMnO4 solution

Potassium permanganate in dilute sulfuric acid oxidizes iron from the ferrous state to the ferric state by changing the electronic configuration of the iron atom. Permanganate ion is the oxidant and the ferrous ion is the reductant in balancing chemical equations.

MnO4 + H+ + Fe+2 → Mn+2 + Fe+3 + H2O

The left-hand side of the ultimate equation will carry MnO4, H+ ion, and Fe+2 ion, and the right-hand side will have Mn+2, water, and Fe+3.

The partial equation represents the reduction of the oxidant,
MnO4 → Mn+2

This reaction occurs in an acid solution, thus we use hydrogen ions to balance the four oxygen atoms in the permanganate ion,
MnO4+ 8 H+ → Mn+2 + 4 H2O

The above partial equation is still unbalanced from the viewpoint of charges. The equation balances by bringing in five electrons,
MnO4 + 8 H+ + 5e → Mn+2 + 4 H2O

The partial equation represents the oxidation of the reductant,
Fe+2 → Fe+3 + e

It is a balanced equation with respect to charges and reacting elements. To balance charges of partial equations, this equation multiplies by 5 and is added to the first equation.

MnO4 + 8 H+ + 5e → Mn+2 + 4 H2O
5 × ( Fe+2 → Fe+3 + e)
MnO4 + 8 H+ + 5 Fe+2 → Mn+2 + 5 Fe+3 + 4 H2O

Balancing potassium iodide in acid solution of dichromate

For balancing oxidation of potassium iodide in an acid solution of dichromate, chromium reduces from +6 state to +3 state, and iodine ion is oxidized to form the elementary iodine molecule.

The balancing partial equations of the dichromate side and iodine side are,
Cr2O7−2 + 14 H+ + 6e → 2 Cr+3 + 7 H2O
2 I → I2 + 2e

For balancing overall equations iodine side multiplies by 3 and is added to the dichromate side.

Cr2O7−2 + 14 H+ + 6e → 2 Cr+3 + 7 H2O
3 (2 I → I2 + 2e)
Cr2O7−2 + 14 H+ + 6 I → 2 Cr+3 + 3 I2 + 7 H2O

Permanganate ion in alkaline solution

Permanganate ion in alkaline solution oxidizes Na2SnO2 to Na2SnO3. The partial equation represents the reduction of oxidants in alkaline solution and balances the chemical equations with a requisite number of OH ions.

MnO4 + 2 H2O → MnO2 + 4 OH

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

MnO4 + 2 H2O + 3e → MnO2 + 4 OH

Again the partial equation represents the oxidation of the reductant.

SnO2−2 + 2 OH → SnO3−2 + H2O + 2e

For balancing overall chemical equations, the first partial equation multiplies 2 and the second partial equation multiplies 3.

2MnO4 + 4 H2O + 6e → 2 MnO2 + 8 OH
3 SnO2−2 + 6 OH → 3 SnO3−2 + 3 H2O + 6e
2 MnO4 + 3 SnO22 + H2O → 2 MnO2 + 3 SnO32 + 2 OH

Problem: How can you balancing the chemical equation of permanganate to manganous by hydrogen peroxide in acid solution?

Solution: Balance partial equations for the above chemical reaction are,
MnO4 + 8 H+ + 5e → Mn+2 + 4 H2O
H2O2 → 2 H+ + O2 + 2e

Balancing the overall charges, the first equation is multiplying by 2 and the second equation multiplying by 5.

2 MnO4 + 16 H+ + 10e → 2 Mn+2 + 8 H2O
5 H2O2 → 10 H+ + 5 O2 + 10e
2 MnO4 + 5 H2O2 + 6 H+ → 2 Mn+2 + 8 H2O + 5O2

Balancing equations practice problems

Problem: How can we balancing equation for oxidation of Mn+2 to MnO4 by acidic NaBiO3?

Solution: NaBiO3 oxidizes Mn+2 to MnO4 in sulfuric acid solution with the formation of BiO+ ion.

Balancing partial equations and overall chemical equation for the above chemical reaction is

2 Mn+2 + 8 H2O → 2 MnO4 + 16 H+ + 10e
5 BiO3 + 20 H+ + 10e → 5 BiO+ + 10 H2O
2 Mn+2 + 5 BiO3 + 4 H+ → MnO4 + 5 BiO+ + 2 H2O

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

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

3 NO3 + 18 H2O + 24 e → 3 NH3 + 27 OH
8 Al + 32 OH → 8 AlO2 + 16 H2O + 24e
3 NO3 + 8 Al + 2 H2O + 5 OH → 3 NH3 + 8AlO2

Balancing chemical equations by oxidation number method

Balancing chemical equations by oxidation number method

Balancing oxidation-reduction reactions by the change of oxidation number goes hand to hand with ion-electron methods for balancing chemical equations.

Let us illustrate this equation balancing method by oxidation of iron in acidic permanganate solution.

MnO4 + Fe+2 → Mn+2 + Fe+3

For balancing, the oxidation number of Mn decreases by 5, and iron increases by 1 in the above equations. Putting the right factor on the above equations,
MnO4 + 5 Fe+2 → Mn+2 + 5 Fe+3

By using hydrogen ions and water molecules, the balancing chemical equation is,
MnO4 + 8 H+ + 5 Fe+2 → Mn+2 + 5 Fe+3 + 4 H2O

Balancing by oxidation number method examples

Balancing iodide ion in acidic dichromate solution

Cr2O72 + I → 2 Cr+3 + I2

The oxidation number of chromium decreases by 3 and the oxidation number of iodine increases by 1. Putting the right factor on the above equations,
Cr2O72 + 6I → 2 Cr+3 + 3 I2

Fourteen hydrogen ions and the requisite number of water molecules are needed to balance the above equation. Therefore, the overall balanced equation is,
Cr2O72 + 6I + 14 H+ → 2 Cr+3 + 3 I2 + 7 H2O

Balancing sodium stannite to stannate in alkaline solution

MnO4 + SnO2−2 → MnO2 + SnO3−2

The oxidation number of manganese decreases by 3 and the oxidation number of tin increases by 2. Equalizing the increase and decrease in oxidation number by putting the right factor,
2 MnO4 + 3 SnO2−2 → 2 MnO2 + 3 SnO3−2

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

Balancing iodide ion and iodate ion in acid solution

The change of oxidation number of iodine to balancing iodide ion and iodate ion in acid solution,
I (−1) + IO3 (+5) → I2 (0)

The oxidation number of iodide increases by 1 and the iodate ion decreases by 5. Putting the right factors for decrease and increase in oxidation number and balancing the above equation,
I + IO3 + 6 H+ → I2 + 3 H2O

Balancing sulfurous acid in dichromate ion

Balancing chemical equations for the oxidation of sulfurous acid to sulfuric acid by potassium dichromate is,

3 SO32 + Cr2O7−2 + 8 H+ → 3 SO4−2 + 2 Cr+3 + 4 H2O

For balancing equations, 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 ions and water to balance the above ionic equations.

The above oxidation-reduction reactions discussion is useful for balancing any chemical equations by ion-electron formula and oxidation number method.

This balancing formula uses only to balance the chemical equations of oxidation-reduction or redox reaction reactions for learning chemistry or chemical science.