Oxidation Number-2

Disproportionation reactions

 “In this type of redox reaction, one element is simultaneously oxidized or reduced.”

This is a special type of redox reaction.


Examples

Balancing of Redox Reactions

 

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.

Half Reaction Method

In this procedure, we split the equation into two halves. Thereafter, we balance both the parts of the reaction separately. Finally, we add them together to give a balanced equation. We will demonstrate this method with an example so as to understand the steps of balancing redox reactions by half-reaction method.

For instance, a reaction is given where Fe2+ ions are converted to Fe3+ ions by dichromate ions in an acidic solution. The dichromate ions (Cr2O72–) are reduced to Cr3+ ions in the reaction. We have to balance the above redox reaction. The steps to balance this equation is as follows.

Step 1

First, we have to write the basic ionic form of the equation.Balancing redox reactions

Step 2

Divide the equation into two separate half reaction-oxidation half and reduction half.

  • ·         Oxidation half

  • ·         Reduction half

Step 3

In the third step of balancing redox reactions by half-reaction method, we will balance the atoms present in each half of the reaction except O and H atoms. In the example question, the oxidation part of the reaction in terms of Fe atoms is already balanced. Therefore, we will just balance the reduction part of the reaction. In this case, we will multiply Cr3+ by 2 in order to balance Chromium atoms.

Step 4

We know that the reaction takes place in an acidic solution. Therefore, we have to add water molecules (H2O) for balancing the O atoms of the equation and H+ for balancing the H atoms in the equation. Now the equation is

Step 5

Now, we will need to balance the charges in both the half reactions. Therefore, we need to multiply the appropriate number to one or both the half reaction and make the number of electrons same. Balancing the oxidation half of the reaction

 

For the reduction half, there are 12 positive charges on the left side of the equation and 6 positive charges on the right side of the equation. Therefore, we need to add 6 more electrons on the left side of the equation to balance the reduction half.

Now, to equate the electrons in two halves of the reactions, we will multiply 6 in the oxidation half reaction. Thus, we get

Step 6

The two halves of the equations are added to complete the overall reactions. After the addition of two reaction halves, cancel the electrons on both sides. The net ionic equation can be written as:

Step 7

Finally, we have to verify whether the equation consists of the same type, number, and charges on both sides of the equation. Moreover, the equation is completely balanced in terms of atoms and charges.

In case of a basic solution, we have to balance the atom similar to acidic solution. Then the equal number of OH ions addition is done for each H+ ion, in both the halves of the equation. When the H+ ions and OH ions will be present on the same side of the equation, we have to combine the ions and write H2O.

Redox Reactions as the Basis for titrations

Redox titrations are based on the transfer of electrons between reactants, which involves oxidation and reduction. In these reactions, one species loses electrons (undergoes oxidation) while the other species gains electrons (undergoes reduction). The reactants involved in redox titrations are usually an oxidizing agent and a reducing agent, and the titration involves determining the amount of one of these species by reacting it with a known amount of the other species until the reaction is complete.

Redox titrations are widely used in analytical chemistry for the determination of a wide range of substances, including metals, organic compounds, and inorganic compounds. The key to performing a successful redox titration is to select the appropriate oxidizing and reducing agents and to control the conditions of the reaction, such as pH and temperature.

Some common redox titrations include:

1.     Permanganate titrations: Permanganate ion  is a strong oxidizing agent that can be used to determine the amount of reducing agents in a sample, such as iron (II) ion or hydrogen peroxide .

2.     Iodine titrations: Iodine is a common oxidizing agent that can be used to determine the amount of reducing agents in a sample, such as thiosulfate ion or ascorbic acid (vitamin C).

3.     Dichromate titrations: Dichromate ion is a strong oxidizing agent that can be used to determine the amount of reducing agents in a sample, such as iron (II) ion  or organic compounds containing double bonds.

4.     Ceric ammonium nitrate (CAN) titrations: Ceric ammonium nitrate (CAN) is a strong oxidizing agent that can be used to determine the amount of reducing agents in a sample, such as ferrous ion.

In each of these titrations, the oxidizing agent is added slowly to the reducing agent until the reaction is complete, and the amount of the reducing agent is determined from the stoichiometry of the reaction. The endpoint of the titration is usually detected using an indicator, such as starch or phenolphthalein, or by monitoring changes in potential using a potentiometer.