Question

how can balance electronic reaction of elements..... with examples

Answer 1

Step -by-step explanation ____

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Balancing Redox Reactions by Oxidation Number Method

As with every other reaction, it is very important to write the correct compositions and formulas. A very important thing to keep in mind while writing oxidation-reduction reactions is to correctly write the compositions and formulas of the substances and products present in the chemical reaction. The steps of the oxidation number method are as follows:

Step 1

Correctly write the formula for the reactants and the products of the chemical reaction.

Step 2

Determine correctly the atoms that undergo oxidation number change in the given reaction by allocating the oxidation number of the individual elements present in the reaction.

Step 3

Calculate the oxidation number on the basis of each atom for the given molecule or ion of the chemical reactions. If the numbers are not equal then multiply it to such a number that overall these numbers become equal. If in a case, two substances are either only oxidized or only reduced then this signifies that something is wrong with the chemical reaction. This signifies that either the formulas of the reactant or the products are incorrect. This can also mean that the allocation of oxidation numbers is incorrect.

Step 4

Keep in mind the involvement of the ions if the reaction occurs in water. Accordingly, add H+ or OH– ions in the appropriate side of the reaction. Overall, the ionic charges of reactant and products will be equal. However, if the reaction takes place in acidic solution then add H+ ions in the chemical equation. Similarly, if the reaction takes place in the basic solution add OH– ions in the chemical equation.

Step 5

It is very important to equate the number of hydrogen atoms on each side of the equation by adding water molecules or H2O molecules. Additionally, it is necessary to check the oxygen atoms present in the equation. It will be a balance reaction if there are equal numbers of oxygen atoms present in both the reactant as well as the product the side.

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