What is the equivalent weight of ${{O}_{2}}$ in the following reaction?

${{H}_{2}}O + \dfrac{1}{2}{{O}_{2}} + 2{{e}^{-}} \to 2O{{H}^{-}}$

To determine the equivalent weight of oxygen ($O_2$) in the given reaction, we first need to understand the concept of equivalent weight and how it relates to the stoichiometry of the reaction. The equivalent weight of a substance is defined as the mass of that substance that will combine with or displace one mole of hydrogen atoms or half a mole of oxygen atoms in a chemical reaction.

Analyzing the Reaction

The reaction provided is:

${{H}_{2}}O + \dfrac{1}{2}{{O}_{2}} + 2{{e}^{-}} \to 2O{{H}^{-}}$

In this reaction, we can see that one mole of $O_2$ is involved, but it is only half a mole ($\frac{1}{2} O_2$) that reacts to produce hydroxide ions ($OH^-$). The key here is to identify how many electrons are transferred per mole of $O_2$ in the reaction.

Identifying the Change in Oxidation State

In this reaction, the oxygen in $O_2$ is reduced to hydroxide ions ($OH^-$). The oxidation state of oxygen in $O_2$ is 0, and in $OH^-$, it is -2. This indicates that each oxygen atom in $O_2$ undergoes a change of 2 units in oxidation state (from 0 to -2).

Calculating the Number of Electrons Transferred

Since one mole of $O_2$ contains two moles of oxygen atoms, and each atom gains 2 electrons, the total number of electrons transferred when half a mole of $O_2$ reacts is:

• 1 mole of $O_2$ = 2 moles of electrons (2 electrons per atom × 1 atom).

Finding the Equivalent Weight

The equivalent weight can be calculated using the formula:

Equivalent Weight = Molar Mass / n

Where:

• Molar Mass of $O_2$ = 32 g/mol (16 g/mol for each oxygen atom × 2).
• n = number of electrons transferred per mole of $O_2$ = 2.

Now, substituting the values into the formula:

Equivalent Weight = 32 g/mol / 2 = 16 g/equiv

Final Thoughts

Thus, the equivalent weight of $O_2$ in the given reaction is 16 g/equiv. This means that 16 grams of oxygen will react with or displace one mole of hydrogen in the context of this specific reaction. Understanding equivalent weight is crucial in stoichiometry and helps in calculating the amounts of reactants and products in chemical reactions.